'■^'■J®*!-"- 
 
IN MEMORIAM 
 FLORIAN CAJORl 
 
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AETTSS Tf 1.4^rr!^ 
 
 A 
 
 COURSE 
 
 OF 
 
 MATHEMATICS. 
 
 IN TWO VOLUMES. 
 
 FOR THE USE OF ACADEMIES, 
 
 AS WELL AS 
 
 PRIVATE TUITION. 
 
 BY 
 
 CHARLES HUTTON, L.L.D. F.R.a 
 
 CATE PROFESSOR OF MATHEMATICS IHJ THE JROltAIiv MILITARY 
 ACADEMY. 
 
 ' '-' 'KZViblJD AND CdKRiSCTEO l/T ' 
 
 ROBERT ADRAIN, A. M. 
 
 FELLOW OF THE AMERICAN PHILOSOPHICAL SOCIETr, 
 
 AND 
 
 PROFESSOR OF MATHEMATICS IN QUEEN'S COLLEGE NEW-SERSEt. 
 
 VOL. I. 
 
 NEW- YORK : 
 
 PUBLISHED BY SAMUEL CAMPBELL, EVERT DUYCKINCl^ 
 
 T. & J. SWORDS, PETER A. MESIER, R. M'DERMUT, 
 
 THOMAS A. RONALDS, JOHN TIEBOUT, 
 
 AND GEORGE LONG. 
 
 1818. 
 

 ADVERTISEMEiNT. 
 
 THE publishers of this third AmericanjedUion of Dr. Hutton'.s Course of 
 Mathfc.natics, were ifTdncdTTo engageT^n't'Ee^work frora a conviction of 
 its utility to private Students, as well as to Colleges and other Seminaries, in 
 vhich Mathematical Science constitutes a branch of education. They also 
 had in view the furnishing of the Military School of their country with a Text 
 Book of high standing, and long in use in the British Military Academy And 
 in order that this edition might derive advantage from the progress of the 
 Science, and thereby become more worthy of the public patronage, they en- 
 gaged a gentleman of acknowledged eminence to revise its pages and super- 
 intend the printing ; and they confidently trust this duty has been performed 
 •\viih some profit to the work generally. To gentlemen, therefore, who study 
 this delightful science in private, and to the literary and military institutions of 
 their country, the publishers and proprietors look for remuneration — and they 
 feel as though they should not look in vain. An increasing taste for Mathe- 
 matical Studies will produce a correspondent increase of purchasers ; whife 
 the preference which an honourable patriotism gives to American editions 
 *when well executed, will receive additional activity from the super-eminence 
 tof the work itself. 
 
 Qfj=* Oiders for this publication will be thankfully received by any of the 
 proprietors; all whose names are printed at the foot of the title-page. 
 
 jSTew-Yorlc, 1818. X* * ' 
 
 DUtrict. nf iJeivTork, ss- \ 
 
 BE IT REMEMBERED, that on the eleventh day of August, in the tliiity-seventli year 
 of the Independence of the United States of America, Samuel Campbell^ of the said district, 
 liath deposited m this office tlie title of a book, the right whereof he claims as proprietor, 
 hi the woi-ds following:, to wit: 
 
 " A Com'se of Mathematics. In two vgjuipjes. For the use of Academies, as well as Pri- 
 vate Tuition. By Charles HiUtoiiI L. L. D. F. R. S. late Professor of Mathematics in the 
 Royal Military Academy. From the fifth and sixth London editions, revised and correctefl 
 by Robert Atlrain, A. M. Fellow of the American Philosophical Society, and Professor of 
 Mathematics in Queen's College. New-Jersey." 
 
 In conformity to the Act of the Congress of the United States, entitled " An act for the 
 encouvaRement of learning, by securing the copies of maps, charts, and books to the' au- 
 thors and iiroprietors ol' such copies^ during the times therein mentioned." And also to an 
 act, entitletl " An act< supplementary to an act, entitled an act for the encouragement of 
 learning, by securing the copies of maps, charts, and books to the authors and proprietors 
 of such coi'ies, during the times therein mentioned, and extending the heaefits thereof to 
 <he arts of designing, engva-y ing, and etching historical and other prints." 
 
 CHARLES CLINTON, 
 Ctcrh of the district of NewTarE;. 
 
 Gem^ge Long, Printer. 
 
 CAJOR? 
 
PREFACE. 
 
 A SHORT and Easy Course of the Mathematical Sciences 
 has long been considered as a desideratum for the use of Stu- 
 dents in the different schools of education : one that should 
 hold a middle rank between the more voluminous and bulky 
 collections of this kind, and the mere abstract and brief com- 
 mon-place forms, of principles and memorandums. 
 
 For long experience, in all Seminaries of Learning, has 
 shown, that such a work was very much wanted, and would 
 prove a great and general benefit ; as, for want of it, recourse 
 has always been obliged to be had to a number of other books 
 hy different authors ; selecting a part from one and a part from 
 another, as seemed most suitable to the purpose in hand, and 
 rejecting the other parts — a practice which occasioned much 
 expense and trouble, in procuring and using such a number 
 of odd volumes, of various forms and modes of composition ; 
 besides wanting the benefit of uniformity and reference, which 
 are found in a regular series of composition. 
 
 To remove these inconveniences, the Author of the present 
 work has been induced, from time to time, to compose various 
 parts of this Course of Mathematics ; which the experience 
 of many years' use in the Academy has enabled him to adapt 
 and improve to the most useful form and quantity, for the be- 
 nefit of instruction there. And, to render that benefit mores 
 eminent and lasting, the Master General of the Ordnance has 
 been pleased to give it its present form, by ordering it to bq 
 enlarged and printed, for the use of the Royal Military Aca- 
 demy. 
 
 As this work has been composed expressly with the inten- 
 tion of adaptmg it to the purposes of academical education, it 
 is not designed to hold out the expectation of an entire new 
 mass of inventions and discoveries : but rather to collect and 
 . arrange the most useful known principles of mathematics, dis- 
 posed in a convenient practical form, demonstrated in a plain 
 and concise way, and illustrated with suitable examples, re- 
 jecting whatever seemed to be matters of mere curiosity, and 
 retaining only such parts and branches, as have a direct tenden- 
 cy and application to some useful purpose in life or profession. 
 It is however expected that much that is new will be found 
 in many parts of these volumes ; as well in the matter, as in 
 the arrangement and manner of demonstration, throughout the 
 whole work, especially in the geometry, which is rendered 
 much more easy and simple thaa heretgfore ; and in the conic 
 
 918242 ""'""' 
 
Vv PREFACE. 
 
 sections, which are here treated in a manner at once new, easy 
 and natural ; so much so indeed, that all the propositions and 
 their demonstrations, in the ellipsis, are the very same, word 
 for word, as those in the hyperbola, using only, in a very few 
 places, the word swm, for the word difference : also in many of 
 the mechanical and philosophical parts which follow, in the 
 second volume. In the conic sections, too, it may bejobserved, 
 that the first theorum of each section*only is proved from the 
 cone itself, and all the rest of the theorems are deduced from 
 the first, or from each other, in a very plain and simple manner. 
 Besides renewing most of the rules, and introducing every 
 where new examples, this edition is much enlarged in several 
 places ; particularly by extending the tables of squares and 
 cubes, square roots and cube roots, to 1000 numbers, which 
 will be found of great use in many calculations ; also by the 
 tables of logarithms, sines, and tangents, at the end of the se- 
 cond volume ; by the addition of Cardan's rules for resolving 
 'cubic equations ; with tables and rules for annuities ; and many 
 other improvements in different parts of the work. 
 
 Though the several parts of this course of mathematics are 
 ranged in the order naturally required by such elements, yet 
 students may omit any of the particulars that may be thought 
 the least necessary to their several purposes ; or they may, 
 study and learn various parts in a different order from their 
 jpresent arrangement in the book, at the discretion of the tutor. 
 So, for instance, all the notes at the foot of the pages may be 
 omitted, as well as many of the rules ; particularly the 1st or 
 Common Rule for the Cube Root, p. 85, may wel\ be omitted, 
 I)eing more tedious than useful. Also the chapters on Surds 
 and Infinite Series, in the Algebra : or these might be learned 
 after Simple Equations. Also Compound Interest and Annui- 
 ties at the end of the Algebra. Also any part of the Geome- 
 try, in vol. 1 ; any of the branches in vol. 2, at the discretion of 
 the preceptor. And, in any of the parts, he may omit some of 
 the examples, or he may give more than are printed in the 
 book ; or he may very profitably vary or change them, by alter- 
 ing the numbers occasionally. — As ta the quantity of writing •,. 
 the author would recommend, that the student copy out into 
 his fair book no more than the chief rules which he is directed 
 to lejarn off by rote, with the work of one example only to. 
 each rule, set down at full length ; omitting to set down the 
 work of all the other examples, how many soever he may be 
 directed to work out upon his slate or waste paper. — In short, 
 a great deal of the business, as to the quantity and order and 
 ^ manner, must depend on the judgment of the discreet and pru- 
 dent tutor or director. 
 
 [Dr. 
 
[Dr. Hutton's Frefaee to the third voiurae of the BoglijOi edition, 
 published in 1811.] 
 
 1 HE beneficial improvements lately made, and still makiog^^ 
 in the plan of the scientific education of the Cadets, in the 
 Royal Military Academy at Woolwich, having rendered a 
 further extension of the Mathematical Course adviseable, I 
 was honoured with the orders of his Lordship the Master 
 General of the Ordnance, to prepare a third volume, in addi- 
 tion to the two former volumes of the Course, to contain 
 such additions to some of the subjects before treated of in 
 those two volumes, with such other new branches of military 
 science, as might appear best adapted to promote the ends of 
 this important institution. From my advanced age, and the 
 precarious state of my health, I was desirous of declining such 
 a task, and pleaded my doubts of being able, in such a state, 
 to answer satisfactorily his lordship's wishes. This difficulty 
 however was obviated by the reply, that, to preserve a uni- 
 formity between the former and the additional parts of the 
 Course, it was requisite that I should undertake the direction 
 of the arrangement, and compose such parts of the work as 
 might be found convenient, or as related to topics in which 
 I had made experiments or improvements ; and for the rest, 
 I might take to my assistance the aid of any other person I 
 might think proper. With this kind indulgence being en- 
 couraged to exert my best endeavours, 1 immediately an- 
 nounced my wish to request the assistance of Dr. Gregory of 
 the Royal Military Academy, than whom, both for his ex- 
 tensive scientific knowledge, and his long experience, I know 
 «f no person more fit to be associated in the due performance 
 ©f such a task. Accordingly, this volume is to be considered 
 as the joint composition of that gentleman and myself, hay- 
 ing each of us taken and prepared, in nearly equal portions^ 
 separate chapters and branches of the work, being such as 
 in the compass of this volume, with the advice and assistance 
 of the Lieut. Governor, were deemed among the most useful 
 additional subjects for the purposes of the education estab- 
 lished in the Academy. 
 
 The several parts of the work, and their arrangement, are 
 as follow. — In the first chapter are contained all the proposi- 
 tions of the course of Conic Sections, first printed for the use 
 of the Academy in the year 1767, which remained, after 
 those that were selected for the second volume of this Course : 
 
 to 
 
yi PREFACE. 
 
 to which is added a tract on the algebraic equations of the 
 several conic sections, serving as a brief introduction to the 
 algebraic properties of curve lines. 
 
 The 2d chapter contains a short geometrical treatise on the 
 elements of Isoperimetry and the maxima and minima of 
 surfaces and solids ; in which several propositions usually in- 
 vestigated by fluxionary processes are effected geometrically ; 
 and in which, indeed, the principal results deduced by Thos. 
 Simpson, Horsley, Legendre, and Lhuillier are thrown into 
 the compass of one short tract. 
 
 The 3d and 4th chapters exhibit a concise but compre- 
 hensive view of the trigonometrical analysis, or that in which 
 the chief theorems of Plane and Spherical Trigonometry are 
 deduced algebraically by means of what is commonly de- 
 nominated the Arithmetic of Sines. A comparison of the 
 modes of investigation adopted in these chapters, and those 
 pursued in that part of the second volume ■ of this course 
 which is devoted to Trigonometry, will enable a student to 
 trace the relative advantages of the algebraical and geome- 
 trical methods of treating this useful branch of science. The 
 fourth chapter includes also a disquisition on the nature and 
 measure of solid angles, in which the theory of that peculiar 
 class of geometrical magnitudes is so represented, as to ren- 
 der their mutual comparison (a thing hitherto supposed im- 
 possible except in one or two Very obvious cases) a matter of 
 perfect ease and simplicity. 
 
 Chapter the fifth relates to Geodesic Operations, and that 
 more extensive kind of Trigonometrical Surveying which is 
 employed with a view to determine the geographical situation 
 «f places, the magnitude of kingdoms, and the figure of the 
 earth. This chapter is divided into two sections ; in the first 
 of which is presented a general account of this kind of survey- 
 ing ; and in the second, solutions of the most important prob- 
 lems connected with these operations. , This portion of the 
 volume it is hoped will be found highly useful ; as there is no 
 work which contains a concise and connected account of this 
 kind of surveying and its dependent problems ; and it cannot 
 fail to be interesting to those who know how much honour re- 
 dounds to this country from the great skill, accuracy, and 
 judgment, with which the trigonometrical survey of England 
 has long been carried on. 
 
 In the 6th and 7th chapters are developed the principles of 
 Pblygonometry.) and those which relate to the Dimsion of lands 
 and other surfaces, both by geoaaetrical coastructiou and by 
 coDM^utatioa. 
 
PREFACE. ^ tii 
 
 The 8th chapter contains a view of the nature and solution 
 of equations in general, with a selection of the best rules for 
 equations of different degrees. Chapter the 9th is devoted to 
 the nature and properties of curves^ and the construction of 
 equations. These chapters are manifestly connected, and 
 show how the mutual relations subsisting between equations of 
 different degrees, and curves of various orders, serve for the 
 reciprocal illustration of the properties of both. 
 
 In the 1 0th chapter the subjects of Fluents and Fluxional 
 equations are concisely treated. The various forms of Fluents 
 comprised in the useful table of them in the 2d volume, are 
 investigated : and several other rules are given ; such as it is 
 believed will tend much to facilitate the progress of students 
 in this interesting department of science, especially those 
 which relate to the mode of finding fluents by continuation. 
 
 The 11th chapter contains solutions of the most useful 
 problems concerning the maximum ejects of machines in mo- 
 tion ; and developes those principles which should constantly 
 be kept in view by those who would labour beneficially for the 
 improvement of machines. 
 
 In the 1 2th chapter will be found the theory of the pres- 
 sure of earth and fluids against walls and fortifications ; and 
 the theory which leads to the best construction of powder 
 magazines with equilibrated roofs. 
 
 The 13th chapter is devoted to that highly interesting sub- 
 ject, as well to the philosopher as to mihtary men, the theory 
 and practice of gunnery. Many of the difficulties attending 
 this abstruse enquiry are surmounted by assuming the results 
 of accurate experiments, as to the resistance experienced by 
 bodies moving through the air, as the basis of the computa- 
 tions. Several of the most useful problems are solved by 
 means of this expedient, with a facility scarcely to be expect- 
 ed, and with an accuracy far beyond our most sanguine ex- 
 pectations. 
 
 The 14th and last chapter contains a promiscuous but ex- 
 tensive collection of problems in statics, dynamics, hydrosta- 
 tics, hydraulics, projectiles, &c. &c. ; serving at once to exer- 
 cise the pupil in the various branches of mathematics com- 
 prised in the course, to demonstrate their utility especially 
 to those devoted to the military profession, to excite a thirst 
 for knowledge, and in several important respects to gratify it. 
 
 This volume. being professedly supplementary to the pre- 
 ceding two volumes of the Course may best be used in tuition 
 by a kind of mutual incorporation of its contents with those 
 of the second volume. The method of effecting this will, of 
 course, vary according to eirctunstances, and the precise em- 
 ployments 
 
viii PREFACE. 
 
 ployments for which the pupils are destined : but in general 
 it is presumed the following may be advantageously adopted. 
 Let the first seven chapters be taught immediately after the 
 Conic Sections in the 2d volume. Then let the substance of 
 
 ^ the 2d volume succeed, as far as the Practical Exercises on 
 Natural Philosophy, inclusive. Let the 8th and 9th chapters 
 in this 3d vol. precede the treatise on Fluxions in the 2d ; and 
 when the pupil has been taught the part relating to fluents in 
 that treatise, let him immediately "he conducted through the 
 10th chapter of the 3d volume. After he has gone over the 
 remainder of the Fluxions with the applications to tangents, 
 Fadii of curvature, rectifications, quadratures, &c. the 1 1th and 
 12th chapters of the 3d vol. should be taught. The prob- 
 lems in the 13th' and 14th chapters must be blended with the 
 practical exercises at the end of the 2d volume, in such manner 
 as shall be found best suited to the capacity of the student, and 
 best calculated to ensure his thorough comprehension of the 
 several curious problems contained in those portions of the work. 
 In the composition of this 3d volume, as weH as in that of 
 the preceding parts of the Course, the great object kept con- 
 stantly in view has been utility, especially to gentlemen in- 
 tended for the Military Profession. To this end, all such in- 
 vestigations as might serve merely to display ingenuity or tal- 
 ent, without any regard to practical benefit, have been carefully 
 excluded. The student has put into his hands the two power- 
 ful instruments of the ancient and the modern or sublime ge- 
 ometry ; he is taught the use of both, and their relative ad- 
 vantages are so exhibited as to guard him, it is hoped, from 
 any undue and exclusive preference for either. Much novel- 
 ty of matter is not to be expected in a work like this ; though, 
 considering its magnitude, and the frequency with which sev- 
 eral of the subjects have been discussed, a candid reader will 
 
 ' not, perhaps, be entirely disappointed in this respect. Per- 
 spicuity and condensation have been uniformly aimed at through 
 the performance : and a small clear type, with a full page, have 
 been chosen for the introduction of a large quantity of matter. 
 A candid public will accept as an apology for any slight dis- 
 order or irregularity that may appear in the composition and 
 arrangement of this Course, the circumstance of the different 
 volumes having been prepared at widely distant times, and 
 with gradually expanding views. But, on the whole, I trust it 
 will be found that, with the assistance of my friend and coad- 
 jutor in this supplementary volume, I have now produced a 
 Course of Mathematics, in which a great variety of useful sub- 
 jects are introduced, and treated with perspicuity and correct-' 
 ness, thanin any three volumes of equal size in any language. 
 
 CHA. HUTXOK. 
 
PREFACE, 
 
 BY THE AMERICAN EDITOR. 
 
 1 HE last English edition of HuttorCs Course of Mathema- 
 tics, in three volumes octavo, may be considered as one of the 
 best systems of Mathematics in the English language. Its 
 great excellence consists in the judicious selection made by 
 the authors of the work, who have constantly aimed at such 
 things as are most necessary in the useful arts of life. To 
 this may be added the easy and perspicuous manner in which 
 the subject is treated — a quality of primary importance in a 
 treatise intended for beginners, and containing the elements 
 of science. 
 
 TheJh[r4js^luaie^^X--lhB^ EBgli»h..€^^^ 
 lately published, is scarcely known at present in this country- 
 ins but justice Itb its excellent authors to state, that they have 
 collected in it a great number of the most interesting sub- 
 jects in Analytical and Mechanical vScience. Analytical Tri- 
 gonometry Plane and Spherical, Trigonometrical Surveying, 
 Maxima and Minima of Geometrical Quantities, Motion of 
 Machines and their Maximum Effects, Practical Gunnery, 
 &c. are among the most important subjects in Mathematics, 
 and are disciissed in the volume just mentioned in such a 
 manner as not only to prove highly useful to pupils, but also 
 to such as are engaged in various departments of Practical 
 Science. 
 
 As the work, after the publication of the third volume, em- 
 braced most subjects of curiosity or utihty in Mathematics, 
 it has been thought unnecessary to enlarge its size by much 
 additional master. The present edition however, differs in 
 several respects from the last English one ; and it is presum- 
 ed, that this difference will be found to consist of improve- 
 ments. These are principally as follows : 
 
 In the first place, it was thought adviseable to pubhsh the 
 work in two volumes instead of three ; the two volumes being 
 still of a convenient size for the use of students. 
 
 Secondly, a new arrangement of various parts of the work 
 has been adopted Several parts of the third volume of the 
 English edition treated of subjects already discussed in the 
 preceding volumes ; in such cases, when it was practicable, 
 the additions in the third volume have been properly incor- 
 porated with the corresponding subjects that preceded them ; 
 
 Vol.1, } and. 
 
X PREFACE. 
 
 and, in general, such a disposition of the various departments 
 of the work has been made as seemed best calculated to pro- 
 mote the improvement of the pupil, and exhibit the respec- 
 tive places of the various branches in the scale of science. 
 
 And thirdly, several notes have been added ; and numerous 
 corrections have been made in various places of the work : 
 it were tedious and unnecessary to enumerate all these at pre- 
 sent ; it may suffice to remark the few following : 
 
 In pages 169, and 263, vol. 1, are given useful notes res- 
 pecting the degree of accuracy resulting from the application 
 of logarithms ; — these notes will appear the more necessary to 
 beginners, when we observe such oversights committed by 
 authors of experience. 
 
 In page 173, vol. 1. a new definition of surds is given, in- 
 stead of that by the author of the work. 
 
 In the Enghsh edition, a surd is defined to be " that which 
 has not an exact root." In Bonnycastle's Algebra, it is " that 
 which has no exact root." And in Emerson's Algebra, it is 
 *■' a quantity that has not a proper root." But notwithstand- 
 ing the weight of authority thus evidently against me, I do not 
 hesitate to assert, that the definition, just stated, is altogether 
 erroneous. According to their definition, the integer 2 is a 
 surd, for it " has not an exact root." 
 
 In the mensuration, page 411, vol. i, a remark is added re- 
 specting the magnitude of the earth. Dr. Hutton has com- 
 monly used a diameter of 7957f English miles, merely be- 
 cause it gives the round number 25,000 for the circumfe- 
 rence : in a few places he has used a diameter of 7930. Hav- 
 ing some years ago discovered the proper method of ascer- 
 taining the most probable magnitude and figure of the earth, 
 from the admeasurement of several degrees of the meridian, 
 I found the ratio of the axis to the equatorial diameter, to be 
 as 320 to 321, and the diameter, when the earth is considered 
 as a globe, to be 79187 English miles. 
 
 In the additions immediately preceding the Table of Loga- 
 rithms in the second volume, a new method is given for ascer- 
 taining the vibrations of a variable pendulum. This problem 
 was solved by Dr. Hutton, in his Select Exercises, 1787, and 
 he has given the same solution in the present work, see page 
 537, vol. 2. The method used by the Doctor appears to me 
 to be erroneous ; but in order that such as would judge for 
 themselves on this abstruse question, may have a fair oppor- 
 tunity of deciding between us, the Doctor's solution is given 
 as well as my own. 
 
 It may be proper to observe, with respect to the new solu- 
 tion, as well as Dr. Hutton's, that the resulting formula does 
 
 not 
 
PREFACE. 
 
 not shew the relation between the time and any number of 
 vibrations ac<wa% performed ; but merely gives the limit to 
 which this relation approaches, when the horizontal velocity 
 is indefinitely diminished. If therefore we would use the 
 new formula as slii approximation in very small finite vibra- 
 tions, the times must not be extended without limitation. 
 
 ROBERT ADRAIN 
 
 JVevf-Brunraick, JVew-Jergei/, 
 July 31, 1812. 
 
 CONTENTS 
 
CONTENTS 
 
 OF VOLUME L 
 
 GEJ^ERAL Pnncipkif .1 . . . . ^ . . i 
 
 ARITHMETIC. 
 
 JSTotation and Mtmeration 4, 
 
 Jtoman JVotation . .' 7 
 
 Addition . . . . 4 . . . . . . . g 
 
 Subtraotion n 
 
 Multiplication 13 
 
 Division » • . .. 18 
 
 jReduction . gS 
 
 Compound Addition . . . . . , , . , \ 33 
 
 >' . Subtraction . . . 36 
 
 — — Multiplication / . , . . 38 
 
 '■■■ Division , . . ,, , , ; . , 41 
 
 Golden Ruley or Rule of Three ....... 4^ 
 
 Compound Proportion . .' 49 
 
 Vulgar Fractions ........... 51 
 
 Reduction of Vulgar Fractions , . , . . . . 52 
 
 Addition of Vuigar Fractions ....... 61 
 
 Subtraction of Vulgar Fractions ^ 62 
 
 Multiplication of Vulgar Fractions ....... 63 
 
 Division of Vulgar Fractions ,64 
 
 Rule of Three in Vulgar Fractions . . . . . . 65 
 
 jpecimal Fractions . . . . . . , . . 66 
 
 Ad^tion of Decimals 67 
 
 l^ufttraction Qf Decimals 68 
 
 Multiplication 
 
CONTENTS. xiii 
 
 Page 
 
 Multiplication of Decimals , . . . ^ . • . 68 
 
 Division of Decimals 70 
 
 Reduction of Decimals 73 
 
 Ilule of Three in Decimals ........ 76 
 
 Duodecimals 77 
 
 Involution ........... 78 
 
 Evolution «0 
 
 To extract the Square Root . 81 
 
 To extract the Cube Root . . . . . . . '85 
 
 To extract any Root -whatever ....... 88 
 
 Table of Squares^ Cubes, Square-Roots and Cube-Roots . 90 
 
 Ratios, Proportions, and Progressions . . , . . ItO 
 
 Arithmetical Progression . . Ill 
 
 Geometrical Progression . 116 
 
 Musical Proportion . , . . r , . . . 119 
 
 Fello-wship, ot Partnership »1>. 
 
 Single Fellowship . ..' . , 120 
 
 Double Felloruship .,.....-. 122 
 
 Simple Interest 124 
 
 Compound Interest . . . - . . ,. . , 127 
 
 AUigation Medial ......... 129 
 
 AlUgation Alternate . . . - . . . . . 131 
 
 Single Position 13S 
 
 Double Position ...,.,.... 137 
 
 Permutations and Combinations ^ 140 
 
 Practical Questions . . . . . . ... 150 
 
 LOGARITHMS. 
 
 Of Logarithms 155 
 
 To compute Logarithms . 159 
 
 Description and use of the Table of Logarithms ... 163' 
 
 MultipUcatim 
 
xiY CONTENTS. 
 
 Page 
 
 Multiplication hy Logarithms ....... 1^7 
 
 Division by Logarithms .....««. 168 
 
 hwolutim by Logarithms . 169 
 
 Evolution by Logarithms 170 
 
 ALGEBRA. 
 
 JDefinitions and Jactation . • • • - . . . 171 
 
 .Addition . . . . 17S 
 
 Subtraction t . . . . 180 
 
 Multiplication 181 
 
 IHvision ... . . . . . . . . t^ 
 
 Jilgebraic Fractions 18S 
 
 Involution 199 
 
 Evolution . . . , . . . ' . . . 202 
 
 Surds '...,., 205 
 
 Infinite Series 213 
 
 Arithmetical Proportion . 218 
 
 PiUngofBaUs , . . . . , 223 
 
 Geometrical Proportion . 4.1^^ '. 1 •. , , 228 
 
 Simple Equations 230 
 
 Quadratic Equations . . . * • • « . . 24d 
 
 JResolution of Cubic and Sigher Equations . , . ♦, • 257 
 
 Simple Interest . . • • ,« • • • 266 
 
 Compound Interest . . ^ ^ • . 267 
 
 Annuities ... 270 
 
 GEOMETRY. 
 
 Definitions . 275 
 
 Axioms— IVieorems 281 
 
 Of Ratios and Proportions .% • • • • - * 319 
 
 Of Planes and SoHds^Dqfimtions . . . . . . SS^ 
 
 Problems . . .353 
 
 Applications 
 
CONTENTS. XV 
 
 Page 
 
 JippKcatioru of Mgebra to Geometry 369 
 
 Plane Trigonometry 377 
 
 Heights and Diatancea 39S 
 
 Mensuraium of Planes 402 
 
 Mensuration of Solids . . > 419 
 
 Land Surveying . • . - - - . . • . 429 
 
 Artijiceri Works . 487 
 
 Timber measuring 466 
 
 Conic Sections . . •' 469 
 
 Of the Ellipse 475 
 
 Of the Hyperbola .......... 49X 
 
 Of the Parabola .......... 514 
 
 Of the Conic Sections as expressed by Algebraic equations, called the 
 
 Equations of the Curve . . 532 
 
 Elements of Isoperimetry , 535 . 
 
 Problems relative to the Division of Surfaces • . . 558 
 
 Construction of Geometrical problems 571 
 
 Practical Exercises fn M^suration . . ... 575 
 
COURSE 
 
 MATHEMATICS, &c. 
 
 GENERAL PRINCIPLES. 
 
 1. Iq^UANTITY, or Magnitude, is any thing that will 
 admit of increase or decrease ; or that is capable of any sort 
 of calculation or mensuration : such as numbers, lines, space, 
 time, motion, weight. 
 
 2. Mathematics is the science which treats of all kinds 
 |of quantity 'whatever, that can be numbered or meaeured. — 
 
 That part which treats of numt)ering is called Arithmetic; 
 and that which concerns measuring, or figured extension, is 
 called Geometry. — These two, which are conversant about 
 multitude and magnitude, being the foundation of all the other 
 parts, arc called Pure or Abstract Mathematics ; because they 
 investigate and demonstrate the properties of abstract num- 
 bers and magnitudes of all sorts. And when these two parts 
 are applied to particular or practical subjects, they constitute 
 the branches or parts called Mixed Mathematics. — Mathematics 
 is also distinguished into Speculative and Practical : viz. Specu- 
 lative, when it is concerned in discovering properties and re- 
 lations ; and Practical, when applied to practice and real use 
 concerning physical objects. ^ 
 
 Vol. L 2 3. In 
 
2 ' GENERAL PRINCIPLES. 
 
 3. In Mathematics are several general terms or principles ; 
 such as, Definitions, Axioms, Propositions, Theorems, Prob- 
 lems, Lemmas, Corollaries, Sclioliums, &c. 
 
 4. A Definition is the explication of any term or word in a 
 science ; showing the sense and meaning in which the term 
 is employed. — Every Definition ought to be clear, and ex- 
 pressed in words that are common and perfectly well under- 
 stood. 
 
 6. A Proposition is something proposed to be proved, or 
 something required to be done ; and is accordingly either a 
 Theorem or a Problem. 
 
 6 A Theorem is a demonstrative proposition ; in which 
 some property is asserted, and the truth of il required to be 
 proved. Thus, when it is said that, The sum of the three 
 angles of any triangle is equal to two right angles, this is a 
 Theorem, the truth of which is demonstrated by Geometry. " 
 — A set or collection of «uch Theorems constitutes a Theory. 
 
 7. A Problem is a proposition or a question requiring 
 something to be done ; either to investigate some truth or 
 property, or to perform some operation. As, to find out the 
 quantity or sum of all the three anglps of any triangle, or to 
 draw one line perpendicular to another. A Limited Prob- 
 lem is that which has but one answer or soluiion. An Un- ' 
 limited Problem is that which has innumerable answers. 
 And a Determinate Problem is that which has a certain num- 
 ber of answers. 
 
 8. Solution of a Problem, is the resolution or answer given 
 to it. A Numerical or Numeral Solution, is the answer©, given 
 in numbers. A Geometrical Solution^ ii the answer given by 
 the principles of Geometry. And a Mechanical Solution, is 
 one which is gained by trials. 
 
 9. A Lemma is a preparatory proposition, laid down in 
 order to shorten the demonstration of the main proposition^ 
 which follows it. ^P 
 
 10. A Corollary, or 'Consectary, is a consequence drawn im- 
 mediately from some proposition or other premises. " 
 
 li. A Scholium is a remark or observation made on some 
 foregoing proposition or premises. - 
 
 12. An Axiom or Maxim^ is a self-evident proposition ; 
 requiring no formal demonstration to prove the truth of it ; 
 but is received and assented to as soon as mentioned. Such 
 as, The whole of any thing is greater than a part of it ; or, 
 The v/hole is equal to all its parts taken together : or. Two 
 quantities that are each of them equal to a third quantity, 
 are equal to each other. 
 
 13.^ 
 
9 GENERAL PRINCIPLES. 3 
 
 13. A Postulate, or Petition, is something required to be 
 done, which is so easy and evident that no person will hesi- 
 tate to allow it. 
 
 14. An Hypothesis is a supposition assumed to be true, in 
 order to argue from, or to found upon it the reasoning^and 
 demonstration of some proposition. 
 
 15. Demonstration is the collecting the several arguments 
 and proofs, and laying them together in proper order, to 
 show the truth of the proposition under consideration. 
 
 16. A Direct, Positive, or Jlffirmative Demonstration, is 
 that which concludes with the direct and certain proof of the 
 proposit^n in hand, — This kind of Demonstration is most 
 satisfacWPy to the mind ; for which- reason it is called some- 
 times an Ostensive Demonstration. 
 
 17. Jin Indirect, or Negative Demonstration, is that which 
 shows a proposition to be true, by proving that some absur- 
 dity would necessarily follow if the proposition advanced were 
 false. This is also sometimes called Reductio ad Ahsiirdum; 
 because it shows the absurdity and falsehood of all supposir 
 tions contrary to that contained in the proposition. 
 
 18. Method is the ait of disposing a train of arguoients in 
 a proper order, to investigate eithsr the truth or falsity of a 
 proposition, or to demonstrate it to others when it has been 
 fcund out. — This is either Analytical or Synthetical. 
 
 19. Analysis, or the Analytic Method, is the art or mode 
 of finding out the truth of a 'proposition, by first supposing 
 the thing to be done, and then reasoning back, step by step, 
 till we arrive at some known truth. — This is also called the 
 Method of Invention, or Resolution. 
 
 20. Synthesis, or • the Synthetic Method, is the searching 
 out truth, by first laying down some simple and easy princi- 
 ples, and pursuing the consequences flowing from them till 
 we arrive at the conclusion. — This is also called the Method 
 of Composition ; and is the reverse of the Anal3^tic method, as 
 this proceeds from known principles to an unknown conclu- 
 sion ; while the other goes in a retrograde order, from the 
 thing sought, considered as if it were true, to some known 
 principle or fact. And therefore, when any truth has been 
 found out by the Analytic method, it may be demonstrated 
 by a process in the contrary order, by Synthesis. 
 
 ARITH- 
 
[4] 
 
 Arithmetic. 
 
 jfl^RITHMETIC is the art or science of numbering; be- 
 ing that branch of Mathematics which treats of the nature 
 and properties of numbers. — ^Wheu it treats of whole num- 
 bers, it is called Vulgar, or Common Arithmetic ; but when of 
 broken numbers, or parts of numbers, it is called FVactions. 
 
 Unity, or an Unit^ is that by which every thinglp called 
 one ; being the beginning of number ; as, one man, one ball, 
 one gun, 
 
 Number is either simply one; or a compound of several 
 units ; as, one man, three men, ten men. 
 
 An Integer^ or Whole Number, is some certain precise 
 quantity of units ; as, one, three, ten. — These are so called as 
 distinguished from Fractions, which are broken numbers, or* 
 parts of numbers ; as, one-half, two-thirds, or three-fourths. 
 
 NOTATION AND NUMERATION. 
 
 Notation, or Numeration, teaches to denote or express 
 any proposed number, either by words or characters ; or to 
 read and write down any sum or number 
 
 The numbers in Arithmetic are expressed by the following 
 ten digits, or Arabic numeral figures, which were introduced 
 into Europe by the Moors, about eight or nine hundred 
 years since ; viz. 1 one, 2 two, 3 three, 4 four, 5 five, 6 six, 
 7 seven, 8 eight, 9 nine, cipher, or nothing. These cha-^ 
 racters or figures were formerly all called by the general 
 name of Ciphers; whence it came to pass that the art of 
 Arithmetic was then often called Ciphering. Also the first 
 nine are called Significant Figures, as distinguished from the 
 cipher, which is of itself quite insignificant. 
 
 Besides this value of thosp figures, they have also another 
 which depends on the place they stand in when joined toge- 
 ther ; as in the following table : 
 
 Units 
 
NOTATION AND NUMERATION. 
 
 
 ^ 
 
 a 
 o 
 
 
 H 
 
 ^ 
 
 
 
 
 
 
 o 
 
 i 
 
 o 
 
 03 
 
 o 
 
 
 O 
 m 
 
 1 
 
 9 
 
 n3 
 
 
 'TO 
 
 
 a 
 
 c 
 
 
 a 
 
 C 
 
 O 
 
 a 
 
 a 
 
 
 u 
 
 9 
 
 <v 
 
 
 a 
 
 V 
 
 ja 
 
 s 
 
 4) 
 
 *2 
 
 -a 
 
 S 
 
 H 
 
 ^ 
 
 K 
 
 H 
 
 H 
 
 S 
 
 H 
 
 t) 
 
 &C. 
 
 9 
 
 8 
 
 7 
 
 6 
 
 6 
 
 4 
 
 3 
 
 2 
 
 1 
 
 
 
 9 
 
 8 
 
 7 
 
 6 
 
 6 
 
 4 
 
 3 
 
 2 
 
 
 
 
 9 
 
 8 
 
 7 
 
 6 
 
 5 
 
 4 
 
 3 
 
 
 
 
 
 9 
 
 8 
 
 7 
 
 6 
 
 6 
 
 4 
 
 
 
 
 
 
 9 
 
 8 
 
 9 
 
 7 
 8 
 9 
 
 6 
 7 
 8 
 9 
 
 6 
 
 6 
 7 
 8 
 9 
 
 Here any figure in the fir«t place, reckoning from right to 
 left, denotes only its own simple value ; but that in the 
 second place, denotes ten times its simple value ; and that in 
 the third place, a hundred times its simple value ; and so on : 
 the value of any figure, in each successive place being always 
 ten times its former value. 
 
 Thus, in the number 1796, the 6 in the first place denotes 
 only six units, or simply six ; 9 in the second place signifies 
 nine tens, or ninety ; 7 in the third place, seven hundred ; 
 and the 1 in the fourth' place, one thousand : so that the 
 whole number is read thus, one thousand seven hundred and 
 ninety-six. 
 
 As to the cipher, 0, though it signify nothing of itself, yet 
 being joined on the right-hand side to other figures, it in- 
 creases their value in the same ten-fold proportion ; thus, 5 , 
 signifies only five ; but 60 denotes 5 tens, or fifty ; and 600 is 
 five hundred ; and so on. 
 
 For the more easily reading of large numbers, they are 
 divided into periods and half-periods, each half-period con- 
 sisting of three figures ; the name of the first period being 
 units ; of the second, millions ; of the third, milUons of 
 millions, or bi-millions, contracted to biUions : of the fourth, 
 millions of millions of millions, or tri-roillions, contracted to 
 trillions, and so on. Also the first part of any period is so 
 many units of it, and the latter part so many thousands. 
 
 The 
 
3 ARITHMETIC. 
 
 The following Table contains a summary of the whole 
 doctrine. 
 
 Periods. 
 
 Quadrill ; Trillions: Billions ; Millions ; Units 
 
 Half-per. 
 
 th. un. th. un. th. un. th. un. th. un. 
 
 Figures. 
 
 123,456 ; 789,098 ; 765,432 ; 101,234 : 567,«90. 
 
 NuMBRATioN is the reading of any number in words 
 that is proposed or set down in figures ; which will be easily 
 done by help of the following rule, deduced from the fore- 
 going tablets and observations — viz. 
 
 Divide the figures in the proposed number, as in the sum- 
 mary above, into periods and half periods ; then begin at the 
 left-hand side, and read the figures with the names set to 
 them in the two foregoing tables. 
 
 EXAMPLES. 
 
 Express in words the following numbers ; viz. 
 
 34 15080 13405670 
 
 96 72003 47050023 
 
 180 109026 309025600 
 
 304 483500 4723507689 
 
 6134 2500639 274866390000 
 
 9028 7523000 6578600307024 
 
 Notation, is the setting down in figures any number pro- 
 posed in words ; which is done by setting down the figures 
 instead of the words or names belonging to them in the sum- 
 mary above ; supplying the vacant places with ciphers where 
 any words do not occur. 
 
 EXAMPLES. 
 
 Set down in figures the following numbers j 
 Fifty-seven. 
 
 Two hundred eighty six. 
 Nine thousand two hundred and ten. 
 Twenty-seven thousand five hundred and ninety-four. 
 Six hundred and forty thousand, four hundred and eighty one. 
 Three millions, two hundred sixty thousand, one hundred 
 
 and six. 
 
 Four 
 
NOTATION AND NUMERATION, 
 
 Four hundred and eight millions, two hundred and fifty-five 
 thousand, one hundred and ninety-two. 
 
 Twenty- seven thousand and eight millions, ninety «8ix thou- 
 sand two hundred and four. 
 
 Two hundred thousand and fire hundred and fifty millions, 
 one hundred and ten thousand, and sixteen. 
 
 Twenty-one billions, eight hundred and ten millions, sixty- 
 four thousand, one hundred and fifty. 
 
 Of the Roman Notation. 
 
 The Romans, like several other nations, expressed their 
 numbers by certain letters of the alphabet. The Romans 
 used only seven numeral letters, being the seven following 
 capitals : viz. I for one ; V for ^^ve ; X for ten ; L for Jifty ; 
 C for an hundred ; D for Jive hundred ; M for a thousand. 
 The other numbers they expressed by various repetitions and 
 combinations of these, after the following manner : 
 1=1 
 
 2 = II As often as any character is re- 
 
 3 = III peated, so many times is its 
 
 value repeated. 
 
 4 = nil or IV A less character before a great- 
 fj = V er diminishes its value. 
 
 6 = VI A less character after a greater 
 
 7 = VII increases its value. 
 - 8 = VIII 
 
 9 = IX 
 
 10 = x 
 
 60 = L 
 100 = C 
 500 = D or 10 
 
 1000 = M or CID 
 2000 = MM 
 
 5000 = V_or lOD 
 6000 = VI 
 10000 = X or CCIOD 
 50000 = Lor 1030 
 60000 = LX 
 
 100000 = 6 or CCCIOOO 
 1000000 = M or CCCCIOOOO 
 2000000 = MM 
 &c. kc. 
 
 For every annexed, this be- 
 comes 10 times as ma^y. 
 
 For every C and 0, placed one 
 at each end, it becomes 10 
 Jimes as much. 
 
 A bar over any number in- 
 creases it 1000 fold. 
 
 BtfLxr- 
 
ARITHMETIC. 
 
 Explanation op certain Characters. 
 
 There are various characters or marks used in Arithmetic, 
 and Algebra, to denote several of the operations and proposi- 
 tions ; the chief of which are as follows : 
 
 -f- signifies plus, or addition. 
 — - - minus, or subtraction, 
 X or . - multipUcation. 
 -f- - - division. 
 : : : : - proportion. 
 = - - equality. 
 ^ - ' square root. 
 ^ - - cube root, &c. 
 
 CO - - diff. between two numbers when it is not known 
 which is the greater. 
 
 Thus, 
 
 6 4-3, denotes that 3 is to be added to 5. 
 
 6 — 2, denotes that 2 is to be taken from 6. 
 
 7 X 3, or 7 . 3, denotes that 7 is to be multiplied by 3.' 
 
 8 -T- 4, denotes that 8 is to be divided by 4. 
 2 : 3 : : 4 : 6, shows that 2 is to 3 as 4 is to 6. 
 
 6 -I- 4 = 10, shows that the sum of 6 and 4 is equal to 10. 
 
 ,J 3, or 32 , denotes the square root of the number 3. 
 
 ^ 6, or 5^, denotes the cube root of the number 5. 
 7s , denotes that the number 7 is to be squared. 
 8 3, denotes that the number 8 is to be cubed. 
 &c. 
 
 OP ADDITION. 
 
 Addition is the collecting or putting of several numberi 
 together, in order to find their sum, or the total amount of the 
 whole. This is done as follows : 
 
 Set or place the numbers under each other, so that each 
 figure may stand exactly under the figures ef the same value, 
 
 that 
 
ADDITION. 9 
 
 that is, units under units, tens under tens, hundreds under 
 hundreds, &c. and draw a line under the lowest number, to 
 separate the given numbers from their sum, when it is found. 
 — Then add up the figures in the column or row of units, 
 and find how many tens are contained in that sum. — Set 
 down exactly below what remains more than those tens, or 
 if nothing remains, a cipher, and carry as many ones to the 
 next row as there are tens. — Next add up the second row, 
 together with the number carried, in the same manner as the 
 first. And thus proceed till "the whole is finished, setting 
 down the total amount of the last row. 
 
 , To PROVE Addition. 
 
 First Method. — Begin at the top, and add together all the 
 rows of numbers downwards ; in the same manner as they 
 were before added upwards ; then if the two sums agree, it 
 may be presumed the work is right. — This method of proof 
 is only doing the same work twice over, a little varied. 
 
 Second Method. — Draw a line below the uppermost number, 
 and suppose it cut oflf.-^Then add all the rest of the numbers 
 together in the usual way, and set their sum under the num- 
 ber to be proved. — Lastly, add this last found number 
 and the uppermost line together ,* then if their sum be the 
 same as that found by the first addition, it may be presumed 
 the work is right. — This method of proof is founded on the 
 plain axiom, that *' The whole is equal to all its parts taken 
 together." 
 
 Third Method. — Add the figures in 
 the uppermost line together, and find 
 how many nines are contained in 
 their sum. — Reject those nines, and 
 set down the remainder towards the 
 right-hand directly even with the 
 figures in the line, as in the annexed 
 example. — Do the same with each 
 of the proposedHnes of numbers, set- 
 ting all these excesses of nines in a co- W 
 lumn on the right-hand, as here 5, 5, 6. Then, if the excess 
 of 9's in this sum, found as before, be equal to the excess 
 of 9's in the total sum 18304, the work is probably right. — 
 Thus, the sum of the right-hand column, 5, 5, 6, is 16, the 
 excess of which above 9 is 7. Also the sum of the figures in 
 
 Vol. L 3 the 
 
 EXAMPLE I, 
 
 
 3497 
 
 S 
 
 5 
 
 6512 
 
 a 
 
 6 
 
 8293 
 
 "a 
 
 6 
 
 18304 
 
 7 
 
 
 O 
 
 
10 
 
 ARITHMETIC. 
 
 the sum total 18304, is 16, the excess of which above 9 is 
 also 7, the same as the former.* . 
 
 OTHER EXAMPLES. 
 
 2. 
 12345 
 
 3. 
 12345 
 
 67890 
 
 9876 
 
 543 
 
 21 
 
 9 
 
 4. 
 12345 
 
 67890 
 98765 
 43210 
 12345 
 67890 
 
 876 
 
 9087 
 
 66 
 
 234 
 1012 
 
 302445 
 
 90684 
 
 23610 
 
 290100 
 
 78339 
 
 11265 " 
 
 302445 ^ 
 
 90684 
 
 23610 
 
 * This method of proof depends on a property of the number 9, 
 ■which except the number 3, belongs to no other digit whatever j 
 namely, that *' any number divided by 9, nvUI leave the same remainder 
 as the sum of its figures or digits divided by 9 :'» -which may be 
 demonstrated in this manner. 
 
 ' Demonstration, Let there be any number proposed, a» 4658. This, 
 separated into its several parts, becomes 4000 •+ 600 -f.50 -^ 8. But 
 4000 = 4 X 1000 =- 4 X (999 + 1) = 4 X 999 + 4. In Uke man- 
 ner 600 = 6 X 99 + 6 i and 50 == 6 X 9 + 5. . herefore the given 
 number 4658 = 4x999-|-4+6x99 + 6 + 5x9 + 5-f. 3'=4 
 X 999 + 6X 99 +5x9+4 + 6+5 + 8; and 4658 -r- 9 = (4 x 
 999 4- 6 X 99 + 5 X 9 + 4 + 6 + 5 + 8) +• 9. But 4 x ^99 + 6 
 X 99 + 5 X 9 is evidently divisible by 9, without a remainder j there- 
 fore if the given number 4658 be divided by nine, it will leave the 
 same remainder as 4 + 6+5 + 8 divided by 9* And the same, it is 
 evident, will htld for any other number whatever. 
 
 In like manner, the san>e property may be shown to belong to the 
 number 3 ; but the preference is usually given lo the number 9, on 
 account of its being more convenient in practice. 
 
 Now, from the demonstration above given, the reason of the rule 
 itself is evident : for the excess of 9*s in two or more number^ being 
 taken separattly, and the excess of 9*s taken also out of the sum of 
 the former excesses, it is plain that this last excess must be equal to 
 the excess of 9's contained in the total sum of all these numbers ; all 
 the parts taken together being equal to the whole.— —This rule 
 was first given by Docter WalUs in his Arithmetici pubUshed in the 
 year 1657. 
 
 Ex- 
 
SUBTRACTION. 11 
 
 Ex. 5. Add 3426 ; 9024 ; 6106 ; 8890 ; 1204, together. 
 
 Ans. 27650. 
 
 6. Add 609267 ; 236809 ; 72920 ; 83^2 ; 420 ; 21 ; and 9, 
 together. Ans 826838. 
 
 7. Add 2 ; 19 ; 817 ; 4298 ; 60916 ; 730206; 91^>0634, 
 together. Ans. 9966891. 
 
 8. Hdw many days are in the twelve calendar months ? 
 
 Ans. 365. 
 
 9. How many days are there from the 1 5th day of April to 
 the 24th day of November, both days included ? Ans 224. 
 
 10. An army consisting of 62714 infantry*, or foot, 6110 
 horse, 6250 dragoons, 3927 light-horse, 928 artillery, or 
 gunners, 1410 pioneers, 250 sappers, and 406 miners : what 
 is the whole number of men ? Ans. 70995. 
 
 OF SUBTRACTION. 
 
 Subtraction teaches to find how much one number 
 exceeds another, called their difference, or the remainder^ by 
 taking the less from the greater. The method of doing which 
 is as follows : 
 
 Place the less number under the greater, in the same man- 
 ner as in Addition, that is, units under units, tens under tens, 
 and so on ; and" draw a line below them. — Begin at the right- 
 hand and take each figure in the lower line, or number, from 
 the figure above it, setting down the remainder below it. — 
 But if the figure in the lower line be greater than that above 
 it, first borrow, or add, 10 to the upper one, and then take 
 the lower figure from that sum, setting down the remainder^ 
 and carrying 1 , for what was borrowed, to the next lower 
 figure, with which proceed as before ; and so on till the 
 whole is finished. 
 
 * The whole body of foot soldiers is denoted by the word Infantry ; 
 and all those that charge on horseback by the word Cavalry. — Some 
 authors conjecture tHat the term infantry is derived from a certain In- 
 fanta of Spain, who finding that the ariry commanded by the k'mg her 
 father had been defeated by the Moors, assembled a body of the peo- 
 ple together on foot, with which she engaged and totally routed the 
 enemy. In honour of this event, and to distinguish the foot soldiers, 
 who were not before held in much estimation, they received the name 
 of Infantrv, from her own title of Infanta. 
 
 To 
 
12 ARITHMETIC. 
 
 i 
 
 To FROVE Subtraction. 
 
 Add the remainder to the less number, or that which is 
 just above it ; and if the sum be equal to the greater or upper- 
 most number, the work is right*. 
 
 EXAMPLES. 
 
 From 5386427 
 Take 2164315 
 
 From 5386427 
 Take 4258792 
 
 Prom 1234567 
 Take 702973 
 
 Rem. 3222112 
 
 Rem. 1127635 
 
 Rem. 531594 
 
 Proof. 5386427 
 
 Proof. 5386427 
 
 Proof. 1234567 
 
 4. From 6331896 take 5073918. 
 6. From 7020974 take 2766809. 
 6. From 8503602 take 574271. 
 
 Ads. 25 
 Ans. 4254165. 
 Ans. 7929131. 
 
 7. Sir Isaac Newton was born in the year 1642, and he 
 clied in 1727 : how old was he at the time of his decease ? 
 
 Ans. 85 years. 
 
 8. Homer was born 2543 years ago, and Christ 1810 years 
 ago : then how long before Christ was the birth of Homer ? 
 
 Ans. 733 years. 
 
 9. , Noah's flood happened about the year of the world 1656, 
 and the birth of Christ about the year 4000 : then how long 
 was the flood before Christ ? Ans. 2344 years. 
 
 10. The Arabian or Indian method of notation was first 
 known in England about the year 1150: then how long is 
 it since to this present year 1810 ? Ans. 660 years. 
 
 11. Gunpowder was invented in the year 1330 : then how 
 long was this before the invention of printing, which was 
 in 1441 ? Ans. Ill years. 
 
 12. The mariner's compass was invented in Europe in the 
 year 1302 : then how long was that before the discovery of 
 America by Columbus, which happened in 1492 ? 
 
 Ans. 190 years. 
 
 * The reason of tlii? method of proof is evident ; for if the differ- 
 ence of two nunrjbers be added to the less, it must manifestly make 
 up a sum equal to the greater. 
 
 OF 
 
MULTIPLICATION. 
 
 l.*^ 
 
 OF MULTIPLICATION. 
 
 Multiplication is a compendious method of Addition, 
 teaching how to find the amount of any given number when 
 repeated a certain number of times; as, 4 times 6, which 
 is 24. 
 
 The number to be multiplied, or repeated, is called the 
 
 Multiplicand. — The number you multiply by, or the number of 
 
 ^^repetitions, is the Multiplier. — -And the number found, being 
 
 ^ the total amount, is called the Product. — Also, both the 
 
 multiplier and multiplicand are, in general, name the Terms 
 
 or Factors. 
 
 Before proceeding to any operations in this rule, it is ne- 
 cessary to learn off very perfectly the following Table, of all 
 the products of the first 12 numbers, commonly called the 
 Multiplication Table,, or sometimes Pythagoras's Table, from 
 its inventor. 
 
 Multiplication Table. 
 
 1 
 
 2 
 3 
 4 
 5 
 6 
 ~ 
 8 
 9 
 
 10 
 
 11 
 
 12 
 
 2 
 
 4 
 
 6 
 
 8 
 
 10 
 
 12 
 
 14 
 
 16 
 
 18 
 
 20 
 
 22 
 
 24 
 
 3 
 6 
 9 
 12 
 15 
 18 
 21 
 24 
 
 4 
 8 
 12 
 16. 
 20 
 24 
 28 
 32 
 36 
 40 
 44 
 48 
 
 10 
 15 
 20 
 25 
 30 
 35 
 40 
 45 
 50 
 55 
 60 
 
 6 
 12 
 18 
 24 
 30 
 36 
 42 
 48 
 54 
 60 
 66 
 72 
 
 7 
 14 
 21 
 28 
 35 
 42 
 49 
 56 
 63 
 70 
 77 
 84 
 
 8 
 16 
 24 
 32 
 40 
 48 
 56 
 64 
 72 
 80 
 S8 
 96 
 
 9 
 18 
 27 
 36 
 45 
 54 
 63 
 72 
 81 
 90 
 99 
 108 
 
 10 
 20 
 30 
 40 
 50 
 60 
 70 
 80 
 90 
 100 
 110 
 120 
 
 11 
 
 22 
 33 
 44 
 5.5 
 66 
 77 
 88 
 99 
 110 
 121 
 132 
 
 12 
 24 
 ^6 
 48 
 60 
 72 
 84 
 96 
 108 
 120 
 
 27 
 30 
 33 
 36 
 
 132 
 
 144 
 
 To 
 
14 ARITHMETIC. 
 
 To multiply any Given Number by a Single Figure^ or by any 
 Number not more than 12. 
 
 * Set the multiplier under the units figure, or right-hand 
 place, of the multiplicand, and draw a line below it. — T hen 
 beginning at. the right hand, multiply every figure in this by 
 the multiplier. — Count how many tens there are in the pro- 
 duct of every single figure, and set ilown the remainder di- 
 rectly under the figure that is multiplied ; and if nothing 
 remains, set down a cipher. — Carry as many units or ones as 
 there are tens couuted, to the product of the next figures ; 
 and proceed in the same manner till the whole is finished. 
 
 EXAMPLE. 
 
 Multiply 9876543210 the Multiplicand. 
 By - - - - 2 the Multiplier. 
 
 19753086420 the Product. 
 
 To multiply by a Number consisting of Several Figures. 
 
 t Set the multiplier below the multiplicand, placing them 
 as in Addition, namely, units under units, tens under tens, &c. 
 drawing a line below it. — Multiply the whole of the multi- 
 plicand by each figure of the multiplier, as in the last article; 
 
 setting 
 
 ♦ The reason of this rule is the same as for 
 the process in Addition, in which 1 is car- 
 ried for every 10, to the next place, gra- 
 dually as the several products are produced 
 one after another, instead of setting them 
 all down one below each other, as in the an- 
 nexed example. 
 
 f After having found the produce of the multiplicand by the first 
 figure of the multiplier, as in the former case, the multiplier is sup- 
 posed to be divided into parts, and the product is found for the second 
 figure in the same manner : but as this figure stands in the place of 
 tens, the product must be ten times its simple v.due ; and therefore the 
 first figure of this product roust be set in the place of tens j or, which 
 
 5678 
 
 
 
 
 4 
 
 
 
 
 32 = 
 
 8 
 
 X 
 
 4 
 
 280 = 
 
 70 
 
 X 
 
 4 
 
 2400 = 
 
 600 
 
 X 
 
 4 
 
 20000 = 
 
 5000 
 
 X 
 
 4 
 
 22712 = 
 
 5678 
 
 X 
 
 4 
 
MULTIPLICATION. 16 
 
 setting down aline of products for each figure in the multi- 
 plier, so as that the first figure of each line may stand straight 
 under the figure multiplying by. — Add all the lines of pro- 
 ducts together, in the order as they stand, and their sum will 
 be the answer or whole product required. 
 
 To PROVE Multiplication. 
 
 There are three different ways of proving Multiplication, 
 which are as below : 
 
 First Method. — Make the multiplicand and multiplier 
 change places, and multiply the latter by the former in the 
 same manner as before. Then if the product found in this 
 way be the same as the former, the number is right. 
 
 Second Method. ■^■*CdiSt all the 9's out of the sum of the 
 figures in each of the two factors, as in Addition, and set 
 down the remainders. Multiply these two remainders 
 together, and cast the 9's out of the product, as also out of 
 
 is the same thing, directly under the figure multiplied by. And pro- 
 ceeding in this manner sepa 
 rately with all the figures of ' 
 
 the multiplier, it is evident 123456r the multiplicand, 
 that we shall multiply all the 4567 
 
 parts of the multiplicand by — — 
 
 all the parts of the multi- 8641969= 7 times the mult, 
 plier, or the whole of the ' 7407402 = 60 times ditto, 
 multiplicand by the whole 6172835 == 500 times ditto- 
 of the multiplier ; therefore 493B268 =4000 times ditto. 
 
 these several products bemg . ^ i. 
 
 added together, will be equal 563826:489 =4567 times ditto, 
 to the whole required pro- i. 
 duct ; as in the example an- 
 nexed. 
 
 * This method of proof is derived fi-om the peculiar property of the 
 number 9, mentioned in the proof of Addition, and the reason for the 
 one may serve for that of the other. Another more ample demonstra- 
 tion of thts rule may be as follows :— Let P and (^denote the number 
 of 9's in the factors to be multiplied, and a and b what remain ; then 9 
 P 4- a and 9 Q_"f 6 will be the numbers themselves, and their product is 
 (9 P X 9 Q) + (9 i" X 6) + (9 ax a) + (a X A), but the first three 
 of these products are each a precise number of 9's, because their fac- 
 tors are so, either one or both : these therefore being cast away, there 
 remains only aX ^ ; and if the 9's also be cast out of this, the excess 
 is the excess of 9's in the total product : but a and b are the ex- 
 cesses in tlie factors themselves, and a X b is their product ; therefore 
 the rule is true. 
 
 the 
 
16 
 
 ARITHMETIC. 
 
 the whole product or answer of the question, reserving the 
 remainders of these last two, which remainders must be equal 
 when the work is right. — Note^ It is common to set the foui-^ 
 remainders within the four angular spaces of a cross, as in the 
 example below. 
 
 Third Method. — Multiplication is also very naturally 
 proved by Division ; for the product divided by either of the 
 factors, will evidently give the other. But this cannot be 
 practised till the rule of Division is learned. 
 
 Mult. 3542 
 by 6196 
 
 EXAMPLES. 
 
 Proof. 
 
 X 
 
 or Mult. 6196 
 by 3542 
 
 21252 
 31878 
 3542 
 2125^ 
 
 12392 
 24784 
 30980 
 18588 
 
 21946232 Product. 
 
 21946232 Proof. 
 
 OTHER EXAMPLES. 
 
 Multiply 123456789 by 3. 
 
 Ans. 
 
 370370367. 
 
 Multiply 123456789 by 4. 
 
 Ans. 
 
 493827156. 
 
 Multiply 123456789 by 5. 
 
 Ans. 
 
 617283945. 
 
 Multiply 123456789 by 6. 
 
 Ans. 
 
 740740734. 
 
 Multiply 123456789 by 7. 
 
 Ans. 
 
 864197523. 
 
 Multiply 123456789 by 8. 
 
 Ans. 
 
 987654312. 
 
 Multiply 123456789 by 9. 
 
 Ans. 
 
 llllIlllOl. 
 
 Multiply 123456789 by 11. 
 
 Ans. 
 
 1358024679. 
 
 Multiply 123456789 by 12. 
 
 Ans. 
 
 1481481468. 
 
 Multiply 302914603 by 16. 
 
 Ans. 
 
 4846633648. 
 
 Multiply 273580961 by 23. 
 
 Ans. 
 
 6292362103. 
 
 Multiply 402097316 by 195. 
 
 Ans. 
 
 78408976620. 
 
 Multiply 82164973 by 3027. 
 
 Ans. 
 
 248713373271. 
 
 Multiply 7564900 by 579. 
 
 Ans. 
 
 4380077100. 
 
 Multiply 8496427 by 874359. 
 
 Ans. 
 
 7428927415293. 
 
 Multiply 2760325 by 37072. 
 
 Ans. 
 
 102330768400. 
 
 CONTRAC- 
 
MULTIPLICATION. 17 
 
 GONTBACTIONS IN MULTIPUCATION. 
 
 L When there are Ciphers in the Factors, 
 
 If the ciphers be at the right-hand of the numbers ; mul- 
 tiply the other figures only, and annex as many ciphers to 
 the right' hand of the whole product, as are in both the fac- 
 tors. — When the ciphers are in the middle parts of the mul- 
 tipher ; neglect them as before, only taking care to place 
 the first figure of every line of products exactly under the 
 figure multiplying with. 
 
 EXAMPLES. 
 
 1. 2. 
 
 Mult. 9001635 Mult. 390720400 
 
 by - gOlOO by - 406000 
 
 9001635 23443224 
 
 63011445 166^:8816 
 
 631014613500 Products 158632482400000 
 
 3. Multiply 81503600 by 7030. Ans. 572970308000. 
 
 4. Multiply 9030100 by 2100. Ans. 189632»0000 
 6. Multiply a057069 by 70050. Ans. 564397683450. 
 
 II. When the multiplier is the Product of two or more Numbers 
 in the Table : then 
 
 * Multiply by each of those parts separately, instead of the 
 whole number at once. 
 
 EXAMPLES, 
 
 1. Multiply 51307298 by 56, or 7 times 8. 
 51307298 
 
 7 
 
 359151086 
 8 
 
 2873208688 
 
 ♦ Tlie reason of this rule is obvious enough ; for any number mul- 
 tiplied by the component parts of another, must give the same pro- 
 duct as if it were multipUed by that number at once. Thus, in the 
 1st example, 7 times the product of 8 by the given number, makes 56 
 times the same number, as plainly as 7 times 8 makes 56. 
 
 Vol. I. , 4 2. Mul- 
 
IB ARITHMETIC* 
 
 2. Multiply 31704592 by 36. Ana. 1141365312. 
 
 3. Multiply 29753804 by 72. Ans. 2142273888. 
 
 4. Multiply 7128368 by 96. Ans. 684323328. 
 6. Multiply 160430800 by 108. Ans. 17326526400. 
 
 6. Multiply 61835720 by 1320. Ans. 81623150400. 
 
 7. There was an army composed of 104 * battalions, eacb 
 consisting of 500 men ; what was the number of men con- 
 tained iR the whole ? Ans. 52000. 
 
 8. A convoy of ammunition f bread, consisting of 250 
 waggons, and each waggon containing 320 loaves, having 
 been intercepted and taken by the enemy ; what is the num- 
 ber of loaves lost ? Ans. 80000. 
 
 OF DIVISION. 
 
 .• 
 
 Division is a kind of compendious method of Subtrac- 
 tion, teaching to find ho\V often one number is contained ia 
 another, or may be taken out of it : which is the same thing. 
 
 The number to be divided is called the Dividend, — 
 The number to divide by, is the Divisor. — And the number 
 of times the dividend contains the divisor, is called the Quo^ 
 tient. — Sometimes there is a Remainder left, after the divi- 
 sion is finished. 
 
 The usual manner of placing the terms, is the dividend in 
 the middle, having the divisor on the left baud, and the quo- 
 tient on the right, each separated by a curve line ; as, td 
 divide 12 by 4, the quotient is 3, 
 
 Dividend 12 
 
 Divisor 4) 12 (3 Quotient ; 4 subtr. 
 
 showing that the number 4 is 3 times — 
 contained in 12, or may be 3 times 8 
 
 subtracted out of it, as in the margin. 4 subtr. 
 
 J Rule — Having placed the divisor — 
 before the dividend, as above direct- 4 
 
 ed, find how often the divisor is con- 4 subtr. 
 
 tained in as many figures of the divi- — 
 dend as are just necessary, and place the . 
 
 number on the right in the quotient. — 
 
 Mul- 
 
 * A battalion is a body of foot, consisting of 500, or 600, or 700 men, 
 more or less. 
 
 \ The ammunition bread, is that which is provided for, and distri- 
 buted to, the soldiers j the usual allowance being a loaf of 6 pounds to 
 every soldier, Once in 4 da)'s. 
 
 * In this way the dividend is resolved into parts, and by trial is 
 
 found 
 
DIVISION. 1» 
 
 Multiply the divisor by this number, and set the product 
 under the figures of the dividend before-mentioned. — Sub- 
 tract this product from that part of the dividend under which 
 it stands, and bring down the next figure of the dividend, or 
 more if necessary, to join on the right of the remainder — Di- 
 vide this number, so increased, in the same manner as before ; 
 and so on till all the figures are brought down and used. 
 
 JV. B. If it be necessary to bring down more figures than 
 one to any remainder, in order to make it as large as the 
 divisor, or larger, a cipher must be set in the quotient for 
 every figure so brought down more than one. 
 
 TO PROVE DIVISION. 
 
 4 
 
 * Multiply the quotient by the divisor ; to this product 
 add the remainder, if there be any ; then the sum will be 
 equal to the dividend when the work is right. 
 
 found how often the divisor is contained in each of those parts, one 
 after another, arranging the several figures of the quotient one after 
 another, into one number. 
 
 When there is no remainder to a division, the quotient is the whole 
 and perfect answer to the question. But when there is a remainder, 
 it goes so much towards another time, as it approaches to the divisor ; 
 so, if the remainder be half thfe divisor, it will go the half of a time 
 more ; if the 4th part of the divisor, it will go one fourth of a time 
 more ; and so on. Therefore, to complete the quotient, set the re- 
 mainder at the end of it, above a small line, and the divisor below it 
 thus forming a fractional part of the whole quotient. 
 
 ♦This method of proof is plain enough: for since the quotient 
 is the number of times the dividend contains the divisor, the quo- 
 tient multiplied by the divisor must evidently be equal to the divi- 
 dend. 
 
 There are also several other methods sometimes used for proving 
 Division, some of the most useful of which are as follow : 
 
 Second ikferAorf— Subtract the remainder from the dividend ; and 
 divide what is left by the quotient ; so shall the new quotient from 
 this last division be equal to the former divisor, when the work is 
 right. 
 
 Third Method— '\ddtogeiher the remainder and, all the products of 
 the several quotient figures by the divisor, according to the order in 
 which they stand in the work ; and the sum Avill be equal to the divi- 
 dend when the work is right. . , 
 
 EXAM 
 
2(^ 
 
 
 ARITHMETIC. 
 
 
 
 
 EXAMPLES. 
 
 
 
 1. 
 
 1234567 
 12 
 
 Quot. 
 (411522 
 mult. 3 
 
 37) 
 
 2. 
 12345678 
 111 
 
 124 
 111 
 
 135 
 111 
 
 246 
 
 222 
 
 Q,aot. 
 
 (333666. 
 
 37 
 
 3 
 3 
 
 1234566 
 add 1 
 
 2335662 
 1000998 
 rem. 36 
 
 4 
 
 1234567 
 
 3 - 
 
 12345678 
 
 ' ■ ■ F 
 
 •roof. 
 
 15 
 35 
 
 Proof. 
 
 6 
 6 
 
 
 
 %41 
 
 222 
 
 
 7 
 6 
 
 
 
 268 
 
 222 
 
 
 Rem. 1 
 
 Rem. 36 
 
 3. Divide 73146085 by 4. 
 
 4. Divide 5317986027 by 7. 
 6. Divide 570196382 by 12. 
 
 6. Divide 7463810,6 "" 
 
 7. Divide 137896264 
 
 8. Divide 35821649 
 
 9. Divide 72091365 
 10. Divide 4637064283 
 
 Ans. 182865211. 
 
 Ans. 759712289f 
 
 Ans 47616366f^. 
 
 Ans. 20 1 7246 3^. 
 
 Ans. 
 
 Ans. 46886lff. 
 Ans. 13861-i-OoV. 
 Ans. 80496^VHV. 
 
 by 37. 
 by 97. 
 by 764. 
 by 5201. 
 by 67606. 
 
 11. Suppose 471 men are formed into ranks of three deep, 
 what is the number in each rank ? Ans. 167. 
 
 12. A party at the distance of 378 miles from the head 
 quarters, receive orders to join the corps in 18 days : what 
 number of miles must they march each day to obey their 
 orders ? Ans. 21. 
 
 13. The annual revenue of a gentleman being 383.^0/; 
 how much per day is that equivalent to, there being 366 d^ys 
 in the year ? Ans. 104/. 
 
 CONTRACTIONS IN DIVISION. 
 
 There are certain contractions in Division, by which the 
 
 operation in particular cases may be performed in a shorter 
 manner as follows : 
 
 * I. Divi- 
 
DIVISION. 
 
 21 
 
 I. Division by any Small Number, not greater than 12, may 
 be expeditiously performed, by multiplying and subtracting 
 mentally, onutting to set down the work, except only the 
 quotient immediately below the dividend. 
 
 3) 66103961 
 Quot. 18701320i 
 
 EXAMPLES. 
 
 4) 52619676 6) 1379192 
 
 6) 38672940 7) 8 1 396627 8) 237 1 8920 
 
 9) 43981962 11) 67614230 12) 27980373 
 
 II. ^When Ciphers are annexed to the Divisor; cut off those 
 ciphers from it, and cut off the same number of figures from 
 the ri<.l<t-hand of the dividend : th^en divide with the remain- 
 ing tigures, as usual. And if there be any thing remaining 
 after this division, place the figures cut off from the dividend 
 to the right of it, and the whole will be the true remainder ; 
 otherwise, the figures cut off only will be the remainder. 
 
 EXAMPLES. 
 
 1. Divide 3704196 by 20. 
 . 2,0) 370419,6 
 
 Quot. 186209^1 
 
 2. Divide 31086901 by 7100. 
 71,00) 310869,01 (4378|i^i.. 
 284 
 
 268 
 213 
 
 656 
 497 
 
 699 
 568 
 
 31 
 
 3. Divide 
 
 * This method is only to avoid a needless repetition of ciphers 
 which would happen in the common way. And the truth of the 
 
 principle 
 
22 ARITHMETIC. 
 
 3. Divide 7380964 by 23000. Ans. 320f||f| 
 
 4. Divide 2304109 by 5800. Ans. 397if^f . 
 III. When the Divisor is the exact Product of two or more 
 
 of the small Numbers not greater than 12 : * Divide by each 
 ©f those numbers separately, instead of the whole divisor at 
 once. 
 
 N. B. There are commonly several remainders in virork- 
 ing by this rule, one to each division ; and to find the true 
 or whole remainder, the same as if the division had been per- 
 formed all at once, proceed as follows : Multiply the last 
 remainder by the preceding divisor, or last but one, and to 
 the product add the preceding remainder ; multiply this sum 
 by the nest preceding divisor, and to the product add the 
 next preceding remainder ; and so on^ till you have gone 
 backward through all the divisors and remainders to the first. 
 As in the example following : 
 
 EXAMPLES. 
 
 1. Divide 31046835 by 56 or 7 times 8. 
 7) 31046835 6 the last rem. 
 
 mult. 7 preced. divisor.^ 
 
 S) 4435262—1 first rem. 
 
 42 
 
 554407 — 6 second rem. add 1 the 1st rem. 
 
 Ans. 554407f| 43 whole rem. 
 
 2. Divide 7014596 by 72. 
 
 3. Divide 5130652 by 132. Ans. 38868/3-. 
 
 4. Divide 83016572 by 240. Ans. 345902^?^-. 
 
 principle on which it is founded is evident: for cutting off the same 
 number of ciphers, or figures, from each, is the same as dividing each 
 of them by 10, or 100. or 1000, &c. according to the number of ciphers 
 cut off; and it is evident, that as often as the whole divisor is contain- 
 ed in the whole dividend, so often must any part of the former be con- 
 taine(J in a like part of the latter. 
 
 * This follows from the second contraction in Multiplication, being 
 only the converse of it ; for the half of the third part of any thing, 
 is evidently the same as the sixth part of the whole ; and so of any 
 other numbers.— The reason of the method of finding the whole re- 
 mainder from the several particular pnes, will best appear from the 
 nature of Vuigiar fractions. Thus in the first example above, the 
 first remainder being 1, when the divisor is 7> makes \ this must be 
 added to the second remainder, 6, making 6\ to the divisor 8, or to be 
 divided by 8. B»\t 6 »6x7 + l 43 
 
 43 43 * _=-; and this divided by a. 
 
 gives— = — - 7 7 
 
 7X8 56 
 
 IV. Common. 
 
REDUCTION. 25 
 
 IV. Common Division may be performed more concisely ^ 
 by omitting the several products, and setting down only the 
 remainders ; namely, multiply the divisor by the quotient 
 figures as before, and without setting down the product, sub- 
 tract each figure of it from the dividend, as it is produced ; 
 always remembering to carry as many to the next figure as 
 were borrowed before. 
 
 ^ EXAMPLES. 
 
 1. Divide 3104G79 by 833. 
 833) 3104679 (^12'7j%^:,. 
 6056 
 2257 
 5919 
 88 
 2. Divide 79165238 by 238, Ans. 332627^^* 
 3 Divide 29137062 by 5317. Ans. 54791114. 
 4. Divide 62016735 by 7803. Ans. 79474ff|. 
 
 OF REDUCTION. 
 
 Redvction is the changing of numbers from one naaoe 
 or denomination to another without altering their -value. — 
 This is chiefly concerned in reducing money, weights, and 
 measures. 
 
 When the numbers are to be reduced from a higher name 
 to a lower, it is called Reduction Descending ; but when, 
 contrarywise, from a lower name to a higher, it is Reduction 
 Ascending. 
 
 Before proceeding to the rules and questions of Reduction, 
 it will be proper to set down the usual Tables of Moneys 
 weights, and measures, which are as follow : 
 
 Of MONEY, WEIGHTS, AND MEASURES. 
 
 TABLES OF MONEY.» 
 
 2 Farthings = 1 Halfpenny \ \ qrs d 
 
 4 Farthings = 1 Penny d\ 4=1 « 
 
 12 Fence = 1 Shilling si 48 = 12 = 1 £ 
 
 20 Shillings = 1 Pound ^ | 960 = 240 =s 20 =^ 1 
 
 PENCE 
 
 ♦^denotes pounds, s shillings, and d denotes pence, 
 i denotes 1 farthing, or oiie quarter of any thing- 
 ■J denotes a halfpenny or the half of any thing. 
 ^ denotes 3 farthings or three quarters of any thing. 
 
 The 
 
24 
 
 ARITHMETIC. 
 
 PENCE TABLE. 
 
 d 
 
 s d 
 
 20 is 
 
 1 8 
 
 30 ~ 
 
 2 6 
 
 40 — 
 
 3 4 
 
 60 — 
 
 4 2 
 
 60 — 
 
 6 
 
 70 ~ 
 
 5 10 
 
 80 — 
 
 6 8 
 
 90 — 
 
 7 6 
 
 100 — 
 
 8 4 
 
 110 — 
 
 9 2 
 
 120 — 
 
 10 
 
 SfflLLINGS TABLE; 
 
 s 
 
 
 d 
 
 1 
 
 is 
 
 12 
 
 ^ 
 
 — 
 
 24 
 
 3 
 
 — 
 
 36 
 
 4 
 
 — 
 
 48 
 
 5 
 
 — 
 
 60 
 
 6 
 
 — 
 
 72 
 
 7 
 
 — 
 
 84 
 
 8 
 
 — 
 
 96 
 
 9 
 
 
 
 108 
 
 10 
 
 — 
 
 120 
 
 11 
 
 
 
 132 
 
 FEDERAL MONEY. 
 
 10 Mills (m)= 1 Cent c 
 10 Cents = 1 Dime d 
 10 Dimes = 1 Dollar D 
 10 Dollars = 1 Eagle E 
 
 Standard Weight, dwt gr^ 
 The Cent weighs 6 23 Copper 
 Dollar 17 ]| Silver 
 
 Eagle 11 4| Gold 
 
 The standard for Federal Money of Gold and Silver is 1 1 
 parts fine, and 1 part alloy. 
 
 A Dollar is equal to 4s and 8d in South Carolina, to 6s in the 
 New-England Spates and Virginia, to 7s and 6d in New- 
 Jersey, Pennsylvania, Delaware, and Maryland, and to 8s in 
 New-York, and North-Carohna. 
 
 TROY 
 
 The full weight and value of the English gold and silver coin, is as 
 
 here below : 
 Gold. 
 
 Vaiue» 
 £, s a 
 110 
 10 6 
 
 A Guinea 
 Half guinea 
 Seven Shillings 7 
 Qiiarter-guinea 5 
 
 W tight. 
 
 diut gr 
 5 9i 
 2 164 
 1 19| 
 1 8^ 
 
 Silver. 
 
 Value. 
 
 Weight. 
 
 
 s d 
 
 dint gr 
 
 A Crown 
 
 5 
 
 19 8§ 
 
 Half-crown 
 
 2 6 
 
 9 16i 
 
 Shilhng 
 
 1 
 
 3 21 
 
 Sixpence 
 
 6 
 
 1 22i 
 
 The usual value of gold is nearly 4? an ounce, or 2d a grain ; and 
 that of Silver is nearly Ss an ounce. Also the value of any quantity 
 of gold, is lo the value of the same weight of standard silver, nearly 
 as 15 to 1, or more nearly as 15 and l-14th to 1. 
 
 Pure gold, free from mixture with other metals, usually called 
 fine gold is of so pure a nature, tliat it will endure the fire 
 
 without 
 
TABLES OP WEIGHTS. 26 
 
 TROYtVElGriT> 
 
 Grains - - marked ^r 
 24 Grains make 1 Penny weight rfa>< 
 20 Pennyweights 1 Ounce oz 
 
 12 Ounces 1 Pound lb 
 
 gr dwt 
 24 = I oz 
 480= 20= 1 lb 
 5760= 24^=12 = 1 
 
 By this weight are Weighed Gold, Silver, and Jewels. 
 APOTHECARIES' WEIGHT. 
 
 Grains - 
 20 Grains make 
 
 3 Scruples - 
 
 8 Drams 
 12 Ounces - 
 
 1 
 
 - 1 
 
 - 1 
 
 - 1 
 
 marked gr 
 Scruple sc or 
 Dram dr or 
 Ounce oz or 
 Pound lb or 
 
 3 
 
 fb 
 
 gr sc 
 20 = 1 
 60 = 3 = 
 480 = 24 = 
 6760 = 988 = 
 
 dr 
 
 1 
 
 8 
 
 96 
 
 oz 
 = 1 
 = 12 
 
 Ih 
 = 1 
 
 
 This is the same as Troy weight, only having some differ- 
 ent divisions. Apothecaries niake use of this weight in com- 
 pounding their Medicines ; but tiiey buy and seH their Drogi 
 by Avoirdupois weight. ' Ak ^ 
 
 -without wasting, though it be kept continually melted. But silver, not 
 having the purity of gold, will not endure the fire like it ; yet fine 
 silver will waste but a very little by being in the fire any moderate 
 time ; whereas copper, tin, lead, &c. will not only waste, but may be 
 calcined, or burnt to a powder. 
 
 Both gold and silver, in their purity, are so very soft and flexible 
 (like new lead, &c), that they are not so useful, either in coin op 
 otherwise (except to beat into leaf gold or silver), as when they are 
 allayed, or mixed and hardened with copper or brass. And though 
 most nations differ, more or less, in the quantity of such allay, as well 
 as in the same place at different times, yet in England the standard for 
 gold and silver coin has been for a long time as follows — viz. That 22 
 parts of fine gold, and 2 parts of copper, being melted together, shall 
 be esteemed the true standard for gold coin : And tliat 11 ounces and 
 2 pennyweights of fine silver, and 18 pennyweights of copper, being 
 melted together, is esteemed the true standard for silver coin, called 
 Sterling silver. 
 
 * The original of all weights used in England, was a grain op 
 com of wheat, gathered out of the middle of the ear, and, being 
 well dried, 32 of them were to make one pennyweight, 20 penny- 
 weights 
 
 Vol. I. 5 
 
26 
 
 ARITHMETIC, 
 
 AVOIRDUPOIS WEIGHI*. 
 
 Drams 
 
 16 Drams 
 
 16 Ounces 
 
 28 Pounds 
 
 4 Quarters 
 
 make 1 Ounce 
 1 Pound 
 1 Quarter 
 1 Hundred Weight 
 
 marked dr 
 oz 
 
 20 Hundred Weight 1 Ton 
 dr 
 
 lb 
 qr 
 
 cwt 
 ton 
 
 16 = 
 
 256 = 
 
 7168 = 
 
 28672 = 
 
 oz 
 1 
 
 16 = 
 448 = 
 1792 
 
 lb 
 
 1 qr 
 28 = 1. 
 112 = 4 = 
 
 cwt 
 
 1 
 
 573440 = 35840 — 2240 = 80 = 20 = 1 
 
 By this weight are weighed all things of a coarse or drossy 
 nature, as Corn, Bread, Butter, Cheese, Flesh, Grocery 
 Wares, and some Liquids j also all Metals, except Silver and 
 Gold. 
 
 oz dwt gr 
 Abfc, that 1Z6 Avoirdupois = 14 11 15i Troy. 
 loz - - = 18 6i 
 Idr - - = 1 3| 
 
 Hence it appears that the pound Avoirdupois contains 6999^ 
 grains, and the pound Troy 5760 ; the former of which aug- 
 mented by half a grain becomes 7000, and its ratio to the latter 
 is therefore very nearly as 760 to 576, that is, as 175 to 144 ; 
 consequently 144 pounds Avoirdupois are very nearly equal to 
 175 pounds Troy : and hence we infer that the ounce Avoirdu- 
 pois is to the ounce Troy as 175 to 192. 
 
 LONG MEASURE. 
 
 
 3 Barley-corns make 
 
 1 Inch 
 
 • In 
 
 12 Inches 
 
 1 Foot 
 
 - Ft 
 
 3 Feet ^ 
 
 1 Yard - 
 
 - Yd 
 
 6 Feet 
 
 1 Fathom - 
 
 ' Fth 
 
 5 Yards and a half 
 
 1 Pole or Rod 
 
 - PI 
 
 40 Poles 
 
 1 Furlong - 
 
 - Fur 
 
 8 Furlongs - 
 
 1 Mile 
 
 - Milt 
 
 3 Miles 
 
 1 League - 
 
 - Lea 
 
 69yV Miles nearly - 
 
 1 Degree - 
 
 - Deg or ^. 
 
 weights one ounce, and 12 ounces one pound. But in latter times, it 
 was thought sufficient to divide the same pennyweight into 24 equal 
 .parts, still called grains, being the least weight now in common use ; 
 »nd from thence the rest are computed, as in the Tables above. 
 
 In 
 
TABLES OF MEASURES. 27 
 
 In Ft 
 
 
 
 12= 1 Yd . 
 
 
 
 36 = 3 = 1 PI 
 
 
 
 198 = 16^ = 6| = . 1 
 
 Fur, 
 
 
 920 = 660 = 220 = 40 = 
 
 1 
 
 MUi 
 
 i360 = 6280 =1760 = 320 = 
 
 8 = 
 
 1 
 
 CLOTH MEASURE. 
 
 2 Inches and a quarter make I Nail ■ - - NX 
 4 Nails - - - 1 Quarter of a Yard (^r 
 
 3 Quarters - - - 1 Ell Flemish - E F 
 
 4 Quarters - - - 1 Yard - - Yd 
 6 Quarters - - - 1 Ell English - E E 
 4 Quarters 1| Inch - 1 Ell Scotch - ES 
 
 SQUARE MEASURE. 
 
 144 Square Inches make 1 Sq Foot - Ft 
 
 9 Square Feet - 1 Sq Yard <- Yd 
 
 30i Square Yards - 1 Sq Pole - PoU 
 
 40 Square Poles - 1 Rood - - Rd 
 
 4 Roods - - 1 Acre - - Acr 
 
 Sq Inch Sq Ft 
 
 144 = 1 SqYd 
 
 1296= 9 = 1 SqPl 
 
 39204 = 2721 = 301 = 1 Rd 
 
 1568160 = 10890 = 1210 = 40 = 1 Acr 
 
 6272640 = 43560 = 4840 = 160 = 4 = 1 
 
 By this measure, Land, and Husbandmen and Gardeners' 
 work are measured ; also Artificers' work, such as Board, 
 Glass, Paveor^ents, Plastering, Wainscoting, Tiling, Flooring, 
 and every dimension of length and breadth only. 
 
 When three dimensions are concerned, namely, length, 
 breadth, and depth or thickness, it is called cubic or solid 
 measure, which is used to measure Timber, Stone, &c. 
 
 The cubic or solid Foot, which is 12 inches in length and 
 breadth and thickness, contains 1728 cubic or solid inches., 
 and 27 solid fe$jt make one solid yard. 
 
 DRY 
 
2« 
 
 2 Pints make 
 
 2 Quarts 
 
 2 Pottles - 
 
 2 Gallons - 
 
 4 P'ecks 
 
 8 Bushels 
 
 5 Quarters - 
 2 Weys 
 
 '^^■--~-.- i 
 
 ARITHMETIC. >5- 
 
 DRY, OR CORN MEASURE. 
 
 1 Quart - - qt 
 
 1 Pottle - - Pot 
 
 1 Gallon - - Qcd 
 
 1 Peck - - Pec 
 
 1 Bushel - - Bu 
 
 1 Quarter - - Qr 
 
 1 Wey, Load, or Ton Wey 
 
 1 Last - - - Last 
 
 Gal 
 1 
 
 Pts 
 
 8 = 
 
 16 = 2 = 
 
 64 = 8 = 
 
 612 = 64 = 
 
 2660 = 320 = 
 
 6120= 640 = 
 
 Pec 
 
 1 Bu 
 
 4=1 Qr 
 
 32 = 8 = 1 Wey 
 160 = 40 = 5 = 1 Last 
 
 320 = 80 = 10 = 2 = 1 
 
 By this are measured all dry wares, as, Corn, Seeds, Roots, 
 Fruit, Salt, Coals, Sand, Oysters, &c. 
 
 The standard Gallon dry measure contains 26 8|- cubic or 
 solid inches, and the Corn or Winchester bushel 2150| cubic 
 inches ; for the dimensions of the Winchester bushel, by the 
 Statute, are 8 inches deep, and 18^ inches wide or in diameter, 
 but the Coal bushel must be 19^ inches in diameter ; and 36 
 bushels, heaped up, make a London chaldron of coals, the 
 weight of which is 31561b Avoirdupois. 
 
 ALE AND BEER MEASURE. ^, *^^ ^ 
 
 A Hi 
 
 2 Pints make 
 
 - 
 
 1 Quart 
 
 - 
 
 Qt 
 
 4 Quarts 
 
 - 
 
 1 Gallon 
 
 - 
 
 Gal 
 
 36 Gallons - 
 
 - 
 
 4 Barrel 
 
 - 
 
 Bar 
 
 1 Barrel and a 
 
 half 
 
 1 Hogshead 
 
 Hhd 
 
 2 Barrels - 
 
 . 
 
 1 Puncheon 
 
 Pun 
 
 2 Hogsheads 
 
 - 
 
 1 Butt 
 
 - 
 
 Butt 
 
 2 Butts 
 
 - 
 
 1 Tun 
 
 - 
 
 Tun 
 
 Pts qt 
 
 
 
 
 
 2= 1 
 
 Gal 
 
 
 
 
 8 = 4 = 
 
 1 
 
 Bar 
 
 
 
 288 = 144 = 
 
 36 = 
 
 1 
 
 ma 
 
 
 432 = 216 = 
 
 54 = 
 
 -- H = 
 
 1 
 
 Butt 
 
 864 = 432 = 
 
 108 = 
 
 = 3 = 
 
 2 = 
 
 1 
 
 JVbfe, The Ale Gallon contains 282 cubic or solid inches. 
 
 WINE 
 
TABLES OF MEASURES AND TIME. 
 
 WINE MEASURE. 
 
 2 Pints make 
 
 4 %iarts 
 
 42 Gallons 
 
 63 Gallons or 1^ Tierces 
 
 2 Tierces 
 
 2 Hogsheads 
 
 2 Pipes or 4 Hhds 
 
 1 Q^art - Q^ 
 
 1 Gallon - G<d 
 
 1 Tierce - Tier 
 
 1 Hogshead - Hhd 
 
 1 Puncheon - Pirn 
 
 1 Pipe or Butt Pi 
 
 I Tun - - Tun 
 
 Pis 
 
 2 = 
 
 8 = 
 
 336 = 
 
 604 = 
 
 672 = 
 
 10Q8 = 
 
 qt 
 
 1 
 
 4 = 
 168 = 
 
 262 = 
 336 = 
 
 Gal 
 
 1 Tier 
 42 = 1 Hhd ^ 
 63 = U = 1 Pun 
 84 = 2" = U = 1 
 604 = 126 = 3 = 2^ ='n 
 
 1 Tmi 
 2016 = 1008 = 252 = 6 = 4 = 3"" = 2 = 1 
 
 JVo/e, By this are measured all Wines, Spirits, Strong- 
 waters, Cyder, Mead, Perry, Vinegar, Oil, Honey, &c. 
 
 The Wine Gallon cantains 231 cubic or solid inches. And 
 it is remarkable, that the Wine and Ale Gallons have the same 
 proportion to each other, as the Troy and Avoirdupois Pounds 
 have J that is, as one Pound Tray is to one Pound Avoirdu- 
 pois, so is one Wine Gallon to one Ale Gallon. 
 
 OF TIME. 
 
 60 Seconds or 60" make 
 
 60 Minutes 
 
 24 Hours - - - - 
 
 7 Days - - - - 
 
 4 Weeks 
 
 13 Months 1 Day 6 Hours, } 
 or 365 Days 6 Hours ^ 
 
 1 Minute 
 
 1 Hour 
 
 1 Day 
 
 1 Week 
 
 1 Month 
 
 M or ' 
 
 Hr 
 
 Day 
 
 JVk 
 
 Mo 
 
 1 Julian Year Yr 
 
 Sec 
 
 60 = 
 
 3600 = 
 
 36400 = 
 
 604800 = 
 
 2419200 = 
 
 31657600 = 
 
 Mill 
 
 1 
 
 60 = 
 
 1440 = 
 
 10080 = 
 
 40320 = 
 
 525960 = 
 
 Hr 
 
 1 
 
 24 = 
 
 168 = 
 
 672 = 
 
 8766 = 366A =r 
 
 Day 
 
 1 Wk 
 
 '7 =^ 1 Mq 
 28 =4=1 
 
 1 Year 
 
 Or 
 
30 ARITHMETIC. 
 
 Wk Da Hr Mo Da Hr, 
 Or 52 1 6=13 1 6=1 Julian Year 
 Da Hr M Sec 
 But 365 5 48 48 = 1 Solar Year. 
 
 RULES FOR REDUCTION. 
 
 I. When the Numhers are to he reduced from a Higher Deno- 
 mination to a Lower : 
 
 Multiply the number in the highest denomination by as 
 many as of the next lower make an integer, or 1, in that 
 higher ; to this product add the number, if any, which was 
 in this lower denomination before, and set down the amount. 
 
 Reduce this amount in like manner, by multiplying it by 
 as many as ©f the next lower, make an integer of this, taking 
 in the odd parts! of this lower, as before. And so proceed 
 through all the denominations to the lowest ; so shall the 
 number last found be the value of all the numbers which 
 were in the higher denominations, taken together.* 
 
 EXAMPLE. 
 
 1. In 1234Z 15s Id^ how many farthings ? 
 I s d 
 1234 15 7 
 20 
 
 24695 Shillings 
 12 
 
 296347 Pence 
 4 
 
 Answer 11 85388 Farthings. 
 
 ♦ The reason of this rule is very evident ; for pounds are brought 
 into shillings by multiplying them by 20 ; shillings into pence, by mul- 
 tiplying them by 12 ; and pence into farthings, by multiplying by 4 ; 
 and the reverse of this rule by Division— And the same, it is evident, 
 vrill be true in the reduction of numbers consisting of any denomina- 
 tions "whatever. 
 
 n. When 
 
RULES FOR REDUCTION. 31 
 
 II. When the Numbers are to be reduced from a Lower Denomi- 
 nation to a Higher : 
 
 DivipE the given number by as many as of that denomina- 
 tion make 1 of the next higher, and set down what remains, 
 as well as the quotient. 
 
 Divide the quotient by as many as of this denomination 
 make 1 of the next higher ; setting down the new quotient, 
 and remainder, as before. 
 
 Proceed in the same manner through all the denomina- 
 tions, to the highest ; and the quotient last found, together 
 with the several remainders, if any, will be of the same value 
 as the first number proposed. 
 
 EXAMPLES. 
 
 2. Reduce 1186388 farthings into pounds, shillings, and 
 pence. 
 
 4) 1186388 
 
 12) 2.96347 ci 
 
 2,0) 2469,6 s—ld 
 Answer 1234/ 15s Id 
 
 3. Reduce 24/ to farthings. Ans. 23040. 
 
 4. Reduce 337687 farthings to pounds, &c. 
 
 Ans. 351/135 Of. 
 
 5. How many farthings are in 36 guineas ? Ans. 36288. 
 
 6. In 36288 farthings how many guineas ? Ans. 36. 
 
 7. In 59 lb 13 dwts 5 gr how many grains ? Ans. 340157. 
 
 8. In 8012131 grains how many pounds, &c. ? 
 
 Ans. 13901b 11 oz 18 dwt 19 gr. 
 
 9. In 35 ton 17 cwt 1 qr 23 lb 7 oz 13 dr how many drams ? 
 
 Ans. 20571006. 
 
 10. How many barley-corns will reach round the earth, 
 supposing it, according to the best calculations, to be 25000 
 miles ? Ans. 4762000000. 
 
 1 1 . How many seconds are in a solar year, or 366 days 
 5 hrs 48 min 48 sec ? Ans. 31556928. 
 
 12. In a lunar month, or 29 ds 12 hrs 44 min 3 sec, how 
 many second* ? • Ans. 2551443. 
 
 COM- 
 
32 ARITHMETIC. 
 
 COMPOUND ADDITION. 
 
 Compound Addition shows how to add or collect several 
 nambers of different denominations into one aafti. 
 
 Rule. — Place the numbers so, that those of the same de- 
 nomination may stand directly under each other, and draw a 
 line below them. Add up the figures in the lowest denomi- 
 nation, and find, by Reduction, how many units, or ones, of 
 the next higher denomination are contained in their sum. — 
 Set down the remainder below its proper column, and carry 
 those units or ones to the next denomination, which add up 
 in the same manner as before. — Proceed thus through all th« 
 denominations, to the highest; whose sum, together with the 
 several remainders, will give the answer sought. 
 
 The method of proof is the same as in Simple Addition. 
 
 EXAMPLES OF MONEY. 
 
 1. 2. 3. 4. 
 
 I s d I s d I s d I s d 
 
 " 7 13 3 14 7 5 15 17 10 53 14 8 
 
 3 5 
 
 lOi 
 
 8*19 
 
 2i 
 
 3 
 
 14 
 
 6 
 
 5 10 
 
 n 
 
 6 18 
 
 7 
 
 7 8 
 
 H 
 
 23 
 
 6 
 
 n 
 
 93 11 
 
 6 
 
 2 
 
 ^ 
 
 21 2 
 
 9 
 
 14 
 
 9 
 
 41 
 
 7 6 
 
 
 
 4 
 
 3 
 
 7 16 
 
 SI 
 
 15 
 
 6 
 
 4 
 
 13 2 
 
 5 
 
 17 15 
 
 H 
 
 4 
 
 3 
 
 6 
 
 12 
 
 9f 
 
 18 
 
 7 
 
 39 15 
 
 n 
 
 
 
 
 
 
 
 
 32 2 
 
 ^ 
 
 
 
 
 
 
 
 
 39 15 
 
 n 
 
 
 
 
 
 
 
 
 6. 
 
 
 6. 
 
 
 
 7. 
 
 
 8. 
 
 
 I s 
 
 d 
 
 I s 
 
 d 
 
 I 
 
 s 
 
 d 
 
 I s 
 
 d 
 
 14 
 
 71" 
 
 37 15 
 
 8 
 
 61 
 
 3^ 
 
 H 
 
 472 15 
 
 3 
 
 8 15 
 
 3 
 
 14 12 
 
 n 
 
 7 
 
 16 
 
 8 
 
 9 2 
 
 n 
 
 62 4 
 
 7 
 
 17 14 
 
 9 
 
 29 
 
 13 
 
 lOf 
 
 27 12 
 
 H 
 
 4 17 
 
 8 
 
 23 10 
 
 n 
 
 12 
 
 16 
 
 '2 
 
 370 16 
 
 H 
 
 23 
 
 4f 
 
 8 6 
 
 
 
 
 
 7 
 
 H 
 
 13 7 
 
 4 
 
 6 6 
 
 7 
 
 14 
 
 5i 
 
 24 
 
 13 
 
 
 
 6 10 
 
 H 
 
 91 
 
 101 
 
 64 2 
 
 ^ 
 
 5 
 
 
 
 lOf 
 
 30 
 
 111 
 
 * 
 
 EXAM- 
 
COMPOUND ADDITION. 
 
 33 
 
 Exam. 9. A nobleman, going out of town, is informed by 
 his steward that his butcher's bill comes to 197/ 13s l^d; his 
 baker's to 69Z 5s 2^d ; his brewer's to 83/ ; his wine mer- 
 chant's to 103/ 13s; to his corn-chandler is due 76/ 3d ; to 
 his tallow-chandler and cheesemonger, 27/ 15s \\\d; and 
 to his tailor 55/ 3s 6|(/ ; also for rent, servants' wages, and 
 other charges, 127/ 3s: Now, supposing he would take 100/ 
 with him to defray his charges on the road, for what sum 
 must he send to his banker ? Ans. 830/ 14s ^d. 
 
 10. The strength of a regiment of foot, of 10 companies, 
 and the amount of their subsistence*, for a month of 30 days, 
 according to the annexed Table, are required ? 
 
 Numb. 
 
 Rank. 
 
 Subsistence for 
 
 a Month. 
 
 
 
 / * 
 
 d 
 
 
 Colonel 
 
 27 
 
 
 
 
 Lieutenant Colonel 
 
 19 10 
 
 
 
 
 Major 
 
 17 6 
 
 
 
 
 Captains 
 
 78 15 
 
 
 
 U 
 
 Lieutenants 
 
 67 16 
 
 
 
 
 Ensigns 
 
 40 10 
 
 
 
 
 Chaplain 
 
 7 10 
 
 
 
 
 Adjutant 
 
 4 10 
 
 
 
 
 Quarter- Master 
 
 6 5 
 
 
 
 
 Surgeon 
 
 4 10 
 
 
 
 
 Surgeon's Mate 
 
 4 10 
 
 
 
 30 
 
 Serjeants 
 
 45 
 
 
 
 30 
 
 Corporals 
 
 to 
 
 
 
 20 
 
 Drummers 
 
 20 
 
 
 
 2 
 
 Fifers 
 
 2 
 
 
 
 390 
 
 Private Men 
 
 292 10 
 
 
 
 607 
 
 Total 
 
 656 10 
 
 
 
 * Subsistence Money, is the nioney paid to the soldiers weekly, 
 •which is short of their full pay, because their clothes, accoutrements, 
 &c. are to be accounted for. It is likewise the money advanced to 
 officers till their accounts are made up, which is commonly once 
 a year, when they are paid their arrears. The following Table shows 
 the full pay and subsistence of each rank on the Eng^Ush establish- 
 ment. 
 
 Vol. L 
 
 BAILY 
 
34 
 
 ARITHMETIC. 
 
 O 
 
 O CO CD CO Q 
 
 05 CO 'it CO t^ 
 
 -> ^ ^ o o 
 
 CO O 
 
 O '-< CO ©< CD 
 T-H ^ O O O 
 
 O O 00 O O C 
 lO '^ u:3 ^ O »0 
 
 o o o o o o 
 
 CO O 
 r}< CO 
 
 o o 
 
 CO CO O 
 I (M l> 10 
 
 OOP 
 
 O 
 
 »o 
 
 ^ 
 
 o o 
 o o o 
 
 -CO 
 
 O CO CO CO CO O CO CO O O 
 
 '— G^»-i COt-* ^[rf COlOiVO 
 
 ^ ^ ^ O O O O OOP 
 
 
 ■P 
 
 p 
 
 OPP P 
 
 p 
 
 p 
 
 1 
 
 3 
 
 
 
 
 
 
 CO 
 
 ;;^ 
 
 CO Tl* CO -^ 
 
 1 1 = 
 
 1 r 1 
 
 (24 
 
 m 
 
 ^^ 
 
 -H ^ O P 
 
 p 
 
 p 
 
 ^ 
 
 o 
 
 CO 
 
 CO p »^ o< 
 
 CO 
 
 p ■ 
 
 S 
 03 
 
 1> 
 
 CO 
 
 Oi CO ©< co» 
 
 P P p o 
 
 1 r 
 
 p 
 
 1 l'^ 1 
 
 p 
 
 
 <'C 
 
 CO ^ '-I 
 
 G^ 
 
 lO 
 
 T-* P 
 
 P P P P P 
 rf Tt P l> lO 
 
 Qj ^ ^ o o 
 
 ^ C» CO 
 
 J O) u:) lo 
 
 p p p 
 
 CO P 
 I rf lO 
 
 P5 
 
 
 CO 
 
 p p p p, 
 p i^ CO io 
 P p p p 
 
 iC Ci j vo 
 
 OP P 
 
 I I I 
 
 ©< I CO I O) 
 
 '^ ^ p 
 
 p p p 
 
 O «0 I -H 
 
 j> p o 
 ^ C5 Ol 
 
 p p p 
 
 I" I 
 
 P P Q P 
 lO P O) CO 
 P P P P 
 
 P 
 
 P "^ P~P~ 
 
 lO »-^ I »o CO 
 
 I I 
 
 J G>l j CO CO 
 
 p p p 
 
 I"" 
 
 oj <u a> :^ 
 a a -^ 
 
 "SScg 
 
 > *^ J s ^ 
 
 
 (^j Q O ^ G^ 
 
 4j . C 
 
 O G T3' 
 
 ■;j LJ < 
 
 P-i Qrc/3 CQ <! > CO 
 
 <s 
 
 *^f, 
 
 ^?. 
 
 ^^ 
 
 
 
 
 1^ 
 
 
 
 
 ^2 
 
 I. 4> 
 
 
 O^-o 
 
 h u 
 
 O J3 
 
 
 ^ S 
 
 la 
 
 "3*^3 
 
 
 11 
 
 et to 
 C.S 
 
 
 
 EXAMPLES 
 
COMPOUND ADDITION. 
 
 35 
 
 EXAMPLES OF WEIGHTS, MEASURES, &c. 
 
 TROY WEIGHT. 
 
 I 
 
 LPOTHE 
 
 ;CAR 
 
 lES' 1 
 
 tVEIGHT. 
 
 1. 
 
 2. 
 
 
 3. 
 
 
 
 4. 
 
 lb oz dwt 
 
 oz dwt gr 
 
 lb 
 
 oz dr 
 
 sc 
 
 OZ 
 
 dr sc gr 
 
 17 3 13 
 
 37 9 3 
 
 3 
 
 6 7 
 
 2 
 
 3 
 
 5 1 17 
 
 7 9 4 
 
 9 6 3 
 
 13 
 
 7 3 
 
 
 
 7 
 
 3 2 5 
 
 10 7 
 
 8 12 12 
 
 19 
 
 10 6 
 
 2 
 
 16 
 
 7 12 
 
 9 5 
 
 17 7 8 
 
 
 
 9 1 
 
 2 
 
 7 
 
 3 2 9 
 
 176 2 17 
 
 6 9 
 
 36 
 
 3 6 
 
 
 
 4 
 
 1 2 18 
 
 23 11 12 
 
 3 19 
 
 5 
 
 8 6 
 
 1 
 
 36 
 
 4 1 14 
 
 
 AVOIRDUPOIS WEIGHT. 
 
 LONG MEASURE. 
 
 5. 
 
 6. 
 
 7. 
 
 8. 
 
 lb oz dr 
 
 cwt qr lb. 
 
 mis fur pis 
 
 j^ds feetinc 
 
 - 17 10 13 
 
 15 2 15 
 
 29 3 14 
 
 127 1 5 
 
 5 14 8 
 
 6 3 24 
 
 19 6 29 
 
 12 2 9 
 
 12 9 18 
 
 9 1 14 
 
 7 24 
 
 10 10 
 
 27 1 6 
 
 9 1 17 
 
 9 1 37 
 
 54 1 11 
 
 4 
 
 10 2 6 
 
 7 3 
 
 5 2 7 
 
 6 14 10 
 
 3 3 
 
 4 5 9 
 
 23 5 
 
 CLOTH MEASURE. 
 
 LAND MEASURE. 
 
 9. 
 
 10. 
 
 11. 
 
 12. 
 
 yds qr nls 
 
 el en qrs nls 
 
 ac ro p 
 
 ac ro p 
 
 26 3 1 
 
 270 I 
 
 225 3 37 
 
 19 16 
 
 13 1 2 
 
 57 4 3 
 
 16 1 25 
 
 270 3 29 
 
 9 1 2 
 
 18 1 2 
 
 7 2 18 
 
 6 3 13 
 
 217 3 
 
 3 2 
 
 4 2 9 
 
 23 34 
 
 9 1 
 
 10 1 
 
 42 1 19 
 
 7 2,16 
 
 a5 3 1 
 
 4 4 1 
 
 7 6 
 
 75 23 
 
 
 
 
 
 WINE MEASURE. 
 
 ALE AND BEER MEASURE. 
 
 13. 
 
 14. 
 
 15. 
 
 16. 
 
 t hds gal 
 
 hds gal pts 
 
 hds gal pts 
 
 hds gal pts 
 
 13 3 16 
 
 15 61 5 
 
 17 37 3 
 
 29 43 5 
 
 8 1 37 
 
 17 14 13 
 
 9 10 15 
 
 12 19 % 
 
 14 1 20 
 
 29 23 7 
 
 3 6 2 
 
 14 16 6 
 
 25 12 
 
 3 15 1 
 
 5 14 
 
 6 8 1 
 
 3 1 9 
 
 16 8 
 
 12 9 6 
 
 57 13 4 
 
 72 3 21 
 
 4 36 6 
 
 8 42 4 
 
 5 6 
 
 
 COM- 
 
i& ARITHMETIC, 
 
 COMPOUND SUBTRACTION; 
 
 Compound Subtraction shows how to find the difference 
 between any two numbers of different denominations. To 
 perform which, observe the following Rule : 
 
 * Place the less number below the greater, so that the 
 parts of the same denomination may stand directly under 
 each other ; and draw a hne below them. — Begin at the 
 right-hand, and subtract each number or part in the lower 
 line, from the one just above it, and set the remainder 
 straight below it. But if any number in the lower line be 
 greater than that above it, add as many to the upper number 
 as make 1 of the next higher denomination ; then take the 
 lower number from the upper one thus increased, and set 
 down the remainder. Carry the unit borrowed to the next 
 number in the lower line ; after which subtract this number 
 from the one above it, as before ; and so proceed till the whole 
 is finished. Then the several remainders, taken together, 
 will be the whole difference sought. 
 
 The method of proof is the same as in Simple Subtraction. 
 
 EXAMPLES OF MONEY. 
 
 1. 
 
 I s 
 From 79 17 
 Take 35 12 
 
 d 
 
 2. 
 
 I s 
 
 103 3 
 
 71 12 
 
 d 
 
 n 
 
 I 
 
 81 
 29 
 
 5. 
 s 
 
 10 
 13 
 
 d 
 11 
 
 I 
 
 254 
 37 
 
 4. 
 
 s d 
 12 
 9 4| 
 
 Rem. 44 5 
 
 "^1 
 
 31 10 
 103 3 
 
 8| 
 
 
 
 
 
 
 Proof. 79 17 
 
 
 
 
 
 
 6. What is the difference between 73Z b^d and 19/135 lOd ? 
 
 Ans. 53/ 65 l\d. 
 
 * The reasoTi of this Rule will easily appear from what has been 
 said in Simple Subtraction ; for the borrowing depends on the same 
 principle, and is only different as the numbers to be subtracted are of 
 different denominations* 
 
 Ex. 6 
 
COMPOUND SUBTRACTION. 
 
 ^.7 
 
 Ei. 6. A lends to B 100/, how much is B in debt after A 
 has taken goods of him to the amount of 73/ I2s 4^d? 
 
 Ans. 26/ 75 7i(?. 
 
 7. Suppose that my rent for half a year is 20/ 125, and that 
 I have laid out for the land-tax 14» 6d, and for several repairs 
 1/ 3s 3i</, what have 1 to pay of my half-year's rent ? 
 
 Ans. Vcl 145 2Srf. 
 
 8. A trader failing, owes to A 35/ 75 6c/, to B 91/ 13s ^d, 
 to C 53/ 7irf, to D 87/ 55, and to E 111/ 35 5^d. When this 
 happened, he had by him in cash 23/ 75 5c/, in wares 53/ 1 Is 
 lO^e/, in household furniture 63/ 175 7|</, and in recoverable 
 book-debts 25/ 7s 5d. What will his creditors lose by him, 
 suppose these things delivered to them ? Ans. 212/ 5s S^d. 
 
 EXAMPLES OF WEIGHTS, MEASURES, &c. 
 
 TROY WEIGHT. APOTHECARIES' WEIGHT. 
 
 1. 
 
 2. 
 
 lb oz dwt gr 
 From 9 2 12 10 
 Take 5 4 6 17 
 
 lb oz dwt gr 
 7 10 4 17 
 3 7 16 12 
 
 lb oz dr scr gr 
 73 4 7 14 
 
 29 5 3 4 19 
 
 Rem. 
 
 Proof 
 
 
 
 AVOIRDUPOIS WEIGHT. 
 4. 5. 
 
 c qrs lb lb oz dr 
 From 6 17 71 5 9 
 Take 2 3 10 17 9 18 
 
 LONG MEASURE. 
 
 6. 7. 
 
 m fu pi yd ft in 
 14 3 17 96 4 
 7 6 11 72 2 9 
 
 Rem. 
 
 
 
 ac 
 
 17 
 16 
 
 
 
 Proof 
 
 
 
 
 
 CLOTH MEASURE. 
 8. 9. 
 yd qr nl yd qr nl 
 From 17 2 1 9 2 
 Take 9 2 7 2 1 
 
 LAND MEASURE. 
 
 10. 11. 
 ro p ac ro p 
 
 1 14 57 1 16 
 
 2 8 22 3 29 
 
 Rem. 
 
 
 
 
 
 
 Proof 
 
 
 
 
 
 ^VINfi 
 
38 ARITHMETIC. 
 
 WINE MEASURE. ALE AND BEER MEASURE. 
 
 12. 
 
 13. 
 
 14. 
 
 15. 
 
 t hdgal 
 
 hd gal pt 
 
 hd gal pt 
 
 hd gal pt 
 
 From 17 2 23 
 
 5 4 
 
 14 29 3 
 
 71 16 6 
 
 Take 9 1 36 
 
 2 12 6 
 
 9 35 7 
 
 19 7 1 
 
 Rem. 
 Proof 
 
 
 
 DRY MEASURE. 
 
 TIME. 
 
 From 
 Take 
 
 la 
 
 9 
 6 
 
 16. 17. 
 
 qr bu bu gal pt 
 4 7 13 7 1 
 3 6 9 2 7 
 
 18. 19. 
 mo we da ds hrs min 
 71 2 5 114 17 26 
 17 1 6 72 10 37 
 
 Rem. 
 Proof 
 
 20. The line of defence in a certain polygon being 236 
 yards, and that part of it which is terminated by the curtain 
 and shoulder being 146 yards 1 foot 4 inches ; what then was 
 the length of the face of the bastion ? Ans. 89 yards 1 ft 8 in. 
 
 COMPOUND MULTIPLICATION. 
 
 Compound Multiplication shows how to find the amount 
 of any given number of different denominations repeated a 
 certain proposed number of times j which is performed by 
 the following rule. 
 
 Set the multiplier ' under the lowest number of the 
 multiplicand, and draw a line below it. — Multiply the num- 
 ber in the lowest denomination by the multiplier, and find 
 how many units of the next higher denomination are con- 
 tained in the product, setting down what remains. — In like 
 manner, multiply the. number in the next denomination, and 
 to the product carry or add the units, before found, and find 
 how many units of the next higher denomination are in this 
 
 amount. 
 
COMPOUND MULTIPLICATION. 39 
 
 amount, which carry in like manner to the next product^ 
 settins; down the overplus. — Proceed thus to the highest de- 
 nonauation proposed : so shall the last product, with the se- 
 Teral remainders, taken as one compound number, be the 
 whole amount required. — The method of Proof, and the 
 reason of the Rule, are the same as in Simple Multiplicatioo. 
 
 EXAMPLES OF MONEY. 
 
 To find the amount of 81b of Tea, at 5s 8^d per lb. 
 s d 
 6 8| 
 8 
 
 £2 5,8 Answ^. 
 
 / s d 
 
 2. 4 lb of Tea, at 7s 8d per lb. Ans. 1 10 8 
 
 3. 6 lb of Butter, at 9^^ per lb. Ans. 4 9 
 
 4. 7 lb of Tobacco, at Is S^d per lb. Ans. 11 lli^ 
 
 5. 9 Stone of Beef, at 2s 1^ per st. Ans. 1 1 
 
 6. 10 cwt of Cheese, at n ifs lOd per cwt. Ans. 28 1 8 4 
 
 7. 12 cwt of Sugar, at 31 Is Ad per cwt. Ans. 40 8 
 
 CONTRACTIONS. 
 
 L If the multiplier exceed 12, multiply successively by its 
 component parts, instead of the whole number at once. 
 
 EXAMPLES. 
 
 IS cwt of Cheese, at 17* 6d per cwt. 
 / s d 
 17 6 
 3 
 
 12 6 
 6 
 
 13 2 6 Answer. 
 
 2. 20 cwt of Hops, at Al Is 2d per cwt. Ans. 87 3 4 
 
 3. 24 tons of Hay, at 31 Is 6rfper ton. Ans. .81 
 
 4. 45 ells of Cloth, at 1« Qd per ell. Ans. 3 7 6 
 
 Ex. 5. 
 
•W 
 
 4© ^ ARITHMETIC. 
 
 / * d 
 
 Ex.6. 63 gallons of Oil, at 2s 3d per gall. Ans. 7 19 
 
 6. 70 barrels of Ale, at 11 45 per barrel Ans. 84 
 
 7. 84 quarters of Oats, at 1/ 12s 8d per qr. Ans. 137 4 
 
 8. 96 quarters of Barley, at 1/ 3s 4d per qr. Ans. 112 
 
 9. 120 days' Wages, at 5s 9d per day. Ans. 34 10 
 
 10. 144 reams of Paper, at 13s 4d per ream. Ans. 96 
 
 11. If the multiplier cannot be exactly produced by the 
 multiplication of simple numbers, take the nearest number to 
 it, either greater or less, which can be so produced, and mul- 
 tiply by its parts, as before. — Then multiply the given mul- 
 tiplicand by the difference between this assumed number and 
 the multiplier, and add the product to that before found, when 
 the assumed number is less than the multiplier, but subtract 
 the same when it is greater. 
 
 % • EXAMPLES. 
 
 1. 26 yards of Cloth, at 3s O^d per yard. 
 I s d 
 3 0| 
 6 
 
 16 3f 
 5 
 
 3 16 6^ 
 3 Of 
 
 £ 3 19 7i Answer. 
 
 I s d 
 
 2. 29quartersofCorn,at2/5s3ic?perqr. Ans. 65 12 loi 
 
 3. 63 loads of Hay, at 3^ 15s 2d per load- Ans. 199 3 10* 
 
 4. 79 bushels ofWheat. at lis 5|rf per bush. Ans. 45 6 10^ 
 
 5. 97 casks of Beer, at 12s 2d per cask. Ans. 69 2 
 
 6. 114stoneofMeat,at 15s3f(/perst0ne. Ans. 87 6 71. 
 
 EXAMPLES OF WEIGHTS AND MEASURES. 
 
 I. 2. 3. 
 
 lb oz dwt gr lb oz dr sc gr cwt qr lb oz 
 
 28 7 14 10 2 6 3 2 10 29 2 16 14 
 
 5 8 12 
 
 ^- 
 
COMPOUND DIVISION. 41 
 
 4. 
 
 mis fu pis yds 
 
 22 5 29 6 
 
 4 
 
 5. 
 yds qrs na 
 126 3 1 
 7 
 
 mo 
 172 
 
 6. 
 ac ro po 
 28 3 27 
 9 
 
 
 
 
 7. 
 tuns hhd gal pts 
 20 2 26 2 
 3 
 
 8. 
 we qr bu pe 
 24 2 5 3 
 6 
 
 9. 
 we da ho min 
 3 6 16 49 
 10 
 
 
 COMPOUND DIVISION. 
 
 Compound Division teaches how to divide a number of 
 several denominations by any given number, or into any num> 
 ber of equal parts ; as follows : 
 
 Place the divisor on the left of the dividend, as in Simple 
 Division. — Begin at the left-hand, and divide the number of 
 the highest denomination by the divisor, setting down the 
 quotient in its proper place. — If there be any remainder after 
 this division, reduce it to the next lower denomination, which 
 add to the number, if any, belonging to that denomination, and 
 divide the sum by the divisor. — Set down again this quotient, 
 reduce its remainder to the next lower denomination again, 
 and so on through all the denominations to the last. 
 
 EXAMPLES OF MONEY. 
 
 1. Divide 237/ 8« Gd by 2. 
 I s d 
 9) 237 8 6 
 
 ig 118 14 3 the Quotient. 
 
 2. Divide 
 
 Vol. I. 
 
4t ARITHMETIC. 
 
 I s d I s d 
 
 2. Divide 432 12 If by 3. Ana. 144 4 Oi 
 
 3. Divide 607 3 5 by 4. Ans. 126 15 lOi 
 
 4. Divide 632 7 6^ by 6. Ans. 126 9 6 
 6. Divide 690 14 3i by 6. Ans. ,115 2 4i 
 
 6. Divide 705 10 2 by 7. Ans. 100 15 8f 
 
 7. Divide 760 5 6 by 8. Ans. 95 8^ 
 S.Divide 761 5 7f by #. Ans. 84 11 ^ 
 9. Divide 829 17 10 by 10. Ans. 82 19 9| 
 
 .10. Divide 937 8 8fbyll. Ans. 86 4 5 
 
 11. Divide H45 11 4i by 12. Ans. 95 9 31 
 
 CONTRACTIONS. 
 
 I. If the divisor exceed 12, find what simple numbers, 
 multiplied together, will produce it, and divide by them sepa- 
 rately, as in Simple Division, as below. 
 
 EXAMPLES. 
 
 1. What is Cheese per cwt, if 16 cwt cost 25f 14s ^d? 
 I s d 
 4)25 14 8 
 
 4) 6 8 8 
 
 £ 1 12 2 the Answer. 
 
 2. If 20 cwt of Tobacco come to } 
 1501 6s M, what is that per cwt ? ^ 
 
 I s d 
 Ans. 7 10 4 
 
 3. Divide 98/ 8s by 36. Ans. 2 14 8 
 
 4. Divide ^1/ 13s lOd by 66. Ans. 15 7i 
 
 5. Divide 44/ 4s by 96. Ans. 9 2^ 
 ^. At 31/ 10s per cwt, how much per lb ? Ans. 5 7^ 
 
 II. If the divisor cannot be produced by the multiplication 
 of small numbers, divide by the whole divisor at once, after 
 the manner of Long Division, as follows. 
 
 ?XAM- 
 
COMPOUND DIVISION. 43 
 
 EXAMPLES. 
 
 1. Divide 59/ 6* 3ft? by 19. 
 I s d I s d 
 
 19) 59 6 3f ( 3 2 ai Ans. 
 57 
 
 2 
 20 
 
 46(2 
 38 
 
 8 
 
 12 
 
 99(5 
 95 
 
 4 
 4 
 
 19(1 
 
 
 / 
 
 9 
 
 d 
 
 Ans. 
 
 
 
 13 
 
 Hi 
 
 Ads. 
 
 2 
 
 18 
 
 3 
 
 Ans. 
 
 6 
 
 11 
 
 10 
 
 Ans. 
 
 
 
 19 
 
 5. 
 
 I s d 
 
 2. Divide 39 14 5i by 67. 
 
 3. Divide 125 4 9 by 43|. 
 
 4. Divide 542 7 10 by 97. 
 
 5. Divide 123 11 2i by 127. 
 
 EXAMPLES OF WEIGHTS AND MEASURES. 
 
 1. Divide 17 lb 9 oz dwts 2 gr by 7. 
 
 Ans. 2 lb 6 oz 8 dvirts 14gr. 
 
 2. Divide 17 lb 5 oz 2 dr 1 scr 4 gr by 12. 
 
 Ans. 1 lb 5 oz 3 dr 1 scr 12 gr. 
 
 3. Divide 178 cwt 3 qrs 14 lb by 53. Ans. 3 cwt 1 qr 14 lb. 
 
 4. Divide 144 mi 4 fur 2 po 1 yd 2 ft in by 39. 
 
 Ans. 3 mi 5 fur 26 po yds 2 ft 8 in. 
 
 5. Divide 534 yds 2 qrs 2 na by 47. Ans. 1 1 yds 1 qr 2 na. 
 
 6. Divide 7 1 ac 1 ro 33 po by 51. Ans. 1 ac 2 ro 3 po. 
 
 7. Divide 7 tu hhds 47 gal 7 pi by 65. Ans. 27 gal 7 pi. 
 
 8. Divide 387 la 9 qr by 72. Ans. 5 la 3 qrs 7 bu. 
 
 9. Divide 206 mo 4 da by 26. Ans. 7 mo 3 we 5 ds. 
 
 THE 
 
44 ARITHMETIC. 
 
 THE GOLDEN RULE, OR RULE OF THREE. 
 
 The Rule of Three teaches how to find a fourth propor- 
 tional to three numbers given : for which reason it is some- 
 times called the Rule of Proportion. It is called the Rule 
 of Three, because three terms or numbers are given, to 
 find a fourth. And because of its great and extensive use- 
 fulness, it is often called the Golden Rule. This Rule is 
 usually considered as of two kinds, namely, Direct, and 
 Inverse. 
 
 The Rule of Three Direct is that in which more requires 
 more, or less requires less. As in this ; if 3 men dig V' 1 yards 
 of trench in a certain time, how much will 6 men dig in the 
 same time ? Here more requires more, that is, 6 men, which 
 are more than 3 men, will also perform more work in the 
 same time. Or when it is thus : if 6 men dig 42 yards, how 
 much will 3 men dig in the same time ? Here then, less re- 
 quires less, or 3 men will perform proportionahly less work 
 than 6 men, in the same time. In both these cases then, the 
 Rule, or the Proportion, is Direct ; and the stating must be 
 
 thus, As 3 : 21 : : 6 : 42, 
 or thus. As 6 : 42 : : 3 : 21, 
 
 But the Rule of Three Inverse, is when more requires less, 
 or less requires more. As in this : if 3 men dig a certain 
 ^ quantity of trench in 14 hours, in how many hours will 6 
 men dig the like quantity'? Here it is evident that 6 men, 
 being more than 3, will perform an equal quantity of work in 
 less time or fewer hours. Or thus : if 6 men perform a 
 certain quantity of work in 7 hours, in how many hours will 
 3 men perform the same ? Here less requires more, for 3 
 men will take more hours than 6 to perform the same work. 
 In both these cases then the Rule, or the Proportion, is In- 
 verse \ and the stating must be 
 
 thus. As 6 ; 14 : J 3 : 7, 
 or thus. As 3 : 7 : : 6 : 14. 
 
 And in all these statings, the fourth term is found, by 
 multiplying the 2d and 3d terms together, and dividing the pro- 
 duct by the 1st term. 
 
 Of the three given numbers ; two of them contain the 
 supposition, and the third a demand. And for stating and 
 working questions of these kind observe the following general 
 R«le : 
 
 State 
 
RULE OF THREE. 45 
 
 State the question, by setting down in a straight line the 
 three given numbers, in the following manner, viz. so that 
 the 2d term be that number of supposition which is of the 
 same kind that the an-awer or fourth terra is to be ; making the 
 other number of supposition the 1st term, and the demanding 
 number the *5d term, when the question is in direct propor- 
 tion ; but contrariwise, the other number of supposition the 
 3d terra, and the demanding number the 1st term, when the 
 question has inverse proportion. * 
 
 Then, in both cases, multiply the 2d and 3d terms together, 
 and divide the product by the 1st, which will give the an- 
 swer, or 4th term sought, viz. of the same denomination as 
 the second term. 
 
 Note^ If the first and third terms consist of different deno- 
 minations, reduce them both to the same : and if the second 
 term be a compound number, it is mostly convenient to re- 
 duce it to the lowest denomination mentioned. — If, after di- 
 vision, there be any remainder, reduce it to the next lower 
 denomination, and divide by the same divisor as before, and 
 the quotient will be of this last denomination. Proceed in 
 the same manner with all the remainders, till they be re- 
 duced to the lowest denomination which the second admits 
 of, and the several quotients taken together will be the an- 
 swer required. 
 
 Note also, The reason for the foregoing Rules will appear, 
 when we come to treat of the nature of proportions. — Some- 
 times two or more statings are necessary, which may always 
 be known from the nature oi the question. 
 
 EXAMPLES. 
 
 I. UB yards of Cloth cost 1/ 45, what will QQ yards cost ? 
 
 yds 1^ yds 1 s 
 As 8 : 1 4 :: 96 : 14 8 the Answer 
 20 
 
 24 ' 
 
 96 
 
 8) 2304 
 2,0) 28,8s 
 
 £U 8 Answer. 
 
 Ex.2. 
 
46 ARITHMETIC. 
 
 Ex. 2. An engineer hailing raised lOO yards of a certain 
 work in 24 days with 5 men ; how many men must he em- 
 ploy to finish a like quantity of work in 15 days ? 
 
 ds men ds men 
 
 As 16 : 6 : : 24 : 8 Ans. 
 5 
 
 • 15) 120 (8 Answer. 
 120 
 
 3. What will 72 yards of cloth cost, at the irate of 9 yards 
 for 6Z 12s ? Ans. 44/ 16s, 
 
 4. A person's annual income being 146Z ; how much is 
 that per day ? Ans. Ss, 
 
 5. If 3 paces or common steps of a certain person be equal 
 to 2 yards, how many yards will 160 of his paces make ? 
 
 Ans. 106 yds 2 ft. 
 
 6. What length must be cut off a board, that is 9 inches 
 broad, to make a square foot, or as much as 12 inches in 
 length and 12 in breadth contains ? Ans. 16 inches. 
 
 7. If 750 men re(^uire 22500 rations of bread for a month ; 
 how many rations will a garrison of 1200 men require ? 
 
 Ans. 36000. 
 
 8. If 7 cwt 1 qr of sugar cost 26Z 10s 4d ; what will be the 
 price of 43 cwt 2 qrs ? Ans. 159/ 25. 
 
 9. The clothing of a regiment of foot of 750 men amount- 
 ing to 2831/ 5s,- what will the clothing of a body of 3500 
 inen amount to ? Ans. 13212/ 10s. 
 
 10. How many yards of matting, that is 3 ft broad, will 
 cover a floor that is 27 feet long and 20 feet broad ? 
 
 Ans. 60 yards. 
 
 11. What is the value of 6 bushels of coals, at the rate of 
 1/ Ms 6f/ the chaldron ? Ans. 5s 9d, 
 
 12. If 6352 stones of 3 feet long complete a certain quan- 
 tity of walling ; how many stones of 2 feet long will raise a 
 like quantity ? Ans. 9528. 
 
 13. What must be given for a piece of silver weighing 
 73 lb 5 oz 15 dwts, at the rate of 5s 9d per ounce ? 
 
 Ans. 253/ 10s Of (^. 
 
 14. A garrison of 536 men having provision for 12 months ; 
 how long will those provisions last, if the garrison be increase* 
 to 1124 men ? Ans. 174 days and xfl?- 
 
 15. What will be the tax upon 763/ 15s, at the rate of 
 3s Qd per pound sterling ? Ans. 133/ 13s l-^rf. 
 
 16. A 
 
RULE OF THREE. 47 
 
 16. A certain work being raised in 12 days, by working 4 
 hours each day ; how long would it have been in raising by 
 working 6 hours per day ? Ans. 8 days. 
 
 17. What quantity of corn can I buy for 90 guineas, at the 
 rate of 6s the bushel ? Ans 39 qrs 3 bu. 
 
 18. A person, failing in trade, owes in all 977/ ; at which 
 time he has, in money, goods, and recorerable debts, .420Z 6s 
 ^id ; now supposing these things delivered to his creditors, 
 how much will they get per pound ? Ans. 85 7Arf. 
 
 19. A plain of a certain extent having supplied a body of 
 3000 horse with forage for 18 days ; then how many days 
 Would the same plain have supplied a body of 2000 horse ? 
 
 Ans 27 days. 
 
 20. Suppose a gentleman's income is 600 guineas a year, 
 and that he spends 25s 6d per day, one day with another ; 
 how much will he have saved at the year's end ? 
 
 Ans. 164/ 125 6d. 
 
 21. What cost 30 pieces of lead, each weighing 1 cwt 121b, 
 at the rate of 16s 4d the cwt ? Ans. 27/ 2s 6c/. 
 
 22. The governor of a besieged place having provision for 
 64 days, at the rate of l^lb of bread ; but being desirous to 
 prolong the siege to 80 days, in expectation of succour, in 
 that case what must the ration of bread be ? Ans, 1 J^ lb. 
 
 23. At half a guinea per week, how long can 1 be boarded 
 for 20 pounds ? Ans. 38 J|^ wks. 
 
 24. How much will 75 chaldrons 7 bushels of coals come 
 to, at the rate of 1/ I3s 6d per chaldron ? 
 
 Ans. 126/ 19s Oi</. 
 
 25. If the penny loaf weigh 8 ounces when the bushel of 
 ' wheat costs 7s 3c/, what ought the penny loaf to weigh when 
 
 the wheat is at 8s 4d ? Ans. 6 oz 15y3_e_ dr. 
 
 26. How much a year will 173 acres 2 roods 14 poles of 
 land givci at the rate of 1/ 7s 8c/ per acre ? 
 
 Ans. 240/ 2s l^^d. 
 
 27. To how much amounts 73 pieces of lead, each weigh- 
 ing 1 cwt 3 qrs 7 lb, at 10/ 4s per fother of I9i cwt ? 
 
 Ans. 69/ 4s 2d l^f q, 
 
 28. How many yards of stuff, of 3 qrs wide, will line a 
 cloak that is If yards in length and 2>\ yards wide ? 
 
 Ans. 8 yds qrs 2| nL 
 
 29. If 5 yards of cloth cost 145 2c/, what must be given for 
 9 pieces, containing each 21 yards 1 quarter ? 
 
 Ans. 27/ Is lOic/, 
 
 30. If a gentleman's estate be worth 2107/ 12s a year ; 
 what may he spend per day, to save 500/ in the year ? 
 
 Ans. 4/.8» I^Vjc/. 
 31. Wanting 
 
48 ARITHMETIC. 
 
 31. Wanting just an acre of land cut off from a piece 
 which is 13i poles in breadth, what length must the piece be ? 
 
 Ans. 11 po ^ yds 2 ft Oif in. 
 
 32. At 7s 9^d per yard, what is the value of a piece of 
 cloth containing 63 ells English 1 qu. Ans. 251 18s Ifrf. 
 
 33. If the carriage of 5 cwt 141b for 96 miles be 1/ 12s 6d; 
 how far may i have 3 cwt 1 qr carried for the same money ? 
 
 Ans. 151 m 3 fur 3J^ pol. 
 
 34. Bought a silver tankard, weighing 1 lb 7 oz 14 dwts ; 
 what did it cost me at 6s 4d the ounce ? Ans. 61 4$ 9^d, 
 
 36. What is the half year's rent of 547 acres of land, at 
 15s 6d the acre ? Ans. 21 U 19s 3d. 
 
 36. A wall that is to be built to the height of 36 feet, was 
 raised 9 feet high by 16 men in 6 days ; then how many men 
 must be employed to finish the wall in 4 days, at the same 
 rate of working ? Ans. 72 men 
 
 37. What will be the charge of keeping 20 horses for a 
 year, at the rate of H^rf per day for each horse ? 
 
 Ans. 441/ Os lOd, 
 
 38. If 18 ells of stuff that is f yard wide, cost 39s 6d ; 
 what will 50 ells, of the same goodness, cost, being yard wide? 
 
 Ans. 71 6s 3f |rf.. 
 
 39. How many yards of paper that is 30 inches wide, will 
 hang a room that is 20 yards in circuit and 9 feet high. 
 
 Ans. 72 yards. 
 
 40. If a gentleman's estate be worth 3«4Z 16s a year, and 
 the land-tax be assessed at 2s d^d per pound', what is his net 
 annual income ? Ans. 3311 Is 9\d, 
 
 41. The circumference of the earth is about 25000 miles ; 
 at what rate per hour is a person at the middle of its surface 
 carried round, one whole rotation being made in 23 hours 
 66 minutes ? Ans. 1044yV3V miles. 
 
 42. If a person drink 20 bottles of wine per month, when 
 it costs 8s a gallon ; how many bottles per month may he 
 drink, without increasing the expence, when wine costs 10s, 
 the gallon? Ans. 16 bottles. 
 
 43. What cost 43 qrs 6 bushels of corn, at IZ 8s &d the 
 quarter. Ans. 62Z 3s Sfd. 
 
 44. How many yards of canvas that is ell wide, will line 
 60 yards of say that is 3 quarters wide ? Ans. 30 yds. 
 
 45. If an ounce of gold cost 4 guineas, what is,the value 
 of a grain ? Ans. '^j'^d, 
 
 46. If 3 cwt of tea cost 401 12s; at how much a pound 
 must it be retailed, to gain 10/ by the whole ? Ans. S^^^s,- 
 
 COMPOUND 
 
f 49] 
 
 COMPOUND PROPORTION. 
 
 ©OMPOUND Proportion shows how to resolve such quesj- 
 tions as require two or more statings by Simple Proportion ; 
 and these may be either Direct or Inverse. 
 
 In these questions, there is always given an odd number of 
 terras, either five, or seven, or nine, kc. These are distinr- 
 guished into terms of supposition, and terms of demand, 
 there being always one term more of the former than of the 
 latter, which is of the same kind with the answer sought; 
 The method is thus : 
 
 Set down in the middle place that term of supposition 
 which is of the same kind with the answer sought. — Take 
 one of the other terms of supposition, and one of the demand- 
 ing terms which is of the same kind with it ; then place one 
 of them for a first term, and the other for a third, according 
 to the directions given in the Rule of Three. — Do the same 
 with another term of supposition, and its corresponding de- 
 manding term ; and so on if there be more terms of each 
 kind ; setting the numbers under each other which fall all on 
 the left-hand side of the middle term, and the same for the 
 others on the right-hand side. — Then, to work 
 
 By several Operations. — Take the two upper terms and 
 the middle term, in the same order as they stand, for the first 
 Rule-of- Three question to be worked, whence will be found 
 a fourth term. Then take this fourth number, so found, for 
 the middle term of a second Rule-of-Three question, and the 
 next two under terms in the general stating in the same 
 order as they stand, finding a fourth term for them. And so 
 on, as far as there are any numbers in the general stating, 
 making always the fourth number, resulting from each simple 
 stating, to be the second term in the next following one. 
 So shall the last resulting number be the answer to the 
 question. 
 
 By one Operation. — Multiply together all the terms stand- 
 ing under each other, on the left-hand side of the middle 
 term ; and in like manner, multiply together all those on the 
 right-hand side of it. Then multiply the middle term by 
 the latter product, and divide the result* by the former pro- 
 duct ; so shall the quotient be the answer sought. 
 
 V©L. 1. 8 ^. EXAMPIiES 
 
50 
 
 ARITHMETIC. 
 EXAMPLES. 
 
 1. How many men can complete a trench of 135 yards 
 long in 8 days, when 16 men can dig 54 yards in 6 days ? 
 
 
 
 General Stating. 
 
 yds. 54 
 days 8 
 
 16 
 
 : : 135 yds 
 6 days 
 
 432 
 
 
 810- 
 16 
 
 
 32) 
 
 4860 
 
 81 men 
 
 4 
 
 12960 (30 Ans. by one operation 
 1296 
 
 The same by two Operations. 
 
 54 
 
 1st. 
 16 : : 135 
 16 
 
 40 
 
 2d. 
 As 8 : 40 : : 6 : 30 
 6 , 
 
 
 810 
 135 
 
 
 8) 240 (30 Ans. 
 24 
 
 
 
 
 
 
 54)2160 
 216 
 
 (40^ 
 
 
 
 2. If 100/ in one year gain bl interest, what will be tb« 
 interest of 750/ for 7 years ? Ans. 262/ 10*. 
 
 3. If a family of 8 persons expend 200/ in 9 months ; how 
 much will serve a family of 18 people 12 mo&tfas ? 
 
 Ans. 300/^ 
 
 4. If 27s be the wages of 4 men for 7 days ; what will be 
 the wages of 14 men for 10 days ? Ans. 6/ 16j. 
 
 6. If a footman travel 130 miles in 3 days, when the days 
 are 12 hours long ; in how many days, of 10 hours each, 
 may he travel 360 miles ? / Ans. 9|f days. 
 
 Ex. 6. 
 
VULGAR FRACTIONS. 61 
 
 Ex. 6. If 120 bushels of corn can serve 14 horses 56 days ; 
 how many days will 94 bushels serve 6 horses ? 
 
 Ans. 102^^ days. 
 
 7. If 3000 lb. of beef serve 340 men 16 days ; how many 
 lbs will serve 120 men for 26 days ? Ans. 1764 lb 1 1 if oz. 
 
 8. If a barrel of beer be sufficient to last a family of 8 per* 
 sons 12 days ; how many barrels will be drank by 16 persons 
 in the space of a year ? Ans. 60f barrels. 
 
 9. If 180 men in 6 days, of 10 hours each, can dig a trench 
 200 yards long, -3 wide, and 2 deep ; in how many days, of 8 
 hours longj will 100 men dig a trench of 360 yards long, 4 wide, 
 and 3 deep ? Ans. 15 days. 
 
 OF VULGAR FRACTIONS. 
 
 A Fraction, or broken number, is an expression of a part, 
 or some parts, of something considered as a whole. 
 
 It is denoted by two numbers, placed one below the other, 
 with a line between them : 
 
 3 numerator ^ . 
 Thus, - ' > which is named 3-fourths. 
 
 4 denominator ) 
 
 The denominator, or number placed below the line, shows 
 how many equal parts the whole quantity is divided into ; 
 and it represents the Divisor in Division. — And the Numera- 
 tor, or number set above the Hne, shows how many of these 
 parts are expressed by the Fraction ; being the remainder 
 after division. — Also, both these numbers, are in general, 
 named the Terms of the Fraction. 
 
 Fractions are either Proper, Improper, Simple, Compound, 
 or Mixed. 
 
 A proper Fraction, is when the numerator is less than the 
 denominator ; as,^, or |, or f , &c. 
 
 An Improper Fraction, is when the numerator is equal to, 
 or exceeds, the denominator ; as, f, or f, or|, &c. 
 
 A Simple Fraction, is a single expression, denoting any num- 
 ber of parts of the integer ; as, |, or #. 
 
 A Compound Fraction, is the fraction of a fraction, or seve- 
 ral fractions connected with the word of between them : as, 
 i of |,orf off of 3, &c. 
 
 A Mixed Number, is composed of a whole number and a 
 fraction together ; as, 3|, or 12f , &c. 
 
 A whole 
 
i^2- ARITHMETIC, 
 
 A whole or integer niioober may be expressed like a frac- 
 tion, by writing I below it, as a denominator ; so 3 is f , or 4 
 is f , &c. 
 
 A fraction denotes division ; and its value is equal to the 
 quotient obtained by dividing the numerator by the denomina- 
 tor ; so *^ is equal to 3, and ^-o is equal to 4. 
 
 Hence then, if the numerator be less than the denominator, 
 the value of the fraction is less than 1. But if the numerator 
 be the same as the denominator, the fraction is just equal to 1. 
 And if the numerator be greater than the denominator, the frac- 
 ^on is greater than 1 . 
 
 REDUCTION OF VULGAR FRACTIONS. 
 
 Reduction of Vulgar Fractions, is the bringing them out of 
 one form or denomination into another ; commonly to prepare 
 them for the operations of Addition, Subtraction, &c. of which 
 there are several cases. 
 
 \ 
 PROBLEM. 
 
 'To find the Greatest Common Measure of Two or mqre Num- 
 bers. 
 
 The Common Measure of two or more numbers, is that 
 number which will divide them both without remainder ; so, 
 3 is a common measure of 18 and 24 ; the quotient of the 
 former being 6, and of the latter 8. And the greatest num- 
 ber that will do this, is the greatest common measure : so 6 
 is the greatest common measure of 18 and 24 ; the quotient of 
 of the former being 3, and of the latter 4, which will not both 
 divide farther. 
 
 nULE: 
 
 If there be two numbers only ; divide the greater by the 
 less ; then divide the divisor by the remainder ; and so on, 
 dividing always the last divisor by the last remainder, till no- 
 thing remains ; so shall the last divisor of all be the greatest 
 common measure sought. 
 
 When there are more than two numbers, find the greatest 
 common measure of two of them, as before ; then do the 
 same for that common measure and another of the numbers ; 
 
 and 
 
REDUCTION OP VULGAR FRACTIONS. 63 
 
 and so on, through all the numbers ; so will the greatest com- 
 mon measure last found be the answer. 
 
 If it happen that the common measure thus found is 1 ; 
 then the numbers are said to be incommensurable, or not 
 having any common measure. 
 
 EXAMPLES. 
 
 1. To find the greatest common measure of 1908, 936, 
 and 630. 
 
 936) 1908 (2 So that 36 is the greatest common 
 
 1872 measure of 1908 and 936. 
 
 36) 936 (26 Hence 36) 630 (17 
 
 72 36 
 
 216 270 
 
 216 252 
 
 18)36 (2 
 
 Hence then 18 is the answer require^. 
 
 2. What is the greatest common measure of 246 and 372 1 
 
 Ans. 6. 
 
 3. What is the greatest common measure of 324, 612, and 
 1'032 ? Ans. 12. 
 
 CASE L 
 
 To Abbreviate or Reduce Fractions to their Lowest Terms. 
 
 * Divide the terms of the given fraction by any number 
 that will divide them without a remainder ; then divide these 
 
 quotients 
 
 * That dividing both the terms of ihe fraction by the same number, 
 whatever it be, will give another fraction equal to Ihe former, is evi- 
 dent. And when these divisions are performed as often as can be 
 done, or when the common divisor is the greatest possible, the terms 
 of the resulting fraction must be the least possible. 
 
 Note 1. Any number ending with an even number, or a cipher, is 
 divisible, or can be divided, by 2. 
 
 2. Any number ending with 5, or 0, is divisible by 5. 
 
 3. If 
 
64 ARITHMETIC. 
 
 quotients again in the same manner ; and so on, till it appears 
 that there is no number greater than 1 which will divide 
 them ; then the fraction will be in its lowest terms. 
 
 Or, divide both the terms of the Fraction by their greatest 
 common measure at once, and the quotients will be the terms 
 of the fraction required, of the same value as at first. 
 
 EXAMPLES. 
 
 . Reduce flf to its least terms. 
 lif = H = f I = If = I = f • the answer. . 
 Or thus : 
 16) 288(1 Therefore 72 is the greatest common 
 
 216 measure; and 72) fif =f the An- 
 
 swer, the same as before. 
 
 72) 216 (3 
 216 
 
 2. Reduce 
 
 3. If the right-hand place of any number be O.tbe whole is divisible 
 by 10 ; if there be two ciphers, it is divisible by 100 ; if three ciphers 
 by 1000 : and so on ; which is only cutting off those ciphers. 
 
 4. If the two right-hand figures of any number be divisible by 4, tlie 
 whole is divisible by 4. And if the three right-hand figures be divisi- 
 ble by 8, the whole is divisible by 8. And so on. 
 
 5. If the sum of the digits in any number be divisible by S, or by 9, 
 the whole is divisible by 3» or by 9. 
 
 6. If the right hand digit be even, and the sum of all tha digits be 
 divisible by 6, then the whole is divisible by 6. 
 
 7. A number is divisible by 11, when the sum of the 1st, 3d, 5th, 
 Sec or all the odd places, is equal to the sum of the 2d, 4th, 6th, &c. or 
 of all the even places of digits. 
 
 8. If a number cannot be divided by some quantity less than, the 
 square root of the same, that number is aprimey or cannot be dindcd 
 by any number whatever. 
 
 9. All prime numbers, except 2 and 5, have either 1, 3, 7, or 9, 
 in the place of units , and all other numbers are composite, or can be 
 divideo. 
 
 iO. When 
 
REDUCTION OF VULGAR FRACTIONS. 65 
 
 2. Reduce Iff to its lowest terms. Ans. a. 
 
 3. Reduce ^f f to its lowest terms. Ans. f . 
 
 4. Reduce |f f to its lowest terms. Ans. |. 
 
 CASE K. 
 
 To Reduce a Mixed Number to its Equivalent improper Fraction, 
 
 * Multiply the integer or whole number by the deno- 
 minator of the fraction, and to the product add the numera- 
 tor ; then set that sum above the denominator for the frac- 
 tion required. 
 
 EXAMPLES. 
 
 1. Reduce 23| to a fraction. 
 23 
 5 
 
 115 Ory 
 
 2 (23X6)4-2 117 
 
 ' = , the Answer. 
 
 117 5 5 
 
 2. Reduce 12^ to a fraction. Ans. ^s, 
 
 3. Reduce 14yV to a fraction. Ana. y/. 
 
 4. Reduce 183^5_ to a fraction. Ans. ^i**. 
 
 10. When narabers, with the sign of addition or subtraction be- 
 tween them, are to be divided by any number, then each of those 
 
 10+8-4 
 numbers must be divided by it Thus =54-4 —2=7. 
 
 2 
 H. But if the numbers have the sign of multiplication between 
 them, only one of them must be divided. Thus, 
 10X8X3 10X4X3 10X4x1 10X2x1 20 
 
 6X2 6X1 . 2X1 1X1 1 
 
 * This is no more than first muhrplying a quantity lyy some num- 
 ber, and then dividing the result back again by the same i which it is 
 evident does not alter the value j for any fraction represents a division 
 of the numerator by the denominator. 
 
 GASE 
 
5<j ARITHMETIC 
 
 CASE m. 
 
 To Reduce an Improper Fraction to it Equivalent Whole oj- 
 Mixed Number. 
 
 * Divide the numerator hy the denominator, and the 
 quotient will be the whole or mixed number sought. 
 
 EXAMPLES. 
 
 1 . Reduce y to its equivalent number. 
 
 Here y or 12 4- 3 = 4, the Answer. 
 
 2. Reduce y to its equivalent number. 
 
 Here y or 15 -f- 7 = 24, the Answer. 
 
 3. Reduce \%^ to its equivalent number* 
 Thus 17) 749 (44yV 
 
 68 
 
 69 So that VV ^ 44xV, the Answer. 
 
 68 
 
 4. Reduce Y to its equivalent number. Ans. 8. 
 
 5. Reduce ^ff^ to its equivalent number. Ans. 54i^f . 
 
 6. Reduce 3^1.8 tQ its equiyaient number. Ang. 171|4. 
 
 * CASE IV. 
 
 To Reduce a Whole Number to an Equivalent Fraction, having a 
 Given denominator, 
 
 t Multiply the whole number by the given denominator : 
 then set the product over the said denominator, and it will 
 form the fraction required. 
 
 * This rule is evidently the reverse of the former ; and the reason 
 of it is manifest from the nature of Common Division, 
 
 f Multiplication and Division being here equally used, the result 
 must b& the same as the quantity first proposed. 
 
 EXAMPLES, 
 
REDUCTION OF VULVAR FRACTIONS, 67 
 
 EXAMPLES. 
 
 1. Reduce 9 to a fractioa whose denominator shall be 7. 
 
 Here 9 X 7 = 63 : then V ^^ the Answer ; 
 For Y = 63 -r- 7 = 9, the Proof. 
 
 2. Reduce 12 to a fraction whose denominator shall be 13. 
 
 Ans. V¥- 
 
 3. Reduce 27 to a fraction whose denominator shall be IK 
 
 Ans. Vt^- 
 
 CASE V. 
 
 To Reduce a Compound Fraction to an Equivalent Simple One. 
 
 * Multiply all the numerators together for a numerator, 
 and all the denominators together for a denominator, and 
 they will form the simple fraction sought. 
 
 When part of the compound fraction is a whole or mixed 
 number, it must first be reduced to a fraction by one of the 
 former cases. 
 
 And, when it can be done, any two terms of the fraction 
 may be divided by the same number, and the quotients used 
 instead of them. Or, when there are terms that are common, 
 they may be ofliitted, or cancelled. 
 
 EXAMPLES. 
 
 1. Reduce x of f of | to a simple fraction. 
 
 1X2X3 6 1 
 Here = — = — , the Answer. 
 
 2 X 3 X 4 24 4 
 
 ix^xgr 1 
 
 Or, = — , by cancelling the 2*s and 3's. 
 
 ^X^X4 4 
 
 * The truth of this Rule may be shown as follows : Let the com- 
 pound fraction be | of 1. Now X of ^ is 1 -f- 3, which is _5_ consequent- 
 ly 2 of s will be -5- X 2 or 4.^ ; that is the numerators are multiplied 
 together, and also the denominatorsi as in the Rule. When the com- 
 pound fraction consists of more than two single ones ; having first re- 
 duced two of them as above, then the resulting fraction and a third 
 will be the same as a compound fraction of two parts j and so on to 
 the last of all. 
 
 2. Reduce 
 Vol, L S 
 
58 ARITHMETIC. 
 
 2. Reduce | of | of \{ to a simple fraction. 
 
 2 X 3 X 10 60 12 4. 
 
 Here — = — = — = — , the Answer. 
 
 3 X 6 X 11 165 33 11 
 
 2 X^Xi:jef '4 " 
 
 Or, = — , the same as before, by cancelling 
 
 ^X^X 11 11 
 the 3's, and dividing by 6's. 
 
 3. Reduce ^ of a to a simple fraction. Ans. if. 
 
 4. Reduce | of | of f to a simple fraction. Ans. f. 
 
 5. Reduce f of | of 3i to a simple fraction. Ans. |. 
 
 6. Reduce f of | of | of 4 to a simple fraction. Ans. |. 
 
 7. Reduce 2 and | of | to a fraction. Ans. |. 
 
 CASE VI. 
 
 To Reduce Fractions of Different Denominators ^ to Equivalent 
 Fractions having a Common Denominator, 
 * Multiply each numerator by all the deriominators ex- 
 cept its own, for the new numerators : and multiply all the 
 denominators together for a common denominator. 
 
 Note, It is evident that in this and several other operations, 
 when any of the proposed quantities are integers, or mixed 
 umbers, or compound fractions, they must first be reduced, 
 ny their proper Rules, to the form of simple fractions, 
 b 
 
 EXAMPLES. 
 
 1. Reduce i, |, and f , to a common denominator. 
 
 1 X3X4 = 12 the new numerator for i 
 
 2 X 2 X 4 = 16 ditto | 
 
 3 X 2 X 3 = 18 ditto | 
 2 X 3 X 4 = 24 the common denominator. 
 
 Therefore the equivalent fractions are ^|, |f , and if. 
 Or the whole operation of multiplying may be best per- 
 formed mentally, only setting down the results and given 
 
 frarfinns thn«5 • JL 3. 3 = it 16 J..8 =::: __6_ _8_ _9_ V^v 
 
 JIdCUUUS IIIUS . 2* 3^» 4J 2ii 24» 24' I2> T2> T2 "J 
 
 abbreviation. 
 
 2. Reduce f and | to fractions of a common denominator. 
 
 Ans. -1^. -3^^ 
 
 ■ 1 • • ■ — ■ ■^ 
 
 * This is evidently no more than multiplying each numerator an'l its 
 denominator by the same quantity, and consequently the value of the 
 fraction is not altered. 
 
 See the note respecting the least common denominator at the end 
 of vol. T. 
 
 3. Reduce 
 
REDUCTION OF VULGAR FRACTIONS. 6& 
 
 3. Reduce ^,4, and |, to a common denominator. 
 
 Ans. a. ih U' 
 
 4. Reduce 4, 24, and 4 to a common denominator. 
 
 Ans.H^if, W- 
 J^ote I. When the denoririnators of two given fractions 
 have a common measure, let them be divided by it ; then 
 multiply the terms of each given fraction by the quotient aris- 
 ing, from the other's denominator. 
 
 Ex, /^, and /^ = jtV and yVs* ^y multiplying the former 
 5 7 , by 7, and the latter by 6. 
 
 2. When the less denominator of two fractions exactly di- 
 vides the greater, multiplyithe terms of that which has the less 
 denominator by the quotient. 
 
 Ex. f and y\ = j\ and y\, by mult, the former by 2. 
 
 2 
 
 3. When more than two fractions are proposed, it is some- 
 times convenient, first to reduce two of them to a common deno- 
 minator ; then these and a third ; and so on till they be all re- 
 duced to their least common denominator. 
 
 Ex, f and f and I = I and f and | = if and || and f f 
 
 CASE vir. 
 
 To Jind the value of a Fraction in Paris of the Integer. 
 
 MuLTiPBY the integer by the numerator, and divide the pro- 
 duct by the denominator, by Compound Multiplication and Di- . 
 vision, if the integer be a compound quantity. 
 
 Or, if it be a single integer, multiply the numerator by the 
 parts in the next inferior denomination, and divide the pro- 
 duct by the denominator. Then, if any thing remains, mul- 
 tiply it by the parts in the next inferior denomination, and di- 
 vide by the denominator as before ; and so on as far as neces- 
 sary ; so shall the quotients, placed in order, be the value of 
 the fraction required*. 
 
 ♦ The numerator of a fraction being considered as a remainder, ia 
 Division, and the denominator as the divisor, this rule is of the same 
 nature as Compound Division, or the valuation of remainders in the 
 Rule «f Three* before explained. 
 
 EXAMPLES. 
 
6i» 
 
 ARITHMETIC: 
 
 EXAMPLES. 
 
 1. What is the f of 2/ 6s ? 
 By the former part of the Rule 
 21 6s 
 4 
 
 Ans. 
 
 6) 9 4 
 1/ 16s9d2|y. 
 
 2. What is the value of | of U ? 
 By the 2d part of the Ruje, 
 2 
 20 
 
 3) 40 (13s 4d Ans. 
 
 1 
 
 12 
 
 3) n{Ad 
 
 3. Find the value of f of a pound sterling. Ans. 7s 6d. 
 
 4. What is the value of | of a guinea ? Ans. 4s 8d: 
 
 5. What is the value of f of a half crown ? Ans. Is 10i(|. 
 
 6. What is the value of f of 4s lOd 1 Ans. Is 1 l^d 
 
 7. What is the value of f lb troy ? Ans. 9 oz 12 dwta. 
 
 8. What is the value of j\ of a cwt ? Ans. 1 qr 7 lb. 
 
 9. What is the value of f of an acre ? Ans. 3 ro. 20 po. 
 10. What is the value of f_ pf a day ? Ans. "^ hrs 12 min. 
 
 CASE VIII. 
 
 To Reduce a Fraction from one Denomination to another. 
 
 * Consider how many of the less denomination make one 
 of the greater ; then multiply the numerator by that number, 
 if the reduction be to a less name, but multiply the denomina- 
 tor, if to a greater. 
 
 EXAMPLES. 
 
 1. Reduce f of a pound to the fraction of a penny, 
 
 I X V X V^ = 4|» = H% the Answer. 
 
 * rhis is the same as the Rule of Reduction in whole numbevs from 
 one denomination to another. 
 
 S. Reduce 
 
ADDITION OF VULGAR FRACTIONS. 61 
 
 2. Reduce ^^ of a penny to the fraction of a pound. 
 
 4 X y'^ X 2^ = ok the Answer. 
 
 3. Reduce ^^l to the fraction of a penny. Ans. ^^d, 
 
 4. Reduce f q to the fraction of a pound. Ans. -^is-o* 
 
 5. Reduce f cwt to the fraction of a lb. Ans. ^^ . 
 
 6. Reduce f dwt to the fraction of a lb troy. Ans ji^. 
 
 7. Reduce f crown to the fraction of a guinea. Ans. /g^. 
 
 8. Reduce | half-crown to the fract. of a shilling. Ans. f f . 
 
 9. Reduce 2s Qd to the fraction of a £. Ans. |. 
 10. Reduce 17s Id ^q to the fraction of a £. 
 
 ADDITION OF VULGAR FRACTIONS. 
 
 If the fractions have a common denominator ; add all the 
 numerators together, then place the sum over the common 
 denominator, and that will be the sum of the fractions 
 required. 
 
 * If the proposed fractions have not a common denomina- 
 tor, they must be reduced to one. Also compound fractions 
 must be reduced to simple ones, and fractions of different 
 denominations to those of the same denomination. Then 
 add the numerators as before. As to mixed numbers, they 
 may either be reduced to Improper fractions, and so added 
 with the others ; or else the fractional parts only added, and 
 the integers united afterwards. 
 
 * Before fractions are reduced to a common denominator, they are 
 quite dissimilar, as much as shillings and pence are, and therefore 
 cannot be incorporated with one another, any more than these can. 
 But when they are reduced to a common denominator, and made parts 
 of the same tiling, their sum, or difference, may then be as properly 
 expressed by the sum or difference of the numerators, as the sum or 
 difference of any two quantities whatever, by the sum or difference of 
 their individuaJs. Whence the reason of the Rule is manifest, both 
 tor Addition and Subtraction. 
 
 When several fractions are to be collected, it is commonly best first 
 to add two of them together that most easily reduce to a common de- 
 nominAtor ; then add their sum and third, and so on. 
 
 EXAMPLES. 
 
g2 ARITHMETIC. 
 
 EXAMPLES. 
 
 1. To add f and f together. 
 
 Here | + j = j = If » the Answer. 
 
 2. To add | and | together. 
 
 3 4- i = ±8 J_ 2 5 = 4 3 :=; 1 1.3 the Answpr 
 
 3. To add f and 7i and | of | together. 
 
 I + 7i + iof f =1 -f y +i =1 4- V +f = ¥=8f 
 
 4. To add ^ and f together. Ans. If. 
 6. To add | and f together. Ans. m. 
 
 6. Add f and j\ together. Ans. i^. 
 
 7. What is the sum of | and f and f ? Ans. 1|^|. 
 
 8. What is the sum of | and f and 2J- ? Ans. Sf-f. 
 
 9. What is the sum of | and |^ of i and 9/^ ? Ans. lOgV- 
 
 10. What is the sum of | of a pound and f of a shilling ? , 
 
 Ans. »|5s or 13s lOd 2^q. 
 
 11. What is the sum of | of a shilling and y*j of a penny 9 
 
 Ans. y-^d or Id lj\q, 
 .12. What is the sum of i of a pound, and | of a shilling, 
 and j% of penny ? Ans. Hof^ or 3* ^^ Hf ?• 
 
 SUBTRACTION OF VULGAR FRACTIONS. 
 
 Prepare the fractions the same as for Addition, when 
 accessary ; then subtract the one numerator from the other, 
 and set the remainder over the common denominator, for the 
 diflerence of the fractions sought. 
 
 EXAMPLES. 
 
 1. To find the difference between | and |. 
 
 Here | — e ~ e ~ l» the Answer. 
 
 2. To find the difference between f and |. 
 I — f = If — fl = A» the Answer. 
 
 3. What 
 
MULTIPLICATION OF VULGAR FRACTIONS. 63 
 
 3. What is the difference between /^ and y'^j ? Ans. |. 
 
 4. What is the difference between j\ and 3V ? Ans. ^^j. 
 $. What is the difference between f'j and x\ ? Ans. ^3^. 
 
 6. What is the diff. between 5f and f of 4} ? Ana. 4/6V- 
 
 7. What is the difference between f of a pound, and | of 
 a of a shilling ? Ans. Vt « or I'Os 7d l^y. 
 
 8. What is the difference between f of 6^ of a pound, 
 and I of a shilling ? Ans. ^^l or 11 ds 11 Jl^. 
 
 MULTIPLICATION OF VULGAR FRACTIONS. 
 
 * Reduce mixed numbers, if there be any, to equivalent 
 fractions ; then multiply all the numerators together for a 
 numerator, and all the denominators to2;ether for a denomi- 
 nator, which will give the product required. 
 
 EXAMPLES; 
 
 1. Required the product of f and f. 
 
 Here f X | = /g = ^' ^^^ Answer. 
 
 2. Required the continual product of |, 3^, 5, and f off. 
 
 g 13 ^ 3 3 13 X 3 39 
 
 Here — X— X— X— X— = = — = 4| Ans. 
 
 ^ 4 1 4,2 JS 4X2 8 
 
 3. Required the product of f and f . Ans. -^j. 
 
 4. Required the product of y*j and /j. Ans. j\, 
 
 5. Required the product of |, f , and ||. Ans. ^\. 
 
 * Multiplication of any tking by a fraction, implies the taking some 
 part or parts of the thing ; it may tJierefore be truly expressed by a 
 compound fraction ; which is resolved by multiplying together the 
 numerators and denominators. 
 
 Noie^ A Fraction is best multiplied by an integer, by dividing the 
 denominator by it ; but if \t will not exactly divide, then multiply the 
 numerator by it. 
 
 6. Required 
 
|S4 ARITHMETIC. 
 
 6. Required the product of i,|, and 3 Ans. 1. 
 
 7. Required the product of i, |, and 4/^ . Ans. 2^V' 
 
 8. Required the product off, and | of ^. Ans. ^f . 
 
 9. Required the product of 6, and f of 5. Ans. 20. 
 10. Required the product off of |, and f of 3f. Ans. ||. 
 U. Required the product of 3f and 4i| Ans. 14i|f. 
 12. Required the product of 5, f, f off, and 4^. Ans. 2-ij. 
 
 DIVISION OF VULGAR FRACTIONS. 
 
 * Prepare the fractions as before in multiplication ; then 
 divide the numerator by the numerator, and the denominator 
 by the denominator, if they will exactly divide : but if not, 
 then invert the terms of the divisor, and multiply the dividend 
 by it, as in multiplication. 
 
 EXAMPLES. 
 
 1. Divide V ^7 f • 
 
 Here V -s- f == f = 1|, by the first method. 
 
 2. Divide! by f 3. 
 
 Heref ^ ^^^ :^ f X »^ = f X f = V = 4^. 
 
 3. It is required to divide if by |. Ans. f . 
 
 4. It is required to divide y*^ by |. Ans. y'g. 
 6. It is required to divide V ^J e • •^'^s* H* 
 
 6. It is required to divide f by y . Ans. y\. 
 
 7. It is required to divide || by |. Ans. ^. 
 
 8. It is required to divide f by |. Ans. |f . 
 
 * Division being the reverse of Multiplication, the reason of the 
 Rule is evident. 
 
 NotCy A fraction is best divided by an intecjer, by dividing the nume- 
 rator by it ; bat if it will not exactly divide, then multiply the deno- 
 minator by it: 
 
 9. rt 
 
RULE OF THREE IN VULGAR FRACTIONS. 61 
 
 9. It is required to divide tV l>y 3. Ans. f^. 
 
 10. It is required to divide ^ by 2. Ans. f^* 
 
 11. It is required to divide 7^ by 9f. . , Ans. ^f. 
 
 12. It is required to divide f of | by 4 of 7|. Ans. j^j. 
 
 RULE OF THREE IN VULGAR FRACTIONS. 
 
 Make the necessary preparations as before directed ; then 
 multiply continually together, the second and the third terms, 
 and the first with its parts inverted as in Division, for th^ 
 
 EXAMPLES. 
 
 1. If I of a yard of velvet cost | of a pound sterling ; what 
 will y\ of a yard cost ? 
 
 3 2 6 8 jgf ^. 
 
 — : — : : — : — X — X — = il = 6s Qd, Answer. 
 
 S 5 16 5 g XJ& 
 
 2. What will 3| oz of silver cost, at 6s 4d an ounce ? 
 
 Ans. 1/ 1* 4^d. 
 
 3. If j\ of a ship be worth 273^ 2* 6d ; what are /^ of het 
 worth ? Ans. 227/ 12s Id, 
 
 4. What is the purchase of 1230/ bank-stock, at 108f per 
 cent. ? Ans. 1336/ Is dd. 
 
 6. What is the interest of 273/ 15s for a year, at 3i per 
 cent. ? Ans. 8/ 17* ll^d. 
 
 6. If I of a ship be worth 73/ Is 3d ; what part of her is 
 worth 250/ 10« ? Ans. |. 
 
 7. What length must be cut off a board that is 7| inches 
 broad, to contain a square foot, or as much as another piege 
 of 12 inches long and 12 broad ? Ans. 18i^ inches. 
 
 8. What quantity of shalloon that is | of a yard wide, wiU 
 line 9i yards of cloth, that is 2^ yards wide ? Ans. 31 J yds. 
 
 * This is only multiplying' the 2d and 3d terms together, and 
 dividing the product by the lirst, as in the Rule of Three in whole 
 nurai>ers. 
 
 Vox.. I. 10 9. If 
 
e.G * ARITHMETIC. 
 
 9. If the penny loaf weighs 6^^ 02, when the price of 
 wheat is 5s the bushel ; what ought it to weigh when the 
 wheat is 8s 6d the bushel ? Ans. 4 y\ oz. 
 
 10. How much in length, of a piece of land that is 11}^ 
 poles broad, will make an acre of land, or as much as 40 
 poles in length and 4 in breadth ? Ads. ISjVt poles. 
 
 11. If a courier perform a certain journey in 35^ days, 
 travelling 13f hours a day; how long would he be in per- 
 forming the same, travelling only 11/^ hours a day ? 
 
 Ans. 40f jf days. 
 
 12. A regiment of soldiers, consisting of 976 men, are to 
 be new cloathed ; each coat to contain 2i yards of cloth 
 that is If yard wide, and lined with shalloon 5. yard wide : 
 how many yards of shalloon will line them ? 
 
 Ans. 4531 yds 1 qr 2^ nails. 
 
 DECIMAL FRACTIONS. 
 
 A Decimal Fraction, is that which has for its deno- 
 minator an unit (1), with as many ciphers anne^xed as the 
 numerator has places ; and it is usually expressed by setting 
 down the numerator only, with a point before it, on the left- 
 band. Thus, j% is -4, and ^-^^ is -24, and j^-^ is -074, and 
 ToWoo is -00124 ; where ciphers are prefixed to make up as 
 many places as are ciphers in the denominator, when there 
 is a deficiency of figures. 
 
 A mixed number is made up of a whole number with some 
 decimal fraction, the one being separated from the other by 
 a point. Thus, 3*25 is the same as Sy^/^, or flf. 
 
 Ciphers on the right-hand of decimals make no alteration 
 in their value ; for -4 or 40, or 400, are decimals having all 
 the same value, each being = yV' ^^ I- ^"^ when they arc 
 placed on the left-hand they decrease the value in a ten-fold 
 proportion : Thus, -4 is ^^, o^ 4 tenths : but -04 is Only y|^j 
 or 4 hundredths, and '004 is only tAo? or 4 thousandths. 
 
 The 1st place of decimals, counted from the left-hand to- 
 wards the right, is called the place of primes, or lOths ; the 
 2d is .the place of seconds, or lOOUi ; the 3d is the place of 
 thirds, or lOOOths ; and so on. For. in decimals, as well as 
 in whole numbers, the values of the places increase towards 
 the left-hand, and decrease towards the right, both in the 
 
 same 
 
ADDITION OF DECIMALS. 61 
 
 same tenfold proportion ; as in the following Scale or Table 
 of Notation. 
 
 1 III I I I I \§ I I i t 
 
 33S33S3333333 
 
 . ADDITION OF DECIMALS. 
 
 Set "the numbers under each other according to the value 
 «f their places, like as in whole numbers ; in which state the 
 decimal separating points will stand all exactly under each 
 other- Then beginning at the right-hand, add up all the 
 columns of numbers as in integers ; and point off as many 
 places, for decimals, as are in the greatest number of decimal 
 places in any of the lines that are added ; or place the point 
 directly below all the other points. 
 
 EXAMPLES. , 
 
 1. To add together 29-0146, and 3148-6, and 2109, and 
 •62417, and 14-16. 
 
 29-0145 
 3146-5 
 210^ 
 
 •62417 
 14-16 
 
 5299-29877 the Sum. 
 
 Ex. 2. What is the sum of 276, 39-213, 72014-d, 417, 
 and 6032 ? . \ . 
 
 3. What is the sum of 7530, 16-201, 3-014^, 957-13, 
 6-72119 and -03014. 
 
 4. What is the sum of 312-09, 3-5711, 7195-6, 71-498, 
 9739-216, 179, and -0027 ? 
 
 SUBTRACTION 
 
SS ARITHMETIC. 
 
 SUBTRACTION OF DECIMALS. 
 
 Place the numbers under each other according to tJbe 
 ualue of their places, as in the last Rule. Then, beginning 
 at the right-hand, subtract, as in whole numbers, and point 
 off the decimals as in Addition. 
 
 IIXAMPLES, 
 
 t. To find the difference between 91-73 and 2-138. 
 91-73 
 2-138 
 
 Ans. 89-592 the Difference. 
 
 2. Find the diff. between 1-9185 and 2-73. Ans. 0-8116 
 
 3. To subtract 4-90142 from 214-81. Ans. 209-90858, 
 
 4. Find the diff. between 2714 and -916. An§. 2713-084. 
 
 MULTIPLICATION OF DECIMALS. 
 
 * Place the factors, and multiply them together the same 
 as if they were whole numbers. — Then point off in the pro» 
 duct just as many places of decimals as there are decimals in 
 both the factors. But if there be not so many figures in the 
 product, then supply the defect by prefixing ciphers. 
 
 * The Rule will be evident from this example : — Let it be re- 
 qyiired to multiply '12 by 361 ; these numbers are eqiiivaknt to 
 J^ and -3_6_L. ; the product of which is -j-f f f f „ = '04332, by the na- 
 ture of Notation, which consists of as many places as there are ciphers, 
 that is, of as many places as there are in both numbers. And in like 
 manner for any other numbers. 
 
 EXAMPLES. 
 
MULTIPLlCATiON OF DECIMALS. 69 
 
 EXAMPLES. 
 
 1. Multiply -321096 
 by -2466 
 
 1605480 
 1926676 
 1284384 
 642192 
 
 Ans. -0791501640 the Product. 
 
 2. Multiply 79-347 by 23-15. Ans. 1836-88306. 
 
 3. Multiply -63478 by -8204. Ans. -520773512. 
 
 4. Multiply -385746 by -00464. Ans. -00178986144. 
 
 CONTRACTION I. 
 
 To multiply Decimals by 1 with any number of Ciphers^ as by 
 10, or 100, or 1000, ^c. 
 
 This is done by only removing the decimal point so many 
 places farther to the right-hand, as there are ciphers in the 
 multiplier : and subjoining ciphers if need be. 
 
 EXAMPLES. 
 
 1. The product of 51-3 and 1000 is 61300. 
 
 2. The product of 2-714 and 100 is 
 
 3. The product of -916 and 1000 is 
 
 4. The product of 21-31 and 10000 is 
 
 CONTRACTION U. 
 
 To Contract the Operation, so as to retain only as many Diecimals 
 in the Product as may be thought Necessary^ when the Pro- 
 duct would naturally contain several more Places. 
 
 Set the units' place of the multiplier under that figure of 
 t-he multiplicand whose place is the same as is to be retained 
 for the last in the product ; and dispose of the rest of the 
 figures in the inverted or contrary order to what they are 
 usually placed in. — Then, in multiplying, reject all the figures 
 that are more to the right-hand than each multiplying figure, 
 and set down the products, so that their right-hand figures 
 
 may 
 
20 ARlTHMETie. r.ttiT I r 
 
 may fall io a column straight below each other ; but observing 
 to increase the first figure df every line with what would 
 •arise from the figures omitted, in this manner, namely 1 from 
 5 to 14, 2 from 15 to 24, 3 from 25 to 34, &c. ; and the sum 
 of all the lines will be the product as required, commonly to 
 the nearest unit in the last figure. 
 
 EXAMPLES. 
 
 1. To multiply 27»14986 by 92-41036, so as to retain oply 
 four places of decimals in the product. 
 
 Contracted Way. Common Way. 
 72-14986 27 14S86 
 
 530 14 -29 92-41035 
 
 24434874 
 
 542997 
 
 108599 
 
 2715 
 
 81 
 
 14 
 
 2508-9280 
 
 13 
 
 574930 
 
 81 
 
 44958 
 
 2714 
 
 986 
 
 103599 
 
 44 V 
 
 642997 
 
 2 
 
 24434874 
 
 
 2608-9280 
 
 650510 
 
 2. Multiply 480-14936 by 2-72416, retaining only four de- 
 cimals in the product. 
 
 3. Multiply 2490-3048 by -573286, retaining only five de- 
 cimals in the product. 
 
 4. Multiply 325-701428 by -7218393, retaining only three 
 ^lecimals in the product. 
 
 DIVISION OF DECIMALS. 
 
 Divide as in whole numbfers ; and point off in the quotient 
 as many places for decimals, as the decimal places in the divi- 
 dend exceed those in the divisor*. 
 
 ♦ The reason of this Rule is evident ; for, since the divisor multiplied 
 by the quotient gives the dividend, therefore the number of decimal 
 places in the dividend, is equal to those in the divisor and quotient, 
 taken together, by the nature of Multiplication ; and consequently the 
 qs>''tient itself must contain as many as the dividend exceeds the di- 
 ^or- 
 
 Another 
 
DIVISION OF DECIMALS. 
 
 71 
 
 Another way to know the .place for the decimal point, is* 
 this : The first figure of the quotient must be made to oc- 
 cupy the same place, of integers or decimals, as doth that 
 figure of the dividend which stands over the unit's figure of the 
 the first product. 
 
 When the places of the quotient are not so many as the 
 Rule requires, the defect is to be supplied by prefixing 
 ciphers. 
 
 When there happens to be a remainder after the division, 
 or when the decimal places in the divisor are more than those 
 in the dividend ; then ciphers may be annexed to the dividend, 
 and the quotient carried on as far as required. 
 
 EXAMPLES. 
 
 1. 
 178) -48520998 (-00372589 
 1292 . 
 460 
 1049 
 1599 
 1768 
 166 
 
 •2639) 27-00000 ( 102-3114 
 6100 
 8220 
 303U 
 3910 
 12710 
 2154 
 
 3. Divide 123-70536 by 64-25. 
 
 4. Divide 12 by 7854. 
 
 5. Divide 4195-68 by 100. 
 
 6. Divide -8297592 by -153. 
 
 Ans. 2-2802. 
 
 Ans. 15-278. 
 
 Ans. 41-9568. 
 
 Ans. 5-4232. 
 
 CONTRACTION 1. 
 
 When the divisor is an integer, with any number of ciphers 
 annexed : cut off those ciphers, and remove the decimal 
 point in the dividend as many places farther to the left as 
 there are ciphers cut off, prefixing ciphers if need be : then 
 proceed as before.* 
 
 * This is no more than dividing both divisor and dividend by the 
 same number, either 10, or 100, or 1000, &c. according to the number 
 of ciphers cut off, which leaving them in the same proportion, does 
 not affect the quotient. And, in the same way, the decimal point may 
 be moved the same number of places in both the divisor and dividend, 
 either to the right or left, whether they have ciphers or not. 
 
 EXAMPLES. 
 
7g ARITHMETIC. 
 
 EXAMPLES^ 
 
 1. Divide 46-5 by 2100. 
 
 21-00 ) -455 ( -0216, mi. 
 35 
 140 
 14 
 
 2. DiTide 41020 by 32000. 
 
 3. Divide 953 by 21600. 
 
 4. Divide 61 by 79000. 
 
 CONTRACTION It 
 
 Hence, if the divisor be 1 with ciphers, as 10, K)0, oj 
 1000, &c. : then the quotient will be found by merely mov- 
 ing the decimal point in the dividend, so many places farther 
 to the left, as the divisor has ciphers ; prefixing ciphers if 
 need be. 
 
 EXAMPLES. 
 
 So, 217-3 -r 100 = 2-173 And 419 ~ 10 = 
 
 And 6-16 -^ 100 = And 21 -f- 1000 = 
 
 CONTRACTION HI. 
 
 When there are many figures in the divisor ; or when only 
 a certain number of decimals are necessary to be retained ia 
 the quotient : then take only as many figures of the divisor 
 as will be equal to the number of figures, both integers and 
 decimals, to be in the quotient, and find how many times 
 they may be contained in the first figures of the dividend, as 
 usual. 
 
 Let each remainder be a new dividend ; and for every such 
 dividend, leave out one figure more on the right-hand side of 
 the divisor ; remembering to carry for the increase of the 
 figures cut off, as in the 2d contraction in Multiplication. 
 
 JVote. When there are not so many figures in the divisor, as 
 are required to be in the quotient, begin the operation with all 
 the figures, and continue it as usual till the number of figures 
 in the divisor be equal to those remaining to be found in the 
 quotient : after which begin the contraction. 
 
 EXAMPLES. 
 
 1. Divide 2508-92806 by 92-41035, so as to have only four 
 decimals in the quotient, in whick case the quotient will con- 
 tain six figures. 
 
 Contracted. 
 
REDUCTION OF DECIMALS. 73 
 
 Contracted. Common, 
 
 92-41Q3,5)2508-928,06(27- 1498 
 660721 
 13849 
 4608 
 
 80 
 ' 6 
 
 92-4103,5)2508-928 06(2ri'498 
 66072106 
 13848610 
 46075760 
 91116100 
 79467850 
 5539570 
 
 2. DivicJe 4109-2351 .by 230-409, so that the quotient may 
 contain only fooT decimals. Aas. 17-8346. 
 
 3. Divide' 37-10438 by 5713-96, that the quotient may con- 
 tail) only live rfecimalap. Ans. -00649. 
 
 4. Divide 913^08 by 2137-2^ that the qijotient may contain 
 •nly three decimals. 
 
 REDUCTION OF DECIMALS. 
 
 CASE I. 
 
 To reduce a Vulgar Fraction to its equivalent Decimal, 
 
 Divide the numerator by the denominator as in Division 
 of Decimals, annexing ciphers to the numerator as far as ne- 
 cessary ,- so »half the qWtient be the decimal required. 
 
 1. Reduce ^\ to at ^cimal. 
 24 = 4 X 6. Then 4) 7- 
 
 6) 1 •750000. 
 •29l6b6 &c. 
 
 t. Reduce i, and ^, and |, to decimals. 
 
 Ans. 't5, afwf '5, arid •'^5. 
 
 3. Redace f to a decimal. Ans. -625. 
 
 4. Reduce 2 J to a decimal. Ans. 12. 
 6. Reduce yfj toadeciaiaL Ans Oj 1350. 
 e^. Reduce ^/Jv to a decimal. Ans. -143155 &c. 
 
 Voc. I. U CASE 
 
74 ARITHMETIC. 
 
 CASE II. 
 
 To find the Value of a Decimal in terms of the Inferior Deno- 
 minations. 
 
 Multiply the decimal by the number of parts in the 
 next lower denomination ; and cut off as many places for a 
 remainder to the right hand, as there are places in the given 
 decimal. 
 
 Mutiply that remainder by the parts in the next lower 
 denomination again, cutting off for another remainder as 
 before. 
 
 Proceed in the same manner through all the parts of the 
 integer ; then the several denominations separated on the left- 
 hand, will make up the answer. 
 
 JVofe, This operation is the same as Reduction Descending 
 in whole Rumbers. 
 
 EXAMPLES. 
 
 i. Required to find the value of '775 pounds sterling. 
 
 •776 
 20 
 
 s 15-600 
 12 
 
 d 6-000 Ans. 16« 6d, 
 
 2. What is the value of -625 shil ? Ans. 7idf. 
 
 3. What is the value of 'S63ol? Ans. 17s 3-24d. 
 
 4. What is the value of -0125 lb troy ? Ans. 3 dwts. 
 6. What is the value of -4694 lb troy ? 
 
 Ans. 6 oz 12 dwts 15-744 gr. 
 
 6. What is the value of -626 cwt ? Ans. 2 qr 14 lb. 
 
 7. What is the value of -009943 miles ? 
 
 Ans. 17 yd 1 ft 5-98848 inc. 
 
 8. What is the value of 6875 yd ? Ans. 2 qr 3 nls. 
 
 9. What is the value of -3375 acr ? Ans. 1 rd 14 poles. 
 10. What is the value of -S^'S hhd of wine ? 
 
 Ans. 13-1229 gal. 
 
 CASE 
 
REDUCTION OF DECIMALS. 75 
 
 CASE ni. 
 
 To reduce Integers or Decimals to Equivalent Decimals of 
 Higher Denominations* 
 
 Divide by the number of parts in the next higher deno- 
 minatioQ ; continuing the operation to as many higher deno- 
 minations as may be necessary, the same as in Reduction 
 Ascending of whole numbers. 
 
 EXAMPLES. 
 1. Reduce 1 dwt to the decimal of a pound troy. 
 
 20 
 12 
 
 1 dwt 
 
 0-06 oz 
 
 0004166 &c. lb. Ans. 
 
 2. Reduce 9d to the decimal of a pound. Ans. '03151. 
 
 3. Reduce 7 drams to the decimal of a pound avoird. 
 
 Ans. •027343751b. 
 
 4. Reduce 26 J to the decimal of a Z. Ans. -0010833 &c. /. 
 
 5. Reduce 2- 15 lb to the decimal of cwt. 
 
 Ans. -oiaise+cwt. 
 
 6. Reduce 24 yards to the decimal of a mile. 
 
 Ans. -013636 &c. mile. 
 
 7. Reduce '056 pole to the decimal of an acre. 
 
 Ans. -00035 ac. 
 €. Reduce 1*2 pint of *vine to the decimal of a hhd. 
 
 Ans. 00238-}-hhd. 
 9. Reduce 14 minutes to the decimal of a day. 
 
 Ans. -009722 &c. da. 
 
 10. Reduce -21 pint to the decimal of a peck. 
 
 Ans. -013125 pec. 
 
 11. Reduce 28" 12'" to the decimal of a minute. 
 
 Note, When there are several numbers ^ to be reduced all to 
 to the decimal of the highest : 
 
 Set the given numbers directly under each other, for divi^ 
 dends, proceeding orderly from the lowest denomination to 
 the highest. 
 
 Opposite to each dividend, on the left-hand, set such a 
 number for a divisor as will bring it to the next higher name ; 
 drawing a perpendicular line between all the divisors and 
 dividends. 
 
 Begin at the uppermost, and perform all the divisions : 
 ©nly observing to set the quotient of each division, as decimal 
 
 parts, 
 
76 ARITHMETIC. 
 
 parts, on the right-han.l of the. dividpnd next below it 
 shall the last quotient be the decimal required. 
 
 EXAMPl^S. 
 
 1. Reduce 17s 9fc? to the decimal of a pound. 
 
 4 
 12 
 
 20 
 
 3- 
 
 9-75 
 17-8125 
 i: ()-89U625 Ans. 
 
 2. Reduce ilH 17^ a^^/ to L Ans. 19-86354166 kc. L 
 
 3. Reduce 15.9 Hrf to the decimal of a L Ans. -775^ 
 
 4. Kedutu^ 7.',c/to the decimal of a shilHng. Ans. '6^55. 
 
 5. Reduce 5 oz 12 dvvts 16 gr to lb. Ans. -46044 kc. lb 
 
 RULE OF THREE IN DECIMALS. 
 
 Prepare the terms by reducing the vulgar fractions to 
 decimals, and any compound numbers either to decimals of the 
 hii:her dcnominationsri. or to integers of the lower, also the 
 first and third terms to the same name : Then multiply and 
 divide as in whole numbers. 
 
 J\'ote, Any of the convenient Examples in the Rule of 
 Three or Rule of Five in Integers, or Vulgar Fractions, may 
 be taken as proper examples to the same rules in Decimals. 
 ' — The following Example, which is the first in Vulgar Frac- 
 tions, is wrought out iiere, to show the method. 
 
 Iff of a yard of velvet cost f/, what will -fj yd cost ? 
 
 yd / yd / s d 
 
 I = -375 -376 : -4 : : -3125 : -333 &c. or 6 8 
 
 •4 
 
 •375) -12500 (-333333 &c, 
 1250 20 
 
 125 
 
 s 6-66666 &c, 
 
 -^^ — 'snr^ 12 
 
 Ans. 6s 8d. d 7-99999 &c.=8a^ 
 
 DUOIDE. 
 
DUODECIMALS. 77 
 
 DUODECIMALS. 
 
 Duodecimals or Cross Multiplication, is a rule used 
 by workmen and artificers, in computing the contents of their 
 works. 
 
 Dimensions are usually taken in feet, inches, and quarters ; 
 any parts smaller than these being neglected as of no conse- 
 quence. And the same in multiplying them together, or cast- 
 ing up the contents. The method is as follows. 
 
 Set down the two dimensions to be multiplied together, one 
 under the other, so that feet may stand under feet, inches un- 
 der inches, &c. 
 
 Multiply each term in the multiplicand, beginning at the 
 lowest, by the feet in the multiplier, and set the result of each 
 straight under its corresponding term, observing to carry 1 for 
 every 1 ^ from he inches to the feet. 
 
 In like manner, multiply all the multiplicand by the inches 
 and parts of the multiplier, and set the result of each term 
 one place removed to the right-hand of those in the multipli- 
 cand ; omitting, however, what is below parts of inches, only 
 carrying to these the proper number of units from the lowest 
 denomination. 
 
 Or, instead of multiplying by the inches, take such parts ol 
 the multiplicand as there are of a foot. • 
 
 Then add the two lines together after the manner of [Com- 
 pound Addition, carr)'ing l.to the feet "for 12 inches, when 
 these come to so many. 
 
 EXAMPLES. 
 
 Multiply 4 f 7 inc 2. Multiply 14 f 9 inc 
 by 6 4 by 4 6 
 
 27 6 
 1 H 
 
 59 
 7 41 
 
 Ans. 29 Oi 
 
 Ans. 66 4-L 
 
 3. Multiply 4 feet 7 inches by 9 f 6 inc. Ans. 43 f 6^ inc. 
 
 4. Multiply 12 f 6 inc by 6 f 8 inc. Ans. 82 9i 
 
 5. Multiply 35 fU inc by 12 f 3 inc. Ans. 433 4} 
 
 6. Multiply 64 f 6 inc by 8 f 9^ inc. Ans. 665 8f 
 
 INVOLUTION. 
 
71 
 
 ARITHMETIC* 
 
 INVOLUTION. 
 
 InvolVtiok is the raising of Powers from any giren number^ 
 as a root. 
 
 A Power is a quantity produced by multiplying any given 
 number, called the Root, a certain number of times continual- 
 ly by itself. Thus, 
 
 2 = 2 is the root, or 1st power of 2. 
 2X2= 4 is the 2d power, or square of 2. 
 2X2X2= 8 is the 3d power, or rube of 2. 
 2 X 2 X 2 X 2 = 16 is the 4th power of 2, &c. 
 
 And in this manner may be calculated the following Table of 
 of the first nine powers of the first 9 numbers. 
 
 TABLE OF THE FIRST NINE POWERS OF NUMBERS. 
 
 1st 
 
 1 
 
 2 
 
 2d 
 1 
 4 
 
 3d 
 
 4th 
 
 5th 
 
 6th 
 
 7th 
 
 8th 
 
 9th 
 
 8 
 
 1 
 
 16 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 32 
 
 64 
 
 128 
 
 256 
 
 ,612 
 
 3 
 
 9 
 
 27 
 
 81 
 
 243 
 
 729 
 
 2187 
 
 6661 
 
 19683 
 
 4 
 
 16 
 
 64 
 
 256 
 
 1024 
 
 4096 
 
 16384 
 
 65536 
 
 262144 
 
 5 
 
 25 
 
 125 
 
 626 
 
 31 £5 
 
 15625 
 
 78125 
 
 390625 
 
 1953126 
 
 •6 
 
 36 
 
 216 
 
 1296 
 
 7776 
 
 466S6 
 
 279936 
 
 1679616 
 
 10077696 
 
 7 
 
 49 
 
 343 
 
 2401 
 
 168U7 
 
 117649 
 
 823543 
 
 5764801 
 
 40353607 
 
 8 
 
 64 
 81 
 
 512 
 
 4096 
 
 32768 
 
 262144 
 
 2097152 
 
 16777216 
 
 134217728 
 
 9 
 
 729 
 
 6561 
 
 59049 
 
 531441 
 
 4782969 
 
 43046721 
 
 387420489 
 
 The 
 
INVOLUTION. 19 
 
 The Index or Exponent of a Power, is the number de- 
 noting the height or degree of that power ; and it is 1 more 
 than the number of multiplications used in producing the 
 same. So 1 is the index or exponent of the first power or 
 root, two of the 2d power or square, 3 of the third power or 
 cube, 4 of the 4th power, and so on. 
 
 Powers, that are to be raised, are usually denoted by placing 
 the index above the root or first power. 
 
 So 23 = 4 is the 2d power of 2. 
 2» = 8 is the 3d power of 2. 
 2* =16 is the 4th power of 2, 
 540* is the 4tb power of 640, &c. 
 
 When two or more powers are multiplied together, their 
 product is that power whose index is the sum of the expo- 
 nents of the factors or powers multiplied. Or the multipli- 
 cation of the powers, answers to the addition of the indices. 
 Thus, in the following powers of 2, 
 
 1st 2d 3d 4th 6th 6th 7th 8th 9th 10th 
 2 4 8 IG 32 64 128 256 512 1024 
 or 2» 22 2» 2* 2* 2^ 2^ 2^ 2* 2»» 
 
 Here, 4 X 4 = 16, and 2 -f 2 = 4 its index ; 
 and 8 X 16 = 128, and 3 -f 4 = 7 its index ; 
 also 16 X 64 = 1024, and 4 -f- 6 = 10 its index. 
 
 OTHER EXAMPLES. 
 
 1. What is the 2d power of 45 ? Ans. 2025. 
 
 2. What is the square of 4-16 ? Ans. 17-3056. 
 
 3. What is the 3d power of 3-5 ? Ans. 42-875. 
 
 4. What is the 6th power of -029 ? Ans. -00000002051 1 149. 
 
 5. What is the square of | ? Ans. f . 
 
 6. What is the 3d power of f ? Ans. iff. 
 
 7. What is the 4th power of f ? Ans. /jV- 
 
 JSVOLUTION. 
 
80 ARITHMETIC. 
 
 EVOLUTION. 
 
 EvottfnoN, or the reverse of Involution, is the extracting 
 or finding the roots of any given powers. 
 
 The root of any number, or power, is such a number, as 
 being multiplied into itself a certain number of times, will 
 produce that power. Thus, 2 is the square root or 2d root 
 of 4, because 23 = 2 X 2 = 4 ; and 3 is the cube root or 3d 
 root of 27, because S^ = 3 X 3 X 3 ~ 27. 
 
 Any power of a given number or root may be found ex- 
 actly, namely, by multiplying the number continually into 
 itself. But there are many numbers of which a proposed root 
 can never be exactly found. Yet, by means of decimals, we 
 may approximate or approach towards the root, to any de^ 
 gr6e of exactness. 
 
 Those roots which only approximate, are called Surd 
 roots ; but those which can be found quite exact, are called 
 Rational Roots. Thus, the square root of 3 is a surd root ; 
 but the square root of 4 is a rational root, being equal to 2 : 
 also the cube root of 8 is rational, being eq.ual to 2 ; but the 
 cube root of 9 is surd or irrational. 
 
 Roots are sometimes denoted by writing the character ^ 
 before the power, with the index of the root against it. 
 Thus, the 3d root of 20 is expressed by ^ 20 ; and the square 
 root or 2d root of it is ^ 20, the index 2 being always omit- 
 ted, when only the square root is designed. 
 
 When the power is expressed by several numbers, with the 
 sign -h or — between them, a line is drawn from the top of 
 the sign over all the parts of it : thus the third root of 
 45 — 12 is ^ 45— 12, or thus ^ (45 --. 12), inclosing the 
 numbers in parentheses. 
 
 But all roots are now often designed like powers, with 
 fractional indices : thus, the square root of 8 is 8^, the cube 
 root of 25 is 253, and the 4th root of 45 — 18 is 45— 18)*, 
 or (45— 18)^. 
 
SQUARE ROOT. 81. 
 
 TO EXTRACT THE SQUARE ROOT. 
 
 * Divide the given number into periods of two figures 
 each, by setting a point over the place of units, another over 
 the place of hundreds, and so on, over every second figure, 
 both to the left-hand in integers, and te the right in deci- 
 mals. 
 
 Find the greatest square in the first period on the left-hand, 
 and set its root on the right hand of the given number, after 
 the manner of a quotient figure in Division. 
 
 * The reason for separating the figures of the dividend into periods 
 or portions of two places each, is, that the square of any single figure 
 never consists of more than two places ; the square of a number of 
 two figures, of not more than four places, and so on. So that there 
 will be as many figures in the root as the given number contains periods 
 so divided or parted off. 
 
 And the reason of the several steps in the operation appears from 
 the algebraic form of the square of any number of terms, whether two 
 or three or more. Thus, 
 
 (a + ^)^=a« + 2ab + b*=a^ + (2a + b) 6, the square of two 
 terms ; where it appears that a is the first term of the root, and b 
 the second term j also a the first divisor, and the new divisor is 
 2a + by or double the first term increased by the second. And hence 
 the manner of extraction is thus : 
 
 . 1st divisor a) az -j- 2ab + b^ ( a + 6 the root. 
 
 2d divisor 2a + 6 I 2(2^ + b- 
 b\2ab ~^b2 
 
 Again, for a root of three parts, a, b, c, thus : 
 
 (a + 6 -f- c) 2 fl2 + 2a6 -f- 62 -f- 2ac -|- 2bc + c^ = 
 
 a2 + {2a + 6) A + (2a 4 26 + c) c, the 
 square of three terms, where a is the first term of the root b, the 
 second, and c the third term ; also a the first divisor, 2a ^b the 
 second, and 2a ■\'2b -f- c the third, each consisting of the double of 
 the root increased by the next term of the same. And the mode of 
 extraction is thus : 
 1st divisor a) a^ -{• 2ab + b"^ + 2ac-f-2dc -f- c' (a + 6 -f- c the root. 
 
 2d divisor 2a'^b\2ab + b^ 
 b\2ab-\- b^ 
 
 od divisor 2<i f 26 -f c\ 2ac -f 2bc + c* 
 c I 2ac + 26c + c2 
 
 Vol. I. }%^ Subtract 
 
82 ARlTHMEtie. 
 
 Subtract the square thus found from the said period, atid 
 to the remaiiidar annex the two figures of the next following 
 period, for a dividend. 
 
 Double the root above mentioned for a divisor ; nnd find 
 how often it is contained in the said dividend, exclusive of 
 its right-hand figure ; and set that quotient figure both in the 
 quotient and divisor. 
 
 Multiply the whole augmented divisor by this last quotient 
 figure, and subtract the product from the said dividend, 
 bringing down to it the next period of the given number, for 
 a new dividend. 
 
 Repeat the same process over again, viz. find another new 
 divisor, by doubling all the figures now found in the root ; 
 from which, and the last dividend, find the next figure of 
 the root as before ; and so on through all the periods, to the. 
 last. 
 
 Note, The best way of doubling the root, to form the new 
 divisors, is by adding the last figure always to the last divisor, 
 as appears in the following examples. — Also, after the figures 
 belonging to the given number are all exhausted, the opera- 
 tion may be continued into decimals at pleasure, by adding any 
 number of periods of ciphers, two in each period. 
 
 EXAMPLES. 
 
 1. To find the square root of 29506624. 
 
 29506624 ( 6432 the root; 
 25 
 
 104 
 
 4 
 
 450 
 416 
 
 1083 
 3 
 
 34G6 
 3249 
 
 10862 
 
 2 
 
 21724 
 21724 
 
 Note, When the root is to he extracted to many places of figures ^ 
 the work may he considerably shortened^ thus • 
 
 Having proceeded in the extraction after the common me- 
 fliod, till there be found half the required number of figures 
 
SQ.UARE ROOT. 
 
 8S 
 
 in the root, or one figure more ; then, for the rest, divide 
 the last remainder by its corresponding diWsor, after the man- 
 ner of the third contraction in Division of Decimals ; thus, 
 2. To find the root of 2 to nine places of figures. 
 
 2 ( 1 '4 142 1366 the root. 
 
 1 
 
 24 
 4 
 
 100 
 96 
 
 281 
 
 1 
 
 400 
 281 
 
 2824 11900 
 4 11296 
 
 28282 
 2 
 
 60400 
 
 6G5G4 
 
 28284 ) 
 
 3836 ( 
 1008 
 160 
 19 
 
 1356 
 
 3. What 
 
 4. What 
 
 5. What 
 
 6. What 
 
 7. What 
 
 8. What 
 
 9. What 
 10. What 
 Ih What 
 12. What 
 
 is the square root of 2025 ? 
 is the square root of 17-3056 ? 
 is the square root of -000729 ? 
 is the square root of 3 ? 
 is the square root of 5 ? 
 is the square root of 6 ? 
 is the square root of 7 ? 
 is the square root of 10 ? 
 is the square root of 11 ? ' 
 is the square root of 12 ? 
 
 Ans. 4S. 
 
 Ans. 416. 
 
 Ans. -027. 
 Ans. 1-732050. 
 Ans. 2-236068. 
 Ans. 2-449489. 
 Ans. 2-643751. 
 Ans. 3-162277. 
 Ans. 3-316624. 
 Ans. 3-464101. 
 
 RULES FOR THE SQUARE HOOTS OF VULGAR FRACTIONS 
 AND MIXED NUMBERS. 
 
 First prepare all vulgar fractions, by reducing them to 
 their least terms, both for this and all other roots. Then 
 
 1. Take the root of the numerator and of the denominator 
 for the respective terms of the root required. And this is the 
 best Way if the denominator be a complete power : but if it 
 be not, then 
 
 2. Multiply the numerator and denominator together ; 
 take the root of the product : this root being made the nume 
 
 rator 
 
84 
 
 ARITHMETIC. 
 
 rator to the denominator of the given fraction, or made the 
 denominator to the numerator of it, will form the fractional 
 root required. 
 
 Thatis,y--- --^^ 
 
 ^ab. 
 
 And this rule will serve, whether the root be finite or infinite. 
 
 3. Or reduce the vulgar fraction to a decimal, and extract 
 its root. 
 
 4. Mixed numbers may be either reduced to improper 
 fractions, and extracted by the first or second rule, or the 
 vulgar fraction may be reduced to a decimal, then joined to 
 tUe integer, and the root of the whole extracted. 
 
 EXAMPLES. 
 
 i. What is the root of |f ? Ans. -f . 
 
 2. What is the root of ^Vt ? Ans. -f. 
 
 3. What is the. root of yV ? Ans. 0-866025. 
 
 4. What is the root of -f^ 1 Ans. 0-645497. 
 
 5. What is the root of 17f ? ' Ans. 4-168333. 
 By means of the square root also may readily be found the 
 
 4th root, or the 8th root, or the 16th root, &c. tlrat is, the 
 root of any power whose index is some power of the number 
 2 ; namely, by extracting so often the square root a»s is de- 
 noted by that power of 2 : that is, two extractions for the 
 4th root, three for the 8th root, and so on. 
 
 So, to find the 4th root of the number 21035'8, extract 
 the square root two times as follows z 
 
 21035 8000 
 1 
 
 (145 037237 (12-0431407 the 4th root. 
 
 X 
 
 24 I 110 
 4 96 
 
 22 I 45 
 21 44 
 
 285 I 
 
 6 ' 
 
 1435 
 1425 
 
 2404 
 
 4 
 
 10372 
 9616 
 
 29003 
 3 
 
 108000 24083 
 87009 3 
 
 75637 
 
 72249 
 
 , 20991 (.7237 3388 (1407 
 687 980 
 
 107 17 
 
 Ex. 2. What is the 4th root of 97-41 ? 
 
 To 
 
CUBE RO0T. B5 
 
 TO EXTRACT THE CUBE ROOT. 
 1. By the Common Rule,* 
 
 1. Having divided the given number into periods of three 
 figures each, (by setting a point over the place of units, and 
 also over every third figure, from thence, to the left hand in 
 whole numbers, and to the right in decimals), find the nearest 
 less cube to the first period ; set its root in the quotient, and 
 subtract the said cube from the first period ; to the remainder 
 brine down the second period, and call this the resolvend. 
 
 2. To three times the square of the root, just found, add 
 three times the root itself, setting this one place more to the 
 right than the former, and call this sum the divisor. Then 
 divide the resolvend, wanting the last figure, by the divisor, 
 for the next figure of the root, which annex to the former ; 
 calling this last figure e, and the part of the root before found 
 let be called a. 
 
 Add all together these three products, namely, thrice a 
 square multiplied by e, thrice a multiplied by e square, and 
 e cube, setting each of them one place more to the right than 
 the former, and call the sum the subtrahend ; which must 
 not exceed the resolvend ; but if it does, then make the last 
 figure e less, and repeat the operation for finding the subtra- 
 hend, till it be less than the resolvend. 
 
 4. From the resolvend take the subtrahend, and to the re- 
 mainder join the next period of the given number for a new 
 resolvend ; to which form a new divisor from the whole root 
 now found ; and from thence another figure of the root, as 
 directed in Article 2, and so on. 
 
 * The reason for pointing the given number into periods of three 
 figures each, is because the cube of one figure never amounts to more 
 than three places. And, for a siinilar reason, a given number is point- 
 ed into periods of four figures for the 4th root, of five figures for 
 the 5tli root, and so on. • 
 
 And the reason for the other parts of the rule depends on the 
 algebraic formation of a cube : for if the root consists of the two 
 parts a + /;, then its cube is as foUow^s : (a + 6)^ =a3 -J- 3a2b-j- 
 3(ibs + ^3 ; where a is the root of the first part u^ ; the resolvend is 
 Sa^b •{• 5ab2 -f- ^^j which is also the same as the three parts of 
 the subtrahend ; also the divisor is 3v3 -|- 3a, by which dividing the 
 first two terms of the resolvend 3a^ b + ab^ , gives b for the second 
 part of the root ; and so on. 
 
 EXAMPLE. 
 
96 ARITHMETIC. 
 
 EXAMPLE. 
 
 To extract the cube root of 48228-544. 
 
 3 X 32 = 27 
 3X3 = 09 
 
 Divisor 279 
 
 48228-544 (36-4 root. 
 27 
 
 21228 resolvend. 
 
 3 X 32 X 6 =162 ) 
 3X3 X 62 = 324 > add 
 
 63= 216' 
 
 3 X 362 = 3888 
 3 X 36 = 108 
 
 38988 
 
 19666 subtrahend. 
 
 1572544 resolvend. 
 
 3 X 362 X 4 = 15552 
 3 X 36 X 42 = 1728) add 
 43 = 64 ^ 
 
 1572544 subtrahend. 
 
 0000000 remainder. 
 
 Ex. 2. Extract the cube root of 571482-19. 
 Ex. 3. Extract the cube root of 1628-1582. 
 Ex. 4. Extract the cube root of 1332. 
 
 II. To extract the Cube Root by a short Way.* 
 
 1. By trials, or by the table of roots at p. 90, &c. take the 
 nearest rational cube to the given number, whether it be 
 greater or less ; and call it the assumed cube. 
 
 2. Then 
 
 * The method usually given for extracting the cube root, is so 
 exceedingly tedious, and difficult to be remembered, that various 
 other approximating rules have been invented; viz. by Newton, 
 Raphson, Halley, De Lagny, Simpson, Emerson, and several other 
 mathematicians ; but no one that I have yet seen, is so simple in 
 its form, or seems so -well adapted for general use, as that above 
 given.. This rule is the same in effect as Dp. Halley's rational 
 
 formula. 
 
CUBE ROOT. 87 
 
 2. Then say, by the Rule of Three, As th6 sum of the givea 
 number and double the assumed cube, is to the sum of the 
 assumed cube and double the given number, so is the root of 
 the assumi'd cube, to the root required, nearly. Or, As the 
 fir^t sum is to the difference of the given and assumed cube, 
 so IS the assumed root to the difference of the roots nearly. 
 
 3. Again, by using, in like manner, the cube of the root 
 last found as a new assumed cube, another root will be obtain- 
 ed still nearer. And so on as far as we please ; using always 
 the cube of the last found root, for the assumed cube. 
 
 EXAMPLE. 
 
 To find the cube root of 21035-8. 
 
 Here we soon find that the root lies between 20 and 30, and 
 then between 27 and 28. Taking therefore 27, its cube is 
 19683, which is the assumed cube. Then 
 19683 21035-8 
 
 2 2 
 
 39366 
 21036-8 
 
 42071-6 
 19683 
 
 60401-8 : 61764-6 : 
 27 
 
 : : 27 : 27-6047. 
 
 4322822 
 1236092 
 
 
 60401-8) 1667374-2 (27-6047 the root nearly. 
 
 469338 
 
 36626 
 
 284 
 
 42 
 
 formula) but more commodiously expressed ; and the first investiga- 
 tion of it was given in my Tracts, p. 49. The algebraic form of it 
 is this : 
 
 As p + 2a : A + 2p : : r : R. Or, 
 
 As p 4" 2-A- : p c/J A : : r ; R c/J r ,- 
 where p is the given number, a the assumed nearest cube, r the cube 
 root of A, and r the root of p sought* 
 
 Again, 
 
88 ARITHMETIC. 
 
 Again, for a second operation, the cube of this root is 
 21035-318645155823, and the process by the latter method 
 will be thus : 
 
 21035-318646, &c. 
 
 42070-637290 21035-8 
 
 21035-8 21035-318G45, &c. 
 
 As 63106-43729 : diff. 481355 : : 27-6047 : 
 
 thediff. -000210560 
 
 conseq. the root req. is 27-604910560. 
 Ex. 2. To extract the cube root of -67. 
 Ex. 3. To extract the cube root of -01. 
 
 TO EXTRACT ANY ROOT WHATEVER. 
 
 Let p be the given power or number, n the index of the 
 power, A the assumed power, r its root, r the required root 
 of p. Then say, 
 
 As the sum of w -f- 1 times a and ?i — 1 times p^ 
 is to the sum of n -{- 1 times p and n — 1 times a ; 
 so is the assumed root r, to the required root R. 
 
 Or, as half the said sum of n -j- 1 times a, and n — 1 times 
 p, is to the difference between the given and assumed powers, 
 so is the assumed root r, to the difference between the true 
 and assumed roots ; which difference, added or subtracted, as 
 the case requires, gives the true root nearly. 
 
 Thati8,n-|-1. A+ n — 1. p : n-fl.p. -f*n — 1. a :: r : r. 
 
 Or, n+ \.\A-\-n — ^^ 1. |p : p c/2 a : ; r : r co r. 
 And the operation may be repeated as often as we please, 
 by using always the last found root for the assumed root, and 
 its nth power for the assumed power a. 
 
 ♦ This is a very g-eneral approximating rule, of which that for the 
 cube root is a particular case, and is the best adapted for practice, 
 and fnr memory, of any that I have yet seen. It was first discoveu-ed 
 in this form by myself, and the investigation and uge of it were given 
 at large in my Tracts^ p. 45, &ۥ 
 
 EXAMPLE. 
 
GENERAL ROOTS, 
 
 89 
 
 EXAMPLE. 
 To extract the 5th root of 21035-8. 
 
 Here it appears that the 5tli root is between 7 3 and 7-4. 
 Takinj?7-3, its 6th power is 20730-71693. Hence we have 
 p = 21036-8, n = 5 r = 7-3 and a = 20730-71593 ; then 
 
 n -f- 1 . i A-f-n — l.^Ptpco A :: r : Rco r, that is, 
 3 X20730-71693 +2 X 21035-8 : 305 084 : : 7-3 : 
 3 2 7-3 
 
 
 62192-14779 42071*6 915252 
 
 
 
 42071-6 2135688 
 
 
 
 
 (04263-74779 ^227-1 132 (-02I3605=R, co 
 
 
 
 7-3= 
 
 :r, add. 
 
 
 7-321360=R,true 
 
 
 
 to the last figure. 
 
 
 OTHER EXAMPLES. 
 
 
 
 1. 
 
 . What is the 3d root of 2 ? 
 
 Ans. 
 
 1-25992L 
 
 2, 
 
 . What is the 3d root of 3214 ? 
 
 Ans. 
 
 14-75758. 
 
 3. 
 
 What is the 4th root of 2 ? 
 
 Ans. 
 
 1-189207. 
 
 4. 
 
 What is the 4th root of 97-41 ?, 
 
 Ans. 
 
 3-1415999. 
 
 5. 
 
 What is the 5th root of 2 ? 
 
 Ans. 
 
 1-148699. 
 
 6. 
 
 What is the 6th root of 21035-8 ? 
 
 Ans. 
 
 5-254037. 
 
 7. 
 
 What is the 6th root of 2 ? 
 
 Ans. 
 
 1-122462. 
 
 8. 
 
 What is the 7th root of 21036-8 ? 
 
 Ans. 
 
 4-145392. 
 
 9. 
 
 What is the 7th root of 2 ? 
 
 Ans. 
 
 1 104089. 
 
 10. 
 
 What is the 8th root of 21035-8 ? 
 
 Ans. 
 
 3 470323. 
 
 11. 
 
 What is the 8th root of 2 ? 
 
 Ans. 
 
 1-090508. 
 
 12. 
 
 What is the 9th root of 21036-8 ? 
 
 Ans. 
 
 3-022239. 
 
 13. 
 
 What is the 9th root of 2 ? 
 
 Ans. 
 
 1-080069. 
 
 The following is a Table of squares and cub^s, as also the 
 'square roots and cube roots, of all numbers from 1 to 1000, 
 which will be found very useful on many occasions, in numeral 
 calculations, when roots or powers are concerned. 
 
 Vol. I, 
 
 13 
 
90 A TABLE OF SQUARES, CUBES, AND ROOTS. 
 
 Number. 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 1 
 
 1 
 
 1 
 
 1 0000000 
 
 i-oooooo 
 
 2 
 
 4 
 
 8 
 
 1-4142136 
 
 1-259921 
 
 3 
 
 9 
 
 27 
 
 1-7320508 
 
 •1-442250 
 
 4 
 
 16 
 
 64 
 
 2-0000000 
 
 1-587401 
 
 5 
 
 25 
 
 125 
 
 2-2360680 
 
 1-709976 
 
 6 
 
 36 
 
 216 
 
 2-4494897 
 
 1-817121 
 
 7 
 
 49 
 
 343 
 
 2-6457513 
 
 1-912933 
 
 8 
 
 64 
 
 512 
 
 2-8284271 
 
 2*000000 
 
 9 
 
 81 
 
 729 
 
 3-0000000 
 
 2*080084 
 
 10 
 
 loo 
 
 1000 
 
 3 1622777 
 
 2-154435 
 
 11 
 
 121 
 
 1331 
 
 3-31G6248 
 
 2223980 
 
 12 
 
 144 
 
 1728 
 
 3-4641016 
 
 2-289428 
 
 13 
 
 169 
 
 2197 
 
 3-6055513 
 
 2-351335 
 
 14 
 
 196 
 
 2744 
 
 3-7416574 
 
 2-410142 
 
 15 
 
 225 
 
 3375 
 
 3-8729833 
 
 2'466212 
 
 16 
 
 256 
 
 4096 
 
 4-0000000 
 
 2-519842 
 
 17 
 
 289 
 
 4913 
 
 4-1231056 
 
 2-571282 
 
 18 
 
 324 
 
 5832 
 
 4-2426407 
 
 2-620741 
 
 19 
 
 361 
 
 6859 
 
 4-3588989 
 
 2 668402 
 
 20 
 
 400 
 
 8000 
 
 4-4721360 
 
 2-714418 
 
 21 
 
 441 
 
 9.261 
 
 4-5825757 
 
 2-758923 
 
 22 
 
 484 
 
 10648 
 
 4-6904158 
 
 2-802039 
 
 23 
 
 529 
 
 12167 
 
 4-7958315 
 
 2-843867 
 
 24 
 
 576 
 
 13824 
 
 4-8989795 
 
 2-884499 
 
 25 
 
 625 
 
 15625 
 
 5-0000000 
 
 2-924018 
 
 26 
 
 676 
 
 17576 
 
 50990195 
 
 2-962496 
 
 27 
 
 729 
 
 19683 
 
 5-1961524 
 
 3-000000 
 
 28 
 
 784 
 
 21952 
 
 5-2915026 
 
 3-036589 
 
 29 
 
 841 
 
 24389 
 
 5-3851648 
 
 3072317 
 
 30 
 
 900 
 
 27000 
 
 5-4772256 
 
 3-107232 
 
 31 
 
 961 
 
 29791 
 
 5-5677644 
 
 3 141381 
 
 32 
 
 1024 
 
 32768 
 
 5-6568642 
 
 3-174802 
 
 33 
 
 1089 
 
 35937 
 
 5-7445626 
 
 3-207534 
 
 34 
 
 1156 
 
 39304 
 
 5-8309519 
 
 3-239612 
 
 35 
 
 1225 
 
 42875 
 
 5-9160798 
 
 3-271066 
 
 36 
 
 1296 
 
 46656 
 
 6-0000000 
 
 3301927 
 
 37 
 
 1369 
 
 50653 
 
 6-0827625 
 
 3-332222 
 
 38 
 
 1444 
 
 54872 
 
 6 1644140 
 
 3-361975 
 
 39 
 
 1521 
 
 59319 
 
 6 2449980 
 
 , 3-39121 1 
 
 40 
 
 1600 
 
 64000 
 
 6-3245553 
 
 3-419952 
 
 41 
 
 1681 
 
 68921 
 
 6-4031242 
 
 3 448217 
 
 42 
 
 1764 
 
 74088 
 
 64807407 
 
 3-476027 
 
 43 
 
 1849 
 
 79507 
 
 6-5574385 
 
 3-503398 
 
 44 
 
 1959 
 
 85184 
 
 6-6332496 
 
 3 530348 
 
 45 
 
 2025 
 
 91125 
 
 6-7082039 
 
 3556893 
 
 46 
 
 2116 
 
 97336 
 
 6-7823300 
 
 3 583048 
 
 47 
 
 2209 
 
 103823 
 
 6-8556546 
 
 3-608826 
 
 48 
 
 2304 
 
 1 10592 
 
 6*9282032 
 
 3-634241 
 
 49 
 
 2401 
 
 117649 
 
 7-0000000 
 
 3659306 
 
 50 
 
 2500 
 
 125000 
 
 7-0710678 
 
 3-684031 
 
SQUARES, CUB^S, AND ROOTS. 
 
 91 
 
 Number. 
 
 Square. 
 
 2601 
 
 . C«be. 
 
 Square Hoot. 
 
 Cube Koot. 
 
 51 
 
 132651 
 
 7 1414284 
 
 3 708430 
 
 52 
 
 2704 
 
 140608 
 
 7-2 111026 
 
 3-732511 
 
 53 
 
 2809 
 
 148877 
 
 7 2801099 
 
 3-756286 
 
 54 
 
 2916 
 
 157464 
 
 7-3484692 
 
 3779763 
 
 55 
 
 3025 
 
 166375 
 
 74161985 
 
 3-802953 
 
 56 
 
 3136 
 
 175616 
 
 7-4833148 > 
 
 3-825862 
 
 57 
 
 3249 
 
 185193 
 
 7-5498344 
 
 3 848501 
 
 58 
 
 3364 
 
 195112 
 
 7-6157731 
 
 3 870877 
 
 59 
 
 3481 
 
 205379 
 
 7-6811457 
 
 3-892996 
 
 60 
 
 3600 
 
 216000 
 
 7-7459667 
 
 3-914867 
 
 61 
 
 3721 
 
 226981 
 
 7 8102497 
 
 3-936497 
 
 62 
 
 3844 
 
 238328 
 
 7-8740079 
 
 3-957892 
 
 63 
 
 3969 
 
 250047 
 
 7-9372539 
 
 3*979057 
 
 64 
 
 •4096 
 
 262144 
 
 8 0000000 
 
 4-000000 
 
 65 
 
 4225 
 
 274625 
 
 8-0622577 
 
 4-020726 
 
 66 
 
 4356 
 
 287496 
 
 8-1240384 
 
 4-041240 
 
 67 
 
 4489 
 
 300763 
 
 8'1853528 
 
 4 061548 
 
 68 
 
 4624 
 
 314432 
 
 8-2462113 
 
 4081656 
 
 69 
 
 4761 
 
 328509 
 
 8-3066239 
 
 4 101566 
 
 70 
 
 4900 
 
 343000 
 
 8-3666003 
 
 4-121285 
 
 71 
 
 5041 
 
 357911 
 
 8-4261498 
 
 4 140818 
 
 72 
 
 5184 
 
 373248 
 
 8-4852814 
 
 4-160168 
 
 73 
 
 5329 
 
 389017 
 
 8-5440037 
 
 4 179339 
 
 74 
 
 5476 
 
 405224 
 
 8 6023253 
 
 4-198336 
 
 75 
 
 5625 
 
 421875 
 
 8-6602540 
 
 4-217183 
 
 76 
 
 5776 
 
 438976 
 
 5 7177979 
 
 4-235824 
 
 77 
 
 5929 
 
 456533 
 
 8-7749644 
 
 4-254321 
 
 78 
 
 6084 
 
 474552 
 
 8-8317609 
 
 4272659 
 
 79 
 
 6241 
 
 493039 
 
 8 8881944 
 
 4 290841 
 
 80 
 
 -6400 
 
 5 1 2000 
 
 8 9442719 
 
 4-308870 
 
 81 
 
 6561 
 
 531441 
 
 9 0000000 
 
 4326749 
 
 82 
 
 6724 
 
 551368 
 
 9 0553851 
 
 4-344481 
 
 83 
 
 6889 
 
 571787 
 
 9-1104336 
 
 4-362071 
 
 &4 
 
 7056 
 
 592704 
 
 9-1651514 
 
 4-379519 
 
 85 
 
 7225 
 
 614125 
 
 9-2195445 
 
 4-396830 . 
 
 86 
 
 7396 
 
 "636056 
 
 9-2736185 
 
 4-414005 
 
 87 
 
 7569 • 
 
 658503 
 
 9 3273791 
 
 4 431047 
 
 88 
 
 7744 
 
 681472 
 
 9-3808315 
 
 4-447960 
 
 89 
 
 7921 
 
 704969 
 
 9-4339811 
 
 4-464745 
 
 90 
 
 8100 
 
 720000 
 
 9 4868330 
 
 4-481405 
 
 91 
 
 8281 
 
 753571 
 
 9o393920 
 
 4-497942 
 
 92 
 
 8464 
 
 778688 
 
 9 5916630 
 
 4514357 
 
 93 
 
 8649 
 
 804357 
 
 9-6436508 
 
 4 530655 
 
 94 
 
 8836' 
 
 830584 
 
 9 6953597 
 
 4 546836 
 
 95 
 
 9025 
 
 857375 
 
 9-7467943 
 
 4-562903 
 
 9*6 
 
 9216 
 
 * 884736 
 
 9 7979590 
 
 4578857 
 
 97 
 
 9409 
 
 912673 
 
 9-8488578 
 
 4-594701 
 
 98 
 
 9604 
 
 941192 
 
 9-8994949 
 
 4-610436 
 
 99 
 
 9801 
 
 970299 
 
 99498744 
 
 4-626055 
 
 100 
 
 10000 
 
 1000000 
 
 10-0000000 
 
 4 641589 
 
92 
 
 ARITHMETIC. 
 
 
 lui 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root 
 
 
 10201 
 
 1030:;oi 
 
 10 0498756 
 
 4 657010 
 
 
 102 
 
 10404 
 
 1061208 
 
 100995049 
 
 4 672330 . 
 
 
 103 
 
 10609' 
 
 1092727 
 
 10-1488916 
 
 4-687548 
 
 
 104 
 
 10816 
 
 1124364 
 
 10 1980390 
 
 4-702669 
 
 
 105 
 
 1 ^9^5 
 
 1157625 
 
 10 2469508 
 
 4717694 
 
 
 106 
 
 1 lf^6 
 
 1191016 
 
 10 2456301 
 
 4-732624 
 
 
 i07 
 
 11449 
 
 1225043 
 
 10-3940804 
 
 4 747459 
 
 
 108 
 
 11664 
 
 1259712 
 
 10-3923048 
 
 - 4 762203 
 
 
 109 
 
 11881 
 
 1295029 
 
 10 4403065 
 
 4-776856 
 
 
 110 
 
 12100 
 
 1331000 
 
 10-4880885 
 
 4-791420 
 
 
 111 
 
 12321 
 
 1367631 
 
 10 5356538 
 
 4-80589.6 
 
 
 112 
 
 12544 
 
 1404928 
 
 10-5830052 
 
 4 820284 
 
 
 113 
 
 12769 
 
 1442897 
 
 I0'6301458 
 
 4-834588 
 
 
 114 
 
 12996 
 
 1481544 
 
 10-6770783 
 
 • 4-848808 
 
 
 115 
 
 13225 
 
 1520875 
 
 107238053 
 
 4-862944 
 
 
 116 
 
 13456 
 
 156<'896 
 
 10-7703296 
 
 4-876999 
 
 
 117 
 
 13689 
 
 16U1613 
 
 10-8166538 
 
 4-890973 
 
 
 118 
 
 13924 
 
 1643032 
 
 108627805 
 
 r4-904868 
 
 
 119 
 
 14161 
 
 1685159 
 
 109087121 
 
 4-918685 
 
 
 120 
 
 14400 
 
 1728000 
 
 10 9544512 
 
 4-932424 
 
 
 121 
 
 14641 
 
 1771561 
 
 11 0000000 
 
 4946088 
 
 
 122 
 
 14884 
 
 1815848 
 
 11-0453610 
 
 4 959675 
 
 
 123 
 
 15129 
 
 1860867 
 
 11-0905365 
 
 4-973190 
 
 
 124 
 
 15376 
 
 1906624 
 
 111355287 
 
 4986631 
 
 
 125 
 
 15625 
 
 1953125 
 
 11-1803399 
 
 5*000000 
 
 
 126 
 
 15876 
 
 2000376 
 
 11 2ia49722 
 
 5013298 
 
 
 127 
 
 16129 
 
 2048383 
 
 11-2694277 
 
 5-026526 
 
 
 128 
 
 16384 
 
 2097152 
 
 11 3137085 
 
 5 039684 
 
 
 129 
 
 16641 
 
 2146689 
 
 11 3578167 
 
 5 052774 
 
 
 150 
 
 16900 
 
 2197000 
 
 11 4017543 
 
 5-065797 
 
 
 131 
 
 17161 
 
 2248091 
 
 11-4455231 
 
 5-078753 
 
 
 132 
 
 17424 
 
 2299968 
 
 11-4891253 
 
 5091643 
 
 
 133 
 
 17689 
 
 2352637 
 
 11 5325626 
 
 5 104469 
 
 
 134 
 
 17956 
 
 2406104 
 
 11-5758369 
 
 5-117230 
 
 
 * 135 
 
 J 8225 
 
 2460375 
 
 ll'6i89500 
 
 5 129928 
 
 
 136 
 
 18496 
 
 2515456 
 
 11 66190;58 
 
 5-142563 
 
 
 13r 
 
 18769 
 
 2571353 
 
 11-7046999 
 
 5-155137 
 
 
 138 
 
 19044 
 
 2628072 
 
 11-7473444 
 
 5 167649 
 
 
 139 
 
 19321 
 
 2685619 
 
 11 7898261 
 
 5-180101 
 
 
 140 
 
 19600 
 
 2744000 
 
 11-8321596 
 
 5 192494 
 
 
 141 
 
 19881 
 
 2803221 
 
 11 '8743421 
 
 5204828 
 
 
 142 
 
 20164 
 
 2863288 
 
 11>9163753 
 
 5-217103 
 
 
 143 
 
 20449 
 
 2924207 
 
 11-9582607 
 
 5229321 
 
 
 144 
 
 20736 
 
 2985984 
 
 12-0000000 
 
 5241482 
 
 
 145 
 
 21025 
 
 3048625 
 
 120415946 
 
 5 25358S 
 
 
 146 
 
 . 21316 
 
 3112136 
 
 12-0830460 
 
 5 265637 
 
 
 147 
 
 2160-^ 
 
 3176523 
 
 12-1243557 
 
 5-277632- 
 
 
 148 
 
 21904 
 
 3241792 
 
 12-1655251 
 
 5-289572 
 
 
 149 
 
 22201 • 
 
 330794a 
 
 12-2065556 
 
 5 301459 
 
 150 1 
 
 22500 
 
 3375000 
 
 12-2474487 
 
 5-313293 
 
SQUARES, CUBES, AND ROOTS. 
 
 93 
 
 Numb. 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 151 
 
 22801 
 
 3442951 
 
 122882057 
 
 5-325074 
 
 152 
 
 23104 
 
 3511808 
 
 123288280 
 
 5-336803 
 
 153 
 
 23409 
 
 3581577 
 
 12 3693169 
 
 5 348481 
 
 154 
 
 23716 
 
 3652264 
 
 12-4096736 
 
 5 360108 
 
 155 
 
 24025 
 
 3723875 
 
 12-4498996 
 
 5-371685 
 
 156 
 
 24336 
 
 3796416 
 
 12 4899960 
 
 5 383213 
 
 157 
 
 24649 
 
 3869893 
 
 12 5299641 
 
 5o94G90 
 
 158 
 
 24964 
 
 3944312 
 
 12-5698051 
 
 5 4U6120 
 
 159 
 
 25281 
 
 4019679 
 
 126095202 
 
 5-417501 
 
 160 
 
 25600 
 
 4096000 
 
 12-6491106 
 
 5-428835 
 
 161 
 
 25921 
 
 4173281 
 
 12 6885775 
 
 5-440122 
 
 162 
 
 26244 
 
 4251528 
 
 12-7279221 
 
 5-451362 
 
 163 
 
 26569 
 
 4330747 
 
 12-7671453 
 
 5-462556 
 
 164 
 
 26896 
 
 4410944 
 
 12-8062485 
 
 5 475703 
 
 165 
 
 27225 
 
 4492125 
 
 12-8452326 
 
 5-484806 
 
 166 
 
 27556 
 
 4574296 
 
 128840987 
 
 5 495865 
 
 167 
 
 27889 
 
 4657463 
 
 12-9228480 
 
 5-506879 
 
 168 
 
 28224 
 
 4741632 
 
 12-9614814 
 
 5-517848 
 
 169 
 
 28561 
 
 4826809 
 
 13-0000000 
 
 5 528775 
 
 170* 
 
 28900 
 
 49 1 3000 
 
 15-0384048 
 
 5-539658 
 
 171 
 
 29241 
 
 5000211 
 
 130766968 
 
 5-550499 
 
 172 
 
 29584 
 
 5088448 
 
 13-1148770 
 
 5-561298 
 
 173 
 
 29929 
 
 5177717 
 
 13-1529464 
 
 5-572054 
 
 174 
 
 30276 
 
 5268024 
 
 13-1909060 
 
 5'582770 
 
 175 
 
 30625 
 
 5359375 
 
 13 2287566 
 
 5-593445 
 
 176 
 
 30976 
 
 545 1776 
 
 13'2664992 
 
 5 '604079 
 
 177 
 
 31329 
 
 5545133 
 
 •13 3041347 
 
 5614673^ 
 
 178 
 
 31684 
 
 5639752 
 
 13 3416641 
 
 5-625226 
 
 179 
 
 3204 1 
 
 5735339 . 
 
 13 3790882 
 
 5-635741 
 
 180 
 
 32400 
 
 5832000 
 
 • 134164079 
 
 5-646216 
 
 181 
 
 32761 
 
 5929741 
 
 13-4536240 
 
 5-656652 
 
 182 
 
 33124 
 
 6028568 
 
 ' 134907376 
 
 5 667051 
 
 183 
 
 33480 
 
 6128487 
 
 13 5277493 
 
 5-677411 
 
 184 
 
 33856 
 
 6229504 
 
 13-5646600 
 
 5-687734 
 
 185 
 
 34225 
 
 6331625 
 
 13-6014705 
 
 5-698019 
 
 186 
 
 34596 
 
 6454856 
 
 13-6381 8 '7 
 
 5708267 
 
 187 
 
 34969 
 
 6539203 
 
 13 6747943 
 
 57 18479 
 
 138 
 
 35344 
 
 6644672 
 
 13 7113092 
 
 •5'728654 
 
 189 
 
 35721 
 
 6751269 
 
 13 7477271 
 
 5-738794 
 
 190 
 
 36100 
 
 6859000 
 
 13-7840488 
 
 5-748897 
 
 191 
 
 36481 
 
 6967871 
 
 13-8202750 
 
 5-7589&5 
 
 192 
 
 36864 
 
 7077888 
 
 13-8564065 
 
 5 768998 
 
 193 
 
 37249 
 
 7189057 
 
 15-8924440 . 
 
 5 778996 
 
 194 
 
 37636 
 
 7301384 
 
 13-9283883 
 
 5-788960 
 
 195 
 
 38025 
 
 7414875 
 
 13'-9642400 
 
 . 5-798890 
 
 196 
 
 38416 
 
 7529536 
 
 14-0000000 
 
 5-808786 
 
 197 
 
 38809 
 
 7645373 
 
 140356688 
 
 5-818648 
 
 198 
 
 39204 
 
 7762392 
 
 14'0712473 
 
 5-828476 
 
 199 
 
 39601 
 
 7880599 
 
 14-1067360 
 
 5-838272 
 
 200 
 
 40000 
 
 8000000 
 
 14-1421356 
 
 5 848035 
 
y4 
 
 
 ARITHMETIC. 
 
 
 Numb. 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 ~ 201 
 
 40401 
 
 8120601 
 
 14M774469 
 
 5-857765 
 
 202 
 
 40804 
 
 8242408 
 
 14-2126704 
 
 5-867464 
 
 203 
 
 41209 
 
 8365427 
 
 14-2478068 
 
 5-8771*30 
 
 204. 
 
 41616 
 
 8489664 
 
 14-2828569 
 
 5-886765 
 
 < 205 
 
 42025 
 
 8615125 
 
 J4-3178211 
 
 5-896368 
 
 206 
 
 42436 
 
 8741816 
 
 143527001 
 
 5905941 
 
 207 
 
 42849 
 
 8869743 
 
 14-3874946 
 
 5-915481 
 
 208 
 
 43264 
 
 8998912 
 
 14-4222051 
 
 5-924991 
 
 209 
 
 43681 
 
 9123329 
 
 14-4568323 
 
 5-934473 
 
 210 
 
 44 i 00 
 
 9261000 
 
 14-4913767 
 
 5-9439 li 
 
 211 
 
 44521 
 
 9393951 
 
 14-5258390 
 
 5-953341 
 
 212 
 
 44944 
 
 9528128 
 
 14-5602198 
 
 5-962731 
 
 213 
 
 45369 
 
 9663597 
 
 14-5945195 
 
 5-972091 
 
 214 
 
 45796 
 
 9800344 
 
 14-6287388 
 
 5981426 
 
 2J5 
 
 46225 
 
 9938375 
 
 14-6628783 
 
 5-909727 
 
 216 
 
 46656 
 
 10077696 
 
 14-6969385 
 
 6-000000 
 
 2ir 
 
 47089 
 
 10218313 
 
 14-7309199 
 
 6-009244 
 
 218 
 
 47524 
 
 10360232 
 
 14-7648231 
 
 6-018463 
 
 219 
 
 47961 
 
 10503459 
 
 14-7986486 
 
 6 027650 
 
 220 
 
 48400 
 
 10648000 
 
 14 8323970 
 
 6036811 
 
 221 
 
 48841 
 
 10793861 
 
 14-8660687 
 
 6-045943 
 
 222 
 
 49284 
 
 10941048 
 
 14-8996644 
 
 6-055048 
 
 223 
 
 49729 
 
 11089567 
 
 14 9331845 
 
 6-064126 
 
 224 
 
 50176 
 
 11239424 
 
 14-9666295 
 
 6-073177 
 
 225 
 
 50625 
 
 11390625 
 
 15 0000000 
 
 6 082201 
 
 226 
 
 51076 
 
 11543176 
 
 , 15 0332964 
 
 6-091199 
 
 227 
 
 51529 
 
 11697083 
 
 15-0665192 
 
 6-100170 
 
 228 
 
 51984 
 
 11852352 
 
 15-0996689 
 
 6-109115 
 
 229 
 
 52441 
 
 12008989 
 
 15 1327460 
 
 6-118032 
 
 230 
 
 52900 
 
 12167000 
 
 15-1657509 
 
 6 126925 
 
 231 
 
 53361 
 
 12326391 
 
 15-1986842 
 
 6-135792 
 
 232 
 
 53824 
 
 12487168 
 
 15-2315462 
 
 6-144634 
 
 233 
 
 54289 
 
 12649337 
 
 15-2643375 
 
 6-153449 
 
 234 
 
 54756 
 
 12812904 
 
 15-2970585 
 
 6-162239 
 
 235 
 
 55225 
 
 12977875 
 
 15-3297097 
 
 6-171005 
 
 236 
 
 55696 
 
 13144256 
 
 15-3622915 
 
 6-179747 
 
 237 
 
 56169 
 
 13312053 
 
 15-3948043 
 
 6-188463 
 
 238 
 
 56644 
 
 13481272 
 
 15-4272486 
 
 6-197154 
 
 239 
 
 57121 
 
 13651919 
 
 15-4596248 
 
 6 205821 
 
 240 
 
 57600 
 
 13824000 
 
 15-4919334 
 
 6 214464 
 
 241 
 
 58081 
 
 13997521 
 
 15-5241747 
 
 6-223083 
 
 242 
 
 58564 
 
 14172488 
 
 15-5563492 
 
 6-231678 
 
 243 . 
 
 59049 
 
 14348907 
 
 155884573 
 
 6-240251 
 
 244 
 
 59536 
 
 14526784 
 
 15-6204994 
 
 6 248800 
 
 245 
 
 60025 
 
 14706125 
 
 15-^524758 
 
 6 257324 
 
 246 
 
 60516 
 
 14886936 
 
 15-6843871 
 
 6 265826 
 
 247 
 
 61009 
 
 15069223 
 
 15-7162336 
 
 6-274304 
 
 248i 
 
 61504 
 
 15252992 
 
 15-7480157 
 
 6-282760 
 
 249 
 
 62001 
 
 15438249 
 
 15-7797338 
 
 6-291194 
 
 250 
 
 62500 
 
 15625000 
 
 15-8113883 
 
 6299604 
 
SQUARES, CUBES, AND ROOTS. 
 
 Numb. 
 ~25T~ 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 63001 
 
 15813251 
 
 15-8429795 
 
 .6-307992 
 
 252 
 
 63504 
 
 16003008 
 
 15*8745079 * 
 
 6316359 
 
 253 
 
 64009 
 
 16194277 
 
 15-9059737 
 
 6-324^04 
 
 '2S4> 
 
 64516 
 
 16387064 
 
 15-9373775 
 
 ^•333C'25 
 
 255 
 
 65025 
 
 16581375 
 
 15-9687194 
 
 6-341355 
 
 256 
 
 65536 
 
 16777216 
 
 J 6 -.0000000 
 
 6-349a02 
 
 25/ 
 
 66049 
 
 16974593 
 
 160312195 
 
 6^357859 
 
 258 
 
 66564 
 
 17173512 
 
 16-0623784 
 
 6 366095 
 
 259 
 
 67081 
 
 17373979 
 
 16'0934769 
 
 6-374310 
 
 260 
 
 67600 
 
 17576000 
 
 16-1245155 
 
 6-382504 
 
 261 
 
 68121 
 
 17779581 
 
 16-1554944 
 
 6-390676 
 
 262 
 
 68644 
 
 17984728 
 
 16-1864141 
 
 6-398827 
 
 263 
 
 69169 
 
 18191447 
 
 16-2172747 
 
 6406958 
 
 264 
 
 69696 
 
 18399744 
 
 16-2480768 
 
 6-415068 
 
 265 
 
 70225 
 
 18609625 
 
 16-2788206 
 
 6-423157 
 
 266 
 
 70756 
 
 18821096 
 
 16-3095064 
 
 6431226 
 
 267 
 
 71289 
 
 19034163 
 
 16-3401346 
 
 6-439275 
 
 268 
 
 71824 
 
 19248832 
 
 16-3707055 
 
 6447305 
 
 269 
 
 72361 
 
 19465109 
 
 164012195 
 
 6 455314 
 
 270 
 
 72900 
 
 19683000 
 
 1.64316767 
 
 6463304 
 
 271 
 
 73441 
 
 19902511 
 
 16 4620776 
 
 6 471274 
 
 272 
 
 73984 
 
 20123648 
 
 16-4924225 
 
 6-479224 
 
 273 
 
 74529 
 
 20346417 
 
 1(5 5227116 
 
 6-487153 
 
 274 
 
 75076 
 
 20570824 
 
 15-5529454 
 
 6-495064 
 
 275 
 
 75625 
 
 20796875 
 
 155831240 
 
 6 502956 
 
 276 
 
 76176 
 
 21024576 
 
 ! 6-6 132477 
 
 6-510829 
 
 277 
 
 76729 
 
 21253933 
 
 16-6433170 
 
 6-518684 
 
 278 
 
 77284 
 
 21484952 
 
 16-6733320 
 
 6-526519 
 
 279 
 
 77841 
 
 21717639 
 
 167032931 
 
 6-534335 
 
 280 
 
 78400 
 
 21952000 
 
 16-7332005 
 
 6 542132 
 
 281 
 
 78961 
 
 22188041 
 
 16-7630546 
 
 6-549911 
 
 282 
 
 79524 
 
 22425768 
 
 16 7928556 
 
 6-557672 
 
 283 
 
 80089 
 
 22665187 
 
 16-8226038 
 
 6-565415 
 
 284 
 
 80656 
 
 22906304 
 
 16-8522995 
 
 6-573139 
 
 285 
 
 81225 
 
 2314912^ 
 
 168819430 
 
 6-580844 
 
 286 
 
 81796 
 
 23393656 
 
 16-9115345 
 
 6-588531 
 
 287 
 
 82369 
 
 23639903 
 
 16 9410743 
 
 6-596202 
 
 288 
 
 82944 
 
 23887872 
 
 16-9705627 
 
 6-603854 
 
 289 
 
 83521 
 
 24137569 
 
 17 0000000 
 
 6 611488 
 
 290 
 
 84100 
 
 24389000 
 
 170293864 
 
 6.619106 
 
 291 
 
 84681 
 
 24642171 
 
 170587221 
 
 , 6 626705 
 
 292 
 
 85264 
 
 24897088 
 
 17 0880075 
 
 6-634287 
 
 293 
 
 85849 
 
 25153757 
 
 171172428 
 
 6-641851 
 
 294 
 
 86436 
 
 25412184 
 
 . 17-1464282 
 
 . 6-649399 
 
 295 
 
 87025 
 
 25672375 
 
 17-1755640 
 
 6656930 
 
 296 
 
 87616 
 
 25934336 
 
 17-2046505 
 
 . 6-664443 
 
 297 
 
 88209 
 
 26198073 
 
 17-2336879 
 
 6-671940 
 
 298 
 
 88804 
 
 26463592 
 
 17-2626765 
 
 6 679419 
 
 299 
 
 89401 
 
 26730899 
 
 17-2916165 
 
 6-686882 
 
 300 
 
 90000 
 
 ^7000000 
 
 17-3205081 
 
 6-694328 
 
96 
 
 ARITHMETIC. 
 
 Numb. 
 
 301 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 90601 
 
 27270901 
 
 17-34935 16 
 
 6 701758 
 
 302 
 
 91204 
 
 27543608 
 
 17'3781472 
 
 6-709172 
 
 303 
 
 91809 
 
 27818127 
 
 17 4068952 
 
 6 716569 
 
 304 
 
 92416 
 
 28094464 
 
 17-4355958 
 
 6-723950 
 
 305 
 
 93025 
 
 28372625 
 
 17-4642492 
 
 6-7313 6 
 
 306 
 
 93636 
 
 28652616 
 
 17-4928557 
 
 6-738665 
 
 307 
 
 94249 
 
 28934443 
 
 17-5214155 
 
 6-745997 
 
 308 
 
 94864 
 
 29218112 
 
 17-5499288 
 
 6 753^13 
 
 309 
 
 95481 
 
 29503629 
 
 17*5783958 
 
 6-7606,4 
 
 310 
 
 96100 
 
 29791000 
 
 17-6068169 
 
 6767899 
 
 311 
 
 96721 
 
 30080231 
 
 17-6351921 
 
 6775168 
 
 312 
 
 97344 
 
 30371328 
 
 17-6635217 
 
 6 782422 
 
 313 
 
 97969 
 
 30664297 
 
 17-6918060 
 
 6 789661 
 
 314 
 
 98596 
 
 30959144 
 
 17 7200451 
 
 6 796884 
 
 ■ 315 
 
 99225 
 
 3:255875 
 
 17-7482393 
 
 6804091 
 
 316 
 
 99856 
 
 31554496 
 
 17 7763888 
 
 6-811284 
 
 317 
 
 100489 
 
 31855013 
 
 17-8044938 
 
 6818461 
 
 318 
 
 101124 
 
 32157432 
 
 178325545 
 
 6-825624 
 
 319 
 
 101761 
 
 32451759 
 
 17-8605711 
 
 6-832771 
 
 320 
 
 102400 
 
 32758000 
 
 17-8885438 
 
 6-839903 
 
 321 
 
 103041 
 
 330r6161 
 
 . 17-9104729 
 
 6-847021 
 
 322 
 
 103684 
 
 33386248 
 
 17-9443584 
 
 6-854124 
 
 323 
 
 104329 
 
 33698267 
 
 17-9722008 
 
 6*861211 
 
 324 
 
 104976 
 
 34012224 
 
 18-0000000 
 
 6*868284 
 
 325 
 
 105625 
 
 34328125- 
 
 18-0277564 
 
 6-875343 
 
 326 
 
 106276 
 
 34645976 
 
 18 0554701 
 
 6-882388 
 
 327 
 
 106929 
 
 34965^83 
 
 18 0831413 
 
 6-889419 
 
 328 
 
 107584 
 
 35287552 
 
 18 1107703 
 
 6-896435 
 
 329 
 
 108241 
 
 35611289 
 
 18-1383571 
 
 6 903436 
 
 330 
 
 108900 
 
 35937000 
 
 18-1659021 
 
 6-910423 
 
 331 
 
 109561 
 
 36264^91 
 
 18-1934054 
 
 6-917396 
 
 332 
 
 110224 
 
 36594568 
 
 18-2208672 
 
 6-924355 
 
 333 
 
 110889 
 
 369260S7 
 
 18-2482876 
 
 6-931300 
 
 334. 
 
 111556 
 
 37259704 
 
 "18-2756669 
 
 6 938232 
 
 335 
 
 112225 
 
 376953r5 
 
 18-3030052 
 
 6 945149 
 
 336 
 
 112896 
 
 •37933056 
 
 18-3303028 
 
 6 952053 
 
 337 
 
 113569 
 
 38272753 
 
 18-3575598 
 
 6-958943 
 
 338 
 
 114244 
 
 58614472 
 
 18-4847763 
 
 6.9658j9 
 
 339 
 
 114921 
 
 38958219 
 
 18 4119526 
 
 6 972682 
 
 340 
 
 115600 
 
 39304000 
 
 184390889 
 
 6.979532 
 
 341 
 
 116281 
 
 39651821 
 
 18^4661853 
 
 6.986369 
 
 342 
 
 116964 
 
 40001688 
 
 18-4932420 
 
 6-993 591 
 
 343 
 
 117649 
 
 40353607 
 
 18-5202592 
 
 7-000000 
 
 344 
 
 118336 
 
 407X)7584 
 
 18*5472370 
 
 7«006796 
 
 345 
 
 119025 
 
 41063625 
 
 18'5741756 
 
 7-013579 
 
 346 
 
 119716 
 
 41421736 
 
 186010752 
 
 7-020349 
 
 347 
 
 120409 
 
 41781923 
 
 18 6279360 
 
 7 027106 
 
 , 348 
 
 121104 
 
 42144192 i 
 
 18'654758l 
 
 r-e33850 
 
 349 
 
 121801 
 
 42508549 
 
 18*6815417 
 
 7-040581 
 
 350 
 
 122500 
 
 43875000 
 
 18*7082869 
 
 7047208 
 
Si^UARES, CUBES, AND ROOTS. 
 
 97 
 
 Numb. 
 
 j Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root 
 
 351 
 
 123201 
 
 43243551 
 
 187349940 
 
 7-054003 
 
 352 
 
 123904 
 
 43614208 
 
 18-7616630 
 
 7-060696 
 
 353 
 
 124609 
 
 43986977 
 
 18-7882942 
 
 7-067376 
 
 354 
 
 125316 
 
 44361864 
 
 18-8148877 
 
 7-074043 
 
 355 
 
 126025 
 
 44738875 
 
 18-8414437 
 
 7 080698 
 
 356 
 
 126736 
 
 45118016 
 
 18-8679623 
 
 7087341 
 
 357 
 
 127449 
 
 45499293 
 
 18-8944436 
 
 7093970 
 
 358 
 
 128164 
 
 45882712 
 
 18 9208879 
 
 7-100588 
 
 359 
 
 128881 
 
 46268279 
 
 18-9472953 
 
 7-107193 
 
 360 
 
 129600 
 
 46656000 
 
 18-9736660 
 
 7 113786 
 
 361 
 
 130321 
 
 47045881 
 
 19'00Q0000 
 
 7*120367 
 
 362 
 
 131044 
 
 47437928 
 
 19-0262976 
 
 7*126935 
 
 363 
 
 131769 
 
 47832147 
 
 19-0525589 
 
 7 133492 
 
 364 
 
 132496 
 
 48228544 
 
 19-0787840 
 
 7*140037 
 
 365 
 
 133225 
 
 48627125 
 
 19-1049732 
 
 7-146569 
 
 366 
 
 133956 
 
 49027896 
 
 19-I3!i265 
 
 7*153090 
 
 367 
 
 134689 
 
 49430863 
 
 19-1572441 
 
 7-159599 
 
 368 
 
 135424 
 
 49836032 
 
 19-1833261 
 
 7-166095 
 
 369 
 
 136161 
 
 50243409 
 
 19*2093727 
 
 7 172580 ' 
 
 370 
 
 136900 
 
 50653000 
 
 19-2353841 
 
 7-179054 
 
 371 
 
 137641 
 
 51064811 
 
 19-2613603 
 
 7-185516 
 
 372 
 
 138384 
 
 51478848 
 
 19-2873015. 
 
 7-191966 
 
 373 
 
 139129 
 
 51895117 
 
 19*3132079 
 
 7-198405 
 
 374* 
 
 139876 
 
 52313624 
 
 19*3390796 
 
 7 204832 
 
 375 
 
 140625 
 
 52734375 
 
 19-3649167 
 
 7-211247 
 
 376 
 
 141376 
 
 53157376 
 
 19-3907194 
 
 7 217652 
 
 377 
 
 142129 
 
 53582633 
 
 19*4164878 
 
 7*224045 
 
 378 
 
 142884 
 
 54010152 
 
 194422221 
 
 7*230427 
 
 379 
 
 143641 
 
 54439939 
 
 194679223 
 
 7-236797 
 
 380 
 
 144400 
 
 54872000 
 
 19-4935887 
 
 7-243156 
 
 381 
 
 145161 
 
 55306341 
 
 19-5192213 
 
 7-249504 
 
 382 
 
 145924 
 
 55742968 
 
 19-5448203 
 
 7 255841 
 
 283 
 
 146689 
 
 56181887 
 
 19-5703858 
 
 7*262167 
 
 384 
 
 147456 
 
 56623104 
 
 19-5959179 
 
 7-268482 
 
 385 
 
 148225 
 
 57066625 
 
 196214169 
 
 7-274786 
 
 386 
 
 148996 
 
 57512456 
 
 19*6468827 
 
 7-281079 
 
 387 
 
 149769 
 
 57960603 
 
 19-6723156 
 
 7^87362 
 
 388 
 
 150544 
 
 58411072 
 
 19 6977156 
 
 7293633 
 
 389 
 
 151321 
 
 58863869 
 
 19*7230829 
 
 7-299S93 
 
 390 
 
 152100 
 
 59319000 
 
 19*7484177 
 
 7-306143 
 
 391 
 
 152881 
 
 59776471 
 
 19-7737199 
 
 7312383 
 
 ^92 
 
 153664 
 
 60236288 
 
 19-7989899 
 
 7-318611 
 
 393 
 
 154449 
 
 60698457 
 
 19-8242276 
 
 7-3248i!9 
 
 394 
 
 155236 
 
 61162984 
 
 19 8494332 
 
 7-331037 
 
 395 
 
 156025 
 
 61629875 
 
 19 8746069 
 
 7-337234 
 
 396 
 
 156816 
 
 62099136 
 
 19-8997487 
 
 7 343420 
 
 397 
 
 157609 
 
 62570773 
 
 19 9248588 
 
 7-349596 
 
 398 
 
 158404 
 
 63044792 
 
 19-9499373 
 
 7 355762 
 
 399 
 
 159201 
 
 63521199 
 
 199749844 
 
 7-361917 
 
 400 
 
 160000 
 
 64000000 
 
 20-.0000000 
 
 7-368063 
 
 Vol. 1 
 
 H 
 
^ 
 
 
 ARITHMETIC. 
 
 
 Numb. 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 401 
 
 1 6080 1 
 
 64481201 
 
 20 0249844 
 
 7-374198 
 
 402 
 
 J 6 1 604 
 
 64964808 
 
 200499377 
 
 7-380322 
 
 403 
 
 162409 
 
 65450827 
 
 20 0748599 
 
 7-386437 
 
 404 
 
 163216 
 
 65939264 
 
 20-0997512 
 
 7-392542 
 
 405 
 
 164025 
 
 66430125 
 
 20-1246118 
 
 7-398636 
 
 406 
 
 164836 
 
 66923416 
 
 20 1494417 
 
 7-404720 
 
 407 
 
 165649 
 
 67419143 
 
 20-1742410 
 
 7-410794 
 
 408 
 
 166464 
 
 67911312 
 
 20-1990099 
 
 7-416859 ' 
 
 409 
 
 167281 
 
 68417929 
 
 20-2237484 
 
 7-422914 
 
 410 
 
 , 168100 
 
 68921000 
 
 20-2484567 
 
 7-428958 
 
 411 
 
 168921 
 
 69426531 
 
 20-2731349 
 
 7 434993 
 
 , 412 
 
 169744 
 
 69934528 
 
 20-2977831 
 
 7-441018 
 
 413 
 
 170569 
 
 70444997 
 
 20-3224014 
 
 7-447033 
 
 414 
 
 171396 
 
 70957944 
 
 20-3469899 
 
 u 7-453039 
 
 415 
 
 172225 
 
 71473375 
 
 203715488 
 
 7-459036 
 
 416 
 
 173056 
 
 71991296 
 
 20-3960781 
 
 7-465022 
 
 417 
 
 173889 
 
 72511713 
 
 . 20-4205779 
 
 7-470999 
 
 418 
 
 174724 
 
 73034632 
 
 204450483 
 
 7-476966 
 
 419 
 
 175561 
 
 73560059 
 
 20-4694895 
 
 7-482924 
 
 420 
 
 176400 
 
 74088000 
 
 20-4939015 
 
 7488872 
 
 421 
 
 177241 
 
 74618461 
 
 20-5182845 
 
 7-494810 
 
 422 
 
 178084 
 
 75151448 
 
 20-5426386 
 
 7-500740 
 
 423 
 
 178929 
 
 75686967 
 
 20-5669638 
 
 7*506660 
 
 424 
 
 179776 
 
 76225024 
 
 20*5912603 
 
 7512571 
 
 425 
 
 180625 
 
 76765625 
 
 20 6155281 
 
 7-518473 
 
 426 
 
 181476 
 
 77308776 
 
 20-6397674 
 
 7-524365 
 
 427 
 
 182329 
 
 77854483 
 
 20-663978^3 
 
 7-530248 
 
 428 
 
 1^5184 
 
 78402752 
 
 20-6881609 
 
 7-536121 
 
 429 
 
 184041 
 
 78953589 
 
 20-7123152 
 
 7-541986 
 
 430 
 
 184900 
 
 79507000 
 
 20-7364114 
 
 7-547841 
 
 431 
 
 185761 
 
 80062991 
 
 20-7605395 
 
 7-553688 
 
 ' 432 
 
 186624 
 
 80621568 
 
 20-7846097 
 
 7-559525 
 
 - 433 
 
 187489 
 
 81182737 
 
 208086520 
 
 7-565353 
 
 434 
 
 ■188356 
 
 81746504 
 
 20-8326667 
 
 7-571173 
 
 435 
 
 189225 
 
 82312875 
 
 20-8566536 
 
 7-576984 
 
 - 436 
 
 190096 
 
 82881856 
 
 20-8806130 
 
 7-582786 
 
 437 
 
 190969 
 
 83453453 
 
 20 9045450 
 
 7*588579 
 
 438 
 
 191844 
 
 84027672 
 
 30 9284495 
 
 7-594363 
 
 439 
 
 192721 
 
 84604519 
 
 209^523268 
 
 7-600138 
 
 440 
 
 193600 
 
 85 1 84t)00 
 
 20-9761770 
 
 7-605905 • 
 
 441 
 
 194481 
 
 S5766121 
 
 210000000 
 
 7-6 11662 
 
 442 
 
 195364 
 
 86350888 
 
 21-0237960 
 
 7-617411 
 
 443 
 
 196249 
 
 86938307 
 
 21 0475652 
 
 7-623151 
 
 444 
 
 197136 
 
 87528384 
 
 210713075 
 
 ^ 7628883 
 
 445 
 
 198025 
 
 88121125 
 
 21-0950231 
 
 7-634606 
 
 446 
 
 198916 
 
 88716536 
 
 2ril87l21 
 
 7-640321 
 
 447 
 
 199809 
 
 89314623 
 
 21-1423745 
 
 7-646027 
 
 448 
 
 200704 
 
 89915392 
 
 21-1660105 
 
 r-651725 
 
 . 449 
 
 201601 
 
 90518849 
 
 21-1896201 
 
 7-657414 
 
 450 
 
 202500 
 
 91125000 
 
 21 2132034 
 
 7-663094 * 
 
SQUARES, CUBES, ANB ROOTS. 
 
 ?5? 
 
 ; Numb. 
 
 Squire. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 451 
 
 203401 
 
 91733851 
 
 21-2367606 
 
 7-668766 
 
 452 
 
 204304 
 
 92345408 
 
 21-2602916 
 
 7-6744:50 
 
 453 
 
 205209 
 
 92959677 
 
 21-2837967 
 
 7.680085 
 
 454 
 
 206116 
 
 93576664, 
 
 21 3072758 
 
 7»685732 
 
 455 
 
 207025 
 
 94196375 
 
 21-3307290 
 
 7-691371 
 
 456 
 
 207936 
 
 94818816 
 
 21-3541565 
 
 7-697002 
 
 457 
 
 208849 
 
 95443993 
 
 21-3775583 
 
 7-702624 
 
 458 
 
 209764 
 
 96071912 
 
 21-4009346 
 
 7-708238 
 
 459 
 
 210681 
 
 96702579 
 
 21-4242853 
 
 7 713844 
 
 460 
 
 2U600 
 
 97336000 
 
 21-4476106 
 
 7-719442 
 
 461 
 
 212521 
 
 97972181 
 
 21-4709106 
 
 7-725032 
 
 462 
 
 213444 
 
 98611128 
 
 21 4941853 
 
 7 730614 
 
 463 
 
 214369 
 
 99252847 
 
 21-5174348 
 
 7-736187 
 
 454 
 
 215296 
 
 99897344 
 
 21-5406592 
 
 7-741753 
 
 465 
 
 216225 
 
 100544625 
 
 21-5638587 
 
 7-747310 
 
 466 
 
 217156 
 
 101194696 
 
 21-5870331 
 
 7-752860 
 
 457 
 
 218089 
 
 101847563 
 
 21-6101828 
 
 7-758402 
 
 468 
 
 219024 
 
 102503232 
 
 21-6333077 
 
 7-763936 
 
 469 
 
 219961 
 
 103161709 
 
 21-6564078 
 
 7-769462 
 
 470 
 
 220900 
 
 103823000 
 
 21-6794834 
 
 7'774980 
 
 471 
 
 221841 
 
 104487111 
 
 21-7025344 
 
 7-780490 
 
 472 
 
 222784 
 
 105154048 
 
 21-7255610 
 
 7-785992 
 
 473 
 
 223729 
 
 105823817 
 
 21-7485632 
 
 7-791487 
 
 474 
 
 224676 
 
 106496424 
 
 21-7715411 
 
 7-796974 
 
 475 
 
 3256'^5 
 
 107171875 
 
 21-7944947 
 
 7-802453 
 
 476 
 
 226576 
 
 107850176 
 
 21-8174242 
 
 7-807925 
 
 477 
 
 227529 
 
 108531333 
 
 21-8403297 
 
 7-813389 
 
 478 
 
 228484 
 
 109215352 
 
 21-8632111 
 
 7.818845 
 
 479 
 
 229441 
 
 109902239 
 
 21-8860686 
 
 7-824294 
 
 480 
 
 230400 
 
 110592000 
 
 21-9089023 
 
 7-829735 
 
 48i 
 
 231361 
 
 Ill284ii41 
 
 21-9317122 
 
 7-835168 
 
 432 
 
 1^32324 
 
 111980168 
 
 21-9544984 
 
 7.840594 
 
 483 
 
 233289 " 
 
 ■112678587 
 
 21-9772610 
 
 7-846013 
 
 484 
 
 234256 
 
 113379904 
 
 22-0600000 
 
 7-851424 
 
 485 
 
 235225 
 
 114084125 
 
 22 0227155 
 
 7-856828 
 
 486 
 
 •236196 
 
 114791256 
 
 22-0454077 
 
 7-862224 
 
 487 
 
 237169 
 
 115501303 
 
 22-0680765 
 
 7-867613 
 
 488 
 
 238144 
 
 116214272. 
 
 22*0907220 
 
 7 872994 
 
 489 
 
 239121 
 
 116930169 
 
 22- ! 133444 
 
 7.878368 
 
 490 
 
 240100 
 
 117849000 
 
 22- 1359436 
 
 7. 883734 
 
 491 
 
 241081 
 
 118370771 
 
 221585198 
 
 7.889094 
 
 492 
 
 242064 
 
 119095488 
 
 221810730 
 
 7-894446 
 
 493 
 
 243049 
 
 119823157 
 
 22-2036033 
 
 7-899791 
 
 494 
 
 244036 
 
 120553784 
 
 22 2261408 
 
 7 905129 
 
 495 
 
 24502 5 
 
 121287375 
 
 22-2485955 
 
 7 910460 
 
 . 496 
 
 246016 
 
 122023936 
 
 22-2710575 
 
 7 915784 
 
 497 
 
 24r009 
 
 122763473 
 
 22-2934968 
 
 7-921100 
 
 498 
 
 248004 
 
 123505992 
 
 22-3159136 
 
 7-926408 
 
 499 
 
 249001 ' 
 
 124251499 
 
 22 338S079 
 
 7 931710 
 
 500 
 
 250000 
 
 125000000 
 
 22*3606798 
 
 7-937005 ' 
 
100 
 
 
 AftlTHMETlC. 
 
 
 
 Numb. 
 
 Square. 
 
 Cube. 
 
 Square Koot. 
 
 22 3830293 
 
 Cube Root. 
 
 
 501 
 
 251001 
 
 125751501 
 
 7-942293 
 
 
 502 
 
 252004 
 
 126506008 
 
 22-4053565 
 
 7-947573 
 
 
 503 
 
 253009 
 
 127263527 
 
 224276615 
 
 7-952847 
 
 
 504 
 
 254016 
 
 128024064 
 
 22-4499443 
 
 7-958114 
 
 
 505 
 
 255025 
 
 128787625 
 
 22-4722051 
 
 r-963374 
 
 
 506 
 
 256036 
 
 129554216 
 
 22.4944438 
 
 7-968627 
 
 
 507 
 
 257049 
 
 130323843 
 
 22-5166605 
 
 7-973873 
 
 
 508 
 
 258064 
 
 131096512 
 
 22-5388553 
 
 7-979 112 
 
 • 
 
 509 
 
 259081 
 
 131872229 
 
 22-5610283 
 
 7-984344 
 
 
 510 
 
 260100 
 
 132651000 
 
 22-5831796 
 
 7-989569 
 
 
 511 
 
 261121 
 
 133432831 
 
 22-6053091 
 
 7-994788 
 
 
 512 
 
 262144 
 
 134217728 
 
 22-6274170 
 
 8-000000 
 
 
 513 
 
 263169 
 
 135005697 
 
 22 6495033 
 
 8005205 
 
 
 514 
 
 264196 
 
 135796744 
 
 22-6715681 
 
 8-010403 
 
 
 515 
 
 265225 
 
 136590875 
 
 22 6936114 
 
 8015595 . 
 
 
 516 
 
 266256 
 
 137388096 
 
 22 7156334 
 
 8 020779 
 
 
 5ir 
 
 267289 
 
 138188413 
 
 22-7376340 
 
 8-025957 
 
 
 518 
 
 268324 
 
 138991832 
 
 22-7596134 
 
 8-031129 
 
 
 519 
 
 269361 
 
 139798359 
 
 22-78 J 5715 
 
 8-036293 
 
 
 520 
 
 270400 
 
 140608000 
 
 228035085 
 
 8 041451 
 
 
 521 
 
 271441 
 
 141420761 
 
 22-8254244 
 
 8046603 
 
 
 522 
 
 272484 
 
 142236648 
 
 22 8473193 
 
 8051748 
 
 
 523 
 
 273529 
 
 143055667 
 
 22-8691933 
 
 8-056886 
 
 
 524 
 
 274576 
 
 143877824 
 
 22 8910463 
 
 8*062018 
 
 
 525 
 
 275625 
 
 144703125 
 
 22-9128785 
 
 8-067143 
 
 
 526 
 
 276676 
 
 145531576 
 
 22-9346899 
 
 8-072262 
 
 
 527 
 
 277729 
 
 146363183 
 
 22 9564806 
 
 8-077374 
 
 
 528 
 
 278784 
 
 147197952 
 
 22-9782506 
 
 8082480 
 
 
 529 
 
 279841 
 
 148035889 
 
 23-0000000 
 
 8-087579 
 
 
 530 
 
 280900 
 
 148877000 
 
 23-0217289 
 
 8-092672 
 
 
 531 
 
 281961 
 
 149721291 
 
 23 0434372 
 
 8-097758 
 
 
 532 
 
 283024 
 
 150568768 
 
 23-0651252 
 
 8-102838 
 
 
 533 
 
 284089 
 
 151419437 
 
 23-0867928 
 
 8-107912 
 
 
 534 
 
 285156 
 
 152273304 
 
 23-1084400 
 
 8-112980 
 
 
 535 
 
 286225 
 
 153130375 
 
 23-1300670 
 
 8-118041 
 
 
 536 
 
 287296 
 
 153990656 
 
 23-1516738 
 
 8-123096 
 
 
 537 
 
 288369 
 
 154854153 
 
 23-1732605 
 
 8-128144 
 
 
 538 
 
 289441 
 
 155720872 
 
 23-1948270 
 
 8-133186 
 
 
 539 
 
 290521 
 
 156590819 
 
 23 2163735 
 
 8*138223 
 
 
 540 
 
 291600 
 
 157464000 
 
 23 2379001 
 
 8*143253 
 
 
 541 
 
 . 292681 
 
 158340421 
 
 23-2594067 
 
 8*148276 
 
 
 542 
 
 293764 
 
 159220088 
 
 23-2808935 
 
 8' 153293 
 
 
 543 
 
 294849 
 
 160103007 
 
 23-3023604 
 
 8-158304 
 
 
 544 
 
 295936 
 
 160989184 
 
 23-3238076 
 
 8 163309 
 
 
 545 
 
 297025 
 
 161878625 
 
 233452351 
 
 8-168308 
 
 
 546 
 
 298116 
 
 162771336 
 
 23-5666429 
 
 8-173302 
 
 
 547 
 
 299209 
 
 163667323 
 
 23 3880311 
 
 8-178289 
 
 
 548 
 
 300304 
 
 164566592 
 
 234093998 
 
 8-183269 
 
 
 549 
 
 301401 
 
 165469149 
 
 23-4307490 
 
 8-188244 
 
 
 550 
 
 302500 
 
 1C6375000 
 
 23*4520788 
 
 8-193212 
 
SQUARES, CUBES, ANt> ROOTS. 
 
 to? 
 
 Numb. 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 " 551 ~ 
 
 303601 
 
 167284151 
 
 23-4733892 
 
 8-198175 
 
 552 
 
 304704 
 
 168196608 
 
 23-4946802 
 
 8-203131 
 
 553 
 
 305809 
 
 169112377 
 
 23-5159520 
 
 8 208082 
 
 554 
 
 306916 
 
 170031464 
 
 23-5372046 
 
 8-213027 
 
 555 
 
 308025 
 
 170953875 
 
 23-5584380 
 
 8-217965 
 
 556 
 
 309136 
 
 171879616 
 
 23-5796522 
 
 8-222898 
 
 557 
 
 310249 
 
 172808693 
 
 23 6008474 
 
 8-227825 
 
 558 
 
 311364 
 
 173741112 
 
 23-6220236 
 
 8-232746 
 
 659 
 
 312481 
 
 174676879 
 
 236431808 
 
 8-237661 
 
 560 
 
 343600 
 
 175616000 
 
 23-6643191 
 
 8-242570 
 
 561 
 
 314721 
 
 176558481 
 
 23-6854386 
 
 8-247474 
 
 562 
 
 315844 
 
 177504328 
 
 23-7065392 
 
 8-252371 
 
 563 
 
 316969 
 
 178453547 
 
 23-7276210 
 
 8-257263 
 
 564 
 
 318096 
 
 179406144 
 
 23-7486842 
 
 8-262149 
 
 ^65 
 
 319225 
 
 180362125 
 
 23-7697286 
 
 8-267029 
 
 566 
 
 320356 
 
 181321496 
 
 23-7907545 
 
 8-271903 
 
 567 
 
 321489 
 
 182284263 
 
 23-8117618 
 
 8-276772 
 
 568 
 
 322624 
 
 183250432 
 
 23-8327506 
 
 8281635 
 
 569 
 
 323761 
 
 184220009 
 
 23-8537209 
 
 8-286493 
 
 570 
 
 324900 
 
 185193000 
 
 23-8746728 
 
 8-291344 
 
 571 
 
 326041 
 
 186169411 
 
 23-8956063 
 
 8-296190 
 
 572 
 
 327184 
 
 187149248 
 
 23-9165215 
 
 8-301030 
 
 573 
 
 328329 
 
 188132517 
 
 23-9374184 
 
 8-305865 
 
 574 
 
 329476 
 
 189119224 
 
 23-9582971 
 
 8'310694 
 
 575 
 
 3306"25 
 
 190109375 
 
 23-9791576 
 
 8*315517 
 
 576 
 
 331776 
 
 191102976 
 
 24*0000000 
 
 8*320335 
 
 ' 577 
 
 332929 
 
 192100033 
 
 24-0208243 
 
 8*325147 
 
 578 
 
 334084 
 
 193100552 
 
 24-0416306 
 
 8-329954 
 
 579 
 
 335241 
 
 194104539 
 
 24-0624188 
 
 8-334755 
 
 580 
 
 336400 
 
 195112000 
 
 24-0831892 
 
 8-339551 
 
 581 
 
 337561 
 
 196122941 
 
 24-1039416 
 
 8-344341 
 
 582 
 
 338724 
 
 197137368 
 
 24 1246762 
 
 8-349125 
 
 583 
 
 339889 
 
 198155287 
 
 241453929 
 
 8-353904 
 
 584 
 
 341056 
 
 199176704 
 
 24-1660919 
 
 8-358678 
 
 585 
 
 342225 
 
 200201625 
 
 24-1867732 
 
 8-363446 
 
 586 
 
 343396 
 
 201230056 
 
 24-2074369 
 
 8-368209 
 
 587 
 
 344569 
 
 202262003 
 
 24-2280829 
 
 8-372966 
 
 538 
 
 345744 
 
 203297472 
 
 24-2487113 
 
 8-377718 
 
 589 
 
 346921 
 
 204336469 
 
 24-2693222 
 
 .8382465 
 
 590 
 
 348100 
 
 205379000 
 
 24-2899156 
 
 8-387206 
 
 591' 
 
 349281 
 
 206125071 
 
 24-3104916 
 
 8-391942 
 
 592 
 
 350464 
 
 207474688 
 
 24-3310501 
 
 8 396673 
 
 593 
 
 351649 
 
 208527857 
 
 243515913 
 
 8-401398 
 
 594 
 
 352836 
 
 209584584 
 
 24-3721152 
 
 8 406118 
 
 595 
 
 354025 
 
 210644875 
 
 24 3926218 
 
 8-410832 
 
 596 
 
 355216 
 
 211708736 
 
 24-4131112 
 
 8-415541 , 
 
 597 
 
 356409 
 
 212776173 
 
 24-4335834 
 
 8 420245 
 
 598 
 
 357604 
 
 213847192 
 
 24-4540385 
 
 8-424944 
 
 599 
 
 358801 
 
 214921799 
 
 24-4744765 
 
 8-429638 
 
 600 
 
 360000 * 
 
 216000000 
 
 24-4948974 
 
 8 434327 
 
■'M'- 
 
 AKITHMETIG. 
 
 Numb. 
 
 Square. 
 
 Cube. 
 
 Square Root 
 
 Cube Root." 
 
 601 
 
 361201 
 
 217081801 
 
 24-5153013 • 
 
 8^439009 
 
 602 
 
 362404 
 
 218167208 
 
 24^5356883 
 
 8-443687 
 
 603 
 
 363609 
 
 219256227 
 
 24-5560583 
 
 8-448360 . 
 
 604 
 
 364816 
 
 220348864 
 
 24'5764115 
 
 8-4*3027 ' 
 
 605 
 
 366023 
 
 221445125 
 
 24-5967478 
 
 8-457689 
 
 606 
 
 367236 
 
 222"545016 
 
 24' 6 170673 
 
 8-462347 
 
 607 
 
 368449 
 
 223643543 
 
 24.637S700 
 
 8-466999 
 
 608 
 
 369664 
 
 224756712 
 
 24-6576560 
 
 8*471647 
 
 609 
 
 37088i 
 
 225866529 
 
 24-6779254 
 
 8*476289 
 
 610 
 
 372100 
 
 226981000 
 
 24-6981781 
 
 81^480926 
 
 611 
 
 373321 
 
 228099131 
 
 24-7184142 
 
 8*485557 
 
 612 
 
 374544 
 
 229220928 
 
 24-7386338 
 
 8-490184 
 
 613 
 
 375769 
 
 230346397 
 
 24-7588368 
 
 8*494806 
 
 614 
 
 376996 
 
 231475544 
 
 24-7790234 
 
 8-499423 
 
 615 
 
 378225 
 
 232.608375 
 
 24-7991935 
 
 8-504034 
 
 616 
 
 379456 
 
 253744896 
 
 24-8193473 
 
 8'608641 
 
 617 
 
 380689 
 
 234885113 
 
 24-8394847 
 
 8-513243 
 
 618 
 
 381924 
 
 236029032 
 
 24-8596058 
 
 8-517840 
 
 619 
 
 383161 
 
 237176659 
 
 24-8797106 
 
 8-5224v32 
 
 620 
 
 384400 
 
 238328000 
 
 24-8997992 
 
 8-527018 
 
 621 
 
 385641 
 
 239483061 
 
 24-9198716 
 
 8-531600 
 
 622 
 
 386884 
 
 240641848 
 
 24-9399278 
 
 8-536177 
 
 623 
 
 388129 
 
 241804367 
 
 24-9599679 
 
 8-540749 
 
 , 624 
 
 389376 
 
 242970624 
 
 24*9799920 
 
 8-545317 
 
 625 
 
 390625 
 
 244140625 
 
 25-0000000 
 
 8-549879 
 
 626 
 
 391876 
 
 245314376 
 
 25-0199920 
 
 8-554437 
 
 627 
 
 393129 
 
 246491883 
 
 ' 25-0399681 
 
 8-558990 
 
 628 
 
 394384 . 
 
 247673152 
 
 25-0599282 
 
 8-563537 
 
 629 
 
 395641 
 
 248858189 
 
 25-0798724 
 
 8-568080 
 
 630 
 
 396900 
 
 250047000 
 
 25 0998008 
 
 8-572618 
 
 631 
 
 398161 
 
 251239591 
 
 25-1197134 
 
 8-577152 
 
 632 
 
 399424 
 
 252435968 
 
 25-1396102 
 
 8-581680 
 
 633 
 
 400689 
 
 253636137 
 
 251594913 
 
 8-586204 
 
 634 
 
 401956 
 
 254840104 
 
 25-1793566 
 
 8-590723 
 
 635 
 
 403225 
 
 256047875 
 
 25-1992063 
 
 8-595238 
 
 636 
 
 404496 
 
 257259456 
 
 25-2190404 
 
 8-599747 
 
 637 
 
 405769 
 
 258474853 
 
 25-2388589 
 
 8 604252 
 
 638 
 
 407044 
 
 259694072 
 
 25-2586619 
 
 8-608753 
 
 639 
 
 408321 
 
 260917119 
 
 25-2784493 
 
 8-613248 
 
 640 
 
 409600 
 
 262144000 
 
 25 2982213 
 
 8617738 
 
 641 
 
 410881 
 
 263374721" 
 
 25-3179778 
 
 8622224 
 
 642 
 
 412164 
 
 264609288 
 
 25-3377189 
 
 8 626706 
 
 643 
 
 413449 
 
 265847707 
 
 25-3574447 
 
 8-631183 
 
 644 
 
 414736 
 
 267089984 
 
 25 3771551 
 
 . 8 635655 
 
 645 
 
 416025 
 
 268336125 
 
 25-3968502 
 
 8 640122 
 
 646 
 
 417316 
 
 269586136 
 
 25-4165301 
 
 8 644585 
 
 647 
 
 418609 
 
 270840023 
 
 25 4361947 
 
 8-649043 
 
 648 
 
 419904 
 
 272097792 
 
 25-4558441 
 
 8 653497 
 
 649 
 
 421201 
 
 273359449 
 
 25-4754784 
 
 8-657946 
 
 650 
 
 422500 
 
 274625000 
 
 25*4950076 
 
 8-662301 
 
SQUARES, CUBES, AND ROOTS. 
 
 io3 
 
 Numb. 
 651 
 
 ^Square. 
 
 Cube. 
 
 Square Hoot. 
 
 Cube Roet. 
 
 433801 
 
 275894451 
 
 25 5147016 
 
 8-666831 
 
 652 
 
 425104 
 
 277167808 
 
 25-5342907 
 
 8-671266 
 
 653 
 
 426409 
 
 278445077 
 
 S5-5538647 
 
 8675697 
 
 654 
 
 427716 
 
 279726264 
 
 25-5734237 
 
 8*680123 
 
 655 
 
 429025 
 
 281011375 
 
 25-5929678 
 
 8 684545 
 
 656 
 
 430336 
 
 282300416 
 
 25-6124969 
 
 8-688963 
 
 657 
 
 431649 
 
 283593393 
 
 25-6320112 
 
 8-693376 
 
 658 
 
 432964 
 
 284890312 
 
 '^5 65\5lOr 
 
 8-697784 
 
 659 
 
 434281 
 
 286191179 
 
 25-^709953 
 
 8-702188 
 
 660 
 
 435600 
 
 287496000 
 
 25-6904652 
 
 8 706587 
 
 661 
 
 436921 
 
 288804781 
 
 25-7099203 
 
 8 710982 
 
 662 
 
 438244 
 
 290117528 
 
 25-7203607 
 
 8 715373 
 
 663 
 
 439569 
 
 291434247 
 
 25-7487864 
 
 8*719759 
 
 664 
 
 440896 
 
 292754944 
 
 25-7681975 
 
 8-724141 
 
 665 
 
 442225 
 
 294079625 
 
 25-7875939 
 
 8-728518 
 
 666 
 
 443556 
 
 295408296 
 
 25-8069758 
 
 8-732891 
 
 667 
 
 444889 
 
 296740963 
 
 25-8263431 
 
 8-737260 
 
 668 
 
 446224 
 
 298077632 
 
 25-8456960 
 
 8-741624 
 
 669 
 
 447561 
 
 299418309 
 
 •25-8650343 
 
 8-745984 
 
 670 
 
 448900 
 
 360763000 
 
 25-8843582 
 
 8-750340 
 
 671 
 
 450241 ■ 
 
 302111711 
 
 25-9036677 
 
 8-754691 
 
 672 
 
 451584 
 
 303464448 
 
 25-9229628 
 
 8-759038 
 
 673 
 
 452929 
 
 304821217 
 
 25-9422435 
 
 8-763380 
 
 674 
 
 454276 
 
 306182024 
 
 25' 96 1 5 100 
 
 8-767719 
 
 675 
 
 455625 
 
 307546875 
 
 25-9807621 
 
 8-772053 
 
 676 
 
 456976 
 
 308915776 
 
 26 0000COO 
 
 8-776382 
 
 677 
 
 458329 
 
 310288733 
 
 26-0192237 
 
 8-780708 
 
 678 
 
 459684 
 
 311665752 
 
 26-0384331 
 
 8-785029 
 
 679 
 
 461041 
 
 313046839 
 
 26-0576284 
 
 8-789346 
 
 680 
 
 462400 
 
 314432000 
 
 260768096 
 
 8-793659 
 
 681 
 
 463761 
 
 315821241 
 
 26-0959767 
 
 8-797967 
 
 682 
 
 465124 
 
 317214568 
 
 26-1151297 
 
 8-802272 
 
 683 
 
 466489 
 
 318611987 
 
 261342687 
 
 8-806572 
 
 684 
 
 467856 
 
 320013504 
 
 26 1535937 
 
 8-810868 
 
 685 
 
 469225 
 
 321419125 
 
 26-1725047 
 
 8-815159 
 
 686 
 
 470596 
 
 322828856 
 
 26.1916017 
 
 8819447 
 
 687 
 
 471969 
 
 324242703 
 
 26-2106848 
 
 8-823730 
 
 688 
 
 473344 
 
 325660672 
 
 26-2297541 
 
 8-828009 
 
 689 
 
 474721 
 
 327082769 
 
 26-2488095 
 
 8-832285. 
 
 690 
 
 476100 
 
 328509000 
 
 26-2678511 
 
 8-836556 
 
 691 
 
 47748 r 
 
 329939371 
 
 26 2868789 
 
 8-840822 
 
 692 
 
 478864 
 
 331373888 
 
 26-3058929 
 
 8-845085 
 
 693 
 
 480249 
 
 332812557 
 
 26-3248932 
 
 8 849344 
 
 694 
 
 481636 
 
 334255384 
 
 26-3438797 
 
 8-853598 
 
 695 
 
 483025 
 
 335702375 
 
 26-3628527 
 
 8-857849 
 
 696 
 
 484416 
 
 337153536 
 
 263818119 . 
 
 8-862095 
 
 697 
 
 485809 
 
 338608873 
 
 26-4007576 
 
 8-866337 
 
 698 
 
 487204 
 
 340068392 
 
 26*4196896 
 
 8-870575 
 
 699 
 
 488601 
 
 341532099 
 
 26-4386081 
 
 8-874809 
 
 700 
 
 490000 
 
 343000000 
 
 26-4575131 
 
 8-879040 
 
104 
 
 
 ARITHMETIC. 
 
 
 Numb. 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Hoot. 
 
 701 
 
 491401 
 
 344472101 
 
 26-4764046 
 
 8-88 3266 
 
 702 
 
 492804 
 
 345948008 
 
 26*4952826 
 
 8-887488 
 
 703 
 
 494209 
 
 347428927 
 
 265141472 
 
 8-891706 
 
 704 
 
 495616 
 
 348913664 
 
 26-5329983 
 
 8'895920 
 
 705 
 
 497025 
 
 350402625 
 
 26-5518361 
 
 8-900130 
 
 706 
 
 498436 
 
 351895816 
 
 26-5706605 
 
 8-904336 
 
 707 
 
 499849 
 
 353393243 
 
 265894716 
 
 8-908538 
 
 708 
 
 501264 
 
 354894912 
 
 26-6082694 
 
 8-912736 
 
 709 
 
 502681 
 
 356400829 
 
 26-6270539 
 
 8-916931 
 
 710 
 
 504100 
 
 357911000 
 
 26*6458252 
 
 8-921121 
 
 711 
 
 505521 
 
 359425431 
 
 26-6645833 
 
 8925307 
 
 712 
 
 506944 
 
 360944128 
 
 26*6833281 
 
 ,8'929490 
 
 713 
 
 508369 
 
 362467097 
 
 26-7020598 
 
 8-933668 
 
 714 
 
 509796 
 
 363994344 
 
 26-7207784 
 
 3'937843 
 
 715 
 
 511225 
 
 365525875 
 
 26-7394839 
 
 8-942014 
 
 716 
 
 512656 
 
 367061696 
 
 26-7581763 
 
 8-946180 
 
 717 
 
 514089 
 
 368601813 
 
 26-7768557 
 
 8-950343 
 
 718 
 
 515524 
 
 370146232 
 
 26-7955220 
 
 8-954502 
 
 719 
 
 516961 
 
 371694959 
 
 26-8141754 
 
 8*958658 
 
 720 
 
 518400 
 
 373248000 
 
 26-8328157 
 
 8-962809 
 
 721 
 
 519841 
 
 374805361 
 
 26-8514432 
 
 8-966957 
 
 722 
 
 521284 
 
 376367048 
 
 26-8700577 
 
 8-971100 
 
 723 
 
 522729 
 
 377933067 
 
 26-8886593 
 
 8-975240 
 
 724 
 
 524176 
 
 379503424 
 
 26-9072481 
 
 8*979376 
 
 725 
 
 525625 
 
 381078125 
 
 26 9258240 
 
 8-983508 
 
 726 
 
 527076 
 
 382657176 
 
 26*9443872 
 
 8'987637 
 
 727 
 
 528529 
 
 384240583 
 
 26-9629375 
 
 8991762 
 
 728 
 
 529984 
 
 385828352 
 
 26-9814751 
 
 8*995883 
 
 729 
 
 531441 
 
 387420489 
 
 27-0000000 
 
 9 000000 
 
 730 
 
 532900 
 
 389017000 
 
 27-0185122 
 
 9*004113 
 
 731 
 
 534361 
 
 390617891 
 
 27-0370117 
 
 9-008222 
 
 732 
 
 535824 
 
 392223168 
 
 27-0554985 
 
 9*012328 
 
 733 
 
 537289 
 
 393832837 
 
 27-0739727 
 
 9-016430 
 
 734 
 
 538756 
 
 396446904 
 
 27-0924344 
 
 9020529 
 
 735 
 
 540225 
 
 397065375 
 
 27-1108834 
 
 9-024623 
 
 736 
 
 541696 
 
 398688256 
 
 27-1293199 
 
 9*028714 
 
 737 
 
 543169 
 
 400315553 
 
 27-1477439 
 
 9 032802 
 
 738 
 
 544644 
 
 401917272 
 
 27-1661554 
 
 9-036885 
 
 739 
 
 546121 
 
 403583419 
 
 27-1845544 
 
 9-040965 
 
 740 
 
 547600 
 
 405224000 
 
 27-2029410 
 
 9 045041 
 
 741 
 
 549081 
 
 406869021 
 
 27-2213152 
 
 9 049114 
 
 742 
 
 550564 
 
 '408518488 
 
 27-2396769 
 
 9-053183 
 
 743 
 
 552049 
 
 410172407 
 
 27-2580263 
 
 9-057248 
 
 744 
 
 553536 
 
 411830784 
 
 27 2763634 
 
 9061309 
 
 745 
 
 555025 
 
 413493625 
 
 27-2946881 
 
 9 065367 
 
 746 
 
 556516 
 
 415160936 
 
 27-3130006 
 
 9-069422 
 
 747 
 
 558009 
 
 416832723 
 
 27-3313007 
 
 9-073472 
 
 748 
 
 559504 
 
 418508992 
 
 27-3495887 
 
 9-077519 
 
 749 
 
 561001 
 
 420189749 
 
 27-3678644 
 
 9-081563 
 
 750 
 
 662500 
 
 421875000 
 
 27-3861279 
 
 9-085603 
 
SQUARES, CUBES, AND ROOTS. 
 
 105 
 
 Numb. 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 751 
 
 564001 
 
 423564751 
 
 27-4043792 
 
 9*089639 
 
 752 
 
 565504 
 
 425259008 
 
 27-4226184 
 
 9093572 
 
 753 
 
 567009 
 
 426957777 
 
 27-4408455 
 
 9097701 
 
 754 
 
 568516 
 
 428661064 
 
 274 590604 
 
 9 101726 
 
 755 
 
 570025 
 
 4303688? 5 
 
 27-4773633 
 
 9 105748 
 
 756 
 
 571536 
 
 432081216 
 
 27 4954542 
 
 9-109766 
 
 757 
 
 573049 
 
 433798093 
 
 27 5136330 
 
 9 113781 
 
 758 
 
 574564 
 
 435519512 
 
 27 5317998 
 
 9.117793 
 
 759 
 
 576081 
 
 437245479 
 
 27-5499546 
 
 9-121801 
 
 760 
 
 577600 
 
 438976000 
 
 27-5680975 . 
 
 9-125805 
 
 761 
 
 579121 
 
 440711081 
 
 275862284 
 
 9-129806 
 
 762 
 
 580644 
 
 442450728 
 
 27 6043475 
 
 9- 133303 
 
 763 
 
 582169 
 
 444194947 
 
 27 6224546 
 
 9 137797- 
 
 764 
 
 583696 
 
 445943744 
 
 27-6405499 
 
 9 141788 
 
 765 
 
 585225 
 
 447697125 
 
 27-6586334 
 
 9-145774 
 
 766 
 
 586756 
 
 449455096 
 
 276767050 
 
 9-149757 
 
 767 , 
 
 588289 
 
 451217663 
 
 27-6947648 
 
 9-153737 
 
 768 
 
 589824 
 
 452984832 
 
 27-7128129 
 
 9 157713 
 
 769 
 
 591361 
 
 454756609 
 
 27-7308492 
 
 9 161686 
 
 770 
 
 592900 
 
 456533000 
 
 27-7488739 - 
 
 9 165656 
 
 771 
 
 594441 
 
 458314011 
 
 27-7668868 
 
 9-169622 
 
 772 
 
 595984 
 
 460099648 
 
 27-7848880 
 
 9 17S585 
 
 773 
 
 597529 
 
 461889917 
 
 27-8028775 
 
 9-177544 
 
 774 
 
 599076 
 
 463684824 
 
 27*8208555 
 
 9 181500 
 
 775 
 
 600625 
 
 465484375 
 
 27-8388218 
 
 9-185452 
 
 776 
 
 602176 
 
 467288576 
 
 27-8567766 
 
 9 189401 
 
 777 
 
 603729 
 
 469097433 
 
 27-8747J97 
 
 9-193347 
 
 778 
 
 605284 
 
 470910952 
 
 27-8926514 
 
 9 197289 
 
 779 
 
 606841 
 
 472729139 
 
 27-9105715 
 
 9 201228 
 
 780 
 
 y608400 
 
 474552000 
 
 27-9284801 
 
 9-205164 
 
 781 
 
 609961 
 
 476379541 
 
 27-9463772 
 
 9 209096 
 
 782 
 
 611524 
 
 478211768 
 
 27-9642629 
 
 9 213025 
 
 783 
 
 613089 
 
 480048687 
 
 27-9821372 
 
 9-216950 
 
 784 
 
 614656 
 
 481890304 
 
 28-0000000 
 
 9220872 
 
 785 
 
 616225 
 
 483736025 
 
 280178515 
 
 9-224791 
 
 786 
 
 617796 
 
 485587656 
 
 28-0356915 
 
 9228706 
 
 787 
 
 619369 
 
 487443403 
 
 28-0535203 
 
 9 232618 
 
 788 
 
 620944 
 
 485303872 
 
 28-0713377 
 
 9237527 
 
 789 
 
 622521 
 
 491169069 
 
 28-0891438 
 
 '9*240433 
 
 790 
 
 624100 
 
 493039000 
 
 28-1069386 
 
 9 244335 
 
 791 
 
 625681 
 
 494913671 
 
 28-1247222 
 
 9-248234 
 
 792 
 
 627264 
 
 496793088 
 
 28 1424946 
 
 9-252130 
 
 793 
 
 628849 
 
 498677257 
 
 28-1602557 
 
 9 256022 
 
 794 
 
 630436 
 
 500566184 
 
 28 1780056 
 
 9 259911 
 
 795 
 
 632025 
 
 502459875 
 
 28-1957444 
 
 9-263797 
 
 796 
 
 633616 
 
 504358336 
 
 28-2134720 
 
 9-267679 
 
 797 
 
 635209 
 
 506261573 
 
 '28-2311884 
 
 9-271559 
 
 798 
 
 636804 
 
 508169592 
 
 28 2488938 
 
 9-275435 
 
 799 
 
 638401 
 
 510082399 
 
 28 2665881 
 
 9-279308 
 
 800 
 
 640000 
 
 512000000 
 
 28-2842712 
 
 9-283177 
 
 Vol. I 
 
 15 
 
306 
 
 ARITHMETIC. 
 
 Numl). 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 ■ 801 
 
 641601 
 
 513922401 
 
 28-3019434 
 
 9-287044 
 
 803' 
 
 643204 
 
 515849608 
 
 28-3196045 
 
 9-290907 
 
 803 
 
 644809 
 
 517781627 
 
 28-3372546 
 
 9-294767 
 
 804 
 
 646416 
 
 519718464 
 
 28-3548938 
 
 9-298623 
 
 805 
 
 64802.5 
 
 521660125 
 
 28-3725219 
 
 9-302477 
 
 806 
 
 649636 
 
 523606616 
 
 28-3901391 
 
 9-S06327 
 
 807 
 
 651249 
 
 525557943 
 
 28-4077454 
 
 9310175 
 
 808 
 
 652864 
 
 527514112 
 
 28-4253408 
 
 9-314019 
 
 809 
 
 654481 
 
 529475129 
 
 28-4429253 
 
 9317859 
 
 810 
 
 656100 
 
 531441000 
 
 28-4604989 
 
 9-321697 
 
 811 
 
 657721 
 
 533411731 
 
 28-4780617 
 
 9-325532 
 
 812 
 
 659344 
 
 535387328 
 
 28-4956137 
 
 9-329363 
 
 813 
 
 660969 
 
 537366797 
 
 28-5131549 
 
 9-333191 
 
 814 
 
 662596 
 
 539353144 
 
 28-5306852 
 
 9-337016 
 
 815 
 
 664225 
 
 541343375 
 
 28-5482048 
 
 9-340838 
 
 816 
 
 665856 
 
 543338496 
 
 28-5657137 
 
 9344657 
 
 817 
 
 667489 
 
 545338513 
 
 28-5832119 
 
 9348473 
 
 818 
 
 669124 
 
 547343432 
 
 28*6006993 . 
 
 9 362285 
 
 819 
 
 670761 
 
 549353259 
 
 28-6181760 
 
 9-356095 
 
 820 
 
 672400 
 
 551368000 
 
 28 6356421 
 
 9-359901 
 
 821 
 
 674041 
 
 553387661 
 
 28-6530976 
 
 9-363704 
 
 822 
 
 675684 
 
 555412248 
 
 28 6705424 
 
 9-367505 
 
 823 
 
 677329 
 
 557441767 
 
 28-6879766 
 
 9*371302 
 
 824 
 
 678976 
 
 559476224 
 
 28*7054002 
 
 9-375096 
 
 825 
 
 680625 
 
 561515625 
 
 28*7228132 
 
 9-378887 
 
 826 
 
 682276 
 
 563559976 
 
 28-7402157 
 
 9 382675 
 
 827 
 
 683929 
 
 565609283 
 
 287576077 
 
 ^-386460 
 
 828 
 
 685584 
 
 567663552 
 
 28-7749891 
 
 9-390241 
 
 829 
 
 687241 
 
 569722789 
 
 28-7923601 
 
 9-394020 
 
 830 
 
 688900 
 
 571787000 
 
 28-8097206 
 
 9 397796 
 
 831 
 
 690561 
 
 573856191 
 
 28-8270706 
 
 9-401569 
 
 832 
 
 692224 
 
 575930368 
 
 28 8444102 
 
 9-405338 . 
 
 833 
 
 693889 
 
 578009537 
 
 28-8617394 
 
 9-409105 
 
 834 
 
 695556 
 
 580093704 
 
 28-8790582 
 
 9 412869 
 
 835 
 
 697225 
 
 582182875 
 
 28-8963666 
 
 9-416630 
 
 836 
 
 698896 
 
 584277056 
 
 28-9136646 
 
 9 420387 
 
 837 
 
 700569 
 
 586376253 
 
 28-9309523 
 
 9-424141 
 
 838 
 
 702244 
 
 588480472 
 
 28 9482297 
 
 9*427893 
 
 839 
 
 703921 
 
 590589719 
 
 28 9654967 
 
 9-431642 
 
 840 
 
 705600 
 
 592704000 
 
 28 9827535 
 
 9-435388 
 
 841 
 
 707-281 
 
 594823321 
 
 29-0000000 
 
 9-439130 
 
 «42 
 
 708964 
 
 596947688 
 
 29 0172363 
 
 9 442870 
 
 843 
 
 710649 
 
 599077107 
 
 29-0344623 
 
 9-446607 
 
 844 
 
 712336 
 
 601211584 
 
 29-0516781 
 
 9-450341 
 
 845 
 
 714025 
 
 603351125 
 
 29-0688837 
 
 9*454071 
 
 846 
 
 715716 
 
 605495736 
 
 29 0860791 
 
 9 457799 
 
 847 
 
 717409 
 
 607645423 
 
 29-1032644 
 
 9-461524 
 
 848 
 
 719104 
 
 609800192 
 
 29-1204396 
 
 9465247 
 
 849 
 
 720801 
 
 611960049 
 
 29-1576046 
 
 9 468966 
 
 I 850 
 
 722500 
 
 614125000 
 
 29*1547505 
 
 9-472682 
 
SQUARES, CUBES, AND ROOTS. 
 
 Wl 
 
 Numb. 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 851 
 
 724201 
 
 616295051 
 
 29-1719043 
 
 9-476395 
 
 852 
 
 725904 
 
 618470208 
 
 29 1890390 
 
 9-480106 
 
 853 
 
 727609 
 
 620650477 
 
 29 2061637 
 
 9 483813 
 
 854 
 
 729316 
 
 622835864 
 
 29 2232784 
 
 9-487518 
 
 855 
 
 731025 
 
 625026375 
 
 292403830 
 
 9 491219 
 
 856 
 
 732736 
 
 627222016 
 
 29-2574777 
 
 9494918 
 
 857 
 
 734449 
 
 629422793 
 
 29-2745623 
 
 9 498614 
 
 858 
 
 736164 
 
 631628712 
 
 29 2916370 
 
 9 502307 
 
 859 
 
 737881 
 
 633839779 
 
 293087018 
 
 9-505998 
 
 860 
 
 739600 
 
 636056000 
 
 29 3257566 
 
 9 509685 
 
 861 
 
 741321 
 
 638277381 
 
 29-3428015 
 
 9 513369 
 
 862 
 
 743044 
 
 640503928 
 
 29 3398365 
 
 9 517051 
 
 863 
 
 744769 
 
 642735647 
 
 29-3768616 
 
 9 520730 
 
 864 
 
 746496 
 
 644972544 
 
 29 3938769 
 
 9 524406 
 
 865 
 
 748225 
 
 647214625 
 
 29 4108823 
 
 9 528079 
 
 866 
 
 749956 
 
 649461896 
 
 29 4278779 
 
 9-531749 
 
 867 
 
 751689 
 
 651714363 
 
 29-4448637 
 
 9 535417 
 
 868 
 
 753424 
 
 653972032 
 
 29-4618397 
 
 9-539081 
 
 869 
 
 755161 
 
 656234909 
 
 29-4788059 
 
 9 542748 
 
 870 
 
 756900 
 
 658503000 
 
 29-4957624 
 
 9-546402 
 
 871 
 
 758641 
 
 660776311 
 
 29 5127091 
 
 9-550058 
 
 872 
 
 760384 
 
 663054848 
 
 29-5296461 
 
 9 553712 
 
 873 
 
 762129 
 
 665338617 
 
 29-5465734 
 
 9-557363 
 
 874 
 
 763876 
 
 667627624 
 
 29 5634910 
 
 9-561010 
 
 875 
 
 765625 
 
 669921875 
 
 29 5803989 
 
 9 564655 
 
 876 
 
 767376 
 
 672221376 
 
 29 5972972 
 
 9 568297 
 
 877 
 
 769129 
 
 674526133 
 
 29-6141858 
 
 9-571937 
 
 878 
 
 770884 
 
 676836152 
 
 29 6310643 
 
 9-575574 
 
 879 
 
 772641 
 
 679151439 
 
 29 6479325 
 
 9 579208 
 
 880 
 
 774400 
 
 681472000 
 
 29-6647939 
 
 9-582839 
 
 881 
 
 776161 
 
 683797841 
 
 29 6816442 
 
 9-586468 
 
 882 
 
 777924 
 
 686128968 
 
 296984848 
 
 9-590093 
 
 883 
 
 779689 
 
 688466387 
 
 29-7153159 
 
 9 593716 
 
 884 
 
 -781456 
 
 690807104 
 
 29-7321375 
 
 9 597337 
 
 885 
 
 783225 
 
 693154125 
 
 29-7489496 
 
 9G00954 
 
 886 
 
 784996 
 
 695506456 
 
 29-7657521 
 
 9-604569 
 
 887 
 
 786769 
 
 698764103 
 
 29-7825452 
 
 9-608181 
 
 888 
 
 788544 
 
 700227072 
 
 29-7993289 
 
 9-611791 ^ 
 
 889 
 
 790321 
 
 702595369 
 
 29 8161030 
 
 9.615397 
 
 890 
 
 792100 
 
 704969000 
 
 29-8328678 
 
 9*619001 
 
 891 
 
 793881 
 
 707347971 
 
 29-8496231 
 
 9-622603 
 
 892 
 
 795664 
 
 709732288 
 
 29 8663690 
 
 9 626201 
 
 893 
 
 797449 
 
 712121957 
 
 29 8831056 
 
 9 629797 
 
 894 
 
 799236 
 
 714516984 
 
 298998328 
 
 9 633390 
 
 895 
 
 801025 
 
 716917375 
 
 29 9165506. 
 
 9-636981 
 
 896 
 
 802816 
 
 719323136 
 
 29 9332591 
 
 9 640569 
 
 897 
 
 804609 
 
 721734273 
 
 29-9499583 
 
 9 644154 
 
 898 
 
 806404 
 
 724150792 
 
 29 9666481 
 
 9-647736 
 
 899 
 
 808201 
 
 726572699 
 
 29 9833287 
 
 9 651316 
 
 900 
 
 810000 
 
 729000000 
 
 30 0000000 
 
 9 654893 
 
lOS 
 
 
 ARITHMETIC. 
 
 
 Numb. 
 
 Square 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 901 
 
 811801 
 
 731432701 
 
 300166620 
 
 9 658468 
 
 902 
 
 813604 
 
 733870808 
 
 300333148 
 
 9 662040 
 
 903 
 
 815409 
 
 736314327 
 
 30-0499584 
 
 9; 665 609 
 
 904 
 
 817216 
 
 738763264 
 
 300665928 
 
 9-669176 
 
 905 
 
 819025 
 
 741217625 
 
 30-0832179 
 
 9 672740 
 
 906 
 
 820836 
 
 743677416 
 
 S-^ 0998339 
 
 9-676301* 
 
 907 
 
 822649 
 
 746142643 
 
 3u 1164407 
 
 9-679860 
 
 908 
 
 824464 
 
 748613312 
 
 30' 1330383 
 
 9-6834! 6 
 
 909 
 
 826281 
 
 751089429 
 
 30- 1496269 
 
 9 686970 
 
 910 
 
 828100 
 
 753571000 
 
 30 1632063 
 
 9-690521 
 
 911 
 
 829921 
 
 756058031 
 
 30 1827765 
 
 9-694069 
 
 912 
 
 831744 
 
 7585o0528 
 
 30 1993377 
 
 9-697615 
 
 913 
 
 833569 
 
 761048497 
 
 302158899 
 
 9 701158 
 
 914 
 
 835396 
 
 763551944 
 
 30-2:>24329 
 
 9 704698 
 
 9)5 
 
 837225 
 
 76G060875 
 
 30*2489669 
 
 9 708236 
 
 916 
 
 839056 
 
 768575296 
 
 30 2654919 
 
 9-711772 
 
 917 
 
 840889 
 
 771095213 
 
 30 2820079 
 
 9-7153Q5 
 
 918 
 
 842724 
 
 773620632 
 
 30-2985148 
 
 9.718835 
 
 919 
 
 844561 
 
 776151559 
 
 30-3150128 
 
 9-722363 
 
 920 
 
 846400 
 
 778688000 
 
 30 3315018 
 
 9-725888 
 
 92t 
 
 8'48241 
 
 781229961 
 
 30 3479818 
 
 9-729410 
 
 922 
 
 850084 
 
 783777448 
 
 30 3644529 
 
 9-732930 
 
 923 
 
 851929 
 
 786330467 
 
 303809151 
 
 9-736448 
 
 •924 
 
 P53776 
 
 7888-89024 
 
 30 3973683 . 
 
 9-739963 
 
 925 
 
 855625 
 
 791453125 
 
 30 4138127 
 
 9-743475 
 
 926 
 
 857476 
 
 794022776 
 
 30'4302481 
 
 9-746985 
 
 927 
 
 859329 
 
 796597983 
 
 30 4466747 
 
 9 750493 
 
 928 
 
 861184 
 
 799178752 
 
 304630924 
 
 9 753998 
 
 929 
 
 & 63041 
 
 801765089 
 
 304795013 
 
 9-757500 
 
 930 
 
 864900 
 
 804357000 
 
 30-4959014 
 
 9-761000 
 
 931 
 
 866761 
 
 806954491 
 
 30 5122926 
 
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 932 
 
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 933 
 
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 935 
 
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 937 
 
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 942 
 
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 946 
 
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 902500 
 
 857375000 
 
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 9-830475 
 
SQUARES, CUBES, AND ROOTS. 
 
 109 
 
 Numb. 
 
 Square. 
 
 Cube. 
 
 Square Root. 
 
 Cube Root. 
 
 951 
 
 9044bT 
 
 " 86008~535r 
 
 30-8382879 
 
 9-833923 
 
 952 
 
 906304 
 
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 30-8544972 
 
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 953 
 
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 954 
 
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 955 
 
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 956 
 
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 958 
 
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 960 
 
 921600 
 
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 961 
 
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 31-0000000 
 
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 96-2 
 
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 963 
 
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 31-0322413 
 
 9 875113 
 
 964 
 
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 965 
 
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 967 
 
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 969 
 
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 970 
 
 ,940900 
 
 912673000 
 
 311448230 
 
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 31-2569992 
 
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 986 
 
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 31*4165561 
 
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 31*4483704 
 
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 980100 
 
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 9 966554 
 
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 31 5119025 
 
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 995 
 
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 9996665 
 
no ARiTHMETiC. 
 
 OF RATIOS, PROPORTIONS, AND PROGRESSIONS. 
 
 Numbers are compared to each other in two different 
 tvays : the one comparison considers the difference of the two 
 numbers, and is named Arithmetical Relation; and the dif- 
 ference sometimes the Arithmetical Ratio : the other consi- 
 ders their quotient, which is called Geometrical Relation ; 
 and the quotient is the Geometrical Ratio. So, of these two 
 numbers 6 and 3, the difference, or arithmetical ratio, is 6—3 
 or 3, but the geometrical ratio is | or 2. 
 
 There must be two numbers to form a comparison : the 
 number which is compared, being placed first, is called the 
 Antecedent ; and that to which it is compared, the Conse- 
 quent. So, in the two numbers above, 6 is the antecedent, 
 and 3 the consequent. 
 
 If two or more couplets of numbers have equal ratios, or 
 equal differences, the equahty is named Proportion, and the 
 terms of the ratios Proportionals. So, the two couplets, 4, 2 
 and 8, 6, are arithmetical proportionals, because 4 — 2 = JJ 
 — 6 = 2; and the two couplets 4, 2 and 6, 3, are geometri- 
 cal proportionals, because |. | = 2, the same ratio. 
 
 To denote numbers as being geometrically proportional, a 
 colon is set between the terms of each couplet, to denote their 
 ratio ; and a double colon, or else a mark of equality between 
 the couplets or ratios. So, the four proportionals, 4, 2, 6, 3 
 are set thus, 4 : 2 : : 6 ; 3, which means that 4 is to 2 as 6 
 is to 3 ; or thus, 4 : 2 = 6 : 3, or thus, ± z=z &^ both 
 which mean, that the ratio of 4 to 2, is equal to the ratio 
 of 6 to 3. 
 
 Proportion is distinguished into Continued and Disconti- 
 nued. When the difference or ratio of the consequent of 
 one couplet, and the antecedent of the next couplet, is not the 
 same as the common difference or ratio of the couplets, the 
 proportion is discontinued. So, 4, 2, 8, 6 are in discontinued 
 arithmetical proportion, because 4 — 2 = 8 — 6 = 2^ where- 
 as 8 — 2 = 6: and 4, 2, 6, 3 are in discontinued geometrical 
 proportion, because f = | = 2, but | = 3, which is not 
 the same. 
 
 But when the difference or ratio of every two succeeding 
 terms is the same quantit}', the proportioa is said to be Conti- 
 nued, and the numbers themselves make a series of Continued 
 
 Proportionals, 
 
ARITHMETICAL PROPORTION. ill 
 
 Proportionals, or a progression. So 2, 4, 6, 8 form an arith- 
 metical progression, because 4—2 = 6—4 = 8—6 = 2, all 
 the same common difference ; and 2, 4, 8, 16 a geometrical 
 progression, because f = | V = 2' »*^ *^^ same ratio. 
 
 When the follov/ing terms of a progression increase, or 
 exceed each other it is called an Ascending Progression, or 
 Series ; but when the terms decrease, it is a descending one. 
 So, 0, f , 2, 3, 4, &c. is an ascending arithmetical progression, 
 but 9, 7, 6, 3, 1 , &c. is a descending arithmetical progression. 
 Also 1, 2, 4, 8, 16, &c. is an ascending geometrical progrebbion, 
 and 16, 8, 4, 2, 1, kc. is a descending geometrical progression. 
 
 ARITHMETICAL PROPORTION and PROGRESSION. 
 
 In Arithmetical Progression, the numbers or terms have all 
 the same common difference. Also, the first and last terms 
 of a Progression, are called the Extremes ; and the other 
 terms, lying between them, the Means. The roost useful 
 part of arithmetical propartions, is contained in the follow- 
 ing theorems : 
 
 Theorem 1. When four quantities are in arithmetical 
 proportion, the sum of the two extremes is equal to the sum 
 of the two means. Thus, of the four 2, 4, 6, 8, here 2 + 
 8 = 4 + 6 = 10. 
 
 Theorem 2. In any continued arithmetical progression, 
 the sum of the two extremes is equal to the sum of any two 
 means that are equally distant from them, or equal to double 
 the middle term when there is an uneven number of terms. 
 
 Thus, in the terms 1, 3, 5, it is 1 -f 5 = 3 -f- 3 = 6. 
 
 And in the series 2, 4, 6, 8, 10, 12, 14, it is 2 -f 14 «= 4 
 -f 12 = 6 4- 10 = 8 + 8 = 16. 
 
 Theorem 3. The difference between the extreme terms 
 of an arithmetical progression is equal to the common dif- 
 ference of the series multiplied by one less than the number 
 of the terms. So, of the ten terms, 2, 4, 6, 8, 10, 12, 14, 
 16, 18, 20, the common difference is 2, and one less than 
 the number of terms 9 ; then the difference of the extremes 
 is 20—2 ~ 18, and 2X9=18 also. 
 
 Consequently, 
 
112 ARITHMETIC. 
 
 Consequently, the greatest term is equal to the least term 
 added to the product of the conunon difference multiplied by 
 1 less than the aumt>er of terms. 
 
 Theorem 4. The sum of all the terms, of any arithme- 
 tical progression, is equal to the sum of the two extremes mul- 
 tiplied by the number of terms, and divided by 2 ; or the sum 
 of the two extremes multiplied by the number of the terms, 
 gires double the sum of all the terms in the series. 
 
 This is made evident by setting the terms of the series in 
 am inverted order, under the same series in a direct order, and 
 adding the corresponding terms together in that order. Thus^ 
 in the series 1, 3, 6, 7, 9, 11, 13, 15. 
 ditto inverted 15, 13, 11, 9, 7, 5, 3, 1 ! 
 . the sums are 16 -f 16 -f 16 -f 16 -f 16 -f 16 -f 16 -f 16, ' 
 which must be double the • sum of the single series, and is 
 equal to the sum of the extremes repeated as often as are the 
 number of the terms. 
 
 From these theorems may readily be found any one of 
 these five parts ; the two extremes, the number of terms, the 
 common difference, and the sum of all the terms, when any 
 three of them are given ; as in the following problems : 
 
 FEODLEM I. 
 
 Given the extremet, and the Number of Tenns ; to find the 
 Sum of all the Terms. 
 
 Add the extremes together, multiply the sum by the num- 
 ber of terms, and divide by 2. 
 
 EXAMPLES. 
 
 1. The extremes being 3 and 19, and the number of 
 terms 9 ; required the sum of the terms ? 
 19 
 3 
 
 19-f3 22 
 
 Or, X9 = -^X9=llX9=i99. 
 
 2 2 
 
 the same answer. 
 
 S. It is required to find the number of all the strokes a 
 common clock strikes in one whole revolution of the index, 
 •r in 12 hours ? Ans. 76. 
 
 El. 
 
 
 22 
 
 
 9 
 
 2) 
 
 198 
 
 Ans 
 
 .99 
 
ARITHMETICAL PROGRESSION. 113 
 
 Ex. 3. How many strokes do the clocks of Venice strike 
 in the compass of the day, which go continually on from 1 
 to 24 o'clock ? Ans. 300. 
 
 4. What debt can be discharged in a year, by weekly pay- 
 ments in arithmetical progression, the first payment being 1», 
 and the last or 52d payment 6/ 3s 1 Ans. 135/ 4«. 
 
 PROBLEM II. 
 
 Given the Extremes, and the Number of Terms ; to find th$ 
 Common Difference. 
 
 Subtract the less extreme from the greater, and divide 
 the remainder by 1 less than the number of terms, for the 
 common difference. 
 
 EXAMPLES. 
 
 1. The extremes being 3 and 19, and the number of terms 
 9 ; required the common difEerence ? 
 19 
 3 19—3 16 
 
 8) 16 ' 9—1 8 
 
 Ans. 2 
 
 2. If the extremes be 10 and 70, and the number of terms 
 21 ; what is the common difference, and the sum of the 
 series ? Ans. the com. diff. is 3, and the sum is 840. 
 
 3. A certain debt can be discharged in one year, by weekly 
 payments in arithmetical progression, the first payment being 
 Is, and the last bl 3s ; what is the common difference of the 
 terms ? Aifs. 2, 
 
 PROBLEM m. 
 
 Given one of the Extrtmes^ the Common Difference, and the 
 JS'umher of Terms : to find the other Extreme, and the sum 
 of the Series. 
 
 Multiply the common difference by 1 less than the num« 
 ber of terms, and the product will be the difference of the 
 extremes : Therefore add the product to the less extreme, to 
 give the greater ; or subtract it from the greater, to give the 
 less extreme. 
 
 Vol. I. 16 
 
 EXAMPI4ES, 
 
lU ARITHMETIC. . 
 
 EXAMPLES. 
 
 1 . Given the least term 3, the common diflference 2, of* ans; 
 arithmt^tkal series of 9 terms ; to find the greatest term, and 
 the sum of the series. 
 
 2 
 
 8 
 
 16 
 
 19 the greatest term> 
 3 the least \ 
 
 22 sum 
 9 number of terms* 
 
 * 2) 198 
 
 99 the sum of the series. 
 
 . 2. If the greatest term be 70, the common diflference 3y 
 and the number of terms 21, what is the least term, and the 
 turn of the series ? 
 
 Ans. The least term is 10, and the sum is 840; 
 
 8. A debt can be discharged in a year, by paying 1 shilling 
 
 the first week, 3 shillings the second, and so on, always 2 
 
 ■hillings more every week ; what is the debt, and what 'will 
 
 ^e last payment be ? 
 
 Ans. The last payment will be 6/ 3s, and the debt is 135/ 45. 
 PROBLEM IV. 
 To find an Arithmetiml Mean Proportional between Two Given 
 Terms. 
 
 Adb the two given extremes or terms together, and take 
 half their sum for the arithmetical mean required. 
 
 EXAMPLE. 
 
 To find an arithmetical mean between the two numbers 4 
 and 14. 
 
 Here 
 14 
 4 
 
 2) 18 
 
 Ana. 9 the mean required. 
 
 PROBLEM 
 
ARITHMETICAL PROGRESSION, lU 
 
 FHOBLEM V. 
 
 To find two Arithmetical Means between Two Given Extremes. 
 
 Subtract the less extreme from the greater, and divide 
 the difference by 3, so vfAl the quotient be the common dif* 
 lerence ; which being continually added to the less extreme^ 
 •r taken from the greater, gives the means. 
 
 EXAMPU5. 
 
 To find two arithmetical means between 2 and 8. 
 Here 8 
 
 3 ) 6 Then 2 -f- 2 = 4 the one mean, 
 and 4 4- 2 = 6 the other mean. 
 
 "^om. dif. 2 
 
 PROBLEM VI. 
 
 To find any Numher of Arithmetical Means between Two Given^ 
 Terms or Extremes. 
 
 Subtract the less extreme from th6 greater, and divide 
 the difference by 1 more than the number of means required 
 to be found, which will give the common difference j then 
 this being added continually to the least term, or subtracted 
 from the greatest, will give the terms required. 
 
 EXAMPLE. 
 
 To find five arithmetical means between 2 and 14. 
 Here 14 
 2 
 
 6) 12 Then by adding this com. dif. continually, 
 - — ^ the means are found 4, 6, 8, 10, 12. 
 com. dif. ^ 
 
 See more of ArithmeticaJ progression in the Algebra. 
 
 GEOMETRICAL 
 
H^ ARITHMETIC. 
 
 GEOMETRICAL PROPORTION AND PROGRESSION, 
 
 In Geometrical Progression the numbers or terms have 
 all the same multiplier or divisor. The most useful part of 
 Geometrical Proportion is eoetaiaed in the following theo- 
 
 Theorem 1. When four quantities are in geometrical pro- 
 portion, the product of the two extremes is equal to the pro- 
 duct of the two means. 
 
 Thus, in the four 2, 4, 3, 6, it is 2 X 6 = 3 X 4 = 12. 
 
 And hence, if the product of the two means be divided by 
 ©ne of the extremes, the quotient will give the other extreme. 
 So, of the a^bove numbers, the product of the means 12-7-2 
 = 6 the one extreme, and 12-^-^ = 2 the other extreme ; 
 and this is the foundation and reason of the practice in the 
 Rule of Three. , 
 
 Theorem 2. In any continued geometrical progression, the 
 product of the two extremes is equal to the product of any 
 two means that are equally distant from them, or equal to the 
 square of the middle term when there is an uneven number 
 
 01 terms. 
 
 Thus, in the terms 2, 4, 8, it is 2 X 8 = 4 X 4 = 16. 
 
 And in the series 2, 4, 8, 16, 32, 64, 128, 
 
 it is 2 X 128 = 4 X 64 = 8 X 32 = 16 X 16 = 256. 
 
 THEORESf 3. The quotient of the extreme terms of a 
 geometrical progression, is equal to the common ratio of the 
 series raised to the power denoted by 1 less than the number 
 of the terms. Consequently the greatest term is equal to 
 the least term multiplied by the said quotient. 
 
 So, of the ten terms, 2, 4, 8, 16,*32, 64, 128, 256, 612, 
 1024, the common ratio is 2, and one less than tbe nuaiber 
 of terms is 9 ; then the quotient of the extremes is 1024-r- 
 
 2 = 612, and 2° = 612 also. 
 
 Theorem 
 
GEOMETRICAL PROGRESSION. 117^ 
 
 Theorem 4. The sum of all the terms, of any geometri- 
 cal progression, is found by adding the greatest term to the 
 difierence of the extremes divided by 1 less than the ratio. 
 
 So, the sum of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 
 1024—2 
 
 (whose ratio is 2), is 1024 -\ = 10244-1022 = 2046. 
 
 2—1 
 
 ^ The focegoing, and several other properties of geometrical 
 proportion, are demonstrated more at large in the Algebraic 
 part of this work. A few examples may here be added of the 
 theorems, just delivered, with some problems concerning mean 
 proportionals. 
 
 EXAMPLES. 
 
 1. The least often terms, in geometrical progression, being 
 1, and the ratio 2 ; what is the greatest term, and the sum of 
 all the terms ? 
 
 Ans. The greatest term is 512, and the sum 1023. 
 
 2. What debt may be discharged in a year or 12 months, by 
 paying 11 the first month, 2/ the second, 4/ the third, and so on, 
 each succeeding payment being doublethe last ; and what will 
 the last payment be ? 
 
 Ans. The debt 4095?, and the last payment 2048/. 
 
 PROBLEM L 
 
 To find One Geometrical Mean Proportional between any Two 
 JVumbers. 
 
 Multiply the two numbers together, and extract the square 
 root of the product, which will give the mean proportional 
 sought. 
 
 EXAMPLE. 
 
 To find a geometrical mean between the two numbers 3 
 and 12. 
 
 12 
 3 
 
 36 (6 the mean. 
 
 3G 
 
 PROBLEM 
 
118 ARITHMETIC, 
 
 PROBLEM H. 
 
 To find Two Geometrical Mean Proportionals between any Tagy 
 Numbers. 
 
 Divide ^he greater number by the less, and extract the 
 cube root of the quotient, which will give the common ratio 
 of the terms. Then multiply the least given term by the ratio 
 for the first mean, and this mean again by the ratio for the se- 
 cond mean : or, divide the greater of the two given terms by 
 the ratio for the greater mean, and divide this again by the 
 ratio for the less mean. 
 
 EXAMPLE. 
 
 To find two geometrical means between 3 and 24. 
 
 Here 3) 24 (8 ; its cube root 2 is the ratio. 
 
 Then 3 X 2 = 6, and 6 X 2 = 12, the two means. 
 
 Or 24 -r 2 = 12, and 12 -f- 2 = 6, the same. 
 
 That is, the two means between 3 and 24, are 6 and 12. 
 
 PROBLEM IIL 
 
 To find any Number of Geometrical Means between Two Nurh" 
 bers. 
 
 Divide the greater number by the less, and extract such 
 root of the quotient whose index is 1 more than the number of 
 means required ; that is, the 2d root for one mean, the 3d root 
 for, two means, the 4th root for three means, and so on ; and 
 that root will be the common ratio of all the terms. Then, 
 with the ratio, multiply continually from the first term, or di- 
 vide continually from the last or greatest term. 
 
 EXAMPLE 
 
 To find four geometrical means between 3 and 96. 
 
 Here 3) 96 ( 32 ; the 5th root of which is 2, the ratio. 
 Then 3 X 2 = 6, & 6 X 2 = 12, & 12 X2 = 24, & 24 X2 = 48- 
 Or 96-^2 =; 48, & 48-f-2 = 24, & 24—2 == 12, & 12-i-2 = 6. 
 
 That is, 6, 12, 24, 48, are fhe four means between 3 and 96. 
 
 OF 
 
MUSICAL PROPORTION. Ud 
 
 OF MUSICAL PROPORTION. 
 
 TvERE is also a third kind of proportion, called Musical, 
 which being but of little or no common use, a very short ac- 
 count of it may here suffice. 
 
 Musical Proportion is when, of three numbers, the first 
 has the same proportion to the third, as the difference between 
 the first and second, has to the difference between the second 
 and third. 
 
 As in these three, 6, 0, 12 ; 
 
 where 6 : 12 : : 8 — 6 : 12 — 8, 
 that is 6 : 12 : : 2 : 4. 
 
 When four numbers are in musical proportion ; then the 
 first has the same ratio to the fourth, as the difference be- 
 tween the first and second has to the difference between the 
 third and fourth. 
 
 As in these, 6, 8, 12, 18 ; 
 where 6:18:: 8-6 : 18 — 12^. 
 that is C : 18 : : 2 : 6. 
 
 When numbers are in musical progression, their recipro- 
 cals are in arithmetical progression ; and the converse, that 
 is, when numbers are in arithmetical progression, their reci- 
 procals are in musical progression. 
 
 So in these musicals 6, 8, 12, their reciprocals }, i, j\, 
 are in arithmetical progression ; for ^ -f- yL = jl. = i j 
 and I 4- i = I = i ; that is, the sum of the extremes is 
 equal to double the mean, which is the property of arithme- 
 ticals. 
 
 The method of finding out numbers in musical proportion 
 is best expressed by letters in Algebra. 
 
 FELLOWSHIP, OR PARTNERSHIP. 
 
 Fellowship is a rule, by which any sum or quantity may 
 be divided into any number of parts, which shall be in any 
 given proportion to one another. 
 
 By this rule are adjusted the gains or loss or charges of 
 
 partners 
 
120 ARITHMETIC. 
 
 partners ia company ; or the effects of bankrupts, or lega- 
 cies in case of a deficiency of assets or effects ; or the shares 
 of prizes ; or the numbers of men to form certain detach- 
 ments ; or the division of waste lands among a number of 
 proprietors. 
 
 Fellowship is either Single or Double. It is Single, when 
 the sharer or portions are to be proportional each to one sin- 
 gle given number only ; as when the stocks of partners are 
 all employed for the same time : And Double, when each 
 portion is to be proportional to two or more numbers ;^as 
 when the stocks of partners are employed for different times.' 
 
 SINGLE FELLOWSHIP. 
 
 GENERAL RULE. 
 
 Add together the numbers that denote the proportion of 
 the shares. Then say, 
 
 As the sum of the said proportional numbers, 
 Is to the whole sum to be parted or divided, 
 So is each several proportional number. 
 To the corresponding share or part. 
 
 Or, as the whole stock, is fo the whole gain or loss, 
 So is each man's particular stock, 
 To his particular share of the gain or loss. 
 
 To yROVE THE Work. Add all the shares or parts to- 
 getlfer, and the sum will be equal to the whole number to 
 be shyed, when the work is right. 
 
 EXAMPLES. 
 
 1. To divide the number 240 into three such parts, as 
 shall be in proportion to each other as the three numbers 1 , 
 2 and 3. 
 
 Here 1 -|- 2 -f- 3 = 6, the sum of the numbers. 
 Then, as 6 : 240 : : 1 : 40 the 1st part, 
 and as 6 : 240 : : 2 : 80 the 2d part, 
 also as 6 : 240 : : 3 : 120 the 3d part, 
 
 Sum of all 240, the prooj^ 
 
 Ex. 2. 
 
SINGLE FELLOWSHIP. 121 
 
 Ex. 2. Three persons, a, b, c, freighted a sliip with 340 tuns 
 of wine ; of which, a loaded 110 tuns, b 97, and c the rest : 
 in a storm the seamen were obliged to throw overboard 85 
 tuns ; how much must each person sustain of the loss ? 
 Here 1 10 -f 97 = 207 tuns, loaded by a and b ; 
 theref. 340— 207 = 133 tuns, loaded by c. 
 Hence, as 340 : 86 : : 1 10 
 
 or as 4 : 1 : : 1 10 : 27i tuns = a's loss ; 
 and as 4 : 1 : : 97 : 24i tuns = b's loss ; 
 also as 4 : 1 : : 133 : 33i tuns = c's loss ; 
 
 Sum 85 tuns, the proof. 
 
 3. Two merchants, c and d, made a stock of 120/, of 
 which c contributed 75/, and d the rest : by trading they 
 gained 30/ ; what must each have of it ? 
 
 Ans. c 18/ 15*, and d 11/ 5s. 
 
 4. Three merchants, e, f, g, made a stock of 700/, of 
 which E contributed 123/, f 358/, and g the rest : by trading 
 they gain 125/ 10s j what must each have of it ? 
 
 Ans. E must have 22/ Is Od 2-^-^q, 
 F - - - 64 3 8 Off. 
 G - - - 39 6 3 ^3V. 
 
 5. A General imposing a contribution* of 700/ on four 
 villages, to be paid in proportion to the number of inhabitants 
 contained in each ; the 1st containing 250, the 2d 350, the 
 3d 400, and the 4th 600 persons ; what part must each vil- 
 lage pay 1 Ans. the 1st to pay 116/ 13s 4rf. 
 
 the 2d - - 163 6 8 
 the 3d - - 186 13 4 
 the 4th - - 233 6 8 
 
 6. A piece of ground, consisting of 37 ac 2 ro 14 ps, is 
 to be divided among three persons, l, m, and n, in propor- 
 tion to their estates : now if l's estate be worth 500/ a year, 
 m's 320/, and k's 75/ ; what quantity of land must each one 
 have I Ans. l must have 2lk ac 3 ro 39if |ps. 
 
 M - - - 13 1 30yY^. 
 N - - - 3 23iif. 
 
 V.^ A person is indebted to o 57/ 16s, to p 108/ 3s 8(/, 
 to Q 22/ lOrf, and to r 73/ ; but at his decease, his effects 
 
 * Contribution is a taj^ paid by provinces, towns, villag-es, &c. to 
 excuse them from being plundered. It is paid in provisions or in 
 money, and sometimes in both. -' 
 
 VoK. r, 17 are 
 
122 ARITHMETIC. 
 
 are found to be worth no more than 170Z 145 ; how must it 
 be divided among his creditors ? 
 
 , Ans. o must have Sll 16s 5d 2^5JL«L2^y, 
 p - - - 70 15 2 ^j\'^-^\. 
 ^ - . - 14 8 4 O/^-'^o^, 
 R - - - 47 14 11 2//j«^V 
 Ex. 8. A ship worth 900/, being entirely lost, of which i be- 
 longed to s, i to T, and the rest to v ; what loss will each 
 sustain, supposing 640Z of her were insured ? 
 
 Ans. s will lose 46/, t 90/, and v 225/. 
 
 9. Four persons, w, x, y, and z, spent among them 26s, 
 and agree that w shall pa}- ^ of it, x i, y i, and z } ; that 
 is, their shares are to be in proportion as ^, ^, i, and i : 
 what are their shares ? Ans. w must pay 9s ^d 3^q. 
 
 X - - 6 5 34^. 
 Y - - 4 10 ^^, 
 z - - 3 10 3^V- 
 
 10. A detachment, consisting of 6 companies, being sent 
 into a garrison, in which the duty required 76 men a day ; 
 what number of men must be furnished by each company, ia 
 proportion to their strength ; the first consisting of 54 men, 
 the 2d of 51 men, the 3d of 48 men, the 4th of 39, and the 
 5th of 36 men ? 
 
 Ans. The 1st must furnish 18, the 2d 17, the 3d 16, the 
 4th 13, and the 5th 12 men.* 
 
 DOUBLE FELLOWSHIP. 
 
 Double Fellowship, as has been said, is concerned im 
 cases in which the stocks of partners are employed or contin- 
 ued for different times. 
 
 * Qiiestions of this nature frequently occurring in military service. 
 General Hsviland, an officer of great merit, contrived an ingenious in- 
 strument, for more expeditiously resolving them ; which is distinguish- 
 ed by the name of the inventor, being called a Haviland* 
 
 Rule. 
 
DOUBLE FELLOWSHIP. 123 
 
 Rule.* — Multiply each person's stock by the time of its 
 continuance ; then flivide the quantity, as in Single Fellow* 
 ship, into shares, in proportion to these products, by saying, 
 
 As the total sum of all the said products, I 
 
 Is to the whole gain or loss, or quantity to be parted, 
 
 So is each particular product, 
 
 To the corre^ondent share of the gain or loss. 
 
 EXAMPLES. 
 
 1. A had in company 60Z for 4 months, and 6 had 601 for 
 B months ; at the end of which time they find 24i gained : 
 how must it be divided between them ? 
 Here 50 60 
 4 5 
 
 200 + 300 = 600 
 
 Then, as 500 : 24 : : 200 : 9| = 9/ 12* = a's share. 
 
 and as 500 : 24 : : 300 : 14| = 14 8 = b's share. 
 
 2. and d hold a piece of ground in common, for which 
 
 they are to pay 54Z. c put in 23 horses for 27 days, and d 
 
 21 horses for 39 days ; how much ought each man to pay of 
 
 the rent ? Ans. c must pay 23/ 5s 9d. 
 
 D must pay 30 14 3 
 
 4. Three persons, e, f, g, hold a pasture in common, 
 for which they are to pay 391 per annum ; into which e put 
 7 oxen for 3 months, f put 9 oxen for 5 ihonths, and g put 
 in 4 oxen for 12 months ; how much must each person pay 
 of the rent ? Ans. e must pay 51 10s 6d lj%q. 
 
 p - - 11 16 lU OyV 
 G - - 12 12 7 2yV 
 
 4. A ship's company take a prize of 1000/, which they 
 agree to divide among them according to their pay and the 
 time they have been on board : now the officers and midship- 
 men have been on board 6 months, and the sailors 3 months ^ 
 
 * The proof of this rule is as follows : When the times are equal 
 the shares of the gain or loss are evidently as the stocks, as in Single 
 Fellowship ; and when the stt^cks are equal, the shares as the times ; 
 therefore, when neither are equal the shares must be as their products. 
 
 the 
 
lU ARITHMETIC. 
 
 the officers have 40s a month, the midshipmen 30*, and the 
 sailors 225 a month ; moreover there are 4 officers, 12 mid- 
 shipmen, and no sailors ; what will each man's share be ? 
 
 Ans. each officer must have 23/ 2s 5d O^^^q. 
 
 each midshipman - 17 6 9 3y\\. 
 
 each seaman - - - 6 7 2 0y|^. 
 
 Ex. 5. H, with a capital of lOOOZ, began trade the first of 
 January, and, meeting with success in business, took in i as a 
 partner, with a capital of 1500/, on the first of March fol- 
 lowing. Three months after that they admit k as a third 
 partner, who brought into stock 2800/. After trading toge- 
 ther till the end of the year, they find there has been gained 
 1776/ 10s; how must this be divided among the partners ? 
 
 Ans. H must have 457/ 9s 4\d. 
 I - - - 571 16 81. 
 K - - - 747 3 Hi. 
 
 6. X, y, and z made a joint-stock for 12 months ; x at 
 first put in 20/, and 4 months afi;er 20/ more ; y put in at 
 first 30/, at the end of 3 months he put in 20/ more, and 2 
 months after he put in 40/ more ; z put in at first 60/, and 
 6 months after he put in 10/ more, 1 month after which he 
 took out 30/ ; during the 12 months they gained 60/ ; how 
 much of it must each have ? 
 
 Ans. X must have 10/ 18s 6d 3f f y. 
 Y - - - 22 8 1 Oif 
 z - - - 16 13 4 0. 
 
 SIMPLE INTEREST. 
 
 Interest is the premium or sum allowed for the loan, or 
 forbearance of money. The money lent, or forborn, is called 
 the Principal. And the sum of the principal and its interest, 
 avlded together, is called the Amount. Interest is allowed 
 at so much per cent, per annum ; which premium per cent 
 per annum, or interest of 100/ for a year, is called the rate of 
 interest : — So, 
 
 When 
 
SIMPLE INTEREST. 125 
 
 When interest is at 3 per cent, the rate is 3 ; 
 
 - 4 per cent. - - 4 ; 
 
 - 6 per cent, - - 5 ; 
 
 - 6 per cent. - - 6 ; 
 
 But, by law in England, interest ought not to be taken higher 
 than at the rate of 5 per cent. 
 
 Interest is of two sorts ; Simple and Compound. 
 
 Simple Interest is that which is allowed for the principal 
 lent or forborn only, for the whole time of forbearance. 
 As the interest of any sum, for any time, is directly propor- 
 tional to the principal sum, and also to the time of continu- 
 ance ; hence arises the following general rule of calcula- 
 tion. 
 
 As 100/ is to the rate of interest, so is any given principal to 
 it interest for one year. And again, 
 
 As 1 year is to any given time, so is the interest for a year, 
 just found, to the' interest of the given sum for that time. 
 
 Otherwise. Take the interest of I pound for a year, 
 which multiply by the given principal, and this product again 
 by the time of loan or forbearance, in years and parts, for the 
 interest of the proposed sum for that time. 
 
 JVbic, When there are certain parts of years in the time, 
 as quarters or months, or days : they may be worked for, 
 either by taking the aliquot or like parts of the interest of a 
 year, or by the Rule of Three, in the usual way. Also to 
 divide "by 100, is done by only pointing off two figures for 
 decimals. 
 
 EXAMPLES. 
 
 1. To find the interest of 230Z IO5, for 1 year, at the rate 6f 
 4 per cent, per annum. 
 
 Here, As 100 : 4 : : 2S01 10s : 91 4s 4^d, 
 4 
 
 100) 9,22 
 20 
 
 4'40 
 12 
 
 4-80 Ans. 9/ 4s 4f df. 
 4 
 
 3-20 
 
 Ex.2. 
 
126 ARITHMETIC. 
 
 Ex. 2. To find the interest of 647/ 15s, for 3 years, at 6 per 
 cent, per annum. 
 
 As 100 : 5 : ; 647-75 : 
 Or 20 : 1 : : 647*75 : 27-3875 interest for 1 year. 
 
 Z 82-1625 ditto for 3 years. 
 20 
 
 8 3-2600 
 12 
 
 d 3-00 Ans. 82/ 5s 3d. 
 
 3. To find the interest of 200 guineas, for 4 years 7 months 
 and 25 days, at 4i per cent, per annum. 
 
 ds I ds 
 
 210/ As 366 : : 9-45 : 25 : / 
 
 4i or 73 : : 9-46 : 5 : -647^ 
 
 5 
 
 840 
 
 105 73) 47-25 (-6472 
 
 345 
 
 9-45 interest for 1 yr. 630 
 
 4 < 19 
 
 37-80 ditto 4 years. 
 6 mo =i 4-725 ditto 6 months. /^ 
 1 mo =i -7875 ditto 1 month. 
 •6472 ditto 25 days. 
 
 / 43-9597 
 20 
 
 s 19-1940 ' 
 
 12 
 
 d 2-3280 
 
 4 Ans. 43/ 19s 2^d, 
 
 q> 1-3120 
 
 4. To find the interest of 450/, for a year at 6 per cent, 
 per annum. Ans. 22/ 10s. 
 
 6. To find the interest of 715/ 12s 6d, for a year, at 4i per 
 cent, per annum. i Ans. 32/ 4s Od. 
 
 6. To find the interest of 720/> for 3 years, at 6 per cent, 
 per annum. Ans. 108/. 
 
 ^ Ex. 7. 
 
COMPOUND INTEREST. 127 
 
 7. To find the interest of 355/ 15s for 4 years, at 4 per 
 cent, per annum. Ans. 66/ 18s 4^d. 
 
 Ex. 8. To find the interest of 32/ 5s Sd, for 7 years, at 4{ 
 per cent, per annum. Ans. 9/ 12s \d, 
 
 9. To find the interest of 170/, for li year, at 5 per cent, 
 per annum. Ans. 12/ 5s. 
 
 10. To find the insurance on 205/ 16s, for i of a year, at 
 4 per cent, per annum. Ans. 2/ Is l|d. 
 
 11. To find the interest of 319/ 6d, for 6f years, at 3| per 
 cent, per annum. Ans. 68/ 15s 9^d. 
 
 12. To find the insurance on 207/, for 117 days, at 4^ per 
 cent, per annum. Ans. 1/ 12s Id. 
 
 13. To find the interest of 17/ 6s, for 117 days, at 4f per 
 cent, per annum. Ans. 5s 3c?. 
 
 14. To find the insurance on 712/ 6s, for 8 months, at 7^ 
 per cent, per annum. Ans. 35/ 12s 3^d. 
 
 Note. The Rules for Simple Interest, serve also to calcu- 
 late Insurances, or the Purchase of Stocks, or any thing else 
 that is rated at so much per cent. 
 
 See also more on the subject of Interest, with the algebraical 
 expression and investigation of the rules at the end of the 
 Algebra, next following. 
 
 COMPOUND INTEREST. 
 
 Compound Interest, called also Interest upon Interest, 
 IS that which arises from the principal and interest, taken 
 together, as it becomes due, at the end of each stated time of 
 payment. Though it be not lawful to lend money at Com- 
 pound Interest, yet in purchasing annuities, pensions, or leases 
 in reversion, it is usual to allow Compound Interest to the 
 purchaser for his ready money. . 
 
 Rules. — 1. F^nd the amount of the given principal, for the 
 time of the first payment, by Simple Interest. Then con- 
 sider this amount as a new principal for the second payment, 
 whose amount calculate as before. And so on through all 
 the payments to the last, always accounting the last amount 
 as a new principal for the next payment. The reason of 
 which is evident from the definition of Compound Interest. 
 Or ehe^ 
 
 2. Find the amount of 1 pound for the time of the first 
 payment, and raise or involve it to the power whose index 
 is denoted by the number of payments. Then that power 
 multiplied by the given principal, will produce the whole 
 
 amount 
 
m ARITHMETIC. 
 
 amount. From which the said principal being subtracted, 
 leaves the Compound Interest of the same. As is evident 
 from the first Rule. 
 
 EXAMP1.ES. 
 
 1. To find the amount of 720/, for 4 years, at 6 per cent, 
 per annum, 
 
 Here 6 is the 20th part of 100, and the interest of 1/ for a 
 • year is -^\ or -05, and its amount 1-05. Therefore, 
 
 1. By the 1st Rule. 2. By the 2d Rule. 
 
 Is d 1'05 amount of 11. 
 
 20)720 1st yr's princip. 1-05 
 
 36 1st yr's interest. 
 
 20)756 2d yr's princip. M025 
 
 37 16 2d yr's interest. 
 
 M 026 2d power of it. 
 
 20)793 16 3d yr's princip. 720 
 
 39 13 91 3d yr's interest. 
 
 /875-1645 
 
 20 ) 833 9 91 4th yr's princip. 20 
 
 41 13 6f 4th yr's interest. 
 
 s 3-2900 
 
 1-21530625 4th pow. of it. 
 
 £815 3 3i the whole amount. 12 
 
 or ans. required. 
 
 d 3-4800 
 
 2. To find the amount of 60/, in 5 years, at 5 per cent, per 
 annum, compound interest. Ans. 631 iGs 3^d. 
 
 3. To find the amount of 501 in 6 years, or 10 half-years, at 
 •5 per cent per annum, compound interest, the interest payable 
 half-yearly. -Ans. 64/ Os Id. 
 
 4. To find the amount of 60/, in 5 years, or 20 quarters, at 
 5 per cent, per annum, compound interest, the interest pay- 
 able quarterly. Ans. 64/ 2s O^d. 
 
 5. To find the compound interest of 370/ forborn for 6 
 years, at 4 per cent, per annum. Ans. 98/ 3s 4}d. 
 
 6. To find the compound interest of 410/ forborn for 2i 
 years, at 4^ per cent, per annum, the interest payable half- 
 yearly. " Ans. 48/ 46 11 ic^. 
 
 7. To find the amount, at compound interest, of 2 17/, for- 
 born for 21 years, at 5 p^r cent per annum, the interest pay- 
 able quarterly. Ans. 242/ 13s 41c/. 
 
 Note. See the Rules for Compound Interest algebraically 
 investigated, at the end of the Algebra. 
 
 ALI.IGATION, 
 
ALLIGATION. 12^ 
 
 ALLIGATION, 
 
 Alligation teaches how to compound or mix together 
 several simples of different qualities, so that the composition 
 may be of some intermediate quality or rate. It is com- 
 monly distinguished into two cases, Alligation Medial, and 
 Alligation Alternate. 
 
 ALLIGATION MEDIAL. 
 
 Alligation Medial is the method of finding the rate or 
 quality of the composition, from having the quantities and 
 rates or qualities of the several simples given. And it is thus 
 performed : 
 
 * Multiply the quantity of each ingredient by its rate or 
 quality ; then add all the products together, and add also all 
 
 * Demonstration. The mle is thus proved by Algebra 
 
 Let a, b, c be the quantities of the ingredients, 
 and «i, w, p their rates, or qualities, or prices ; 
 then am, bn^ cp are their several values, 
 and am -f- 6n 4- <:p the sum of their vahies, 
 also a -^ b ■{• CIS the sum of the quantities, 
 and if r denote the rate of. the whole composition, 
 
 then « -f- 6 -f c X '' will be the value of the whole, 
 
 conseq. a + 6 -f- c X ^ ==• aw + A/i -f- c/&, 
 
 and r z= am ■{- bn + cp -r- a + b + c, which is the Rule. 
 
 Kote, If an ounce or any other quantity of pure gold be reduced in- 
 to 24 equal parts, these parts are catled Caracts j but ,^old is often 
 mixed with some base metal, which is called the Alloy, and the mix- 
 ture 15 said to be of so many caracts fine, according to the proportion 
 of pure gold contained in it ; thus, if 22 caracts of pure gold, and 2 of 
 alloy be mixed together, it is said to be 22 caracts fine. 
 
 If any one of the s-mples be of little or no value with respect to the 
 rest, its rate is supposed to be nothing ; as water mixed with wine, 
 and alloy with gold and silver. 
 
 Vol. L 18 tlic 
 
130 ARITHMETIC. 
 
 the quantities together into another sum ; then divide the 
 former sum by the latter, that is, the sum of the products 
 by the sum of the quantities, and the quotient will be the 
 rate or quality of the composition required. 
 
 EXAMPLES. 
 
 1. If three sorts of gunpowder be mixed together, viz. 
 601b at 12rf a pound, 44lb at 9d!, and 261b at td a pound ; how 
 much a pound is the composition worth ? 
 
 Here 50, 44, 26 are the quantities, 
 and 12, 9, 8 the rates or qualities ; 
 then ^60 X 12 = 600 
 
 •44 X 9 = 396 
 
 26 X 8 = 208 
 
 120) 1204 (10^f^=10J^. 
 
 Ans. The rate or price is 10 ^^d the pound. 
 
 2. A composition being made of 51b of tea at Is per lb, 
 91^ at 8s 6d per lb, and I4ilb at 5s lOd per lb ; what is a 
 lb of it worth ? Ans. 6s iO^d. 
 
 3. Mixed 4 gallons of wine at 4s lOd per gall, with 7 gal- 
 lons at 5s 3d per gall, and 9f gallons at 5s 8d per gall ; 
 what is a gallon of this composition worth ? Ans. 5s 4^^. 
 
 4. A mealman would mix 3 bushels of flour at 3s Bd 
 per bushel, 4 bushels at 5s 6d per bushel, and 5 bushels at 
 4s 8rf per bushel ; what is the worth of a bushel of this 
 mixture ? Ans. 4s l^d. 
 
 5. A farmer mixes 10 busJhels of wheat at 5s the bushel, 
 with 18 bushels of rye at 3s the bushel, and 20 bushels of 
 barley at 2s per bushel : how much is a bushel of the mixture 
 worth ? Ans. 3s. 
 
 6. Having melted together 7 oz of gold of 22 caracts fine, 
 12i oz of 21 caracts fine, and 17 oz of 1^ caracts fine : I 
 would know the fineness of the composition ? 
 
 Ans. 20 41 caracts fine. 
 
 7. Of what fineness is that composition, which is made by 
 mixing 31b of silver of 9 oz fine, with 51b 8 oz of 10 oz 
 fine, and lib 10 oz of alloy ? Ans. 7|i oz fine. 
 
 ALLIGATION 
 
[131 j 
 
 ALLIGATION ALTERNATE. 
 
 Alligation Alternate is the method of finding what quan- 
 tity of aiiy number of simples, whose rates are given, will 
 compose a mixture of a given rate. So that it is the reverse 
 of AHigation Medial, and may be proved by it. 
 
 1. Set the rates of the simples in a column under each 
 other. — 2. Connect, or link with a continued line, the rate of 
 each simple, which is less than that of the compound, with one, 
 or any number, of those that are greater than the compound ; 
 and each greater rate with one or any number of the less. — 
 3. Write t-he difference between the mixture rate, and that of 
 each of the simples, opposite the rate with which they are 
 linked. — 4. Then if only one difference stand against, any 
 rate, it will be the quantity belonging to that rate ; but if there 
 be several, their sura will be the quantity. 
 
 The examples may be proved by the rule for Alligation 
 Medial. 
 
 ♦ Ikmonst. By connecting the less rate to the gre^Her, and placing 
 the difference between then) and the rate alteitiately, the quantities 
 resulting are such, that there is precisely as much gained by one 
 quantity as is lost by the other, and therefoie the gain and lo&s upon 
 the whole is equal, and is exactly the pioposed rate : and the same 
 will be true of any other two simples mfin.ged according to ihe ilule 
 
 In like manner, whatever the number of simples may be, and with - 
 how many soever every one is linked, since il; is always a less with a 
 greater than the mean price, there will be an e'iu:il balance of loss 
 and gain betweeri every two, and consequently an equal balance on the . 
 
 whole. Q; E. D. 
 
 It is obvious, from this Rule, that questions of this sort admit of a 
 great variety of answers ; for, having f i»md one answer , we may find 
 as many more as we please, by only multiplying or dividing each of the 
 quantities found, by 2, or 3, or 4, &c. : the reason of which is 
 evident ; for, if two quantities, of two simples, make a balance of loss 
 and gain, with respect to the mean price, so must also the double or 
 treble, the i or ^ part, or any other ratio of these quantities, and so on 
 ad infinitum* 
 
 These kinds of questions are called by algebraists indeterminate or 
 wn?«m<cd problems ; and by an analytical process, theorems may be 
 raised that will give all the possible answers. 
 
 EXAMPLES. 
 
132 ARITHMETIC. 
 
 EXAMPLES. 
 
 1. A merchant would mix wines at 165, at 18*, and at 
 22s per gallon, so as that the mixture may be worth 205 the 
 gallon : what quantity of each must be taken ? 
 
 r 16-^ 2 at 165 
 
 Here 20^18->^^ \2 at 18s 
 
 - (22^^/4 + 2= 6 at 225. 
 Ans. 2 gallons at lGs2 gallons at I85, and 6 at 22*. 
 
 2. How much wine at 6s per gallon, and at 4s per gallon 
 must be mi\ed together, that the composition may be worth 
 5s per gallon ? Ans. 1 qt or'l gall, &c. 
 
 3. How much sugar at 4c?, at 6^d and at llrf per lb, must be 
 mixed together, so that the composition formed by them may 
 be worth 7c? per lb ? 
 
 Ans. 1 lb, or 1 stone, or 1 cwt, or any other equal quantity 
 of each sort. 
 
 4. How much corn at 2s 6c?, 8s 8d, 45. and 45 8c? per bushel, 
 must be mixed together, that the compound may be worth 
 3s 10c? per bushel ? 
 
 Ans. 2 at 2s 6c?, 2 at 3s 8rf, 3 at 45, and 3 at 4s 8df. 
 
 5. A goldsmith has gold of 16, of 18, of 23, and of 24 ca- 
 racts fine : how much must he take of each, to make it 21 
 <!aracts fine ? Ans. 3 of 16^2 of 18, 3 of 23, and 5 of 24. 
 
 6. It is required to mix brandy at 125, wine at 10s, cyder 
 at Is, and water at per gallon together, so that the mixture 
 inay be worth 8s per gallon ? 
 
 Ans. 8 galh of brandy, 7 of wine, 2 of cyder, and 4 of water. 
 
 RULE II. 
 
 When the whole composition is limited to a certain quantity : 
 Find an answer as before by linkinjj; ; then say, as the sum of 
 the .quantities, or diiferences thus determined, is to the given 
 quantity ; so is each ingredient found by linking, to the re- 
 quired quantity of each. 
 
 , EXAMPLES. 
 
 1. How much gold of 15, 17, 18, and 22 caracts fine, must 
 be mixed together, to form a composition of 40 oz of 20 ca- 
 racts fine ? . , 
 
 Here 
 
ALLIGATION ALTERNATE. 133 
 
 Here 20 
 
 54.3-1-2 = 10 
 
 16 
 Then, as 16 : 40 : : 2:6 
 and 16 : 40 .: : 10 : 25 
 Ana. 5 oz of 15, of 17, and of 18 caracts fine, and 25 oz of 22 
 caracts fine*. 
 
 Ex. 2. A vintner has wine at 4s, at os, at 5s 6rf, and at 6s 
 a gallon ; and he would make a mixture of 18 gallons, so 
 that it might be afforded as 6s 4d per gallon ; how much of 
 each sort must he take ? 
 
 Ans. 3 gal. at 4s, 3 at 5s, 6 at 5s 6d, and 6 at 6s, 
 
 • A great number of questions might be here given relating to the 
 specific gravities of metals, &c. but one of the most curious may here 
 suffice. 
 
 Hiaro, kii% of Syracuse, gave orders for a crown to be made entirely 
 of pure gold j but suspecting- the workmen had debased it by mixing 
 it with silver ot- copper, he recommended the discovery of the fraud 
 to the famous Archimedes, and desired to know the exact quantity of 
 alloy in the crown. 
 
 Archimedes, in order to detect the imposition, procured two otlier 
 masseS; tlie one of pure gold, the other of silver or copper, and each 
 of the same weight with the former j and by putting eack%eparately 
 into a vessel full of water, the quantity of water expelled b/them de- 
 termined their specific gravities ; from which, and their given weights, 
 the exact quantities of gold and alloy in the crown may be determined. 
 
 Suppose the weight of each crown to bfe lOlb, and that the water 
 expelled by the copper or silver was 92lb, by the gold 52lb, and by the 
 compound crown 641b ; what will be the quantities of gold and alloy in 
 the crown ? 
 
 The rates of the simples are 92 and 52, and of the compound 64 ; 
 therefore 
 
 g, 192-^ 12 of copper 
 °*|52-^ 28 of gold 
 And the sum of these is 12 + 28 = 40, which should have been 
 but 10 ; therefore by the Rule, 
 
 40 : 10 : : 12 : 31b of coppcr7 ^. 
 
 40 : 10 : : 28 : ;^lb of gpld i ^^^ ^"^w^^' 
 
 RULE 
 
134 ARITHMETIC. 
 
 RULE m». 
 
 When one of the ingredients is limited to a certain quan- 
 tity ; Take the difference between each price, and the mean 
 rate as before ; then say, As the difference of that simple 
 whose quantity is given, is to the rest of the differences se- 
 verally ; so is the quantity given, to the several quantities re- 
 quired. 
 
 EXAMPLES. 
 
 1. How much wine at 6s, at 5s 6d, and 6s the gallon, 
 must be mixed with 3 gallons at 4s per gallon, so that the 
 mixture may be worth 5s 4d per gallon ; 
 
 l-^sN. 8+2 = 10 
 
 , A\) \ 8+2= 10 
 Here 64 <^ rx ) 
 
 '^^\ yi6+4 = 2o 
 
 >^^ 16+4 = 20 
 Then 10 - 10 : : 3 : 3 
 10 : 20 : : 3 : 6 
 10 : 20 : : 3 : 6 
 Ans. 3 gallons at 5s, 6 at 5s 6d, and 6 at 6s. 
 
 2. A grocer would mix teas at 12s, 10s, and 6s per lb, with 
 201b at 4s per lb. how much of each sort must he take to make 
 the composition worth 8s per lb *? 
 
 Ans. 201b at 4s, 101b at Gs, lOlb at 10s and 20lb at 12s, 
 
 3. How much gold of 15, of 17, and of 22 caracts fine, 
 must be mixed with 5 oz of 18 caracts fine, so that the com- 
 position may be 20 caracts fine ? 
 
 Ans. 5 oz of 15 caracts fine, 5 oz of 17, and 25 of 22. 
 
 * In the very same manner questions may be wrought when several 
 of the ingredients are limited to certain quantities, by finding first for 
 one limit, and then for another. The two last Rules can need no de- 
 monstration, as they evidently result fi-om the first, the reason of which 
 has been already explained. 
 
 POSITION. 
 
SINGLE POSITION. 135 
 
 POSITION. 
 
 Position is a method of performing certain questions, 
 which cannot be resolved by the common direct rules. It is 
 sometimes called False Position, or False Supposition, because 
 it makes a supposition of false numbers, to work with the 
 same as if they were the true ones, and by their means dis- 
 covers the true numbers sought. It is sometimes also called 
 Trial-and-Error, because it proceeds by trials of false num- 
 bers, and thence finds out the true ones by a comparison of 
 the errors. — Position is either Single op Double. 
 
 SINGLE POSITION. 
 
 Single Position is that by which a question is resolved 
 by means of one supposition only. Questions which have 
 their result proportional to their suppositions belong to 
 Single Position : such as those which require the multipli- 
 cation or division of the number seught by any proposed num^, 
 ber ; or when it is to be increased or diminished by itself, or 
 any parts of itself, a certain proposed number of times. The 
 rule is as follows : 
 
 Take or assume any number for that which is required, 
 and perform the same operations with it, as are described or 
 performed in the question. Then say. As the result of the 
 said operation, is to the position, or number assumed ; so is 
 the result in the question, to a fourth term, which will be the 
 number sought*. 
 
 ♦ The reason of this Rule is evident, because it is supposed that the 
 results are proportional to the suppositions- 
 Thus, rM i a : : nz : Z, 
 
 a z 
 
 or — la I I — I z, 
 
 n n 
 
 a a z z 
 
 or — ± — &c. : a : : — ± — S;c. : 2r, 
 
 n m n m 
 
 and so on. 
 
 EXAMPLES. 
 
136 ARITHMETIC, 
 
 EXAMPLES. 
 
 1. A person after spending i and ^ of his money, has yet 
 remaining 60/; what had he at first ? 
 
 ijuppose he had at first 120/. Proof. 
 
 Now i of 120 is 40 i of 144 is 48 
 
 i of it is 30 xof 144.is 36 
 
 their sum is 70 their sum 84 
 
 which taken from 120 taken from 144 
 
 leaves 60 leaves 60 as 
 
 Then, 50 : 120 : : 60 : 144, the Answer. per question. 
 
 2. What number is that, which being multiplied by 7, and 
 the product divided by 6, the quotient may be 21 ? Ads. 18. 
 
 3. What number is that, which being increased by i, i, 
 and ^ of itself, the sum shall be 75 ? Ans. 36. 
 
 4. A general, after sending out a foraging ^ and i of his 
 men, had yet remftining 1000 ; what number had he in com- 
 mand ? Ans. 6000. 
 
 5. A gentleman distributed 52 pence among a number of 
 poor people, consisting of men, women, and children : to 
 each man he gave 6d, to each woman, 4d, and to each child 
 2d : moreover there were twice as many women as men, and 
 thrice as many children as women. How many were there 
 ®f each ? Ans. 2 men, 4 women, and 12 children ? 
 
 6. One being asked his age, said, iff of the years I have 
 lived, be multiplied by 7, and | of them be added to the pro- 
 dust, the sum will be 219. What was his age ? 
 
 Ans. 45 years. 
 
 •■11? 
 
 DOUBLE 
 
[ 137 ] 
 
 DOUBLE POSITION. 
 
 Double Position is the method of resolving certain ques- 
 tions by means of two suppositions of false numbers. 
 
 To the Double Rule of Position belong such questions as 
 ha^re their results not proportional to their positions : such are 
 those, in which the numbers sought, or their parts, or their 
 multiples, are increased or diminished by some given absolute 
 number, which is no known part of the number sought. 
 
 RtJLE I*. 
 
 Tare or assume any two convenient numbers, and proceed 
 with each of them separately, according to the conditions of 
 the question, as in Single Position ; and find how much each 
 result is diiTerent from the result mentioned in the question, 
 calling these differences the errors, noting also whether the 
 results are too great or too little. 
 
 • Demonstr. The Rule is founded on this supposition, namely, that 
 the first error is to the second, as the difference between the true and 
 first supposed numbeis is to the difference between the true and se- 
 cond sup[)Osed number ; when that is not the case, the exact answer 
 to the question cannot be found by this Rule.— That the Rule is true, 
 according to that supposition, may be thus proved. 
 
 Let a and b be the two suppositions, and A and b their results, 
 produced by similar operation ; also r and j their errors, or the 
 differences between the results A and B from the true result n ; 
 and let x denote the number sought, answering to the true result n of 
 the question. 
 
 Then is n — a = r, and n — B — «. And, according to the 
 supposition on which the Rule is founded, r :s : : x— a : x — bi 
 hence, by multiplying extremes" and meaps, rx — rb => sx — sa ,• 
 then, by transposition, rx -^ sx *^ rb — sa / and, by division, 
 
 rb-~8a 
 X = ■ the number sought, which is the rule when the 
 
 • r—s 
 results are both too little. 
 
 If the results be both too great, so that a and b are both greater 
 than N i then n — a = — r, and n — b = — », or r and « are both 
 negative ; hence — r : — s :: x — a : x '- b^ but — r : — j : : -f. r 
 : 4f *, therefore r ■ s i : x ^^a: x ^ b ; and the rest will be exactly 
 as in the former case. 
 
 But if one result A only be too little, and the other b too great, 
 or one error r positive, and the other 3 negative, then the theorem be- 
 
 rb-{.3a 
 comes X = — . — , which is the Rule in this case, or when the errors 
 
 are unlike. 
 Vol. I. 19 Then 
 
131 
 
 ARITHMETie. 
 
 Then multiply each of the said errors by the contrary sup- 
 position, namely, the first position by the second error, and 
 the second position by the first error. Then, 
 
 If the errors are alike, divide the difference of the products 
 by the difference of the errors, and the quotient will be the 
 answer. 
 
 But if the errors are unlike, divide the sum of the products 
 by the sum of the errors, for the answers. 
 
 JVbfe, The errors are said to be ahke, when they are either 
 both too great or both too little ; and unlike, when one is too 
 great and the other too little. 
 
 EXAMPLES. 
 
 1. What number is that, which being multiplied by 6, 
 the product increased by 18, and the sum divided by 9, the 
 quotient shall be 20 ? 
 
 Suppose the two numbers 18 and 30. Then, 
 
 First Position. Second Position. Proof. 
 
 18 Suppose - 30 27 
 
 6 mult. .6 6 
 
 
 108 
 
 
 
 180 
 
 -162 
 
 
 18 
 
 add 
 
 
 18 
 
 18 
 
 
 9) 126 
 
 div. 
 
 9) 
 
 198 
 
 9) 180 
 
 
 
 
 
 
 
 
 14 
 
 results 
 
 
 22 
 
 20 
 
 
 20 
 
 true res 
 
 • 
 
 20 
 
 
 
 + 6 
 
 errors unlike 
 
 —2 
 
 
 2d pos. 
 
 30 
 
 mult. 
 
 
 18 1st 
 
 po^. 
 
 Er- ^2 
 
 180 
 
 
 
 36 
 
 
 rors jle 
 
 36 
 
 
 
 — -— 
 
 , 
 
 sum 8) 
 
 216 
 
 sum of 
 
 products. 
 
 
 
 
 27 
 
 Answer 
 
 sought. 
 
 
 
 RULE II. 
 Find, by trial, two numbers, as near the true number as 
 convenient, and work with them as in the question ; marking 
 the errors which arise from each of them. 
 
 Multiply the difference of the two numbers assumed, or 
 found by trial, by one of the errors, and divide the product by 
 the difference of the errors, when they are alike, but by their 
 sum when they are unlike. 
 
 Add 
 
DOUBLE POSITION. 139 
 
 Add the quotient, last found, to the number belonging to 
 the said error, when that number is too little, but subtract it 
 when too great, and the result will give the true quantity 
 souffht*. 
 
 EXAMPLES. 
 
 1. So, the foregoiog example, worked by this 2d rule will 
 be as follows : • 
 
 30 positions 18 ; their dif. 12 
 
 — 2 errors -{- 6 ; least error 2 
 
 sum of errors 8) 24 (3 subtr. 
 from the position 30 
 
 leaves the answer 27 
 
 Ex* 2. A son asking his father how old he was, received 
 this answer : Your age is now one-third of mine ; but 5 years 
 ago, your age was only one-fourth of mine. What then are 
 their two ages ? • Ans. 16 and 45. 
 
 3. A workman was hired for 20 days, at 3s per day, for 
 every day he worked ; but with this condition, that for every 
 day he played, he should forfeit Is. Now it so happened, that 
 upon the whole he had 21 4s to receive. How many days did 
 he work ? Ans. 16 
 
 4. A and e began to play together with equal sums of money : 
 A first won 20 guineas, but afterwards lost back f of what he 
 then had ; aft«- which, b had 4 times as much as a. What 
 sum did each begin with ? ^ Ans. 100 guineas. 
 
 5. Two persons, a and b, have both the same inconie, a 
 saves i of his ; but b, by spending 50t per annum more than a, 
 at the end of 4 years finds himself 100/ in debt. What does 
 each receive and spend per annum ? 
 
 Ans. They receive 125/ per annum ; also a spends 100/, 
 and B spends 150/ per annum. 
 
 * Fop since, by the supposition, r : s :'. x — a:x -^6, therefore by 
 ^vision, r - « : j ; : d — a : ?c — 6, which is the 2d Rule. 
 
 PERMUTATIONS 
 
140 ARITHMETIC. 
 
 PERMUTATIONS AND COMBINATIONS. 
 
 Permutation is the altering, changing or varying the 
 position or order of things ; or the showing how many diffe- 
 rent ways they may be placed. — This is otherwise called Al- 
 ternation, Changes, or Variation ; and the only thing to be re- 
 garded here, is the order they stand in ; for no two parcels 
 are to have all their quantities placed in the same situation : 
 as, how many changes may be rung on a number of bells, or 
 how man}' different ways any number of persons may be 
 placed, or how many several variations may be made of any 
 number of letters, or any other things proposed to be varied. 
 
 Combination is the showing how often a less number of 
 things can be taken out of a greater, and combined together, 
 without considering their places, or the order they stand in. 
 This is somtjtimes called Election or Choice ; and here every 
 parcel must be different from all the rest, and no two are to 
 have precisely the same quantities or things. 
 
 Combinations of the same Form, are those in which there are 
 the same number of quantities, and the same repetitions : thus, 
 aabc, bbcdf ccde^ are of the same form ; aabc, abbb\ aabb, are 
 of different forms. 
 
 Composition of Quantities, is the taking a given number of 
 quantities out of as many equal rows of different quantities, 
 one out of every row, and combining them together. 
 
 Some illustration of these definitions are in the following 
 Problems : 
 
 PROBLEM r. ^ 
 
 7^0 assig)i the number of Permutations, or Changei, that can be 
 7nade of any Given Number of things, all different from each 
 other. 
 
 EULE*. 
 
 Multiply all the terms of the natural series of numbers^ 
 from 1 up to the given number, continually together, and the 
 last product will be the answer required. 
 
 EXAMPLES. 
 
 ♦ The reason of the Rule may be shovm thus ; any one thlog a is 
 capable only of one position, as a* 
 
 Any two things a and b, are only capable of two variations > as a^, 
 6a i whose number is expressed by 1 X 2' 
 
I 
 PERMUTATIONS AND COMBINATIONS. 141 
 
 EXAMPLES. 
 
 1. How many changes may be rung on 6 bells? 
 
 1 
 2 
 
 2 
 
 6 
 4 
 
 24 
 
 120 
 6 
 
 720 the Answer. 
 Or 1 X 2 X 3 X 4 X 6 X 6 = 720 the Answer. 
 
 2. How many days can 7 persons be placed in a .different 
 position at dinner ? Ans. 5040 days. 
 
 3. How many changes may be rung on 12 bells, and what 
 time would it require, supposing 10 changes to be rung in 1 
 minute, and the year to consist of 366 days, 6 hours, and 49 
 minutes ? 
 
 Ans. 479001600 changes, and 91 years, 26 days, 22 hours, 
 41 minutes. 
 
 4. How many changes may be made of the words in the 
 following verse : Tot tibi sunt dotes, virgo, quot sidera ccelo ? 
 
 Ans. 40320 changes,. 
 
 If there be three things, a, b, and c / then any two of them, leaving; 
 out the 3d, will have 1X2 variations j and consequently when the 3d 
 is taken in, there will be 1 X 2 X 3 variations. 
 
 In the same manner, when there are 4 things, every three> leaving 
 out the 4th, will have i X '^ X 3 variations ; consequently by taking 
 in successively the 4 left out, there will be 1X2X3x4 variations. 
 And s© on as far as we please- 
 
 PROl- 
 
142 ARITHMETIC, 
 
 PRQBLEM II. 
 
 Any Numher of different Things being given ; to find how many 
 Changes can he made out of them, by taking a Given Kum- 
 ber of Quantities at a Time. 
 
 RULE.* 
 Take a series of numbeis, beginning at the number of 
 things given, and decreasing by 1 tc the number of qoanti- 
 ties to be taken at a time, and the product of all the terms 
 will be the answer required. 
 
 EXAMPLES. 
 1. How many changes may be rung with 3 bells out gf 8 ? 
 8 
 
 7 
 
 66 
 6 
 
 336 the Answer. 
 
 Qr, 8 X 7 X 6 (= 3 terms) = 336 the Answer. 
 !2. How many words can be made with 5 letters of the 
 alphabet, supposing 24 letters in all, and that a number of con- 
 sonants alone will make a word. Ans, 5100480. 
 3. How many words can be made with 5 letters of the alpha- 
 bet in each word, there being 26 letters in all, and 6 vowels, 
 admitting that a number of consonants alone will not make a 
 word ? Ans. 137858400. 
 
 FROB- 
 
 * This Rule, expressed in algebraic termsj is as follows ; 
 
 m X wi — l X M— 2 X w— 3 &c. to n terms : where m ;= the num- 
 ber of things given, and n => the quantities to be taken at a time. 
 
 In order to demonstrate the Rule, it will be proper to premise the 
 following Lemma ; 
 
 Lemma. The number of changes of m ihing.e, taken n at a time, Is . 
 equal to m changes of m— 1 things, tiken n — 1 at a lime. 
 
 DemonstV' Let any five quantities a h c d ehe given 
 
 First, leave out the a, and let v *• the number of all the variations 
 of every two, 6c, bd^ Sic. that can be taken out of the four remaining 
 quantities bade. 
 
 Now, let a be pat in the first place of each of them, a, 6, c, a, *, d, 8m:. 
 and the number of changes which still remain the same ; that i??, v -«= 
 the number of variations of every 3 out of the 5, a, 6, r, </, <?, when a 
 is first. 
 
 In like manner, if b, c, rf. e be successively left out, the number of 
 vai:iations of all the two's will also be ^ f ; and putting t, c, d, e re- 
 spectively 
 
PERMUTATIONS AND COMBINATIONS. 143 
 
 proALem iir. 
 
 Any Numher of Things being given ; of •which there are several 
 given Thiii.gs of one Sort^ and several of another, ^c. ; to 
 Find how tnany Changes can he made out of them all. 
 
 BULB*. 
 
 Take the series 1 X 2 X 3 X 4, &;c. up to the number 
 of things given, aod find the product of all the terms. 
 
 Take the series 1 X 2 X 3 X 4, Sic. up to the number of 
 given things of the first sort, and the series 1 X 2 X 3 X 4, 
 &c. up to the number of given things of the second sort, &c. 
 
 Divide 
 
 spectively in the first place, to make 3 quantities out of 5, there will 
 still be V variations, as before. 
 
 But these are all the variations that can happen of 3 things out of 
 5, when a, b. c, dt e, are successively put first j and therefore the suifta 
 of all these is the sum of all the changes of 3 things out of 5. 
 
 But :he sum of these is so many times t», as is the number of things j 
 that IS 5t>> or mv, = all the changes of 3 times out of 5. 
 
 And the same way of reasoning may be applied to any numbers 
 whatever. 
 
 Demon, of the Rule. Let any 7 things, ab cdef g^he. given, and let 
 3 be the number of quantities to be taken,. 
 
 Then m z=>7^ and n =3 3. 
 
 Now, It is evident, that the number of changes that can be made by 
 taking 1 by 1 out of 5 things, will be 5, which let = v. 
 
 Then, by the Lemma, when m = 6, and n = 2, the number of 
 changes will be = mv ==6X5; which let be = t> a second time. 
 
 Again, by the Lemma, when in = 7 and n = 3, the number of 
 changes h mv = 7 X 6 X o ; that is wit; = w X (m-1) X 
 (m—2), continued to 3, or n terms. 
 
 And the same may be shown for any othfer numbers, 
 
 * This Ilule is expressed in terms thus : 
 
 1 X 2 X 3 X 4 X 5, &c. to m 
 
 1 X 2 X 3, &c. to p X 1 X 2 X 3. &c. to 7 &c. 
 where w =:the number of things given, j&= the number of things 
 of the first sort, g = the number of things of the second sort &c. 
 The Demonstration may be shown as follows ; 
 
 Any two quantities, <7, b, both different, admit of 2 changes ; but If 
 the quantities are the same, or a b becomes a o, there will be only 
 
 1X2 
 one position ; which may be expressed by ■ ■ = 1. 
 
 1 X 2 
 Any 3 quantities, a, b, c, all different from each other, afford 6 varia- 
 tions; but if the quantities be all alike, ova be becomes a a a, then 
 
 the 
 
144 ARITHMETIC. 
 
 piride the product of all the terms of the first series by the 
 joint product of all the terms of the remaining ones, and the 
 i|Uotient will be the answer required. 
 EXAMPLES. 
 
 1. How many variations can be made of the letters in the 
 word Bacchanalia ? 
 
 1 X 2 (= number of c's) = 2 
 1 X 2 X 3 X 4 (= number of a's) = 24 
 1X2X3X4X5X6X7X8X9X10X11 
 ( = number of letters in the word) = 39916800 
 2 X 24 = 48) 39916800 (831600 the Answer. 
 151 
 76 
 288 
 
 2. How many different numbers can be made of the follow- 
 ing figures, 1220005555 ? Ans. 12600. 
 
 3. How many varieties will take place in the successior. of 
 the following musical notes, fa, fa, fa, sol, sol, la, mi, fa ? 
 
 Ans. 3360. 
 
 the 6 variations will be reduced to 1 j which may be expressed by 
 1X2X3 
 
 — = 1. Again, if two of the quantities only are alike, or a ^ c 
 
 1X2x3 
 
 becomes, aac ; then the 6 variations will be reduced to these 3j a a c, 
 
 1X2X3 
 c aa^ and a c a ; which may be expressed by- — — — == 3. 
 
 1X2 
 Any 4 quantities, abed, all different from each other, will admit 
 of 24 variations. But if the quantities be the same, or a b c d be- 
 comes a a a Of the number of variations will be reduced to one ; 
 1X2X3X4 
 
 which is =s ■ ■■ — — = 1. 
 
 1x2x3x4 
 Again, if three of the quantities only be the same, or a b c d 
 becomes a a a b, the number of variations will be reduced to 
 these Ataaabtaabayab a a, and b a a a; which is =: 
 1x2x3x4 
 
 1=4. • 
 
 1X2X3 
 
 And thus it may be shewn, that if two of the quantities be alike, 
 or the 4 quantities be n a 6 c, the number of variations will be re- 
 
 1X2X3 X 4 
 duced to 12 ; which may be expressed by ■ = 12- 
 
 1X2 
 And by reasoning in the same manner, it will appear, that the 
 number of changes which can be made of the quantities a * ^ c c, is 
 
 1X2X3X4X5X6 
 equal to 60 s which may be expressed by ----.----—-——---—-----«----- 
 
 1X2X1X2X3 
 = 60. And so on for any other quantities whateveri 
 
 PBOa- 
 
PERMUTATIONS AND COMBINATIONS. 146 
 
 PROBLEM IV. 
 
 To Jind the Changes of any Criven JSfumber of Things^ taking a 
 Given Number at a Time : in which there art several Given 
 Things of one Sort^ several of another, ^c. 
 
 RULE*. 
 
 Find all the different forms of combination of all the given 
 thinajs, taken as many at a time as in the question. 
 
 Find the number of changes in any form, and multiply it by 
 the number of combinations in that form. 
 
 Do the same for every distinct form, and the sum of all the 
 products will give the whole number of changes required. 
 
 EXAMPLES. 
 
 1. How many alterations, or chanses, can be made of 
 every four letters out of these 8, aaabbbcc ? 
 
 No. of forms. No. of changes. 
 
 a^h^a'^c^b^a, b^c 4 
 
 o2 62,a2c2, 62^3 6 
 
 a^bCyb^ac.c^ ab 12 
 
 /4 X 4 = 16 
 
 Therefore <3X 6=18 
 
 ^3 X 12 = 36 
 
 70 == number of changes 
 — required. 
 
 2. How many changes can be made of every 8 letters out 
 ef these 10 ; aaaabbccde ? Ans. 22260. 
 
 3. How many different numbers can be made out of 1 unit, 
 
 * The reason of this Rule is plain from what has been shown be- 
 fore, and the nature of the problem. 
 
 A Rule for finding the Number of Fofms. 
 
 1. Place the things so, that the greatest indices may be first, and 
 the rest in brder 
 
 2. Begin with the first letter, and join it to the second) third* fourth, 
 &c. to i he last. 
 
 3. Then take the second letter, and join it to the third, fourtht &c, 
 to the last. And so on, till they are entirely exhausted* always remem- 
 bering to reject such combinations as have occurred before ; and this 
 wiU give the combinations of all the twos. 
 
 4. Join the first letter to every one of the twos, and the second, 
 third, &c. as before ; and it will give the combinations of all the threes. 
 
 5. Proceed n the same manner to get the combinations of all the 
 fours, &c. and you will at last get all the several forms of combina- 
 tions, and the number in each form. 
 
 Vol. L 2I> 2 twos, 
 
14G ARITHMETIC. 
 
 2 twos, 3 threes, 4 fours, and 5 fives ; taken 6 at a time ? 
 
 Ans, 2111, 
 PROBLEM V. 
 
 To find the JVurnber of Pombinations of any Given JVufnber of 
 things all different from each other ^ taken any Given Number 
 at a time. 
 
 RULE.* 
 
 Take the series 1, 2, 3, 4, &c. up to the number to be 
 taken at a time, and find the product of all the terms. 
 
 Take a series of as many terms, decreasing by 1, from the 
 given number, out of which the election ia to be made, and 
 find the product of all the terms. 
 
 Divide the last product by the former, and the quotient- 
 will be the number sought. 
 
 EXAMPLES. 
 
 1. How many combinations can be made of 6 letters out 
 of ten ? 
 
 * This Rule> expressed algebraically, is, 
 tn m — 1 m — 2 m — 3 
 
 — X — — X " . X Oc. to n terms ; where m is the 
 
 12 3 4 
 
 number of given quantities, and n those to be taken at a time. 
 
 Demonstr. of the Mule. 1. Let the number of tilings to be taken at 
 a time be 2, and the things to be combined = m. 
 
 Now, when m, or the number of things to be combined, is only two, 
 as a and Z>, it is evident that there can be but one combination, as ab ; 
 but if m be increaseti by one, or the letters to be combined be 3, as a, 
 b, c ; then it is plain that the nr.mber of conibinations will be increas- 
 ed by 2, since with each of the former letters a and Z>, the new letter 
 c may be joined. In this case therefore, it is evident that the whole 
 number of combinations will be truly expressed by 1 -f-2. 
 
 Againt if m be increased by one letter more, or \he whole number of 
 letters be four, as o, b, c, d ; then it will appear that the whole num- 
 ber of combinations must be increased by 3, since with each of the 
 preceding letters the new letter d may be combined. Tue combina- 
 tions therefore, in this case will be truly expressed by 1+2+3. 
 
 And in Ihe same manner it may be shown that the whole number of 
 combinations pf 2, in 54hings, will be 1+2 +3-+ 4; of 2 in 6 things, 
 1 + 2 + 3 + 4 + 5 ; and of 2» in /things, 1+2 + 3-+4 + 5 + 
 6, &c. ; whence, universally, the number of combinations of m things, 
 taken 2 by-2, is = 1 + 2 •+ 3 + 4 + 5 +- C, &c. to (7n— 1) terms, 
 m m — 1 
 
 But the sum of this series is = ~ X ■; which is the same as 
 
 12 
 tl^e rule. 
 
 2. Let now the number of quantities in each combination be sup- 
 posed to be three. 
 
 Then 
 
PERMUTATIONS AND COMBINATIONS. 147 
 
 1 X2X3X4X5X6( = the number to be taken 
 at a time ) = 720. 
 
 10 X 9X8 X7X6 X6(= same number frdna 10) 
 = 151200. 
 
 Then 720 ) 151200 (210 the Answer. 
 1440 
 
 720 
 720 
 
 2. How many combinations can be made of 2 letters out of 
 the 24 letters of the alphabet ? Ans. 276. 
 
 3. A general, who had often been successful in war, was 
 asked by his king what reward he should confer upon him for 
 his services ; the general only desired a farthing for every file, 
 of 10 men in a file, which he could make with a body of 100 
 men ; what is the amount in pounds sterling ? 
 
 Ans. 18031572350Z95 2(;. 
 
 Then it is plain that when m ==» 3, or the things to be combined arc 
 <j, 6> c, there cun be only one combination. But if mbe increased by 
 1, or the things to be combined are 4, as a, 6, c, d, then will the num- 
 ber of combinations be increased by 3 ; since 3 is the number of com- 
 binations of 2 in all the preceding letters, a, b, c, and with each two of 
 these the new letter d may be combined. 
 
 The number of combinations, ih-^refore in this case, is 1 -f- 3. 
 
 Again, if m be increased by one more, or the number of letters be 
 supposed 5 ; then the former number of combinations will be in- 
 creased by 6, that is, by all the combinations of 2 in thti 4 preceding 
 letters, at b, c, d : since, as before, with each two of these the new 
 letter c may be combined. 
 
 The number of combinations, therefore, in this case, is 1 -4- 3 + 6. 
 
 "VVhence, universally, the number of combinations of »a things, taken 
 3 by 3, is 1 + 3 4- 6 -f 10 &c. to m — 2 terms. 
 
 VI m — 1 m — 2 
 
 But the sum of this series is = — X — — — X— — — ; which is 
 12 3 
 
 the same as the rule. 
 
 And the same thing will hold, let the number of things to be taken 
 at a time be what it will ; therefore tlie number of combinations of m 
 things, taken n at a time, will be =*• ^ 
 
 w m — 1 7n— 2 wj — 3 
 •r- X — — X X —— -, &c. to n termS' a* e. d. 
 
 1 2 r. 4 
 
 I'RQBv 
 
148 ARITHMETIC. 
 
 PROBLEM VI. 
 
 To find the Number of Combinations of any Given Number of 
 Things^ by taking any Given Number at a time ; in which there 
 are several Things of one Sort, several of another, ^c. 
 
 RULE. 
 
 Find, by trial, the number of different forms which the 
 things to be taken at a time will admit of, and the number of 
 combinations there are in each. 
 
 Add all the combinations, thus found together, and the sum 
 will t>e the number required. i 
 
 EXAMPLES. 
 
 1 . Let the things proposed be a a a b b c ; it is required to 
 find the number of combinations made of every 3 of these 
 quantities ? 
 
 Forms. Combinations.. 
 
 a3 1 
 
 fl26, a^c, b^a, b^c - - - - - 4 
 ab c - -- - - - -J 
 
 Number of combinations required = 6 
 
 2. Let aaabbbcche proposed ; it is required to find the 
 number of combinations of these quantities, taken 4 at a time ? 
 
 Ans. 10. 
 
 3. How many combinations are there in a a a a b b c c d e, 
 taking 8 at a time ? Aps. 13. 
 
 4. How many coipbinations are there in a a a aa bb b b h 
 ccccddddeeee fff gy taking 10 at a time ? Ans, 2619. 
 
 PROBLEM yiL 
 
 To find the Compositions of any Number, in an equal Number of 
 ^ets, the things themselves being all different, 
 
 RULE*. 
 
 Multiply the number of things in every set continually to- 
 gether, and the product will be the answer required. 
 
 ♦ Demonatr. Suppose there are only two sets ; then, it is plain, that 
 every quantity of the one set being combined with every quantity of the 
 ^Ithej", will make all the compositions, of two things in these two sets; 
 
 and 
 
PERMUTATIONS AND COMBINATIONS. 14^ 
 
 EXAMPLE 
 
 1. Suppose there are four companies, in each of which 
 there are 9 men ; it is required to find how many ways 9 
 men may be chosen, one out of each company ? 
 
 9 
 
 9 
 
 6561 the Answer. 
 Op, 9 X 9 X 9 X 9 = 6561 the Answer. 
 
 2. Suppose there are 4 companies ; in one of which there 
 are 6 men, in another 8, and in each of the other two 9 ; what 
 are the choices, by a composition of 4 men, one out of each 
 company ? Ans. 3888. 
 
 3. How many changes are there in throwing 6 dice ? 
 
 Ans. 7776. 
 
 and the number of these compositions is evidently the product of the 
 number of quantities in one set by that in the other. 
 
 AgaiTif suppose there are three sets ; then the composition of two, 
 in any two of the sets, being combined with every quantity of the third, 
 will make all the compositions of three in the three sets. That is, the 
 coMtpositions of two in any two of the sets, being multiplied by the 
 number >f quantities in the remaining' set, will produce the composi- 
 tions r)f three in the three sets ; which is evidently the continual pro- 
 duct of all the three numb* rs in the three sets. 
 
 And the same manner of reasoning will hold, let the number of sets 
 be what it will. q. e. d. 
 
 Ihe doctrine of permutations, combinations, &c. is of very exten- 
 siv:- use in different parts of the Mathematics ; particularly in the 
 calculat.on of annuities and chances. The subject might have been 
 pursued to a much greater length ; but what is here done, will be 
 found sufficient for most of the purposes to which things of this nature 
 are applicable* 
 
 PRACTICAL 
 
150 ARITHMETIC. 
 
 PRACTICAL QUESTIONS iN ARITHMETIC. 
 
 Quest. 1. The swiftest velocity of a cannon-ball, is about 
 2000 feet in a second of time. Then in what time, at that 
 rate, would such a ball be in moving from the eartlf to the 
 sun, admitting the distance to be 100 millions of miles, and 
 the year to contain 365 days 6 hours. 
 
 Ans. SyV/A years. 
 
 Quest. 2. What is the ratio of the velocity of light to that 
 of a cannon-ball, which issues from the gun with a velocity of 
 1500 feet per second ; light passing from the sun to the earth 
 in 1} minutes ? Ans. the ratio of 782222| to 1. 
 
 Quest. 3. The slow or parade-step being 70 paces per 
 minute, at 28 inches each pace, it is required to determine 
 at what rate per hour that movement is ? Ans. l|if miles. 
 
 Quest. 4. The quick-time or step, in marching, being 2 
 paces per second, or 120 per minute, at 28 inches each ; then 
 at what rate per hour does a troop march on a route, and how 
 long will they be in arriving at a garrison 20 miles distant, al- 
 lowing a halt of one liour by the way to refresh ? 
 
 ^ ( the rate is 3/y miles an hour. 
 °** ^ and the time T^- hr. or 7 h. 17i min. 
 
 Quest. 5. A wall was to be built 700 yards long in 29 days. 
 Now, after 12 men had been employed on it for 11 days, it 
 was found that they had completed only 220 yards of the wall. 
 It is required then to determine how many men must be added 
 to the former, that the whole number of them may jusf finish 
 the wall in the time proposed, at the same rate of working ? 
 
 Ans. 4 men to be added. 
 
 Quest. 6. To determine how far 500 millions of guineas 
 will reach, when laid down in a straight line touching one 
 another ; siipposing each guinea to be an inch in diameter, 
 as it is very nearly. Ans. 71'91 miles, 728 yds. 2ft, 8 in. 
 
 Quest. 7. Two persons, a and b, being on opposite sides 
 of a wood, which is 536 yards about, they begin to go round 
 it, both the same way, at the same instant of time ; a goes at 
 the rate of 1 1 yards per minute, and b 34 yards in 3 minutes ; 
 and the question is, how many times will the wood be gone 
 round before the quicker overtake the slower ? 
 
 Ans. 17 times. 
 
 Quest. 
 
PtlACTICAL QUESTIONS. 151 
 
 Quest. 8. a cab do a piece of work alone in 12 days, 
 and B alone in 14 ; in what time will they both together per- 
 form a like quantity of work ? Ans. 6 -^.j days. 
 
 Quest. 9. A person who was possessed of a | share of 
 a copper mine, sold | of his interest in it for 1800/ ; what was 
 the reputed value of the whole at the same rate ? Ans. 4000/. 
 
 Qhest. 10. A person after spending 20/ more than i of 
 his yearly income, had then remaining 30/ more than the half 
 of it ; what was his income ? Ans. 200/. 
 
 Quest. 11. The hour and minute hand of a clock arc 
 
 exactly together at 12 o'clock ; when are they next together ? 
 
 Ans. at 1 yV hr or 1 hr, /y min. 
 
 Quest. 12. If a gentleman whose annual income is 1500/, 
 spends 20 guineas a week ; whether will he save or run in 
 debt, and how much in the year ? Ans. save 408/. 
 
 Quest. 13. A person bought 180 oranges at 2 a pennj^ 
 and 180 more at 3 a penny ; after which, selling them out 
 again at 5 for 2 pence, whether did he gain or lose by the 
 bargain ? ' Ans. he lost 6 pence. 
 
 Quest. 14. If a quantity of provisions serves 1500 men 
 12 weeks, at the rate of 20 ounces a day for each man ; how 
 many men will the same provisions maintain for 20 weeks, at 
 the rate of 8 ounces a day for each man ? Ans. 2250 men. 
 
 Quest. 15. In the latitude of London, the distance round 
 the earth, measured on the parallel of latitude, is about 15550 
 miles ; now as the earth turns round in 23 hours 56 minutes, 
 at what rate per hour is the city of London carried by this 
 motion from west to east ? Ans. 64&||f miles an hour. 
 
 Quest. 16. A father left his son a fortune, ^ of which he 
 ran through in 8. months ; | of the remainder lasted him 12 
 months longer ; after which he had bare 820/ left. What 
 sum did the father bequeath his son ? Ans. 1913/65 8c/. 
 
 Quest. 17. If 1000 men, besieged in a town with pro- 
 visions for 5 weeks, allowing each man 16 ounces a day, be 
 reinforced with 500 men more ; and supposing that they can- 
 not be relieved till the end of 8 weeks, how many ounces a 
 day must each man have, that the provision may last that 
 time ? * Ans. 6| ounces. 
 
 Quest. 13. A younger brother received 8400/, whicli 
 was just I of his elder brother's fortune : What was the 
 father worth at his death ? Ans. 19200/. 
 
 Quest. 
 
132 ARITHMETIC. 
 
 Quest. 19. A person, looking on his watch, was askedl 
 what was the time of the day, who answered, It is between 
 6 and 6 ; but a more particular answer being required, he 
 said that the hour and minute hands were then exactly toge- 
 ther : What was the time ? Ans. 27 y3_ min. past 5. 
 
 Quest. 20. If 20 men can perform a piece of work in 
 12 days, how many men will accomphsh another thrice as 
 large in one-fifth of the time ? Ans 300. 
 
 Quest. 21. A father devised y\ of his estate to one of his 
 sons, and /j of the residue to another, and the surplus, to his 
 relict for life. The children's legacies were found to be 
 614/ 6s Bd diflferent : Then what money did he leave the 
 widow the use of? * Ans. 1270/ Is d^^d. 
 
 Quest. 22. A person, making his will, gave to one child 
 if of his estate, and the rest to another. When these legacies 
 came to be paid the one turned out 1200/ more than the 
 other : What did the testator die worth ? Ans. 4000/. 
 
 Quest. 23. Two persons, a and b, travel between London 
 and Lincoln, distant 100 miles, a from London, and b from 
 Lincoln, at the same instant. After 7 hours they meet on the 
 road, when it appeared that a had rode 1^ miles an hour more 
 than B. At what rate per hour then did each of the travellers 
 ride ? Ans. a. 7|f , and b 6^^ miles. 
 
 Quest. 24. Two persons, a and b, travel between Lon- 
 don and Exeter, a leaves Exeter at 8 o'clock in the morn- 
 ing, and walks at the rate of 3 miles an hour, without inter- 
 mission ; and b sets out from London at 4 o'clock the S3me 
 evening, and walks for Exeter at the rate of 4 miles an hour 
 constantly. Now, supposing the distance between the two 
 cities to be_ 130 miles, whereabouts on the road will they 
 meet ? Ans. 69f miles from f^xeter. 
 
 Quest. 25. Ope hundred eggs being placed on the 
 ground in a straight line, at the distance of a yard from each 
 ©ther : How far will' a person travel who shall bring them 
 ©ne by one to a basket, which is placed at one yard from the 
 first egg ? Ans. 10100 yards, or 5 miles and 1300 yds. 
 
 Quest. 26. The clocks of Italy go on to 24 hours : 
 Then how many strokes do they strike in one complete re- 
 volution of the index ? • Ans. 300. 
 
 Quest. 27. One Sessa, an Indian, having invented the 
 game of chess, shewed it to bis prince, who was so delighted 
 
 with 
 
PRACTICAL QUESTIONS. 153 
 
 t^ith it, that he promised him any rewai^d he should ask ; on 
 which Sessa requested that he might be allowed one grain of 
 wheat for the first square on the chess board, 2 for the second, 
 4 for the third, and so on, doubling continually, to 64, the 
 whole number of squares. Now, supposing, a pint to contain 
 7680 of these grains, and one quarter or 8 bushels to be worth 
 27s 6dy it is required to compute the value of all the corn ? 
 
 Ans. 6460468216285/ 17s 3d ^^lq. 
 
 Quest. 28. A person increased his estate annually by 
 100/ more than the a part of it ; and at the end of 4 'years 
 found that his estate amounted to 10342/ 3s 9d. What had he 
 at first? ^ Ans. 4000/. 
 
 Quest. 29. Paid 1012/ 10« for a principal of 750/, taken 
 in 7 years before : at what rate per cent, per annum did I pay 
 interest ? Ans. 5 per cent. 
 
 Quest. 30. Divide 1000/ among a, b> c ; so as to give 
 A 120 more, and b 95 less than c 
 
 Ansi A 445, b 230, c 325. 
 
 Quest. 31. A person being asked the hour of the day, 
 said, the time past noon is equal to f ths of the time till mid- 
 night. What was the time ? Ans. 20 min. past 6. 
 
 Quest. 32. Suppose that I have fV of a ship worth 1200/ ; 
 what part of her have I left after selhng f of f of my share, 
 and what is it worth ? Ans. -^W* worth 185/. 
 
 Quest. 33. Part 1200 acres of land among a, b, c ; so 
 that B may have 100 more than a, and c 64 more than b. 
 
 Ans. A 312, B 412, c 476. 
 
 Quest. 34. What number is that, from which if there be 
 taken f of f, and to the remainder be added y j of y5_, the 
 sum will be 10 ? Ans. 9|f . 
 
 Quest. 35. There is a number which if multiplied by | 
 ef 1^ of 1 ^, will produce 1 : what is the square of that number Z 
 
 Ans. IV^. 
 
 Quest. 36. What length must be cut off a board, 8^^ inches 
 broad, to contain a square foot, or as much as 12 inches in 
 length and 12 in breadth ? Ans. 16 |f inches. 
 
 Quest. 37. What sum of money will amount to 138/ 2s 
 6d, in 15 months, at 5 per cent, per annum simple interest ? 
 
 Ans. 130/. 
 
 Quest. 38. A father divided his fortune among his three 
 sons, A, B, c, giving a 4 as often as b 3, and c 5 as often as 
 
 Vol. I. 21 
 
164 ARITHMETIC. 
 
 B 6 ; what was the whole legacy, supposing a's share was 
 4000Z. Ans. 9600^. 
 
 Quest. 39. A young hare starts 40 yards before a grey- 
 bound, and is not perceived by him till she has been up 40 
 seconds ; she scuds away at the rate of 10 miles an honr, and 
 the dog, on view, makes after her at the rate of 18 ; ho,w long 
 will the course hold, and what ground will be run over, count- 
 ing from the outsetting of the dog ? 
 
 Ans. eOj'gSec. and 630 yards run. 
 
 Quest. 40. Two young gentlemen, without private for- 
 tune, obtain commissions at the same time, and at the age of 
 18. One thoughtlessly spends 10/ a year more than his pay ; 
 but, shocked at the idea of not paying his debts, gives his 
 creditor a bond for the money, at the end of every year, and 
 also insures his hfe for the amonnt ; each bond costs him 30 
 shilliijgs, besides the lawful interest of 6 per cent, and to in- 
 sure his hfe costs him 6 per cent. 
 
 The other, having a proper pride, is determined never to 
 run in debt ; and, that he may assist a friend in need, perse- 
 veres in saving IQl every year, for which he obtains an inter- 
 est of 5 per cent, which interest is every year added to his 
 savings, and laid out, so as to answer the effect of compound 
 interest. 
 
 Suppose these two officers to meet at the age of 50, when 
 each receives from Government 400/ per annum ; that the 
 one, seeing his past errors, is resolved in future to spend no 
 more than he actually has, after paying the interest for what 
 he owes, and the insurance on his life. 
 
 The other, having now something before hand, means in 
 future, to spend his full income, without increasing his stock. 
 
 It is desirable to know how much each has to spend per 
 annum, and what money the latter has by him to assist the 
 distressed, or leave to those who deserve it ? 
 
 Ans. The reformed officer has to spend 661 19s 1| •6389c?. 
 per annum. 
 The prudent officer has to spend 437Z 12s 1 If •4379c?. 
 
 per annum. 
 And the latter has saved, to dispose of, 762Z 19s 9-1896(1 
 
 END OP THE ARITHMETIC 
 
[ 15g] 
 
 OF LOGARITHMS*. 
 
 JLjOGARITHMS are made to facilitate troublesome calcu- 
 lations in numbers. This they do, because they perform 
 multiplication by only addition, and division by only subtrac- 
 tion, and raising of powers by multiplying the logarithai by 
 the index of the power, and extracting of roots by dividing 
 the logarithm of the number by the index of the root. For, 
 logarithms are numbers so contrived, and adapted to other 
 numbers, that the sums and differences of the former shall 
 correspond to, and show, the products} and quotients of the 
 latter, &c. 
 
 Or, more generally, logarithms are the rinmerical expo- 
 nents of ratios ; or they are a series of numbers in arith- 
 metical 
 
 * The invention of Logaritliras is due to Lord Napier, Baron of 
 Merchiston, in Scotland, and is properly considered as one of the 
 most useful inventions of modern times. A. table of these numbers 
 was first published by the inventor at Edinbui'gh, in the year 1614, in 
 a treatise entitled Canon Mirijiciim Logarithmornm ; which was eager- 
 ly received by all the learned J hroughout Enrope. Mr. Henry Briggs, 
 then professor of geometry at Gresham College, soon after the dis- 
 covery, went to visit the noble inventor j after which, they jointly un- 
 dertook the arduous task of computing new tables on this subject, and 
 reducing them to a more convenient form than that which was at first 
 thought of. But Lord Napier dying soon after, the whole burden fell 
 upon Mr. Briggs, who, with prodigious laboiur and great skill, made an 
 entire Canon, according to the new form, for all numbers from 1 to 
 20000, and from 900u0 to 10100, to 14 places of figures, and published 
 it at London, in the year 1624, in a treatise entitled Arithmetica Loga- 
 rithmica, with directions for supplying the intermediate parts. 
 
 This 
 
166 LOGARITHMS. 
 
 metical progression, answering to another series of numbers 
 in geometrical progression. i 
 
 rp, JO, 1, 2, 3, 4, 6, 6, Indices, or logarithms, 
 
 inus, ^ j^ 2^ ^^ g^ 16, 32, 64, Geometric progression. 
 
 Q JO, 1, 2, 3, 4, 5, 6, Indices, or logarithms. 
 ^1, 3, 9, 27, 81, 243, 729, Geometric progression. 
 
 ^ Jo, 1, 2, 3, 4, 6, Indices, or logs. 
 
 ^^ ^1, 10, 100, 
 
 1000, 10000, 100000, Geom. progress. 
 
 Where it is evident, that the same indices serve equally 
 for any geometric series ; and consequently there may be an 
 
 This Canon was again published in Holland by Adrian Vlacq, in the 
 year 1628, together with the Logarithms of all the numbers which 
 Mv. Briggs had omitted ; but he contracted them down to 10 places 
 of decimals. Mr. Briggs also computed the Logarithms of the sines, 
 tangents, and secants, to every degree, and centesm, or 100th part of a 
 degree, of the whole quadrant ; and annexed them to the natural 
 sines, tangents, and secants, which he had before computed, to fifteen 
 places of figures. These Tables, with their construction and use, 
 ■were first published in the year 1633, after Mr. Brigg*s death, by Mr. 
 Henry Gellibrand, under the title of Trigonometria Britannica. 
 
 Benjamin Ursinus also gave a Table of Napier's Logs, and of sines, 
 to every 10 seconds. And Chr. Wolf, in his Mathematical Lexicon, 
 says that one Van Loser had computed them to every single second, 
 but his untimely death prevented their publication. Many other 
 authors have treated on this subject ; but as their numbers are fie- 
 quenliy inaccurate and incommodiously disposed, they are now gene- 
 rally neglected. The Tables in most repute at present, are those of 
 Gardiner in 4to, first published in the year 1742 ; and my own Tables 
 in 8vo, first printed in the year 1785, where the Logarithms of ail num- 
 bers may be easily found from 1 to lOuOOOOO ; and those of the sines, 
 tangents, and secants, to any degree of accuracy required. 
 
 Also, Mr. Michael Taylor's Tables in large 4to, containing the com- 
 mon logarithms, and the logarithmic sines and tangents to every se- 
 cond of the quadrant. And, in France, the new book of logarithms 
 by Callet ; the 2d edition of winch, in 1795, has the tables stil) far- 
 ther extended, and are printed with what are called stereotypes, tlie 
 types in each page being soldered together into a solid mass or block. 
 
 Dodson's Antilogarithmic Canon is likewise a very elaborate work, 
 and used for finding the numbers answering to any given logarithm. 
 
 endless 
 
LOGARITHMS. 157 
 
 endless variety of systems of logarithms, to the same common 
 numbers, by only changing the second term, 2, 3, or 10, &c. 
 of the geometrical series of whole numbers ; and by interpo- 
 lation the whole system of numbers may be made to enter the 
 eometric series, and receive their proportional logarithms, 
 whether integers or decimals. 
 
 It is also apparent, from the nature of these series, that if 
 any two indices be added together, their sum will be the index 
 of that number which is equal to the product of the two terms, 
 in the geometric progression, to which those indices belong. 
 Thus, the indices 2 and 3, being added together, make 6 ; and 
 the numbers 4 and 8, or the terms corresponding to those indi- 
 ces, being multiplied together, make 32, which is the number 
 answering to the index o. 
 
 In like manner, if any one index be subtracted from another, 
 the difference will be the index of that number which is 
 equal to the quotient of the two terras to which those indi- 
 ces belong. Thus, the index 6, minus the index 4, is = 2 ; 
 and the terms corresponding to those indices are 64 and 16, 
 whose quotient is = 4, which is the number answering to the 
 index 2. 
 
 For the same reason, if the logarithm of any number 
 be multiplied by the index of its power, the product will 
 be equal to the logarithm of that power. Thus, the index or 
 logarithm of 4, in the above series, is 2 ; and if this number be 
 multiplied by 3, the product will be = 6 ; which is the loga- 
 rithm of 64, or the third power of 4. 
 
 And, if the logarithm of any number be divided by the 
 index of its root, the quotient will be equal to the logarithm 
 of that root. Thus, the index or logarithm of 64 is 6 ; and if 
 this number be divided by 2, the quotient will be = 3 ; which 
 is the logarithm of 8, or the square root of 64. 
 
 The logarithms most convenient for practice, are such as 
 are adapted to a geometric series increasing in a tenfold pro- 
 portion, as in the last of the above forms ; and are those 
 which are to be found, at present, in most of the common 
 tables on this subject. The distinguishing mark of this 
 system of logarithms is, that the index or logarithm of 10 
 is 1 ; th'at of 100 is 2 ; that of 1000 is 3 ; &c. And, in 
 
 decimals. 
 
168 LOGARITHMS. 
 
 decimals, the logarithm of •! is — 1 ; that of •01 is — 2 ; that 
 of -001 is — 3 ; &c. The log. of 1 being in every system. 
 Whence it follows, that the logarithm of any number between 
 1 and 10, must be and some fractional parts ; ?nd that of a 
 number between 10 and 100, will be 1 and some fractional 
 parts ; and so on, for any other number whatever. And 
 since the integral part of a logarithm, usually called the Index, 
 or Characteristic, is always thus readily found, it is commonly 
 omitted in the tables ; being left to be supplied by the opera- 
 tor himself, as occasion requires. 
 
 Another Definition of Logarithms is, that the logarithm of 
 any number is the index of that power of some other num- 
 ber, which is equal to the given number. So, if there be 
 N = r", then n is the log. of N ; where n may be either posi- 
 tive or negative, or nothing, and the root r any number 
 "whatever, according to the different systems of logarithms. 
 When n is = 0, then N is = 1, whatever the vahie of r is ; 
 which shows, that the log. of 1 is always 0, in every system of 
 logarithms. When n is = 1, then N is = r ; so that the 
 radix r is always that number whose log is 1, in every sys- 
 tem. When the radix r is = 2-718281828469, &c. the in- 
 dices n are the hyperbolic or Napier's log. of the numbers N ; 
 so that n is always the hyp. log. of the number N or (2*718 
 &c.)n . 
 
 But when the radix r is = 10, then the index n becomes 
 the common or Briggs's log. of the number N : so that the 
 common log. of any number 10" or N, is n the index of 
 that power of 10 which is equal to the said number. Thus 
 100, being the second power of 10 will have 2 for its loga- 
 rithm : and 1000, being the third power of 10, will have 3 
 for its logarithm: hence also, if 60 be = lO'-eosoT^ then 
 is 1-69897 the common log. of 50. And, in general, the fol- 
 lowing decuple series of terms, 
 
 viz. lOS 103, 102, 101, 100^ lO-S 10-2, 10-3^ jo_4, 
 oV 10000, 1000, 300, 10, 1, -1, -01, -001, -0001, 
 have 4, 3, 2, 1, 0,-1, —2, —3, —4, 
 
 for their logarithms, respectively. And from this scale of 
 numbers and logarithms, the same properties easily follow, 
 as above mentioned* 
 
 PROBLEM, 
 
LOGARITHMS. 169 
 
 PROBLEM. 
 
 Po compute the Logarithm to any of the Natural Numbers 
 1, 2, 3, 4, 6, ^c. 
 
 RULE I*. 
 
 Take the geometric series, 1, 10, IGO, 1000, 10000, &c. 
 and apply to it the arithmetic series, 0, 1, 2, 3, 4, &c. as 
 logarithms. — Find a geometric mean between 1 and 10, or 
 between 10 and 100, or any other two adjacent terms of the 
 series, between which the number proposed lies. — In like 
 manner, between the mean, thus found, and the nearest ex- 
 treme, find another geometrical mean ; and so on, till you 
 arrive, within the proposed limit of the number whose loga- 
 rithm is sought. — Find also as many arithmetical means, in 
 the same order as you found the geometrical ones, and these 
 will be the logarithms answering to the said geometrical 
 means. 
 
 EXAMPLE. 
 
 Let it be required to find the logarithm of 9. 
 Here the proposed number lies between 1 and 10. 
 
 First, then, the log. of 10 is 1, and the log. of 1 is ; 
 theref. 14-0-5-2=1^= -6 is the arithmetical mean, 
 and ^ f X 1 = y/ 10 = 3-1622777 the geom. mean ; 
 hence the log. of 3- 1622777 is -5. 
 
 "Secondly, th e log, of 10 is I, and the log. of 3-1622777 is -5 : 
 theref. 1 + '5 -=- 2 = -7 5 is the arithmetical mean, 
 and ^ io X 3-1622777 = 5-6234132 is the geom. mean ; 
 hence the log. of 3-6234132 is -76. 
 
 Thirdly, the log, of 10 is 1, and the log. of 5-6234132 is -76 ; 
 theref. \ -\- -15 -^ 2 = -875 is the arithmetical mean, 
 and y/ 10 X 5-6236132 = 7-4989422 the geom. mean ; 
 hence the log. of 7-4989422 is -875. 
 
 Fourthly, th e log, of 1 is 1, and the log. of 7-4989422 is -875} 
 theref. 1 -j- '875 -r 2 = -9375 is the arithmetical mean, 
 and ^ 10 X 7-4989422 = 8-6596431 the geom. mean j 
 hence the log. oS 8-6596431 is -9375. 
 
 * The reader who wishes to inform himself more particularly con- 
 cerning the history, nature, and construction of Logarithms, may con- 
 sult the Introduction to my Mathematical Tables, lately published, 
 where he will find his curiosity amply gratified. 
 
 Fifthly, 
 
160 LOGARITHMS. 
 
 Fifthly, the l og, of 10 is 1, and the log. of 8-6596431 is -9876; 
 theref. 14-9375 -j- 2 = -96875 is the arithmetical mean, 
 and ^ 10 X 8-6596431 = 9-3057204 the geom. mean ; 
 hence the log. of 9-3057204 is -96875. 
 
 Sixthly, the log. of 8-6596431 is -9373, and the log. of 
 9-30 57204 is -96875 ; 
 
 theref. -9375 + -96875^ 2 = - 953125 is the arith. mean, 
 and ^ 8-6596431 X 9-3057204 = 8-9768713 the geo- 
 metric mean ; 
 hence the log. of 8-9768713 is -953125. 
 
 And proceeding in this manner, after 25 extractions, it 
 will be found that the logarithm of 8-9999998 is -9542425; - 
 which may be taken for the logarithm of 9, as it differs so. 
 little from it, that it is sufficiently exact for all practical pur- 
 poses. And in this manner were the logarithms of almost all 
 the prime numbers at first computed. 
 
 RULE U*. 
 
 Let b be the number whose logarithm is required to be 
 found ;.and a the number next less than b, so that b — a = 1, 
 the logarithm of a being known ; and let s denote the sum 
 of the two numbers a -f 6. Then 
 
 1. Divide the constant decimal -8685889638 &c. by s, and 
 reserve the quotient : divide the reserved quotient by the 
 square of s, and reserve this quotient : divide this last quo- 
 tient also by the square of s, and again reserve the quotient : 
 and thus proceed, continually dividing the last quotient by the 
 square of s, as long as division can be made. 
 
 2. Then write these quotients orderly under one another, 
 the first uppermost, and divide them respectively by the odd 
 numbers, 1, 3, 5, 7, 9, &;c. as long as division can be made; 
 that is, divide the first reserved quotient by 1, the second by 
 3, the third by 5, the fourth by 7, and so on. 
 
 3. Add all these last quotients together, and the sum will 
 be the logarithm of 6 -7- a ; therefore to this logarithm add 
 also the given logarithm of the said next less number a, so 
 will the last sum be the logarithm of the number 6 proposed. 
 
 * For tke demonstration of this rule, see my Mathematical Tables, 
 p. 109. &c. 
 
 That 
 
LOGARITHMS. 
 
 161 
 
 That is, 
 
 n 111 
 
 Log. of 6 is log. a'\ X (IH 1 1 +&c.) 
 
 8 . 352 5s4 7s« 
 
 where n denotes the constant given decimal '8686889638 &g. 
 
 EXAMPLES. 
 
 Ex. 1. Let it be required to find the log. of number 2. 
 Here the given number b is 2, and the next less number a 
 is 1 , whose log. is ; also the sum 2 -{- 1 = 3 = s, and it« 
 square s^ = 9. Then the operation will be as follows : 
 
 
 •868588964 
 
 289529664 
 
 32169962 
 
 3574440 
 
 397160 
 
 44129 
 
 4903 
 
 645 
 
 61 
 
 •289629664 ( 
 32169962 ( 
 
 •289629654 
 
 10723321 
 
 3574440 ( 
 
 714888 
 
 397J60 ( 
 
 66737 
 
 44129 ( 
 
 4903 
 
 4903 ( 
 
 446 
 
 645 ( 
 
 42 
 
 61 ( 
 
 4 
 
 log. off - 
 
 •301029995 
 
 add log. 1 - 
 
 •000000000 
 
 log. of 2 - -301029995 
 
 Ex. 2. To eomputa the logarithm of the number 3.. 
 Here 6 = 3, the next less number a =: 2, and the sum 
 a -f 6 = 5 = 5, whose square s^ is 25, to divide by which, 
 always multiply by '04. Then the operation is as follows : 
 5 ) -868588964 1 ) -173717793 ( -173717793 
 
 25 ) 
 25 ) 
 25 ) 
 25 ) 
 25) 
 
 -868588964 
 
 1) 
 
 -1737 J 7793 
 
 3 ) 
 
 5) 
 
 6948712 
 
 277948 
 
 7 ) 
 
 11118 
 
 9) 
 
 445 
 
 11 ) 
 
 18 
 
 
 6948712 ( 
 
 2316237 
 
 277948 ( 
 
 55590 
 
 11118 ( 
 
 1588 
 
 445 ( 
 
 50 
 
 18 C 
 
 2 
 
 log. off - -176091260 
 log. of 2 add -301029995 
 
 log. of 3 sought -477121255 
 
 Then, because the sum of the logarithms of numbers, 
 gives the logarithm' of their product ; and the difference of 
 the logarithms, gives the logarithm of the quotient of the 
 
 Vol. I. 22 numbers : 
 
162 
 
 LOGARITHMS. 
 
 numbers ; from the above two logarithms, and the logarithm 
 of 10, which is 1, we may raise a great many logarithms, as 
 in the following examples : 
 
 EXAMPLE 3. 
 Because ^X2 = 4, therefore 
 to log. 2 - -3010299951 
 add log. 2 - ^3010299961 
 
 sum is log. 4 -602059991^ 
 
 EXAMPLE 4. 
 
 Because 2X3 = 6, therefore 
 
 to log. 2 - -301029995 
 
 add log. 3. - -477121255 
 
 s«m is log. 6 -778151250 
 
 EXAMPLE 5. 
 Because 2 = 8, therefore 
 log. 2 - -3010299961 
 mult, by 3 3 
 
 gives log. 8 -903089987 
 
 EXAMPLE 6. 
 Because 3^ = 9, iht refore 
 log. 3 - •477r^l254y'?^ 
 mult, by 2 '2 
 
 gives log. 9 -954242509 
 
 EXAMPLE 7. 
 Because lp = 5, therefore 
 from log. 10 1-000000000 
 take log. 2 •301029995| 
 
 leaves log. 6 •69897«004i 
 
 EXAMPLE 8. 
 Because 3X4 = 12, therefore 
 to log. 3 - -477121265 
 add log. 4 - -602059991 
 
 gives log. 12 1-079181246 
 
 And thus, computing, by this general rule, the logarithms 
 to the other prime numbers, 7, 11, 13, 17, 19, 23, &c and 
 then using composition and division, we may easily find as 
 many logarithms as we please, or may speedily examine any 
 jlojgarithm in the table*. 
 
 ^ There are, besides these, many other ingenious methods, which 
 later writers hate discovered for findr;g the logarithms of numbers, 
 in a much easier way than by the original inventor ; but, as they 
 cannot Vve understood without a knowledge of some of the higher 
 b/anches ^ the mathematics, it is thought proper to omit them, and 
 to refer the reader to those works which are Wi itten expressly on the 
 subject. It vould likewise much exceed tlie limits of this compen- 
 dium, to point cut all the particular artifices that are made use of for 
 constructing an entire table of these numbers ; but any information of 
 this kind, which the leamer may wish to obtain, may be found in my 
 Tables, before mentioned. 
 
 Description 
 
I^OGARITHMS, 165 
 
 Description and Vie of the TABLE of LOGARITHMS. 
 
 Having explained the manner oi forming a table of the log- 
 arithms of numbers, greater than unity ; the next thaig to 
 be done is, to show how the logarithms of fractional quanti- 
 ties may be found. In order to this, it may be observed, 
 that as in the former case a geometric series is supposed to 
 increase towards the left, from unity, so in the latter case 
 it is supposed to decrease towards the right hand, still be- 
 ginning with unit ; as exhibited in the general description, 
 page 148, where the indices being made negative, still show 
 the logarithms to which they belong. Whence it appears, 
 that as -f- 1 is the log. of 10, so — 1 is the log. of yV or -1 ; 
 and as 4- 2 is the log. of 100, so — 2 is the log. of y^^ or 
 •01 : and so on. 
 
 Hence it appears in general, that all numbers, which consist 
 of the same figures, whether they be integral, or fractional, or 
 mixed, will have the decimal parts of their logarithms the same» 
 but diifuring only in the in lex, which will be more or less, and 
 positive or negative, according to the place of the first figure 
 of the number. 
 
 Thus, the logarithm of 2651 being 3-423410, the log. of 
 tV> or To J or r oVo 5 ^c- P*"^* ^^ ** > ^^^^ ^® ^s follows : 
 
 Numbers. 
 
 Logarithms. 
 
 2 6 6 1 
 
 3-423410 
 
 2 6 5-1 
 
 2-423410 
 
 2 6-51 
 
 1-423410 
 
 2-651 
 
 0-423410 
 
 •2651 
 
 •—1423410 
 
 2 6 5 1 
 
 —2-423410 
 
 2 6 5 1 
 
 —3 -423410 
 
 Hence it also appears, that the index of any logarithm, n 
 always less by 1 than the number of integer figures which the 
 natural number consists of ; or it is equal to the distaare of 
 the first figure from the place of units, or first place of inte- 
 gers, whether on the left, or on the right, of it : and this index 
 is constantly to be placed on the left hand side of the decimal 
 part of the logarithm. 
 
 When there are integers in the given number, the index is 
 always affirmative ; but when there no integers, the index is 
 negative, and is to be marked by a shoit line drawn before it^ 
 or else above it. Thus, 
 
 A number having 1, 2, 3, 4, 5, &c. integer places, 
 
 the index of its log. is 0, 1, 2, 3, 4, &c. or 1 less than those 
 places. 
 
 And 
 
164 LOGARITHMS. 
 
 And a decimal fraction having its first figure in the 
 
 1st, 2d, 3d, 4th, &c. place of the decimals, has always 
 
 — 1, — 2, — 3, — .4, &;c. for the index of its logarithm. 
 
 It may also be observed, that though the indices of fractional 
 quantities are negative, yet the decimal parts of their loga- 
 rithms are always affirmative. And the negative mark ( — ) 
 may be set either before the index or over it. 
 
 1. TO FIND, IN THE TABLE, THE LOGARITHM TO ANY 
 NUMBER*. 
 
 1. If the given Number he less than 100, or consist of 
 only two figures ; its log is immediately found by inspection 
 in the first page- of the table, which contains all numbers from 
 1 to 100, with their logs, and the index immediately annexed in 
 the next column. 
 
 So the log. of 6 is 0-698970. The log. of 23 is 1-361728. ^ 
 The log. of 60 is 1 -698970. And so on. 
 
 2. ff^ihe Number he more than 100 hut less than 10000 ; 
 that is, consisting of either three or four figures ; the decimal 
 part of the logarithm is found by inspection in the other pages 
 of the table, standing against the given number, in this manner; 
 viz. the first three figures of the given number in the first 
 column of the page, and the fourth figure one of those along 
 the top line of it ; then in the angle of meeting are the last 
 four figures of the logarithm, and the first two figures of the 
 same at the beginning of the same line in the second column 
 of the page : to which is to be prefixed the proper index, 
 which is always 1 less than the number of integer figures. 
 
 So the logarithm of 251 is 2-399674, that is, the decimal 
 •399674 found in the table, with the index 2 prefixed, because 
 the given number contains three integers. And the log. of 
 34-09 is. 1-632627, :that is, the decimal -632627 found in the 
 table, with the index 1 prefixed, because the given number 
 contains two integers. 
 
 3. But if the eiven Number contain more than four figures ; 
 take out the logarithm of the first four figures by inspection 
 in the table, as before, as also the next greater logarithm, sub- 
 tracting the one logarithm from the other, as also their cor- 
 responding numbers the one from the other. Then say. 
 
 As the difference between the two numbers, 
 Is lo the difference of*their logarithms. 
 So is the remaining part of the given number, 
 To the proportional part of the logarithm. 
 
 ♦ See the table of Loffarithms at the end of the 2d volume. 
 
 Which 
 
LOGARITHMS. 16o 
 
 Which part being added to the less logarithm, before taken 
 •ut, gives the whole logarithms sought very nearly. 
 
 EXAMPLE. 
 
 To find the logarithm of the number 340926. 
 The log. of 340*500, as before, is 532627. 
 
 And log. of 341000 - - ^ is 532754. 
 The diffs. are 100 and 127 
 
 Then as 100 : 127 :: 26 : 33, the proportional part. 
 
 This added to - - 532627, the first log. 
 
 Gives, with the index, 1-532660, for the log. of 34-0926. 
 
 4. If the number consist both of integers and fractions, or 
 IS entirely fractional : find the decimal part of the logarithm 
 the same as if all its figures were integral ; then this, having 
 prefixed to it the proper index, will give the logarithm re- 
 quired. 
 
 5. And if the given number be a proper vulgar fraction ; 
 subtract the logarithm of the denominator from the loga- 
 rithm of the numerator, and the remainder will be the loga- 
 rithm sought ; which, being that of a decimal fraction, must 
 always have a negative index. 
 
 6. But if it be a mixed number ; reduce it to an improper 
 fraction, and find the difference of the logarithms of the nu- 
 merator and denominator, in the same manner as before. 
 
 EXAMPLES. 
 
 1. To find the log. of f f 
 Log. of 37 - 1-568202 
 
 Log. of 94 - 1-973128 
 
 Dif. log. of |i ■-- 1-595074 
 
 Where the index 1 is negative. 
 
 2 To find the log. of ll^. 
 First, I7J^ = o/. ThenV 
 Log. of 405 - 2-607455 
 Log. of 23 - 1-361728 
 
 Dif. log. of 17i| 1-245727 
 
 n. TO FIND THE NATURAL NUMBER TO ANY GIVEN 
 - LOGARITHM. 
 
 This is to be found in the tables by the reverse method 
 to tlie former, namely, by searching for the proposed loga- 
 rithm pmong those in the table, and taking out the corres- 
 ponding number by inspection, in which the proper number 
 of integers are to be pointed, off, viz. 1 more than the 
 index. For, in finding the number answering to any given 
 logarithm, the index always shows how far the first figure 
 
 must. 
 
166 LOGARITHMS. 
 
 must be removed from the place of units, viz. to the left hand, 
 or integers, when the index is affirmative ; but to the right 
 hand, or decimals, when it is negative. 
 
 EXAMPLES. 
 
 So, the number to the log. 1 -632882 is 34-11. 
 
 And the number of the log. 1-632882 is '3411. 
 
 But if the logarithm cannot be exactly found in the table ; 
 take out the next greater and the next less, subtracting the 
 one of these logarithms from the other, as also their natural 
 numbers the one from the other, and the less logarithm from 
 the logarithm proposed. Then say, 
 
 As the difference of the first or tabular logarithms. 
 Is to the difference of their natural numbers. 
 So is the differ, of the given log. and the least tabular log. 
 To their corresponding numeral difference. 
 Which being annexed to the least natural number above taken, 
 gives the natural number sought, corresponding to the pro- 
 posed logarithm. 
 
 EXAMPLE. 
 
 So, to find the natural number answering to the given 
 
 logarithm 1-532708. 
 Here the next greater and next less tabular logarithms, 
 with their corresponding numbers, are as below : 
 
 Next greater 632744 its num. 341000 ; given log. 632708 
 Next less 632627 its num. 340900 ; next less 632627 
 
 Differences 127 — 100 — 81 
 
 Then, as 127 : 100 : i 81 : 64 nearly, the numeral differ. 
 Therefore 34-0964 is the number sought, marking off two 
 integers, because the index of the given logarithm is 1. 
 
 Had the index been negative, thus 1*632708, its corres- 
 ponding number would have be^n -340966, wholly dc' 
 cimal. 
 
 MULTIPU- 
 
[ 167 ] 
 
 MULTIPLICATION BY LOGARITHMS. 
 
 RULE. 
 
 Take out the logarithms of the factors from the table, 
 then add them together, and their sum will be the logarithm 
 of the product required. Then, by means of the table, take 
 out the natural number, answering to the sum, for the pro- 
 duct sought. 
 
 Observing to add what is to be carried from the decimal 
 part of the logarithm to the affirmative index or indices, or 
 else subtract it from the negative. 
 
 Also, adding the indices together when they are of the 
 same kind, both affirmative or both negative ; but subtracting 
 the less from the greater, when the one is affirmative and the 
 other negative, and prefixing the sign of the greater to the 
 remainder. 
 
 EXA> 
 
 . To Multiply 23-14 by 
 5-062. 
 Numbers. Logs. 
 23-14 - 1-364363 
 5-062 - 0-704322 
 
 IPLES. 
 
 2. To multiply 2-58192C 
 by 3-457291, 
 Numbers. Logs. 
 2-581926 - 0-411944 
 3-457291 - 0-538736 
 
 )duct 117-1347 2-068685 
 
 Prod. 8-92648 - 0-950680 
 
 3. To mult. 3-902 and 597-16 
 and -0314728 all together. 
 Numbers. , Logs. 
 3-902 - 0-591287 
 597-16 - 2-776091 
 •0314728—2-497935 
 
 Prod. 73-3333 
 
 I 865313 
 
 Here the — 2 cancels the 2, 
 and the 1 to carry from the 
 decimals is set down. 
 
 4. To mult. 3-586, and 2-1046, 
 and 0-8372, and 0-0294 all 
 together. 
 
 Numbers. Logs. 
 3-586 - 0-654610 
 2-1046 - 0-323170 
 0-8372—1-922829 
 00294 — 2-468347 
 
 Prod. 0-1057618-1-26895G 
 
 Here the 2 to carry cancels 
 the — 2, and there remains the 
 — 1 to set down. 
 
 DIVISION 
 
[168] 
 
 DIVISION BY LOGARITHMS. 
 RULE. 
 
 From the logarithm of the dividend subtract the logarithm 
 of the divisor, and the. number answering to the remainder 
 will be the quotient required. 
 
 Observing to change the sign of the index of the divisor, 
 from affirmative to negative, or fioni negative to affirmative ; 
 then take the sum of the indices if they be of the same name, 
 or their difference when of different signs, with the sign of 
 the greater, for the index to the logarithm of the quotient. 
 
 And also, when 1 is borrowed, in the left-hand place of 
 the decimal part of the logarithm, add it to the index of the 
 divisor when that index is affirmative, but subtract it when 
 negative ; then let the sign of the index arising from hence 
 fee changed, and worked with as before. 
 
 EXAMPLES. 
 
 1. To divide 24163 by -4567. 
 
 Numbers. Logs. 
 Dividend 24163 - 4-383151 
 Divisor - 4567 - 3-659631 
 
 quot. 5-29078 0-723520 
 
 2. To divide 37-149 by 623-76 
 
 Numbers. Logs. 
 Dividend 37-149 - 1-569947 
 Divisor 523-76 - 2-719132 
 
 Quot. -0709275— 2-850816 
 
 3. Divide -06314 by -007241 
 Numbers. Logs. 
 Divid. -06314 — 2-800305 
 Divisor -007241 — ■ 3-859799 
 
 Quot. 
 
 8-71979 0-940506 
 
 Here 1 carried from the 
 decimals to the — 3, makes it 
 become — 2, which taken from 
 the other — 2, leaves re- 
 maining. 
 
 4. To divide -7438 by 12-9476 
 
 Numbers. Logs. 
 Divid. -7438 — 1-871456 
 Divisor 12-9476 1-112189 
 
 Q.uot. 
 
 •057447 — 2-759267 
 
 Here the 1 taken from the 
 — 1, makes it become — 2, t® 
 set down. 
 
 Note. As to the Rule-of-Three, or Rule of Proportion, it 
 is performed by adding the logarithms of the 2d and 3d terms, 
 and subtracting that o? the first term from their sum. 
 
 INVOLUTION 
 
[ 169 J 
 INVOLUTION BY LOGARITHMS. 
 
 RULft. 
 
 Take out the logaVithm of the given number from the 
 table. Multiply the log. thus found, by the index of the 
 power proposed. Find the number answering to the pro- 
 duct, and it will be the power required. 
 
 Note. In multiplying a logarithm with a negative index, 
 by an affirmative number, the product will be negative. 
 But what is to be carried from the decimal part of the loga- 
 rithm, will always be affirmative. And therefore their dif- 
 ference will be the index of the product, and is always to be 
 made of the same kind with the greater. 
 EXAMPLES. 
 
 , 1 . To square the number 
 
 2-6791. 
 
 Numb. Log. 
 
 Root 2-6791 - - 0-411468 
 
 The index - - 2 
 
 Power 6-65174 
 
 0-822936 
 
 2. To find the cube of 
 3-07146. 
 Numb. Log, 
 
 Root 307146 - - 0-487346 
 The index - - 3 
 
 Power 28-9758 
 
 1-462035 
 
 3. To raise -09163 to the 4th 
 power. 
 Numb. Log. 
 
 Root -09163 —2-962038 
 
 The index - - 4 
 
 Pow. -000070494— 5-848152 
 
 Here 4 times the negative 
 index being — 8 and 3 to carry, 
 the difference — 5 is the index 
 of the product. 
 
 4. To raise 1-0045 to the 
 365th power. 
 Numb. Log. 
 
 Root 1-0045 - - 0-001950 
 The index - - 365 
 
 9750 
 11700 
 5850 
 
 Power 5-14932* 711750 
 
 * This answer 5'14932 though found strictly according to the gene- 
 ral rule, is not correct in the last two figures 32 j nor can the answers 
 to such questions relating to very high powers be generally found true 
 to 6 places of figures by the table of logarithms in this work : if any 
 power above the hundred thousandth were required, not one figure of 
 the answer found by the table of logarithms here given could be de- 
 pended on. 
 
 The logarithm of 1-0045 is 00194994108 true to eleven places, which 
 multiplied by 365 gives -7117285 true to 7 places, and the correspond- 
 ing numb c true to 7 places is 5* 149067. 
 
 Vol. I. 23 EVOLUTION 
 
[ 1.70 3 
 
 EVOLUTION BY LOGARITHMS. 
 
 Take the log. of the given number out of the table. 
 Divide the log. thus found by the index of the root. Then 
 the number answering to the quotient, will be the root. 
 
 Note. When the index of the logarithm, to be divided, is 
 negative, and does not exactly contain the divisor, without 
 some remainder, increase the index by such a number as will 
 make it exactly divisible by the index, carrying the units bor- 
 rowed, as so many tens, to the left-hand place of the decimal, 
 and then divide as in whole numbers. 
 
 Ex. 1. To find the square root 
 of 365 
 Numb. Log. 
 
 Power 365 2)2-562293 
 
 Root 19-10496 l-281146i 
 
 Ex. 2. To find the 3d root of 
 
 12345. 
 
 Numb. Log. 
 
 Power 12345 3) 4-091491 
 
 Root 23-1116 1-363830^ 
 
 Ex. 3. To find the 10th root 
 
 of 2. 
 
 Numb. Log. 
 
 Power 2 - - 10) 0-301030 
 
 Root 1-071773 0-030103 
 
 Ex. 4. To find the 365th root 
 
 of 1-045. 
 
 Numb. Log. 
 
 Power 1-045 365) 0-019116 
 
 Root 1000121 0-000052-1 
 
 EiL. 5. To find ^ -093. 
 Numb. Log. 
 
 Power -093 2) — 2-968483 
 Root -304959 — 1 -4842411 
 
 Here the divisor 2 is con- 
 tained exactly once in the ne- 
 gative index — 2, and there- 
 fore the index of the quotient 
 is — L 
 
 Ex. 6. To find the %/ -00048. 
 
 Numb. Log. , 
 
 Power -00048 3) — 4-681241 
 
 Root -0782973 — 2-893747 
 
 Here the divisor 3 not being ex- 
 actly contained in — 4, it is augment- 
 ed by 2,to make up 6, in which the 
 divisor is contained just 2 times ; 
 then the 2, thus borrowed, being 
 carried to the decimal figure 6, 
 makes 26, which divided by 3, 
 gives 8, &c. 
 
 Ek. 7. To find 3-1416 X 82 X ff. 
 Ex. 8. To find -02916 X 751-3 X ^fy. 
 Ex. 9. As 7^41 : 3-58 : : 20-46 : ? 
 Ex. 10. As y/ 724 : v'ff : : 6-927 : ? 
 
 ALGEBRA. 
 
[ 171 ] 
 
 ALGEBRA. 
 
 DEFINITIONS AND NOTATION. 
 
 l.Al 
 
 lLGEBRA is the science of computing by symbols. 
 It is sometimes also called Analysis ; and is a general kind of 
 arithmetic, or universal way of computation. 
 
 2. In this science, quantities of all kinds are represented 
 by the letters of the alphabet. And the operation to be per- 
 formed with them, as addition or subtraction, Sac. are denoted 
 by certain simple characters, instead of being expressed by 
 words at length. 
 
 3. In algebraical questions, some quantities are known or 
 given, viz. those whose values are known : and others un- 
 known, or are to be found out, viz. those whose values are not 
 known. The former of these are represented by the leading 
 letters of the alphabet, a, fc, c, d, &c. ; and the latter, or un- 
 known quantities, by the final letters, Zf y, x, u, &c. %- 
 
 4. The characters used to denote the operations, are chiefly 
 the following : 
 
 -f- signifies addition, dnd is named vhts, 
 
 — signifies subtraction, and is named minus. 
 
 X or . signifies multiplication, and is named into. 
 
 -f- signifies division, and is named hy. 
 
 y/ signifies the square root ; %/ the cube root ; \/ th« 4tit 
 root, &c. ; and ^ the nth root. 
 
 : : : : signifies proportion. 
 
 = signifies equality, and is named equal to. 
 
 And so on for other operations. 
 
 Thus a H- 6 denotes that the number represented by h is to 
 be added to that represented by a. 
 
 a — 6 denotes, that the number represented by 6 is to be 
 subtracted from that represented by a. 
 
 o OD 6 denotes the difference of a and 6, when it is not known 
 which is the greater. 
 
 a6, or 
 
172 ALGEBRA. 
 
 ,<z6, or a X h, or a.h, expre^es the product, by multiplicatioD, 
 «f the numbers represented by a and h. 
 a 
 a -f- 6, or — , denotes, that the number represented by a 
 b 
 is to be divided by that which is expressed by h. 
 
 a : b : : c : d, signifies that a is in the same proportion to 6, 
 as c is to d. 
 
 X = a — 6 -{- c is an equation, expressing that x is equal to 
 the difference of a and 6, added to the quantity c. 
 
 i J. 
 
 ^ a, or a^, denotes the square root of a ; ^ a, or o^, the 
 
 cube root of a ; and ^/a^ or a^ the cube root of the square of a; 
 
 also ^ a, or a»», is the mth root, of a ; and "^ a" or a ^ is the 
 
 n 
 
 nth power of the mth root of a, or it is a to the ^ power. 
 
 a2 denotes the square of a ; a^ the cube of a ; a* the fourth 
 power of a ; and a" the nth power of a. 
 
 a-\-b X c, or («+6) c, denotes the product of the compound 
 quantity a -{- 6 multiply by the simple quantity c. Using the 
 bar , or the parenthesis ( ) as a vinculum, to connect seve- 
 ral simple quantities into one compound. 
 
 ft-f.6 
 
 a.-\-b -j-a-hoT , expressed like a fraction, means 
 
 a — b 
 the quotient of a + ^ divided by a — b. 
 
 ^ ab -\- cd, or [ab -\- cd)^j is the square root of the com- 
 pound quantity ab + cd. And c ^ ab -\- cd, or c {ab -\- cd)'^j 
 denotes the product of c into the square root of th^ compound 
 quantity ab -\- cd. 
 
 3 
 
 a-{-b — c, or (<x + 6 — c)^, denotes the cube, or third 
 power, of the compound quantity a -\- b — c. 
 
 3a denotes that the quantity a is to be taken 3 times, and 
 4 {a -\- b) is 4 times a -^ b. And these numbers, 3 or 4, 
 showing how often th^e quantities are to be taken, or multiplied, 
 are called Co-efficients. 
 
 Also fa; denotes that a; is multiplied by | ; thus | X a; or 
 ^x 
 
 5. Like Quantities, are those which consist of the same let- 
 ters, and powers. As a and 3a; or 2ab and 4ab ; or 3a^bc 
 and — ba^ be. 
 
 6. Unlike Quantities, are those which consist of different 
 letters, or different powers. As a and 6 ; or 2a and a^ ; or 
 ^al- and 3abc. 
 
 7. Simple 
 
DEFINITIONS AND NOTATION. 173 
 
 7. Simple Quantities, are those wliich consist of one term 
 only. As 3a, or 5ab, or Qabc^ . 
 
 8. Compound Quantities are those which consist of two or 
 more terms. As a + 6, or 2a — 3c, or a -f 26 — 3c. 
 
 9. And when the compound quantity consists of two terms, 
 it is called a Binomial, as a + 6 ; when of three terms, it is a 
 Trinomial, as a 4- 26 — 3c ; when of four terms, aQuadrino- 
 mial, as 2a — 36 + c — 4d ; and so on. Also, a Multinomial 
 or Polynomial, consists of many terms. 
 
 10. A Residual Quantity, is a binomial having one of the 
 terms negative. As a — 26. 
 
 11. Positive or Affirmative Quantities, are those which are 
 to be added, or have tl^ sign -f- As a or -}- «> or a6 ; for 
 when a quantity is found without a sign, it is understood to be 
 positive, or have the sign + prefixed. 
 
 12. Negative Quantities, are those which are to be subtrac- 
 ted. As — a, or — 2a6, or — 3a62 . 
 
 13. Like Signs, are either all positive ( -f- ), or all negative 
 
 ( - )• 
 
 14. Unlike Signs, are when some are positive ( + ), and 
 ethers negative ( — ). 
 
 16. The Co-efficient of any quantity, as shown above, is the 
 number prefixed to it. As 3, in the quantity 3a6. 
 
 16. The Power of a quantity (a), is its square (a^), or 
 cube (a 3), or biquadrate (a*), &c. ; called also, the 2d power, 
 or 3d power, or 4th power, &c. 
 
 17. The Index or Exponent, is the number which denotes 
 the power or root of a quantity. So 2 is the exponent of 
 the square or second power a^ ; and 3 is the index of the 
 
 cube or 3d power ; and i is the index of the square root, a^ 
 
 or ^ a; and ^ is the index of the cube root, a^, or ^ a. 
 
 18. A Rational Quantity, is that which has no radical sign 
 ( ^ ) or index annexed to it. As a, or 3a6. 
 
 19. An Irrational Quantity, or Surd, is that of which the 
 value cannot be accurately expressed in numbers, as the square 
 roots of 2, 3, 6. Surds are commonly expressed by means of 
 
 the radical sign ^, as y'S, ^ a, ^a^ , or ab^, 
 
 20. The Reciprocal of any quantity, is that quantity in- 
 verted, or unity divided by it. So, the reciprocal of a, or 
 a 1 a b 
 
 — , is — , and the reciprocal of — is — . 
 la ha 
 
 ^ 21. The 
 
174 ALGEBRA. 
 
 21. The letters by which any simple quantity is expressed, 
 may be ranged according to any order at pleasure. So the 
 product of a and 6, may l)e either expressed by a6, or ba ; 
 and the product of a, b, and c, by either a6c, or acfc, or bac, 
 or bca, or cab. or c6a , .as it matters not which quantities are 
 placed or multiplied first. But it will be sometimes found 
 convenient in long operations, to place the several letters 
 according to their order in the alphabet, as abc, which order 
 also occurs most easily or naturally to the mind. 
 
 22. Likewise, the several taembers, or terms, of which a 
 compound quantity is compos^id, may be disposed in any or- 
 der at pleasure, without altering the value of the signification 
 of the whole. Thus, 3a — tab -}- 4fl6c may also be written 
 3a -f 4a6c — 2a6, or Aabc -f 3a — Saj^or — 2a6 -f- 3a-|-4afec, 
 &c. ; for all these represent the sain6 thing, namely, the 
 quantity which remains, when the quantity or term 2a6 is sub- 
 tracted from the sum of the terms or quantities 3a aftd 4afec. 
 But it is most usual and natural, to begin with a positive term, 
 and with the first letters of the alphabet. 
 
 SOME EXAMPLES FOR PRACTICE. 
 
 fing the numeral values of various expressions, or 
 combinations, of quantities. 
 
 Supposing a = 6, and 6 = 6, and c =a 4, and ci = 1, and 
 « = 0. Then 
 
 1. Will a2 4- Sab — c^ — 36 + 90 — 16 = llO. 
 
 2. And 2a3 — 3aH -f c^ = 432 — 640 -f 64 = — 44. 
 
 3. And a2 X a+T— 2aic =36 X 11 — 240 = 166. 
 
 a3 216 
 
 4. And f- c2 = {-16 = 12+16= 28. 
 
 a -f 3c 18 
 
 5. And ^2ac -\- c^ or 2ac + c^ ^ " = ^ 64 = 8. 
 
 26c 40 
 
 6. And^cHh-;^^-^ = 2 + - = 7. 
 
 .^h^—ac 36—1 35 
 --/fea -ra^'^12 — 7~ 6 
 8. And v' fc2— ac4-^2flc + c2 = 1 + 8 = 9. 
 
 9. And ^62— ac + ^2ac+c3 =^26 — ^4 + 8 = 3. 
 
 10. Anda26 + c — d= 183. 
 
 11. And 9a6— 106^ + c = 24. 
 
 *^ . 12. And 
 
 ^ 
 
12. 
 
 ADDITION, 
 And X d = 45. 
 
 
 c 
 
 13. 
 
 a+b b 
 
 And X-^- 13|. 
 
 c d 
 
 14. 
 
 a-\'b a—b 
 
 And = If. 
 
 c • d 
 
 15. 
 
 a^b 
 And h c = 43. 
 
 
 c 
 
 16. 
 
 And X c = 0. 
 
 
 c 
 
 175 
 
 17. And6— cX(^-e = 1. 
 
 18. Anda-f6— c -(^ = 8. 
 
 19. Anda-f6-.c-"tZ = 6. 
 
 20. AndaStX-is = 144. 
 
 21. And ac«j(-r£ = 23. 
 
 22. Anda2c-ffe2e4.f^=: i. 
 
 23. And X = 18^. 
 
 d— c c — d 
 
 24. And y'a^b^ -^ a^ —6a = 4-4936249. 
 
 25. And 3ac2 -{- 3/^3 -P"= 292-497942. 
 
 26. And 4a2—3a ^/a^ — fa6 = 72. 
 
 ADDITION. 
 
 Addition, in Algebra, is the connecting the quantities 
 together by their proper signs, and incorporating or uniting 
 into one term or sum, such as are simiUr, and can be united. 
 As, 3a 4- 26 - 2a = a 4- 26, the sum. 
 
 The rule of addition in algeJjra, may be divided into three 
 cases : one when the quantities are like, and their signs like 
 also ; a second, when the quantities are like, but their signs 
 unlike ; and the third, when the quantities are unlike. 
 Which are performed as follows. 
 
 CASE 
 
 • * The reasons on which these operations are founded, will rea- 
 dily appear, by a little reflection on the nature of the quantities to 
 
176 ALGEBRA. 
 
 CASE L 
 
 When the Quantities are Like, and have Like Signs. 
 
 Add the co-efficients together, and set down the sum 5 
 after which set the common letter or letters of the like quan- 
 tities, and prefix the common sign -f- or — . 
 
 be added or collected together. For, with regard to the first ex- 
 ample, where the quantities are 3a and 5a whatever a represents 
 in the one term, it will represent the same thing, in the other ; so 
 that 3 times any thing and 5 times the same thing, collected 
 together, must needs make 8 times that thing. As if a denote a 
 shilling ; then 3a is 3 shillings ; and 5a is 5 shillings* and their sum 
 8 shillings. In like manner, — 2ad apd — 7a6, or — 2 times any 
 thing, and — 7 times the same thing, make — 9 times that thing. 
 
 As to the second case, in which the quantities are like, but the 
 signs unlike ; the reason of its operation will easily appear, by re- 
 flecting, that addition means only the uniting of quantities together 
 by means of the arithmetical operations denoted by their signs 4- »nd 
 — , or of addition and subtraction ; which being of contrary or oppo- 
 site natures, the one co-efficient must be subtracted from the other, to 
 obtain the incorporated or united mass. 
 
 As to the third case, where the quantities are unlike, it is plain 
 that such quantities cannot be united into one, or otherwise added, 
 than by means of their signs ; thus, for example, if a be supposed 
 to represent a crown, and 6 a shilling ; then the sum of a and b 
 can be neither 2a nor 2b, that is neithei- 2 crowns nor 2 shillings, 
 hut only 1 crown plus 1 shilling, that is a + 6. 
 
 In this rule, the word addition is not very properly used ; being much 
 too limited to express the operation here performed. The business of 
 this operation is to incorporate into one mass, or algebraic expression, 
 different algebraic quantities, as far as an actual incorporation or 
 union is possible ; and to retain the algebraic marks for doing it, in 
 eases where the former is not possible. When we have several quanti- 
 ties, some affirmative and some negati^ e ; and the relation of these 
 quantities can in th€ whole or in part be discovered ; such incorpora- 
 tion of two or more quantities into one, is plainly efiected by the fore- 
 going rules. 
 
 It may seem a paradox, that what is called addition in algebra, 
 should sometimes mean addition, and sometimes subtraction. But 
 the paradox wholly arises from the scantiness of the name given to 
 the algebraic process ; from eihploying an old term in a new and more 
 enlarged sense. Instead of addition, call it, incorporation, or union, 
 or striking a balance, or any name to which a mote extensive idea may 
 be annexed, than that which is usually implied by the word addition ; 
 and the paradox vanishes. 
 
 Thus, 
 
ADDITION. 177 
 
 Thus, 3a added to 5a, makes 8 a. 
 
 And — 2ab added to — 7a6, makes — 9ab. 
 
 And 5a -{- 76 added to 7a + 36, makes I2a -{- 106. 
 
 OTHER EXAMPLES FOR PRACTICE. 
 
 I 
 
 3a 
 
 — 36x 
 
 
 bxy 
 
 9a 
 
 — 56a; 
 
 
 2bxy 
 
 5a 
 
 — 46a; 
 
 
 bbxy 
 
 12a 
 
 — 262; 
 
 
 bxy 
 
 a 
 
 — 76a; 
 
 
 3bxy 
 
 2a 
 
 — bx 
 
 
 6bxy 
 
 32a 
 
 —226a; 
 
 
 ISbxy 
 
 3z 
 
 3^2 ^ sxy 
 
 
 2ax — 4y 
 
 2z 
 
 j?2 4. ry 
 
 
 4ax — y 
 
 4z 
 
 2x^ -f 4xy 
 
 
 ax — 3y 
 
 z 
 
 5x2 ^ 2xy 
 
 
 5ax — 5y 
 
 hz 
 
 4x'i + 3xy 
 
 
 lax — 2y 
 
 \bz 
 
 15a:3 -f 15^2/ 
 
 19aj? — 15y 
 
 5xy 
 
 — 12?/2 
 
 
 4a — 46 
 
 Uxy 
 
 - V 
 
 
 5a — 56 
 
 22xy 
 
 - 22/^ 
 
 
 6a— 6 
 
 llxy 
 
 - 4y2 
 
 
 3a — 26 
 
 H^y 
 
 - r 
 
 
 2a — 76 
 
 i^y 
 
 - 3y2 
 
 
 8a— 6 
 
 30 — 
 
 13a;i — 3:1-^ 
 
 5xy 
 
 — 3:r -f- 4a6 
 
 28 — 
 
 lOxi — 4xy 
 
 Sxy 
 
 — 4x-\- 3ab 
 
 14 — 
 
 Ux^ — 7^:?/ 
 
 3xy 
 
 — 5x -f- 5a6 
 
 10 — 
 
 16a;^ — 5xy 
 
 xy 
 
 — 2x + ab 
 
 16 — 
 
 20a;i— j:^/ 
 
 Axy 
 
 — a; 4- 7a6 
 
 
 
 
 Vot. I. 2^ CASE 
 
17B ALGEBRA. 
 
 CASE 11. 
 
 fVfien the Quantities are Likcy but have Unlike Signs ; 
 
 Add alf the affirmative co-efficients into one suno, and all 
 rtie negative ones into another, when there are several of a 
 kind. Then subtract the less sum, or the less co-efficient, 
 from the greater, and to the remainder prefix the sign of the 
 greater, and subjoin the common quantity or letters. 
 
 So -|- 5a, and — 3a, united, make -{- 2a. 
 And — 5a, and + 3a, united, make — 2a, 
 
 OTHER EXAMPLES FOR PRACTICE. 
 
 — 5a 
 + 4a 
 + 6a 
 —.3a 
 + a 
 
 + Saxs 
 4- 4aa:2 
 
 — 8ax2 
 
 — 6ax2 
 -i- 5aa;2 
 
 + 8a;3 4- 3y 
 
 — 6a;» 4 4y 
 
 — 16a;3 4- 5^ 
 4- 3x^ — ly 
 + 2a;3 — 2y 
 
 + 3a 
 
 — 2aa;2 
 
 — 8x3 _{. sy 
 
 — . 3a2 
 
 — 5a2 
 
 — 10a2 
 + 10a2 
 + 14a2 
 
 H- 3621/3 
 
 4- 9622,3 
 
 — . 1062^3 
 
 — 1962i/s 
 
 — 2621/3 
 
 -f 4a6 4- 4 
 — 4a6 -\- 12 
 4- 7a6 — 14 
 4- a6 -f- 3 
 _ 6a6 — 10 
 
 
 
 ^ 3a:«:2 
 4- ax± 
 + bax.^ 
 — 6axi 
 
 4- 10^ ax 
 
 — 3^ax 
 4- 4^a:c 
 
 — 12^aa: 
 
 -1- 3?/ + 4ax^ 
 
 — 2/ "— ^(^^\ 
 + 4y'-\' 2ax^ 
 
 — 22/4- 6aar| 
 
 
 
 
 CASE 
 
ADDITION. 119 
 
 CASE UI. 
 
 Whenjhe Quantities are Unlike, 
 
 Having collected together all the like quantities, as in the 
 two foregoing cases, set down those that are unlike, one after 
 another, with their proper signs. 
 
 EXAMPLES. 
 
 I 
 
 3xy Sjry— 12^2 4a:r — 1 30 + Sx^ 
 
 2aj? — 4^2 4- 3xy >5x^ -f Sax + 9x^ 
 
 — 5xy + 4:r2 — 2xy Ixy — Ax^ -{- 90 
 
 6ax — 3xy -f- 4ar3 ^a: + 40 — 6a:« 
 
 — ^xy + 8a J7 Axy — ^x^ lax + \St^ •\-lry 
 
 9 '\- \0^ ax — 6y 
 2r 4- 1 y/xy -f 5y 
 % H- 3 ^ ax — 4y 
 10 — 4 ^ ax + 4y 
 
 9x^y 
 
 14aa; 
 
 — 2a;2 
 
 -^Ix^y 
 
 Saa; •{- 3xy 
 
 + Saxy 
 
 8y3 
 
 — 4aa; 
 
 —4x2 2/ 
 
 3a:2 
 
 + 26 
 
 
 
 
 4a;2y 
 
 4y :r 
 
 - 3y 
 
 — 6xy2 
 
 2^ary 
 
 -{- 14a: 
 
 + 3^2^ 
 
 3x + 
 
 22/ 
 
 -7a:2y 
 
 — 9 + 2^.Ty 
 
 3a2 + 9 -f X2-.4 
 2a — 8 -I- 2a2 — Sx 
 4x^ — 2a2 4- 18 — 7 
 -12 4- a — 3x2— 2y 
 
 Add a + 6 and 3a — 56 together. 
 Add 6a — 8x and 3a — 4x together. 
 Add 6a: -66 + a + 8 to - 5a — 4jc + 46 - 3. 
 Add a 4- 26 — 3c - 10 to 36 — 4a H- 5c 4- 10 and 56 — c. 
 Add a 4- 6 and a - 6 together. 
 
 Add 3a + 6- 10 to c — rf - a and — 4c -I- ^a — 3Z> - 7. 
 Add 3a2 -f 62 — c to 2a6 - Sa^ -f- 6c - 6. 
 Add a3 -I- 62c — 62 to a62 — a6c -f 62. 
 
 Add 9a - 86 + lOx - 6<^ - 7c -f 60 to 2x - 3a - 5c -{- 4fc 
 4- 6d - 10. 
 
 BUBTRAC- 
 
180 
 
 ALGEBRA. 
 
 SUBTRACTION. 
 
 Set down in one line the first quantities from which the 
 subtraction is to be made ; .and underneath them place all the 
 other quantities composing the subtrahend : ranging the like 
 quantities under each other, as in Addition. 
 
 Then change all the signs (-f- and — ) of the lower line, or 
 conceive them to be changed ; alter which, collect all the 
 terms together as in.the cases of Addition*. 
 
 From 7a- — 36 
 Take 3a^ — 86 
 
 EXAMPLES. 
 
 6x^ -f 5?/ --• 4 
 
 80:^ — 3 + 60;— y j 
 4xy — 7 — 6a; — 4y 
 
 Rem. 4a2 -f 56 Sx^ — 61/+ 12 4xy + 4-{-na: -{-3y 
 
 From 5xy — 6 
 Take— 2x2/ + ^ 
 
 Rem. '7xy~\2 2y^ ^ by — S -^2S-i'3x—8xy-{'2ay 
 
 4y2^Sy-^4 — 20 — 6a;— So;?; 
 
 22/2 4- 2?/ -I- 4 ' oxy — 9x -\- S ^2ay 
 
 From Sx^y-^G 5^xy -}- 2x^xy 7a;2-f2^ or— 18+36 
 Take— 2x^y + 2 l^xy -\- 3-^^xy 9x2—12 + 66+0:2 
 
 Rem. 
 
 * This rule is founded on the consideration, that addition and sub- 
 traction are opposite to each other in their nature and operation, as 
 are the signs + and — , by which they are expressed and represent- 
 ed. So that, since to unite a negative quantity with a positive 
 one of the same kind, has the effect of diminishing it, or subducting 
 an equal positive one from it, tlierefore to subtract a positive (which 
 is the opposite of uniting or adding) is to add the equal negative 
 quantity. In like manner, to subtract a negative quantity, is the same 
 in effect as to add or unite an equal positive one. So that, by chang- 
 ing the sign of a quantity from -f to — , or from — to -J- , changes 
 its nature from a subdiictive quantity to an additive one ; and any 
 quantity is in effect subtracted, by barely changing its sign. 
 
 Bxv 
 
MULTIPLICATION. 181 
 
 bxy - 30 7x3 ^ 2(a-\'b) 3xy^ -}" 20 a^ (xy+10) 
 Ixy — 60 2x^ - 4 (a 4- i) ^a;^?/^ -f \2a^ (xy+10) 
 
 From a -f- ^» take a— 6. 
 
 From 4a + 46, take b -{■ a. 
 
 From 4a — 4b, take 3a + 56. 
 
 From 8a — 12x, take 4a — 3^:. 
 
 From 2x — 4a— 26 -f 5 take 8 - 66 -f a + 6j:. 
 
 From 3a 4- 6 4- c — J — 10, take c -{- 2a — rf. 
 
 From 3a -f 6 -I- c — d — 10, take 6 — 19 + 3a. 
 
 From 2a6-f-62 -.4c + 6c- 6, take 3a2 — . c + 6^. 
 
 From a3 -f 363c -f a63 — a6c, take 6^ -f- 06^ — a6c. 
 
 From nx + 6a— 46 + 40, take 46 — 3a + 4r -f 6rf— 10. 
 
 From 2ar— 3a -f 46 -f 6c-50,take 9a -f- x + 66-6c— 40. 
 
 From 6a— 46— 12c + 12x, take 2a; — 8a + 46- 5c. 
 
 MULTIPLICATION. 
 
 This consists of several cases, according as the factors are 
 simple or compound quantities. 
 
 CASE 1. When both the Factors are Simple Quantities : 
 
 First multiply the co-efficients of the two terms together, 
 then to the product annex all the letters in those terms, which 
 will give the whole product required. 
 
 Note*. Like signs, in the factors, produce + and unlike 
 signs — , in the products. 
 
 EXAMPLES. 
 
 ♦ That this rule for the signs is true, may be thus shown, 
 1. When + a is to be moltiplied by + c / the meaning is, that^- a 
 is to be taken as many times asthere are units in c; and since the sum 
 of any number of positive terms is positive, it follows that + a x +c 
 makes + ac 
 
 2. When 
 
3 Oct 
 2b 
 
 ALGEBRA 
 
 EXAMPLES. 
 
 —3a , 
 4-26 
 
 7« 
 — 4c 
 
 — 4a 
 
 20ab 
 
 ^6ab 
 
 ~-28ac 
 
 -f 24aa7 
 
 4ac 
 
 Ax 
 
 — ^x^y 
 6x^y^ 
 
 •^4xy 
 
 — l2aHc 
 
 ' 36fl2x2 — 
 
 + 4x^y^ 
 
 4x 
 
 — ax 
 
 + 3xy 
 — 4 
 
 — dxyz 
 
 — 5ajr 
 
 
 CASE n. 
 
 When one of the Factors is a Compound Quantity. 
 
 Multiply every term of the multiplicand, or compound 
 ^quantity, separately, by the multiplier, as in the former 
 case ; placing the products one after another, with the 
 proper signs ; and the result will be the whole product re- 
 quired. 
 
 2- When two quantities are to be multiplied together, the result will 
 be exactly the same, in whatever order they are placed ; for a times 
 tT 13 the same as c times a, and therefore, when — a is to be multiplied 
 by -f" <^» or -|- c bv— .a ; this is the same thing as taking — a as many 
 times as there are units in 4. c ; and as the sum of any number of 
 negative terms is negative, it follows that «-. a X +c, or -f.*^ X •- 
 c make or produce— ac. * 
 
 3. When— a is to be multiplied by — c : here —a is to be subtract- 
 ed as often as there are units in c .• but subtracting negatives is 
 the same thing as adding affirmatives, by the demonstration of the 
 rule for subtraction ; consequently the product is c times a, or -|- ac. 
 
 Otherwise. Since a — a = 0, therefore (a — a)X — ■ c is also = 0, 
 because 0, multiplied by any quantity, is still but ; and since the first 
 term of the product, or a X -• c is ■= — ac, by the second case j 
 therefore the last term of the product, 01 — a x — c, must be-4-ac, 
 to make the sum =ts o» or — ac 4- ac=^0 ; that is, — a x - c «= 
 + ««• 
 
 EXAMPLES. 
 
MULTIPLICATION'. im 
 
 EXAMPLES. 
 
 5a — Sc 
 2a 
 
 3ac — 46 
 3a 
 
 2a2 — 3c + 5 
 6c 
 
 10a3 — 6ac 
 
 9a3c ^ \2ab 
 
 2a2 6c — 36c4-56c 
 
 12x — 2ac 
 4a 
 
 25c — 76 
 
 — 2a 
 
 4x — 6 + 3a6 
 2a6 
 
 
 3c3 4-^ 
 
 4xy 
 
 10x2—3^2 
 — ■4x2 
 
 3a3— 2x2 — 66 
 
 2(1X3 
 
 
 
 
 
 CASE nr. 
 
 
 When both the Factors are Compound Quantities ; 
 
 Multiply every term of the multiplier by every term of 
 the multiplicand, separately ; setting down the products one 
 after or under another, with their proper signs ; and add the 
 several lines of products all together for the whole product 
 required. 
 
 a -f 6 3x 4- 2y 2a:2 4- xy— 2y^ 
 
 « 4- 6 4x — 5y 3x — 3y 
 
 «a + a6 
 
 + ab + b^ 
 
 12x3 ^ Qjry 
 
 -Ux2/— 10i/2 
 
 12:r3- Ixy-^ \0y^ 
 x^ +y 
 
 x* 4" y^^ 
 
 6x3 43a-3 2/-6x2/3 
 
 -6a;2 2/-3a;2/3 46?/3 
 
 «2 4- 2a6 4 b^ 
 
 6x 3 -3x2 y - exy^ +6y^ 
 
 a-{-6 
 « — 6 
 
 a3 4a6+63 
 a —6 
 
 a2 4-a6 
 -a6-63 
 
 a3 4-«^6+a63 
 -a26-a62 — 6- 
 
 a2 * —63 
 
 flS * * _ ^r. 
 
 J^nte. 
 
T84 ALGEBRA. 
 
 Note, In the multiplication of compound quantities, it i» 
 the best way to set them down in order, according to the 
 powers and the letters of the alphabet. And in multiplying 
 them, begin at the left-hand side, and multiply from the left 
 hand towards the right, in the manner that we write, which is 
 contrary to the way of multiplying numbers. But in setting 
 down the several products, as they arise, in the second and 
 following lines, range them under the like terms in the lines 
 above, when there are such like quantities ; which is the 
 easiest way for adding them up together. 
 
 In many cases, the multiplication of compound quantities 
 is only to be performed by setting them down one after 
 another, each within or under a vinculum with a sign of 
 multiplication between them. As'(a-f 6) X (a — t) X 3aK 
 or a-f"^ . — 6 . 3a6. 
 
 EXAMPLES FOR PRACTICE. ' 
 
 1. Multiply lOac by 2a. Ans. 20a2C. 
 
 2. Multiply 3aa — 26 by 36. Ans. ^a^-Qh^, 
 
 3. Multiply 3a + 26 by 3a — 26. Ans. 9a2— 462. 
 
 4. Multiply x^ — xy •\"y^ hy X -{- y. Ans. x^ -f y^. 
 
 5. Multiply a3 -f a^b\-ah^ 4" 6^ by a -6. Ans. a*- 64. 
 
 6. Multiply a2 4. a6 -f- 6^ by a2 - a6 + 62 . 
 
 7. Multiply 3a;2 ^2a;«/ + 5 by x^ -\- 2a;i/— 6. 
 
 * 8. Multiply 3a2 -2aa; -f bx^ by 3a2 — Aax - Ix^, 
 9. Multiply 3x3 4. 2^72^2 -f- 3^3 by ^x^^dx^y'^ -J- Sy^. 
 10. Multiply a2 + a6 -f 62 by a- 26. 
 
 division- 
 
 Division in Algebra, like that in numbers, is the converse 
 of multiplication ; and it is performed like that of numbers 
 also, by beginning at the left-hand side, and dividing all the 
 parts of the dividend by the divisor, when they can be so 
 divided ; or else by setting them down like a fraction, the 
 dividend over the divisor, and then abbreviating the fraction 
 as much as can be done. This will naturally divide into the 
 following particular cases. 
 
 CASE 
 
DIVISION. 18i 
 
 CASE I. 
 
 When the Divisor and Dividend are both Simple Quantitiei ; 
 
 Set the terms both down as in division of numbers, either 
 the divisor before the diridend, or below it, hke the deno- 
 minator of a fraction. Tbew abbreviate these terms as 
 much as can be done, by canceUinjo; or striking out all the 
 letters that are common to them both, and also dividing 
 the one co-efficient by the other, or abbreviating them after 
 the manner of a fraction, by dividing them by their commoQ 
 measure. 
 
 Note. Like signs jn the two factors make 4- in the quo- 
 tient ; and unlike signs make — ; the same as in multipli- 
 eation*. 
 
 EXAMPLES. , 
 
 1. To divide 6a6 by 3a, 
 
 Qah 
 
 Here Qah -r- 3a or 3a) Qah ( or = 2t. 
 
 3a 
 
 c ahx a 
 2. Also c -r- c =^^ — = 1 ; and ahx -r- hxy = = — . 
 
 c bxy y 
 
 •3. Divide 16^2 by 8j:. Ans. 2x. 
 
 4. Divide l^a-x^ by-*- 3a^x. Ans. — 4x. 
 
 5. Divide — 1 bay^ by 3ay. * Ans. — by, 
 
 9xy 
 
 6. Divide— ISax^y hy^Qaxz. Ans. — '-, 
 
 4z 
 
 * Because the divisor multiplied by the quotient, must produce the 
 dividend. Therefore, 
 
 1. When both the terms are -f-, the quotient must be -f ; because 
 + in the divisor X 4- in the quotient, produces -|- in the dividend. 
 
 2. When the terms are both — , the quotient is also + ; because 
 — in the divisor X + in the quotient, produces — in the dividend. 
 
 3. When one term is -f. and the other — , the quotient must be — i 
 because -^ m the divisor X — in the quotient produces — in the divi- 
 dend, or,- in the divisor x + in the quotient gives — in the. dividend. 
 
 So that the rule is general ; viz. that like signs give -|- , and unlike 
 signs give — , in the quotient. 
 
 Vol. I. 2i 
 
 CASE 
 
186 ALGEBRA. 
 
 CASE II. 
 
 When the Dividend is a Compound Quantity, and the Divisor 
 a Simple one : 
 
 PirjDF eypry term of the dividend by the divisor, as in 
 the former case. 
 
 EXAMPLES. 
 
 ab-^-b^ a-\-b 
 
 1. (afe+fca ) -J- 2b, or = = ^a + ^b, 
 
 2b 2» 
 
 10a6 4- ISao: 
 
 2. {lOab -f I5ax) -j- 5o, or — = 26 + 3ar. 
 
 6a 
 
 3. (30a2r - 482:) ^ z, or = 30a — 48. 
 
 z 
 
 4. Divide 6a6 — 'Sajr -f- a by 2a. 
 
 5. Divide 3x2 __ 15 ^ g^ ij. g^ |jy 3_j.^ 
 
 6. Divide 6a6c -{- 12a6a? — 90^6 by 3a6. 
 
 7. Divide 10a2x — 16x2 _ 25x by hx. 
 
 8. Divide 16a^ic — Xbacx^ -}- 5aa* by — Sac. 
 
 9. Divide 15a -f- 3ay — 18y by 21a. 
 18. Divide — 20a6 -f 60a63 by — 6a6. 
 
 CASE in. 
 
 When the Divisor and Dividend are both Compound Quantities . 
 
 1. Set them down as in common division of numbers, 
 the divisor before the dividend, with a small curved line 
 between them, and ranging the terms according to the 
 powers of some one of the letters in both, the higher 
 powers before the lower. 
 
 2. Divide the first term of the dividend by the first term 
 of the divisor, as in the first case, and set the result in the 
 quotient. 
 
 3. Multiply the whole divisor by the term thus found, and 
 subtract the result from the dividend. 
 
 4. To this remainder bring down as many terms of the 
 dividesd as are requisite for the next operation, dividing as 
 befgre ; and so on to the end, as in common arithmetic. 
 
 JVoit. 
 
DIVISIO.V. 187 
 
 Note. If the divisor be not exactly contaiaed in the divi- 
 dend, the quantity which remains after the operation is fin- 
 ished, may be placed over the divisor, like a vulgar fraction, 
 and set down at the end of the quotient as in common aritn-* 
 metic. 
 
 EXAMPLES. 
 
 o3— a6 
 
 -a6 -f- h^ 
 
 -c) a^ — 4a3c 4- ^ac^ ^-c^ (a* —Sac -^ c^ 
 
 -3a2c4 3oc2 
 
 ac^ — c' 
 
 «— 2) a»-6a3 -f- 12a— 8 (a^ -4a -{- 4 
 a3~2a3 
 
 -4a3 4- 12a 
 -4fl3 -{- 8a 
 
 4a- 8 
 4a-8 
 
 ct -|- 2-) a3 4. 2r3 ^ci2 ^a^ -{- ^2 
 a3 4- a22 
 
 — a^z — az2 
 
 a2r2 ^2^3 
 a2r2 4.^3 
 
 ^ + ar) 
 
188 ALGEBRA. 
 
 <» -f x) (X* — 5x4 .(a3 — ($2 X -{- aa:^ —^ x^ -— 
 
 o-i-x 
 «4 4- a^x 
 
 — asx 
 
 — 3x4 
 
 — a^xs 
 
 — 3x4 
 
 + ax3 
 
 
 
 a2x2 
 
 
 
 
 — ax3 — 
 
 — ax3 — 
 
 3x4 
 
 X4 
 
 
 — 
 
 2x4 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Divide a^ + 4ax -f- 4x2 by a + 2ar. Ans. a + 2x. 
 
 2. Divide a^ — 3a2^ -|- 3az^ — z^ by a — z. 
 
 Ans. a2_2ct2r4- z^ . 
 
 3. Divide 1 by 1 + a. Ans. I — a + a^ -^ a^ -\- kc. 
 
 4. Divide 12a:4 — 192 by 3a; — 6. 
 
 Ans. 4x3 -\- 8x^ + 16ar -f 32. 
 
 6. Divide a^ — da'^b + lOa^fea^iOasfes -|- 5a64 — 6^ by 
 
 a2 _ 2a6 4- ^2. Ans. a^ -_3a2 6 + 3a62 —63. 
 
 6. -Divide 48^3 __ 960^2 _ 64aaz + 150a 3 by 2z — 3a. 
 
 7. Divide 6^ — 364a-2 +362x4 ^^^ by 63 - 362a;+36a:2 — a;3, 
 
 8. Divide a'^ ^x'^ by a — x. 
 
 9. Divide a^ -\- ba^x -\- 5a x" -}- x^ by a -^ x. 
 .10. Divide a^ + 4^262 — 326^ by a + 26. 
 
 11. Divide 24a4 — 64 by 3a — 26. 
 
 ALGEBRAIC FRACTIONS. 
 
 Algebraic Fractions have the same names and rules of 
 operation, as numeral tractions in common arithmetic; as ap- 
 pears in the following Rules and Cases. 
 
 CASE 
 
FRACTIONS. 189 
 
 CASE L 
 
 To reduce a Mixed Quantity to an Improper Fraction, 
 Multiply the integer by the denoniinator of the fraction, 
 and to the product add the numerator, or connect it with its 
 proper sign, -f- or — ; then the denominator being set under 
 this sum, will give the improper fraction required. 
 EXAMPLES. 
 b 
 
 1. Reduce 3f and a - — to improper fractions. 
 
 j: 
 3X64-4 164-4 19 
 
 First, 3f = = = — the Answer. 
 
 5 6 6 
 b a X or " b ax — b 
 And, a — — = = the Answer. 
 
 2. Reduce a -\ and a to improper fractions. 
 
 b a 
 
 a2 a X 6 + a3 ab-^-a^ 
 
 First, a -\ = = the Answer. 
 
 b b b 
 
 z^ — a2 a3 — 2-2 -f- a2 2(x2 —2-2 
 
 And, a — = = ^ the Answer. 
 
 a a a 
 
 3. Reduce 6^ to an improper fraction. Ans. y* 
 
 3a X — 3a 
 
 4. Reduce 1 — — to an improper fraction. Ans. 
 
 X X 
 
 3ax + a2 
 
 6, Reduce 2a to an improper fraction. 
 
 4x 
 4j:— 18 
 
 6. Reduce 12 -| to an improper fraction. 
 
 5x 
 1 - 3a - c 
 
 7. Reduce x -\ to an improper fraction. 
 
 c 
 2x3 — 3^ 
 
 8. Reduce 4 -f 2a: to an improper fraction. 
 
 5a 
 CASE U. 
 To Reduce an Improper Fraction to a •whole or Mixed Quantity. 
 Divide the numerator by the denominator, for the integral 
 part ; and set the remainder, if any, over the denominator, 
 for the fractional part ; the two joined together will be the 
 mixed quantity required. 
 
 EXAMPLES 
 
190 ALGEBRA. 
 
 EXAMPLES. 
 
 16 a6 + a2 
 
 1. To reduce — and to mixed quantities. 
 
 3 b 
 
 First, L6 = 16 ^ 3 = 6^, the Answer required. 
 
 ab 4- 0.^ a2 
 
 And, = a6-{-a*-r-6=a-{ . Answer. 
 
 b b 
 
 2ac-3a2 3aj:+4x2 
 
 2. To reduce and to mixed quantities. 
 
 c a -\- a; 
 
 2ac-3a2 3a2 
 
 First, = 2a(?— 3a2 -j- c = 2a . Answer. 
 
 c c 
 
 3ax + 4j:* or^ 
 
 And, = 3ax + 4x^ -i- a-f x = 3a; H . Ans. 
 
 a -f- ar a+rc 
 
 33 2ax--3x2 
 
 3. Reduce — and • to mixed quantities. 
 
 6 a 
 
 3x2 
 
 Ans. 6|, and 2x — . 
 
 a 
 4a^x 2a2 4-2&2 
 
 4. Reduce and' to whole or mixed quan- 
 
 2a a — b 
 
 tities. 
 
 3x^ — 32/2 2x3 — 22/3 
 
 6. Reduce , and to whole tr mixed 
 
 x + y X —y 
 
 quantities. 
 
 10a2_4a-f 6 
 
 6. Reduce . to a mixed quantity. 
 
 ba 
 15a3-f6a2 
 
 7. Reduce to a mixed quantity. 
 
 3a3 + 2a2 —2a -4 
 
 CASE III. 
 To Reduce Fractions to a Common Denominator. 
 Multiply every numerator, separately, by all the denomi- 
 nators except its own, for the new numerators ; and all the 
 denomincitors together, for the common denominator 
 
 When il>e denominators have a common divisor, it will be 
 better, instead of multiplying by the whole denominators, to 
 multiply only by those parts which arise from dividing by the 
 common divisor. And observing also the several rules and 
 directions as in Fractions iti the Arithmetic. 
 
 EXAMPLES. 
 
FRACTIONS. l&l 
 
 EXAMPLES. 
 
 a b 
 
 1. Reduce — and — to a commoo denominator. 
 
 X z 
 
 a h az hx 
 
 Here — and — = — and — , by multiplying the terms of the 
 
 X z xz xz 
 
 first fraction by z, and the terms of the 2d by x. 
 a X b 
 
 2. Reduce — , — and — to a common denominator. 
 
 X b c 
 
 ax b abc cx^ b^x 
 
 Here — , — , and — = , , and , by muhiplying the 
 
 X b c bcx bcx bcx 
 
 terms of the 1st fraction by 6c, of the 2d by ex, and of the 
 3d by bx, 
 
 2a 36 
 
 3. Reduce — and — to a common denominator. 
 
 X 2c 
 
 4ac 3bx 
 Ans. and . 
 
 2ca: 2cx 
 2a 3a4-26 
 
 4. Reduce — and ■ to a common denominator. 
 
 6 2c 4ac Sab -f 26« 
 
 Ans. and , 
 
 26c 26c 
 
 5a 36 
 
 5. Reduce — and — , and 4d, to a common denominator. 
 
 3x 2c 
 
 lOac 9bx 24cdx 
 
 Ans. and and . 
 
 Sex 6cx 6cx 
 
 5 3a 3a 
 
 6. Reduce — and — and 26 -j , to fractions having a 
 
 6 4 6 
 
 206 18tt6 48624.72a 
 
 common denominator. Ans. and and — _ 
 
 246 246 . 246 
 
 1 2a2 2a2 4-62 
 
 7. Reduce — and and to a common denomi- 
 
 3 4 a-i-6 
 
 nator. 
 
 3b 2c d . 
 
 8. Reduce — and — and — to a common denominator. 
 
 4aa 3a 2a 
 
 CASE 
 
192 ALGEBRA. 
 
 CASE IV. 
 
 To find the Greatest Common Measure of the Terms of a 
 Fraction. 
 
 Divide the greater term by the less, and the last divi^sor by 
 the last remainder, and so on till nothing remains ; then the 
 divisor last used will be the common measure required ; just 
 the same as in common numbers. 
 
 But note, that it is proper to range the quantities according 
 to the dimensions of some letters, as is shown in division. 
 And note also, that all the letters or figures which are com- 
 mon to each term of the divisors, must be thrown out of 
 them, or must divide them, before they are used in the 
 •pe ration. 
 
 EXAMPLES. 
 
 ab-^-b' 
 
 1. To find the greatest common measure of - 
 
 ac^ -f 6c2 
 ab -\- b^) ac^ -f- bc^ 
 OT a 4- b ) ac2 4- 6c2 (c2 
 
 Therefore the greatest common measure is a + ^» 
 
 as — ab^ 
 2. To find the greatest common measure of - 
 
 «a -f2a6-f62)a3 -. ab^{a 
 
 a3 4- 2a^b -^ ab^ 
 
 ^2a^b — 2a63^ a^ + 2ab -f b^ 
 er « + 6 ) a2 4- 2a6 -f fc^ (a 4- * 
 a^ + ab 
 
 ab -f b' 
 a6 + b^ 
 
 Therefore a -|- ft is the greatest common divisor. 
 
 a2 — 4 
 3. To find the greatest common divisor of- 
 
 ab+2b 
 
 Ans. a— 2. 
 
 4. Tt 
 
FRACTIONS. m 
 
 4. To find the greatest common divisor of - 
 
 .6* 
 Ans. a2 —b^, 
 
 5. Find the greatest com. measure of . 
 
 6a5-{-lUa*x+5a*;c3 
 
 CASE V. 
 
 To Reduce a Fraction to its Lowest Terms, 
 
 Find the greatest common measure, as in the last pro- 
 blem Then divide both the terms of the fraction by the 
 common measure thus found, and it will reduce it to its lowest 
 terms at once, as was required. Or divide the terms by any 
 quantity which it may appear will divide them both, as in 
 arithmetical fractions. 
 
 EXAMPLES. 
 
 ab + b^ 
 
 1. Reduce to its lowest terms. 
 
 ac2 -I- 6c2 
 
 ab-i-b^) «c2 -|-6c2 
 or a -f 6 ) «c3 4- 6c3 (c* 
 
 Here ab -f- b^ is divided by the common factor b. 
 
 Therefore a -f- ^ is the greatest common measure, am! 
 ab + b^ b 
 
 hence a -}- b) = — , is the fraction required. 
 
 ac^ -f 6c2 c2 
 c3 — 62c 
 
 2. To reduce : — to its least terms. 
 
 c2-f-26c+63 
 €2 ^2be'\-b^)c^ — b^c{c 
 
 c3 -f-26c2-{-63c 
 
 -^^bc^^n^c) c2 -f- 26c -f 62 
 
 or c 4- fe) c2 -f- 26c + 62 (c -f 6 
 02-1- be 
 
 6c + 62 
 6c + 62 
 
 Vol.,!. 2$ Therefore 
 
194 ALGEBRA. 
 
 Therefore c -f fc is the greatest common measure, and 
 c3 — 62c c^—bc 
 
 hence c -f- 6) = is the fraction required. 
 
 C'-f 5^6c-f 6* c + 6 
 
 c3— 63 c3 4- 6c-f h" 
 
 3. Reduce to its lowest terms. Ans. . 
 
 C4~62c2 c3 4- 6c2 
 
 a2 — 62 1 
 
 4. Reduce to its lowest terms. Ans. • 
 
 a* —64 a« 4-6* 
 
 a4--6* 
 
 6. Reduce to its lowest terms. 
 
 a3-3a264-iia6*— fc=* 
 3a« + 6a*c + 3a^c^ 
 
 6. Reduce ■ — - to its lowest terms. 
 
 a^c-^ 3a2c«-|-3ac3-|-c* 
 a3— «62 
 
 7. Reduce — — to its lowest terms. 
 
 o2 4-2a6 4-6« 
 
 CASE VI. 
 To add Fractional Q;uantities together, 
 
 Ip the fractions have a common denominator, add all the 
 numerators together ; then under their sum set the common 
 denominator, and it is done. 
 
 If they have not a common denominator, reduce them t& 
 one, and then add them as before. 
 
 EXAMPLES. 
 
 1. Let — and — be given, to find their sum. 
 
 3 4 
 
 a a 4a Sa la ) - 
 
 Here 1 = 1 = — is the sum required. 
 
 3 4 12 12 12 
 ah c 
 
 2. Given — , — , and — , to find their sum. 
 
 be d 
 
 a b c acd bbd bcc acd'{-bbd'{-bcc 
 
 Here — + — + — = + + = — 
 
 bed bed bed bed bed 
 
 the sum roauired. 
 
 ^ 3. Let 
 
FRACTIONS. 196 
 
 *3 . Let « — — and b H be added together: 
 
 b e 
 
 3x2 Zaa: Scx^ 2a6jr 
 
 Here a h * H == « \- b -\ — — 
 
 b c be be 
 
 ^bx—Scx^ 
 
 ' = a -f- fc H the sum required. 
 
 be 
 4x 2x 206x -f- 6fl^ 
 
 4. Add — and — together. Ans. . 
 
 3a 5b 16a6 
 
 a a a 
 
 5. Add — , — and — together. Ans. f^a. 
 
 3 4 6 
 
 2a- 3 5a 9a^6 
 
 6. Add and — together. Ans. . 
 
 4 8 .8 
 
 a4-3 2a— 5 14a-13 
 
 7. Add 2a + -: — to 4a H . Ans. Oa + -: . 
 
 6 4 20 V 
 
 3a2 a 4- 6 
 
 8. Add Sttf and and together. 
 
 46 36 
 6a 6a 3a + 2 
 ' Q, Add *- , and — and together. 
 
 4 6 7 
 3a a, 
 
 10. Add 2a, and — and 3 -f- - together. 
 
 8 6' 
 
 3a 6a 
 
 U. Add Ba-^ and 2a— — together. 
 
 4 8 
 
 CASE VII. 
 
 * To Subtraet one Fractional Quantiiif from another. 
 
 Reduce the fractions to a common denominator, as in addi- 
 tion, if they have not a common denominator. 
 
 Subtract the numerators from each other, and under their 
 difference set the common denominator, and the work is done. 
 
 * In the addition of mixed quantities, it is best to bring the frac- 
 tional parts only to a coromon denominator^ and to annex their sr.ni to 
 the sum of the integers* w'th the proper sign. And the same rule 
 may be observed for mix§d quantities in subtraction ^Iso. 
 
 EXAMPLES. 
 
i^ ALGEBKA. 
 
 EXAMPLES. 
 
 3g 4a 
 
 1. To find the difference of — and — . 
 4 7 
 
 3a 4a 21a 16a 5a 
 
 Here = : — = — is the difference requireel. 
 
 4 7 28 28 28 
 
 2a— 6 3a--4fe 
 
 :2. To find the difference of and . 
 
 4c 36 
 
 2a— A 3a — 46 6a6 -366 12ac— 166c 
 
 Here = = 
 
 4c 36 126c 126c 
 
 6a6— . 366- 12ac + 166c 
 
 —- is the difference required. 
 
 126c 
 
 10a 4a 
 
 3. Required the difference of and — . 
 
 9 7 
 
 3a 
 
 4. Required the difference of 6a and — .. 
 
 4 
 
 f 5a 2a 
 
 5. Required the difference of — and ^^. 
 
 4 3 
 
 26 3a -f e 
 
 6. Subtract — from . 
 
 c 6 
 
 2a -f 6 4a + 8 
 
 7. Take from -. 
 
 9 ' 6 
 
 a-^36 2a 
 
 8. Take 2a from 4a -f — 
 
 CASE VIII. 
 
 To Multiply Fractional Quantities together. 
 
 Multiply the numerators together for a new numerator, 
 and the denominators for ^ new denominator.* 
 
 * 1. When the rumera'tor of one fraction, and the denominator of 
 th^ other, can be divided by some quantity, which is common to both, 
 the quotients may be used instead of them. 
 
 2. When a fraction is to be multiplied by an integer, theproduct is 
 found either by multiplying the numerator, or dividing the denomina- 
 tor by it ; and if the integer be the same with the denominator, the 
 nwmerator may be taken for the product. 
 
 IPAMPLES^ 
 
FRACTIONS. 1^ 
 
 EXAMPLES. 
 
 a ta 
 
 1 . Required to find the product of — and — . 
 
 8 5 
 
 a X 2a 2a2 a^ 
 
 Here = = — the product required. 
 
 8 X 5 40 20 
 
 a 3a 6a 
 
 2. Required the product of — , — , and — . 
 
 3 4 7 
 a X 3a X 6a ISa^ So^ 
 
 = = the product required. 
 
 3X4X7 84 14 
 
 2a a 4- 6 
 
 3. Required the product of — and . 
 
 b 2a-{-c 
 
 2a X (a-f-6) 2aa -f 2a6 
 
 Here = the product re(luired. 
 
 b X (2a+c) 2ab -f be 
 
 4a Ga 
 
 4. Required the product of — and — . 
 
 3 5c 
 3a 4b^ 
 6. Required the product of — aad . 
 
 4 3a 
 3a 8ac 4ab 
 
 6. To multiply — and and together. 
 
 6 6 3c 
 
 ab 3a2 
 
 7. Required the product of 2a + — and . 
 
 2c b 
 
 2a2 —262 4a2 + 26- 
 
 8. Required the product of and— . 
 
 36c a -I- 6 
 
 2a + 1 2a - 1 
 
 9. Required the product of 3a, and and . 
 
 a 2a -f 6 
 
 X x^ a a^ 
 
 10. Multiply a -h by a; {- 
 
 2a 4a2 2x Ax^ 
 
 CASE IX. 
 To Divide one Fractional (Quantity by another. 
 Divide the numerators by each other, and the denomina- 
 tors by each other, if they will exactly divide. But, if not, 
 then invert the terms of the divisor, and multiply by it exact- 
 ly as in muItipHcation, EXAMPLES. 
 
 * I . I f the fractions to be divided have a coniaion denominator, take the numerator of the 
 dividend for a new numerator, and the numerator of the divisor iVir the new denominator. 
 
 2. When a fraction is t , be divided hy any quantttr. it is the stme thing whether the 
 munemiot be diYnied t»y it, or the deooiaiaator multiplied by it. 3. When 
 
198 ALGEBRA. 
 
 EXAMPLES. ' 
 
 a .3a. 
 
 1. Required to divide — by — . 
 
 4 8 
 a 3a a 8 8a 2 
 
 Here r — = — X — — = — the quotient. . 
 
 4 8 4 3a I2a 3 
 3a 5c 
 
 2. Required to divide — by — . 
 
 26 4d 
 3a 6c 3a 4a VZad 6ad 
 
 Here — -j = — X — = — ^ — =: the quotient, 
 
 26 4d 26 5c 106c 56c 
 2a + b 3a + 26 
 
 3. To divide by . Here, 
 
 3a— 26 4a + 6 
 2a + 6 4a -f- 6 8a2 -|- 6a6 -f 62 
 
 X — : = the quotient required. 
 
 3a— 26 3a+26 9a^ - 46^ 
 
 3a2 2a 
 
 4. To divide — by — . 
 
 a2 4- 65» 2a -f- 26 
 3a2 a-f6 3a2X(a + 6) 3a 
 
 Here, X • 
 
 a2-f-63 a (a3-f63)Xa a^ — a6 -f 6^ 
 
 is the quotient required 
 
 3a; U 
 
 5. To divide — by — . ^ 
 
 4 12 
 
 6. To divide by 3x, 
 
 5 
 3a; -f- 1 4x 
 
 7. To divide by — . 
 
 9 3^ 
 
 4x X 
 
 8. To divide ^ — by—. 
 
 2x- 1 3 
 
 4x 3a 
 
 9. To divide — by —. 
 
 6 66 
 2a— 6 5ac 
 
 10. To divide by . 
 
 4cd 6d 
 
 6a* -56* 6a3-f6ai 
 
 11 . ©ivide by . 
 
 2a2 -4a6 + 26^ 4a — 46 INVOLU- 
 
 3. When the two numerators, or t)ie two denominators, can be divided by jome commvn 
 quantity, let that be doae, ami the quotients qsqd uute^id of the fractiosa fir«t prepoted. 
 
[ m ] 
 
 INVOLUTION. 
 
 iNVOLtJTioN is the raising of powers from any proposed 
 root ; such as findin/j; the square, cube, biquadrate, &,c. of any 
 given quantity. '1 he n>ethod is as follows : 
 
 * Multiply the root or given quantity by itself, as m^ny 
 tiiues a.* there are units in the index less one, and the last pro- 
 duct will be the power required. — Or, in literals, multiply 
 the index of the root by the index of the power, and the result 
 will be the power, the same as before. 
 
 JVote. When the sign of the root is -f, all the powers of it 
 will be -f ; but when the sign is — , all the even powers will be 
 -i-, and all the odd powers — ; as is evident from multrplication. 
 EXAMPLES. 
 
 a, the root 
 
 a2, the root 
 
 a2 = square* 
 
 a* = square 
 
 a3 = cube 
 
 a6 = )cnhe 
 
 a* = 4th power 
 
 a a = 4th power 
 
 a^ = 5th power 
 
 ftio = 5th power 
 
 &c. 
 
 &c. 
 
 — 2a, the root 
 
 — 3a63 , the root 
 
 -|- 4a2 = square 
 
 -|- 9a2 6* = square 
 
 — 8a 3 = cube 
 
 — 27a3 66 = cube 
 
 -f 16a* =4th power 
 
 "--f Sla46« = 4th power ■ 
 
 — 32a5 = 5th power 
 
 — 243as6»o= 5th power 
 
 2ax^ 
 
 a 
 
 , the root 
 
 — , the root 
 
 36 
 
 26 
 
 4a3a;* 
 
 a2 
 
 -| = square 
 
 — = Square 
 
 96^ 
 
 463^ 
 
 Sa^x^ 
 
 a3 
 
 — cube 
 
 — == cube 
 
 863 
 
 2763 
 
 16a4x» 
 
 , fl4 
 
 -\ — = 4th power. 
 
 
 816^ 
 
 166* 
 
 ♦ A ay power of the product of two or moie quantities, is equal t© 
 the same power of each of the factors, multiplied together. 
 
 And any power of a fi-action, is equal to the san^e power of the nu- 
 me ator, divided by the I ke power of the denominator. 
 
 Also, powers or roots of the same quantity, are multiplied by one 
 another, by adding their exponents j or divided, by subtracting their 
 exponents. 
 
 a3 3-2 
 
 Thus,a3 X «2 = a8*2 ^ a5. And a3 -r a^ or - = a =«. 
 
200 ALGEBRA. 
 
 X — a = root ^ -|- o = root 
 
 X — a X -\- a 
 
 X* —ax x^-\- ax 
 
 —ax -[- a2 -\' ax '\- a^ 
 
 x^^ 2ax + a2 square x^-\- 2ax -{- a^ 
 X — a JT -j- a 
 
 
 :r3— 3ax2 '{-Sa'^x—a^ x^-}- 3ax^ -{-Sa^x-^-a^ 
 
 the cubes, or third powers, of x — a and x -{- a. 
 EXAMPLES FOR PRACTICE. 
 
 1. Required the cube or 3d power of 3a^. 
 
 2. Required the 4th power of 2a^b. 
 
 3. Required the 3d power of — ■ 4a^b^. 
 
 a^x 
 
 4. To find the biquadrate of . 
 
 263 ' 
 6. Required the 6th power of a — 2x, 
 
 6. To find the 6th power of 2a2 . 
 
 Sir Isaac Newton's Rule for raising a Binomial to any 
 Power whatever*. 
 
 1. To find the terms mthout the Co-efficients. The index of 
 the first, or leading quantity, begins with the index of the 
 given power, and in the succeeding terms decreases conti- 
 nually by 1, in every term to the last ; and in the 2d or 
 following quantity, the indices of the terms are 0, 1,2, 3, 4, 
 &c. increasing always by 1 . That is, the first term will con- 
 tain only the first part of the root with the same index, or of 
 
 * This rule, expressed in general terms, is as follows ; 
 
 fifl\x)^ — a" + n . a^-^x + «.— — ao~2x2-^M.— . — an-^x^ &c. 
 2 2 3 
 
 n— 1 n— 1 n — 2 
 
 ^a— x)» «■ a" - n . an-lx+ n . — -^a^-^^x^ - n — . -^ a^-^x^ &c. 
 2 2 3 
 
 J^Tote. The sum of the co-efficients, in every power, is equal to the 
 number 2, when raised to that power. Thus 1 + 1 = 2 in the first 
 power ; 1 + 2 + 1 = 4 -= 22 in the sljaare ;l+3+3 + l«'8=s 
 23 in the«ube, or third power ; and s« on* 
 
 the 
 
INVOLUTION. 201 
 
 the same height as the intended power : and the last term of 
 the series will contain only the 2d part of the given root, 
 when raised also to the same height of the intended power : 
 hut all the other or intermediate terms will contain the pro- 
 ducts of some powers of both the members of the root, in 
 such sort, that the powers or indices of the 1st or leading 
 member will always decrease by 1, while those of the 2d 
 member always increase by 1. 
 
 2. To find the Co-efficients. The first co- efficient is always 
 1 , and the second is the same as the index of the intended 
 power ; to find the 3d co-efficient, multiply that of the 2d 
 term by the index of the leading letter in the same term, and 
 divide the product by 2 ; and so on, that is, multiply the co- 
 efficient of the term last found by the index of the leading 
 quantity in that term, and divide the product by the numl)er 
 of terms to that place, and it will give the co-efficient of the 
 term next following ; which rule will find all the co-efficients, 
 one after another. 
 
 JVote. The whole number of terms will be 1 more than the 
 index of the given power : and when both terms of the root 
 are +, all the terms of the power will be -{- ; /but if the se- 
 cond term be — , all the odd terms will be -4-, and all the 
 even terms — , which causes the terms to be 4* and — alter- 
 nately. Also the sum of the two indices, in each term, is 
 always the same number, viz. the index of the required 
 power : and counting from the middle of the series, both 
 ways, or towards the right and left, the indices of the two 
 terms are the same figures at equal distances, but mutually 
 changed places. Moreover, the co-efficients are the same 
 numbers at equal distances from the middle of the series, 
 towards the right and left ; so by whatever numbers they 
 increase to the middle, by the same in the reverse order they 
 decrease to the end. 
 
 EXAMPLES. 
 
 1. Let a -{- jc he involved to the 5th power. 
 
 The terms without the co-efficients, by the 1st rule, 
 will be 
 
 and the co -efficients, by the 2d rule, will be 
 
 ..5X4 10X3 10x25x1 
 1, 6, -y-, — ^ — , — - — , -J- ; 
 
 or 1, 5, 10, 10, 5, 1 ; 
 
 Therefore the 5th power altogether is 
 a5 4- 5a*x -I- lOa^jps ^ lOa^x^ -f- dax* -f ««. 
 Vol. I. 27 But, 
 
202 ALGEBRA. 
 
 But it is best to set down both tbe co-efficients and the 
 powers of the letters at once, in one line, without the inter- 
 mediate lines in the above example, as in the example here 
 below. 
 
 2. Let o — X be involved to the 6th power. 
 
 The terms with the co-efficients will be 
 a^ ~ ea^x -f- 15a4:r2 — 20a^x^ + Iba'^x'^ - 6aa;« -f x*. 
 
 3. Required the 4th power of a — x. 
 
 Ans. a* — 4a^x -f Ga^x^ — 4ax^ -f ^*. 
 And thus any other powers may be set down at once, is 
 the same manner ; which is the best way. 
 
 EVOLUTION. 
 
 Evolution is the reverse of Involution, being the method 
 of finding the square root, cube i^oot, &c. of any given 
 quantity whether simple or compound. 
 
 CASE L 
 
 To find the Roots of Simple Quantities, 
 
 Extract the root of the co-efficient for the numeral 
 part ; and divide the index of tl.e letter or letters, by the 
 index of the power, and it will give the root of the literal 
 part ; then annex this to the former, for the whole root 
 sought*. 
 
 * Any even root of an affirmative qiiantity, may be either + or — : 
 thus the square root of + as is either -f- a, or — as because -j- aX 
 -f. a = -f. a, and — aX — a = -^ ais also. 
 
 But an odd root of any quantity will have the same sign as the 
 quantity itself: thus the cube root of + a3 is + a and the cube 
 i-cot of a3 is - a ; for -f a X + a X -f* a =r + a3, and — rt X — 
 a X — a = — «3, 
 
 Any even root of a negative quantity is impossible ; for neither 
 -f a X + fl, nor — a X — a can produce — flS. 
 
 An . root of a product, is equal to the like root of each of the fac- 
 tors multiplied togetlier. And for the root of a fraction, take tlie root 
 of the numerator, and the root of the denominator. 
 
 EXAMPUIS. 
 
EVOLUTION. 20S 
 
 EXAMPLES. 
 
 1. The square root of 4a^ , is 2a. 
 
 2. The cube root of Sa^, is 2a^ or 2a. 
 
 Sii^b^ 5j2b2 . ab 
 
 3. The square root of -^, or v/"^^ > ^^Yc ^ ^' 
 
 4. The cube root of ^n^^^' sT ^ 
 
 6. To find the square root of ^a^bK Ans. ab^ ^2. 
 ^ 6. To find the cube root of — 64a^b^ Ans. — 4ab^. 
 
 7. To find the square root of Arj-' ^^^' 2a^ >/ -3^- 
 
 8. To find the 4th root of Sla^ft^. Ans. Sab ^ b. 
 
 9. To find the 6th root of - 32a^b\ Ans--2a6 ^ ^• 
 
 CASE II. 
 
 To find the Square Root of a Compound Quantity . 
 
 This is performed like as in numbers, thus : 
 
 1. Range the quantities according to the dimensions of 
 one of the letters, and set the root of the first term in the 
 quotient. 
 
 2. Subtract the square of the root, thus found, from the 
 first term, and brin^ down the next two terms to the re- 
 mainder for a dividend ; and take double the root for a 
 divisor. 
 
 3. Divide the dividend by the divisor, and annex the re- 
 sult both to the quotient and to the divisor. 
 
 4. IVIultiply the divisor thus increased, by the term last 
 set in the quotient, and subtract the product from the divi- 
 dend. 
 
 And so on, always the same, as in common arithmetic. 
 
 EXAMPLES. 
 
 p 1. Extract the square root of a4—4a3i-f-6a2 62— 4a63-f 6*. 
 a* — 4a36 -f ^a^b^ — 4ab^ -f- b*{a^ — 2ab + 6^ the root. 
 
 2a3 — 2ab) - 4a^b -}- 6a^b^ 
 - 4a^b 4- 4a362 
 
 2a2 — 4a&-i-62) 2an^ - 4ab^ -i- b^ 
 2a2 62.-4a63 4-64 
 
 2. Find 
 
204 ALGEBRA. 
 
 2. Find the root of a* -f 4a^b + 10a3 62 4- i2ab^ -f 96". 
 a* -f 4a^b + 10a2 62 + 12ab^ -f 96* (aa+ 2a6 -f 3fe2. 
 
 2a2 + 2a6 ) 4aH + lOaafe^ 
 4a36 4- 4a2 62 
 
 2a2 + 4a6-|-363 ) 60262 + 1206 3 + 9^* 
 6a2 62-f 12a63-|-96* 
 
 3. To find the square root of a* + 4a^ + 6a^ -|- 4a + 1. 
 
 Ans. a2 -|-2a +1. 
 4« Extract the square root of a* — 2a^ + 2a2 — a + a. 
 
 Ans. x^ — X -f" i- 
 
 5. It is required to find the square root of a^ — ab. 
 
 6 62 d3 „ 
 
 Ans. a — - — &c. 
 
 2 8a Jba2 
 
 To find the Roots of any Powers in General. 
 
 This is ako done like the same roots in numbers, thus ; 
 
 Find the root of the first term, and set it in the quotient. 
 —Subtract its power from that term, and bring down the* 
 second term for a dividend. — Involve the root, last found, to 
 the next lower power, and multiply it by the index of the 
 given power, for a divisor. — l/ivide the dividend by the di- 
 visor, and set the quotient as the next term of the root. — 
 Involve now the whole root to the power to be extracted ; 
 then subtract the power thus arising from the given power, 
 and divide the first term of the remainder by the divisor first 
 found ; and so on till the whole is finished*. 
 
 EXAMPLES. 
 
 ♦ As this method, in high powers, may be thought too laborious, it 
 ■will not be improper to observe, that the roots of compound quantities 
 may sometimes be easily discovered, thvis : 
 
 Extract th- roots of some of the most simple terms, and connect 
 them together by the sign + or — , as may be judged most suitable 
 for the purpose Involve the compound root, thus found, to the pro- 
 per power i then, if this bf the same wivh the given quantity, it is the 
 
 root required But if it be found to difrer only in some of the signs, 
 
 change them from -f to — , or from — to +, till its power agrees 
 witii the given one throughout. 
 
 , Thus 
 
EV6LUT101I5J. . 20b 
 
 EXAMPLES. 
 
 1. To find the square root of a* -Sa^fc+Sasfcz— 2a6a+fe<. 
 a* — 2a^b -f Sa^fc^ — Saft^ 4-6* (a3~ 06 -f b^ 
 
 • 2a^b 
 
 
 2. Find the cube root of a^ — 6a« + 21a* — 44a=» + 63a« 
 
 ~54a -f- 27. 
 a6-6a5 4-21a*-44a3 + 63a3— 54a + 27 (a3-.2a 4- 3. 
 a6 
 
 3a4 ) — 6a5 
 
 a6_6a5 -f 12a4— 8a3 = (a^ -.2a)» 
 
 3a4 ) -j- na* 
 a6_6a:5-|-21a4— 44a34-63a2~54a + 27 = (a2— 2a— 3)^-. 
 
 3. To find the square root of a^ — 2ab + 2aa; + 62 — 
 2bx -\~ x^- Ans. a — b -^ x, 
 
 4. Find the cube root of a6-3a5 -{- 9a4 — iSa^ -f 18a2 — 
 12a + 8. Ans. a2-.a -f- 2. 
 
 5. Find the 4th root of 81a* — 216a36 -f- 216a2 62 _ geab'-' 
 + 166*. Ans. 3a — 26. 
 
 6. Find the 5th root of a« — 10a* -{- 40a3 - SOa^ -f 80ft 
 — 32. Ans. a — 2. 
 
 7. Required the square root of 1 — x^ . 
 
 8. Required the cube root of 1 — x^. 
 
 Thus, in the 5th exninple, the root 3a — 26, is the difference of 
 the roots of the first and last terms ; and in the third example, the 
 root a ~ 6 4- >r, is the sOrri of the roots of the 1st, 4th, aiid 6lh terms. 
 The same may ulso be observed of the 6th example, where the root is 
 foiyjd from the first and last terms. 
 
 SURDS. 
 
2e6 ALGEBRA. 
 
 SURDS. 
 
 Surds are such quantities as have not exact values in num- 
 bers ; and are usually expressed hy fractional indices, or by 
 
 means of the radical sign y/. Thus, 3^, or ^ 3, denotes the 
 
 square root of 3 ; and 2^ or »/ 22, or ^ 4, the cube root of 
 the square of 2 ; where the numerator shows the power to 
 which the quantity is to be raised, and the denominator its root, 
 
 PROBLEM I. 
 
 To Reduce a Rational Quantity to the Form of a Surd, 
 
 Raise the given quantity to the power denoted by the index 
 •f the surd ; then over or before this new quantity set the 
 radical sign, and it will be the form required. 
 
 EXAMPLES, 
 
 1. To reduce 4 to the form of the square root. 
 First, 42 = 4 X 4 = 16 ; then ^ 16 is the answer. 
 
 2. To reduce 3a^ to the form of the cube root. 
 First 3a2 X 3a2. x 3a2 = (3a2)3 = 27a« j 
 
 then 3/ 27a6 or (27a6) 3- is the answer, 
 
 3. Reduce 6 to the form of the cube root. 
 
 Ans. (216)2 or ^216, 
 
 4. Reduce ^ab to the form of the square root. 
 
 Ans. ^ ^a^b''^ 
 
 5. Reduce 2 to the form of the 4th root. Ans. (16)*. 
 
 6. Reduce a^ to the form of the 5th root. 
 
 7. Reduce a -j- a; to the form of the square root. 
 
 8. Reduce a — a; to the form of the cube root. 
 
 PROBLEM IL 
 
 To Reduce Quantities to a Common Index. 
 
 1. Repuce the indices of the a;iv<^n quantities to a common 
 denominator, and involve each of them to the power denoted 
 by its numerator ; then 1 set over the common denominator 
 will form the common index. Or, 
 
 2. Ijf 
 
SUMS. 207 
 
 ^ f . If tbe comnion index be gives, divide the indices of the 
 quantities by the given index, and the quotients will be the 
 new indices for those quantities. Then over the said quan- 
 tities, with their new indices, set the given index, and thejr 
 ^ill make the equivalent quantities sought. 
 
 EXAMPLES. 
 
 1. Reduce 3* and 6* to a common index. 
 Here 1 and } = j\ and j\. 
 
 Therefore 3^" and 5^^° = (3s)^^znA (62)t'»=^ 3^ and v' 6« 
 
 = ^ 243 and ^ 26. 
 
 2. Reduce a^ and b^ to the same common index |. 
 Here, f -j- i = i X f = f the 1st index. 
 
 and ^ 4- i = i X f = I the 2d index. 
 
 I a. L 2 
 
 Therefore(a«)^ and (63) 2^ or ^^ a^ and y 6^ are the quaa- 
 
 tities. 
 
 3. Reduce 4^ and 5^ to the common index i. 
 
 Ans. 2563 )^ and 25 . 
 
 4. Reduce o^ and x* to the common index a. 
 
 Ans. (a3)« and(x^)^. 
 
 5. Reduce a^ and x^ to the same radical sign. 
 
 Ans. -^ a* and y/ x^. 
 
 6. Reduce (a -|- ^)^ *°<^ (a — x)^ to a common index. 
 
 7. Reduce (a -f- 6)^ and (0—6/ to a common index, 
 
 PROBLEM ra. 
 
 To Reduce Surds to more Simple Terms^ 
 
 ^iND out the greatest power contained in, or to divide the 
 given surd ; take its root, and set it betore the quotient or the 
 remaining quantities, with the proper radical sign between 
 them. 
 
 EXAMPLES. 
 
 1. To reduce ^ 32 to simpler terms* 
 
 Here y 32 = y 16X2 = y 16 Xy2 = 4 X^2 = 4^2^ 
 
 2. To reduce ^ 320 to simpler terms. 
 
 3/320= ^64X5 = ^64X V'^ = 4X \/ 5^4 1/5. 
 
 3. Reduce 
 
208 AL,GEBRA. 
 
 3. Reduce -^ 75 to it3 simplest terms. Ans. 5 ^ S. 
 
 4. Reduce ^ f | to simpler terms. Ans. fj y/ 33. 
 
 5. Reduce ^ 109 to its simplest terms. Ans. ^ 37. 
 
 6. Reduce ^ *^^ to its simplest terms. Ans. -|- ^ 10. 
 
 7. Reduce y/ Iba^b to its simplest terms. Ans. ba ^ 36. 
 jVyic, There are other cases of reducing algebraic surds 
 
 to simpler forms, that are practised on several occie^ions ; one 
 instance of which, on account of its simplicity and nsetulness, 
 may be here noticed, viz. in fractional forms having com- 
 pound surds in the denominator, multiply both numerator and 
 denominator by the same terms of tht denominator, hut hav- 
 ing one sign changed, from + to— or from — to -}■» which will 
 reduce the fraction to a rational denominator. 
 
 Ex. To reduce , multiply it by , ^nd 
 
 V'ft — v/3 a/54--v/3 
 16 -f 2 ^ 15 Sy'lS- V5 
 it becomes == 8 -f ^ 16. Also, if ; 
 
 ^15—^5 65 — 7^/76 
 
 multiply it by , and it becomes = 
 
 ^\b^^b 15 — 5 
 
 65 — 35^3 13 — 7^3 
 
 10 2 
 
 PROBLEM IV. 
 
 To add Surd (Quantities together. 
 
 1. Bring all fractions to a common denominator, and reduce 
 the quantities to their simplest terms, as in the last problem. 
 — 2. Reduce also such quantities ds have unhke indices to 
 other equivalent ones having a common index. — 3. Then, if 
 the surd part be the same in them all, annex it to the sum of 
 the rational parts, with the sign of multiplication, and it will 
 give the total sum required. 
 
 But '}i the surd part be not the same in all the quantities, 
 they can only Jm? added by the signs -j- and — . 
 : EXAMPLES. 
 
 1 . Required to add ^ 1 8 and ^ 32 together. 
 
 First, y 18 = v^yX2 = 3^2; and ^^32 = ^^16X2= 4^2 : 
 Then, 3^2 + 4^2 = (3 -f 4)^2 = 7^2 — sum required. 
 
 2. It is required to add \/ 375, and y 192 together. 
 
 First, ^3^75=1/ 125X3=53/3; and^/ 192=3/64X3=43/3; 
 Then, 5 ^3 -f 4^' 3 — (5 4- 4)^3 -}-9 ^'3 = sum required. 
 
 3. Required! 
 
SURDS. 20p 
 
 3. Required the sum of y 27 and ^ 48. Ans. 7 ^^ 3. 
 
 4 Required the sum of ^ 50 and ^ 72. Ans. 11^2. 
 
 5. Required the sum of ^ ^ and y/ ^s- 
 
 Ans. 4 ^ yV or A \/ 1^- 
 
 6. Required the sum of %/ 56 and %/ 189. Ans. 5 3/7. 
 
 7. Required the sum of y ^ and ^ -i-. Ans. \ %/ 2. 
 
 8. Required the sum of 3^a^b and 5^ \6a^b. » 
 
 PROBLEM V. 
 
 To find the Difference of Surd Quantities. 
 
 Prepare the quantities the same way as in the last rule ; 
 then subtract the rational parts, and to the remainder annex 
 the common surd, for the difference of the surds required. 
 
 But if the quantities have no common surd, they can only 
 be subtracted by means of the sign — . 
 
 EXAMPLES. 
 
 1. To find the difference between ^ 320 and y 80. 
 
 First, ^ 320 = ^64 X 5=8^5 ; and^80=y 16X5=4^5, 
 Then 8^5 — 4-^5 = 4^5 the difference sought 
 
 2. To find the difference between %/ 128 and y 54. 
 First, ^ 128 = ^64X2=43/ 2 ; and^/ 4 = 3/27X2~= »/ 2. 
 Then 4 ^ 2 — 3 3/ 2 = y 2, the difference required. 
 
 3. Required the difference of ^75 and ^ 48. Ans. ^ 3. 
 
 4. Required the difference of 1/ 256 and 1/ 32. Ans. 23/4. 
 
 5. Required the difference of ^ ^ and ^ |. Ans. i 1/ 6. 
 
 6. Required the difference of 3/ | and y %'. Ans. ^j^Td, 
 
 7. Find the difference of ^24a3 62 and ^54ab*. 
 
 Ans. (a- 26) ^ (362_2a6) ^ 6a,. 
 
 PROBLEM VL 
 
 To Multiply Surd Quantities together. 
 
 Reduce the surds to the same index, if necessary ; next 
 multiply the rational quantities together, and the surds too-e- 
 ther ; then annex the one product to the other for the whole 
 product required ; which may be reduced to more simple 
 terms if necessary. 
 
 ,^ ' " EXAMPLES. 
 
210 AI^GEBRA. 
 
 EXAMPLES. 
 
 1. Required to find the product of 4 ^ 12, and 3^2, 
 Here, 4 X 3 X v' 12X^/2=12 ^12 x'2=12^24=12y'4X6 
 
 = 12 X 2 X <v/ 6 = 24 ^ 6, the product required. 
 
 2. Require to multiply i 3/ i by | »/ f . 
 HereiXiVIXVI=TVXy3%-TVXl/if=TVX-}X^18 
 
 = jV V ^^j *^^ product required. 
 
 3. Required the product of 3 y 2 and 2 ^ 8. Ans, 24. 
 
 4. Required the product of J-^4 and 1^12. Ans. i^6. 
 
 5. To find the product of f^f and yVv^l- Ans. /o\/l^- 
 
 6. Required the product of 23/14 and 33/4. Ans. 123/7. 
 
 ,2. i. 
 
 7. Required the product of 2a^ an4 c^. Ans. 2o2. 
 
 i 3 
 
 8. Required the product of (a -f 6)3 and (a -|- 6)*. 
 
 9. Required the product of 2x -{- y/ b and 2x — ^ b. 
 
 10. Required the product of (a 4-2v^6)2 and (a — 2^6)-. 
 
 11. Required the product of 2x » and 3x"*. 
 
 12. Required the product of 4j;« and 2^'» . 
 
 PROBLEM tn. 
 
 To Divide one Surd Quantity by another. 
 
 Reduce the surds to the same index, if necessary ; then 
 take the quotient of the rational quantities, and annex it to 
 the quotient of the surds, and it will give the wholf^ quotient 
 required ; which may be reduced to more simple terms if 
 requisite, 
 
 EXAMPLES. 
 
 1. Required to dinde 6 v^ 96 by 3 y 8. 
 
 Here 6 — 3 . ^ (96 ~ 8) =2^ 12=2^^(4X3) =2X2^/3. 
 = 4^3, the quotient required. 
 
 2. Required to divide 123/280 by 33/6. 
 
 Pere 12 ^ 3 = 4, and 280 -j- 5 = 66 = 8X7 = 23 . 7 ; 
 Therefore 4 X 2 X ^7 = 8^7, is the quotient required. 
 
 3. Let 
 
SURDS. «11 
 
 3. Let 4 ;;/ 50 be divided by 2 v' 6. Ans. 2 ^ 10, 
 
 4. Let 6 3/ 100 be divided by 3 V' ^- Aas. 2 y 20. 
 
 5. Let f y J^- be divided by f ^ f . Ans. yV v^ ^• 
 
 6. Let f y ^j be divided by f ^ f . Ans, ^ ^ 30. 
 •7. Let A ^a, or f^t-, be divided by §a^. Ans. f a^. 
 
 i. 3l 
 
 8. Let a" be divided by a^. 
 
 9. To divide 3a« by 4a"». 
 
 PROBLEM VUI. 
 
 To Involve or Raise Surd Quantities to any Power, 
 
 Raise botb the rational part and the surd part. Or mul 
 tiply the index of the quantity by the index of the power to 
 which it is to be raised, and to the result annex the power of 
 the rational parts, which will give the power'required. 
 
 EXAMPLES. 
 
 1. Required to find tbe square of |a2. 
 
 thirst, (f )2 = I XJ = tV' and {a^y == a^ ^ ^ = a^ =.a. 
 Therefore (fa^)^ = i\a, is the square required. 
 
 3. 
 
 2. Required to find the square of ^a^ 
 First, i X i = i and (a^ )3 = a» = a 3/ « ; 
 Therefore (ict^)^ = la^/a is the square required. 
 
 3. Required to find the cube of | ^ 6 or | X 6^' 
 First, (1)3 = I X f X I = ^\ and (6^)3 =6^ = 6^6. 
 Theref (1^6)3 = /^ x6y/6= V a/ 6, the cube required. 
 
 4. Required the square of 2 ^ 2. Ans 4^4. 
 
 6. Required the cube of 3^, or ^ 3. Ans. 3 ^ 3. 
 G, Required. the 3d power of i ^ 3. Ans. } ^ 3. 
 
 7. Required to find th© 4th power of ^ ^2. Ans. }. 
 
 8. Required 
 
212 ALGEBRA. 
 
 i. 
 
 8. Required to find the mth power of a". 
 
 9. Required to find the square of 2 -f ^ 3. 
 
 PROBLEM IX. 
 
 To Evolve or Extract the Roots of Surd Quantities*. 
 
 Extract both the rational part and the surd part. Or di- 
 vide the index of the given quantity by the index of the root 
 to be extracted ; then to the result annex the root of the ra- 
 laonal part, which will give the root required. 
 
 EXAMPLES. 
 
 1. Kequired to find the square root of 16-^6. 
 irirst, ^ 16 = 4, and {f)^ =ze^^^ = 6^ ; 
 
 theref. (16 y 6)2 = 4 . 6*v= 4 y 6, is the sq. root required^. 
 
 2. Required to find the cube root of -^-^ y/ 3. 
 First, %/ ^V = h and W.^f = 32 -^ ^ = 36 ; 
 
 theref. (2V ^ 3)^ = i . 3^ = i ^ 3, is the cube root required. 
 
 5. Required the square root of G^. Ans. 6 y^ 6. 
 4. Required the cube root of \a'^h. Ans. \a ^6. 
 
 6. Required the 4th root of 16a2. Ans. 2 ^ a, 
 
 6. Required to find the mth root of x^. 
 
 7. Required the square root of a^ ^ ^a ^h ■\- 96. 
 
 * The square root of a binomial or residual surd, a + *» <>i' ^ -~ ^ 
 Hiay be found thus : Take ^ a^ -b'^ ^ c; 
 — — . rt + C « — c 
 
 then v/ a -}- 6 x= x/ ' + \/ > 
 
 2 2 
 — — rt-fc-c a—c 
 and v^ a — A = v/—— — \/ • 
 
 Thus the square root of4 + 2v^3 = l-f-v^3; 
 
 and the square root of 6 •— 2 ^^ 5 — ^ 5 — 1. 
 
 But for the cube, or any higher root, no general rule is known. 
 
 INFINITE 
 
SURDS. 213 
 
 INFINITE SERIES. 
 
 An Infinite Series is formed either from division, dividing 
 hy a comj)ound divisor, or by extracting the root of a com- 
 pound surd quantity ; and is such as, being continued, would 
 run on infinitely, in the manner of a continued decimal frac- 
 tion. 
 
 But, by obtaining a few of the first terms, the law of thfr^ 
 pros^ression will be manifest ; so that the series may thence be 
 continued, without actually performing the whole operation. 
 
 PROBLEM I. 
 To Reduce Fractional Quantities into Infinite Series by Division. 
 
 Divide the numerator by the denominator, as in common 
 division ; then the operation, continued as far as may be 
 thought necessary, will give the infinite series required. 
 
 EXAMPLES. 
 
 2ab 
 
 1. To change into an infinite series. 
 
 a 4- 6 
 
 « + 6) 2a6 . . ( 26 1 + &c. 
 
 a a2 a3 
 2o6 + 262 
 
 .- 262 
 
 26» 
 « 263 
 
 268 
 
 263 26* 
 a a3 
 
 a3 
 26* 26s 
 
 a^ o3 
 
 , &c. 
 
 a' 
 
 ' " 2. Let 
 
214 ALGEBRA. 
 
 2. Let be changed into an infinite series. 
 
 1 — a 
 
 1 — a) 1 ( 1 + a + a2 + a3 4- o* + &c. 
 
 1 -- a 
 
 a2 
 
 a2 
 
 a2 -C3 
 
 a3«a4 
 
 
 h 
 
 
 3. 
 
 Expand ■ into an infinite series. 
 
 
 
 h c 
 
 C2 C3 
 
 
 Ans. — X<1 1 
 
 + &C.) 
 
 
 a a 
 
 a2 a» 
 
 
 a 
 
 
 4. 
 
 Expand into an infinite series. 
 
 a — b 
 
 
 
 h 
 
 is fc3 
 
 
 Ans. 1-1 1- 
 
 - + - + &C. 
 
 
 a . 
 
 a2 a3 
 
 
 1 - a; 
 
 
 6. 
 
 Expand into an infinite series. 
 
 
 1 -f re 
 
 Ans. 1 — 2a; -f 2a;2— .gx^ 4- 2x*y &c. 
 
 6. Expand ■ " into an infinite series. 
 
 26 362 463 
 
 ' Ans. 1 1 ,&c 
 
 a aa o=» 
 1 
 
 7. Expand =^, into an infinite series. 
 
 l-fl 
 
 PJROBLEM II. 
 
 To Reduce a Compound Sufd into an Infinite Series. 
 Extract the root as in common arithmetic ; then the ope- 
 ration, continued as far as may be thought necessary, will give 
 the series required. But this method is chiefly of use in ex- 
 tracting the square root, the operation being too tedious for 
 the higher powers. EXAMPLES. 
 
INFINITE SERIES. f If 
 
 EXAMPLES. 
 
 K Extract the root of a^ — x' in an infinite series. 
 x^ X* X* 6x3 
 
 a» — ar2 (^ — &c. 
 
 ia 8o3 I6a« 1280' 
 a2 
 
 
 x^ 
 
 
 
 
 
 
 2a 
 
 2a 
 
 ^x» 
 
 4a3 
 
 X* 
 
 
 
 
 X8 
 
 X* 
 
 
 
 2a < 
 
 a 
 
 8o3 
 
 )- 
 
 4a3 
 
 X9 
 
 x» 
 
 
 
 
 
 ' + 
 
 — + 
 
 
 
 JC-3 
 
 X4 
 
 
 4a2 
 
 Sa* 
 
 64a» 
 
 
 ore 
 
 a;' 
 
 B 
 
 2a- 
 
 -. . 
 
 . - — 
 
 &c.) 
 
 .; 
 
 _ —— 
 
 — 
 
 
 a 
 
 4o3 
 
 
 8a* 
 
 64a< 
 
 x^ 
 
 S 
 
 
 
 
 
 -f 
 
 &c. 
 
 
 
 
 
 8a* 
 
 16a« 
 
 
 
 
 6a:« 
 
 
 
 %t 
 
 
 
 ' 
 
 64a6 
 
 &c. 
 
 2. Expand ^Z 1 + 1 = \/ ^> into an infinite series. 
 
 Ans 1 + i-i + i^^jh &c. 
 
 3. Expand y^ 1 — 1 into an infinite series., 
 
 Ans. 1-±-.i-J--t|3. &c. 
 
 4. Expand ^ a'^ + x into an infinite series. 
 
 5. Expand ^ a^ — 26a; - x^ to an infinite series. 
 
 PROBLEM III. 
 
 To Extract any Root of a BinomifU : or to Reduce a Binomial 
 
 Surd into an Infinite Series. 
 
 This will be.doue by substituting the particular letters of 
 the binomial, with their proper signs, in the following general 
 theorem or formula, viz. 
 
 m mm m — n m—2n 
 
 (P + Pft) *» = P n -I A^ -I Bft H C€l 4- &C. 
 
 n 2n 3» and 
 
216 ALGEBRA. 
 
 and it will give the root required : observing that p denotei 
 the first term, q the second term divided by the first, ^ the 
 index of the power or root ; and a, b, c, d, &,c., denote the 
 several foregoing terms with their proper signs. 
 
 EXAMPLES. 
 
 1. To extract the sq. root of a^ + ^", in an infinite series. 
 
 6« m 1 
 
 Here p = a^ , ci = — and — = — : therefore 
 
 m m X 
 
 jp - = (o^) - = (aa) 2 = a = A, the 1st term of the series. 
 
 m b'' b^ 
 
 — AQ=|^ X aX — = — =B, the 2d term: 
 
 n a^ 2a 
 
 m—n 1—2 62 fc2 54 
 
 B^ X X— X— = — = c, the 3d term, 
 
 2n 4 2a eta 2.4a3 
 
 m-^2n 1—4 b^ b^ 3b^ 
 
 c^ = X X— = = D the 4tli. 
 
 3fi 6 2.4a3 a3 2.4.6a5 
 
 .62 h 3.6 e 
 
 Hence a -f- j — &c. or 
 
 2a 2.4a3 2.4 6a^ 
 62 5* 66 568 
 
 a -{ 1 &c. is the series required. 
 
 2a 8a3 16a« 1280^ 
 
 1 -^ 
 
 2. To find the value of , or its equal (a - a;)-2 , in an 
 
 infinite series*. (a~x) 
 
 ♦ N'ote, To facilitate the af plication of the rule to fractional ex- 
 amples, it is proper to observe, that any surd may be taken from the 
 denominator of a fraction and placed in the numerator, and vice versa^ 
 by only changing the sign of its index. Thus, 
 1 1 
 
 — =IX:j^' or only jr 2 ; apd . ■» 1 X (o + 6) ~* or 
 
 ^2 (a + ^)2 ^ 
 
 aa ^* , I 
 
 (a +6)~2 ; and— ^a^{<i-\-xy~^; and — —a:^x^^ ; als» 
 
 , {a -f :r)2 x^ 
 
 (fl2 4. a?2)2 ^ -X 
 
 , = (a2 + J»)2 X (a2 -,ar2) ^ , g^^, 
 
 ga2 _ x2 )2 
 
 Here 
 
INFINITE SERIES. 2f7 
 
 —a: m —2 
 
 Herep = a, a = = — a"'x, and— = =—2 ; theref. 
 
 a n 1 
 
 P ^ = ro^""2 r= = A, th e 1st term of the series. 
 
 « ^ ' o2 
 
 — A€i = — 2X--X = — = 2a 3 ;i^ = ^^ the 2d term. 
 
 n a* a a=* 
 m — n 2x "X 3x2 
 Ba = — I X — X = = 3a""*a:2 = c, the 3d. 
 
 2ra a3 a a* 
 
 m— 2n 3j;2 —o: ^x"^ 
 
 cft = — f X X = = 4a~5 j:3 = ©. 
 
 3n a* a a* 
 
 Hence ar^-\'^a"^x-\-'^a~'^x^ -f- ^a~^x^ -{- &c, or 
 
 1 2a: 3x3 4j;3 5^4 
 
 1 1 1 1 &c, is the series required. 
 
 a2 a^ a* a^ a^ 
 a2 
 
 3. To find the value of , in an infinite series. 
 
 a*^x 
 
 x^ x^ x^ 
 
 Ans. o + x-l 1 1 &c. 
 
 a a2 a» 
 
 4. To expand v^ , . . or , . . J" in a series. 
 
 1 ar2 3x4 5^6 
 
 Ans. — — 1 &c. 
 
 a 2a3 8a5 16a'' 
 a2 
 
 5. To expand — ' in an infinite series. 
 
 (a-6)2 
 
 26 362 453 5J4 
 
 Ans. 1 -I 1 h 1 &c, 
 
 a a^ a2 a* 
 
 ^. To expand -^a2 — x2 or (a2 — x^y in a series. 
 
 x^ X* x« 6z« 
 
 Ans. a &c. 
 
 2a 8a 3 ISa^ 128a7 
 
 7. Find the value of y (a^- 63) or (a3 — 63)3 in a series. 
 
 63 6« 669 
 
 Ans. a ' — —1 < &.C. 
 
 3a2 9a« 81a» 
 
 5. To find the value of ly (a^ -|-x* ) or (a* 4-^*)^ in a series. 
 x^ 2x»» 6x»« 
 
 Ans. a H [- . &c. 
 
 5a* SSa" 125a»* 
 Vol. I. 29 . 9. To 
 
218 ALGEBRA. 
 
 9. To find the square root of in an infinite series. 
 
 b X^ 073 
 
 Ans. 1 1 &c. 
 
 a 2a2 2a3 
 
 ^ 10. Find the cube root of , in a series. 
 
 3a3 9a^ 81«» 
 
 ARITHMETICAL PROPORTION^. 
 
 Arithmetical Proportion is the relation between tw# 
 »auHibers with respect to their difference. 
 
 Four quantities are in Arithmetical Proportion, when the 
 difference between the first and second is equal to the dif- 
 ference, between the third and fourth. Thus, 4, 6, 7, 9, and 
 a, a -{- d^btb X d are in arithmetical proportion. 
 
 Arithmetical Progression is when a series of quantities 
 have all the same common difference, or when they either 
 increase or decrease by the same common difference. Thus 
 2, 4, 6, 8, 10, 12, &c. are in arithmetical progression having 
 the common difference 2 ; and a, a -{- c?, a -f- 2<f, a -f 3d, 
 a -\- 4d, a -{- Bd, &c. are series in arithmetical progression, 
 the common difference being d. 
 
 The most useful part of arithmetical proportion is contained 
 in the following theorems : 
 
 1. When four quantities are in Arithmetical Proportion, 
 the sum of the two extremes is equal to the sum of the two 
 means. Thus, in the arithmetical 4, 6, 7, 9, the sum 4 + 
 9 = 64-7=13: and in the arithmeticals a, a -{- d, fc, 6 -{- </, 
 the sum a + 6 -|- d = a -f- 6 -{- d« 
 
 2. In any continued arithmetical progression, the sum of 
 the two extremes is equal to the sum of any two terms at an 
 equal distance from them. 
 
 Thus, 
 
ARITHMETICAL PROPORTION. 219 
 
 Thus, if the series be 1, 3, 6, 7, 9, 11, &c. 
 Then 1 + 11 = 3 -f 9 = 5 -f 7 = 12. 
 
 3. The last term of any increasing arithmetical series, is 
 equal to the first term increased by the product of the com- 
 mon difference multiplied by the number of terms less one ; 
 but m a decreasing series, the last term is equal to the first 
 term lessened by the said product. 
 
 Thus, the 20th term of the series, 1, 3, 6, 7, 9, &c. is ~ 
 1 4- 2 (20-1) = 1 +2 X 19 = 1 +38 = 39. 
 
 And the nth term of c, a — c?, a-- 2d, a— 3rf, a - 4d, &c. is 
 = a-(n— 1) X«? = a— (n — l)d. 
 
 4. The sum of all the terras in any series in arithmetical 
 progression, is equal to half the sum of the two extremes 
 multiplied by the number of terms. 
 
 Thus, the sum of 1,3, 6, 7, 9, &c. continued to the 10th 
 (1 -I- 19) X 10 20 X 10 
 term, is = ^^ ^' — ^ — = 10 X 10 = 100. 
 
 And the sum of n terms of a, a -f- d, a 4* 2c?, a -f- 3d, to 
 n 
 a -{- md, is = (a -|- a -{- tnd) . — = (^ + i md)n. 
 
 ' 2 
 
 EXAMPLES FOR PRACTICE. 
 
 1. The first term of an increasing arithmetical series is I, 
 the common difference 2, and the number of terms 21 ; re- 
 quired the sum of the series ? 
 
 First, 1 -}- 2 X 20 = 1 -f 40 = 41, is the last term. 
 1 + 41 
 ' Then — z — X 20 = 21 X 20 = 420, the sum required. 
 
 2. The first term of a decreasing arithmetical series is 199, 
 the common difference 3, and the number of terms 67 ; re- 
 quired the sum of the series ? 
 
 First, 199— 3 . 66 = 199—198 = 1, is the last term. 
 Then i?^ X 67 = 100 X 67 = 6700, the sum re- 
 quired. 
 
 3. To find the sum of 100 terms of the natural numbers 
 1, 2, 3, 4, 6, 6, &c. Ans 6050. 
 
 4. Required 
 
22^ ALGEBRA. 
 
 4. * Required the sum of 99 terms of the odd numbers 
 1,3, 6,7, 9, &c. Ans. 9811. 
 
 6. The first term of a decreasing arithmetical series is 10, 
 the common difiference ^, and the number of terms 21 ; re- 
 quired the sum of the series ? Ans. 140. 
 
 6. One hundred stones being placed on the ground, in a 
 straight line, at the distance of 2 yards Irom each other ; how 
 far will a person travel, who shall bring them one by one to 
 a basket, which is placed 2 yards from the first stone ? 
 
 Ans 1 1 miles and 840 yards. 
 
 APPLICATION OF ARITHMETICAL PROGRESSION 
 TO MILITARY AFFAIRS. 
 
 QUESTION I. 
 
 A Triangular Battalion,! consisting of thirty ranks, in 
 which the first rank is formed of one man only, the second 
 
 of 3 
 
 * The sum of any number (n) of terms of the arithmetical series of 
 odd number 1, 3, 5, 7, 9, &c. is equal to the square («3 ) of that num- 
 ber. 1 hat is, 
 If 1, 3, 5, 7, 9, &c. be the numbers, then will 
 
 12,2^, 33, 42, 53, &c. be the sums of 1,2, 3, &c. terms. 
 Thus, -|- 1 = 1 or 12 , the sum of 1 term, 
 14-3= 4 or 22 , the sum of 2 terms, 
 4 + 5 = 9 or 32 , the sum of 3 terms, 
 9 Hh 7 = 16 or 42 , the sum of 4 terms, &ۥ 
 For, by the 3d theorem, 1+2 (n— 1) = 1 + 2n— 2 = 2n— 1 is the 
 last term, when the number of terms is n ,- to this last term 2n — 1, 
 add the first term 1, gives 2n the sum of the extremes, or n half the 
 sum of the extremes ; then, by the 4ih theorem, ?iXn t= w2 is the sum 
 of all the terms. Hence it ai)pears in general, that half the sum of 
 the extremes, is always the same as the number of the terms n ; and 
 that the sum of all the terms, is the same as the square of the same 
 number, n2. 
 
 See more on Arithmetical Proportion in the Arithmetic, p. Ill' 
 
 f By triangular battalion, is to be understood, a body of troops, 
 ranged in the form of a triangle, in which the ranks exceed each 
 
 other 
 
ARITHMETICAL PROGRESSION. 221 
 
 of 3 the'third of 5 and so on : What is the strength of such a 
 triaigular battalion ? Answer, 900 men. 
 
 QUESTION n. 
 
 A detachment having 12 successive days to march, with or- 
 ders to advance the first day only 2 leagues, the second S^, and 
 so on increasing 1^ league each day's march : What is the 
 length of the whole march, and what is the last day's march ? 
 
 Answer, the last day's march is 18^ leagues, and 123 leagu€fs 
 is the length of the whole march. 
 
 QUESTION HL 
 
 A brigade of sappers,* having carried on 15 yards of sap 
 the first night, the second only 13 yards, and so on, decreasing 
 2 yards every night, till at last they carried on in one night 
 only 3 yards : What is the number of nights they were em- 
 ployed ; and what is the whole length of the sap ? 
 
 Answer, they were employed 7 nights, and the length of the 
 whole sap was 63 yards. 
 
 other by an equal number of men ; if the first rank consist of one man, 
 only, and the diiference between the ranks be also l,then its form is 
 that of an equilateral triangle ; and when the difference between the 
 ranks is more than 1, its form may then be an isosceles or scalene tri- 
 angle. The practice of forming troops in this order, which is now 
 laid aside, was formerly held in greater esteem than forming them in 
 a solid square as admitting of a greater front, especially when the 
 troops were to make simply a stand on all sides. 
 
 * A brigade of sappers, consists generally of 8 men divided equally 
 into two parties. While one of these parties is advancing the sap, the 
 other is furnishing the gabions, fascines, and other necessary imple- 
 ments, and when the first party is tired, the second takes its place 
 and soon, till each man in turn has been at the head of the sap. A sap 
 is a small ditch, between 3 and 4 feet in breadth and depth ; and is 
 distinguished from the trench by its breadth only, the trench having 
 between 10 and 15 feet breadth. As an encouragement to sappers, 
 the pay for all the work carried on by the whole brigade, is given to 
 the survivors. 
 
 QUESTIOX 
 
222 ALGEBRA. 
 
 QUESTION rV. 
 
 A number of gabions * being given to be placed in six 
 ranks, one above the other, in such a manner as that each 
 rank exceeding one another equally, the first may consist of 
 4 gabions, and the last of 9 : What is the number of gabions 
 in the six ranks ; and what is the difference between each 
 rank ? 
 
 Answer, the difference between the ranks will be 1 , and 
 the number of gabions in the six ranks will be 39, 
 
 QUESTION V. 
 
 Two detachments, distant from each other 37 leagues, and 
 both designing to occupy an advantageous pest equi-distant 
 from each other's camp, set out at different times ; the first 
 detachment increasing every day's march I league and a half, 
 and the second detachment increasing each day's march 2 
 leagues : both the detachments arrive at the same time ; the 
 first after 5 days' march, and the second after 4 days' march : 
 V/hat is the number of leagues marched by each detachment 
 each day ? 
 
 The progression ^V, 2^3^, Sf^, 5y\, 6j\y answers the con* 
 ditions of the first detachment and the progression If, 3f, £|, 
 7f , answers the conditions of the second detachment. 
 
 QUESTION VI. 
 
 A deserter, in his flight, travelling at the rate of 8 leagues a 
 day ; and a detachment of dragoons being sent after him with 
 orders to march the first day only 2 leagues, the second 5 
 leagues, the third 8 leagues, and so on : What is the number of 
 days necessary for the detachment to overtake the deserter, 
 and what will be the number of leagues marched before he is 
 overtaken? 
 
 Answer, 5 days are necessary to overtake him ; and conse- 
 quently 40 leilgues will be the extent of the march. 
 
 * Gabions are baskets, open at both ends, made of ozier twigs, and 
 of a cylindrical form ; those made use of at the trenches are 2 feet 
 wide, and about 3 feet high ; which, being filled with earth, serve as 
 a shelter from the enemy's fire ; and those made use of to construct 
 batteries, are generally higher and broader. There is another sort of 
 gabion made use of to raise a low parapet ; its height is from 1 to 2 
 feet, and I foot wide at top, but somewhat less at bottom, to give room 
 for placing the muzzel of a firelock between them j these gabions 
 serve instead of sand bags. A sand bag is generally made to contain 
 about a cubical foot of earth. 
 
 QUESTION 
 
PILING OP BALLS, 
 
 22? 
 
 QXJESTION Vn, 
 
 A convoy* distant 35 leagues, having orders to join its 
 camp, and to march at the rate of 5 leagues per day ; its 
 escort departing at the same time, with orders to march the 
 first day only half a league, and the last day 9^ leagues ; and 
 both the escort and convoy arriving at the same time : At 
 what distance is the escort from the convoy at the end of each 
 march ? 
 
 OF COMPUTING SHOT OR SHELLS IN A FINISHED PILE. 
 
 Shot and Shells are generally piled in three different 
 forms, called triangular, square, or oblong piles, according 
 as their base is either a triangle, a square, or a rectangle. 
 Fig, 1. C G^ Fig, 2, 
 
 H 
 
 
 B E 
 
 ABCD, fig. 1, is a triangular pile, 
 EFGH, fig. 2, is a square pile. . 
 .. E 
 
 
 F D ' 
 
 ABCDEF, fig. 3, is an oblong pile. 
 
 * By convoy is generally meant a supply of ammunition or provi- 
 sions, conveyed to a town or army. The body of men that guard 
 this supply is called escort. 
 
 A triangular 
 
224' ALGEBRA. 
 
 A triangular pile is formed by the continual laying of trian- 
 gular horizontal courses of shot one above another, in such a 
 manner, as that the sides of these courses, called rows, 
 decrease by unity from the bottom row to the top row, which 
 ends always in 1 shot. 
 
 A square pile is formed by the continual laying of square 
 horizontal courses of shot one above another, in such a man- 
 ner, as that the sides of these courses decrease by unity from 
 the bottom to the top row, which ends also in 1 shot. 
 
 In the triangular and the square piles, the sides or faces 
 being equilateral triangles, the shot contained in those faces 
 form an arithmetical progression, having for first term unity^ 
 and for last term and number of terms, the shot contained in 
 the bottom row ; for the number of horizontal rows, or the 
 number counted on one of the angles from the bottom to the 
 top, is always equal to those counted on one side in the bot- 
 tom : the sides or faces in either the triangular or square piles, 
 are called arithmetical triangles ; and the numbers contained 
 in these, are called triangular numbers : abc, fig. 1, efg, fig. 
 2, are arithmetical triangles. 
 
 The oblong pile may be conceived as formed from the 
 isquare pile abcd : to one side or face of which, as ad, a 
 number of arithmetical triangles equal to the face have beea 
 added : and the number of arithmetical triangles added to the 
 square pile, by means of which the oblong pile is formed, is 
 always one less than the shot in the top row ; or, which is the 
 same, equal to the difierence between the bottom row of the 
 greater side and that of the lesser. 
 
 QUESTION VIIL 
 
 To find the shot in the triangular pile abcd, fig. 1, the 
 bottom row ab consisting of 8 shot. 
 
 SOLUTION. 
 
 'The proposed pile consisting of 8 horizontal courses, each 
 of which forms an equilateral triangle ; that is, the shot con- 
 tained in these being in an arithmetical progression, of which 
 the first and last term, as also the number of terms, are 
 known ; it follows, that the sum of these particular courses, 
 or of the 8 progressions, will be the shot contained in the 
 proposed pile ; then 
 
 The 
 
/; ^ > ^ 
 
 PILING OF BALLS. 
 
 225 
 
 The shot of the first or lower } 
 triangular course will be ^ 
 
 the second _ - 
 
 the third - . - . 
 
 the fourth - - - • 
 
 the fifth . - . . 
 
 the sixth - - - - - 
 
 the seventh . - 
 
 the eighth - - - - 
 
 8 + 1 >4^4b = 36 
 7 -f 1 X 3J^ = 2« 
 
 6 -f 1 X 3 =21 
 
 6-1-1 X 2i = !• 
 4 4-1X2 =1© 
 
 3 4- 1 X U 
 
 2 -f 1 X 1 =3 
 
 1 -1- 1 X 
 
 = 1 
 
 Total - 120 shot 
 in the pile proposed. 
 QUESTION IX. 
 
 To find the shot of the square pile efgh, fig. 2, the bottom 
 row EF consisting of 8 shot. 
 
 SOLUTION. 
 
 The bottom row containing 8 shot, the second only 7 ; 
 that is, the rows forming the progression, 8, 7, 6, 5, 4, 3, 2, 1, 
 in which each of the terms being the square root of Ihe shot 
 contained in each separate square course employed in forming 
 the square pile ; it follows, that the sum of the squares of 
 these roots will be the shot required : and the sum of the 
 squares of 8, 7, 6, 6, 4, 3, 2, 1, being 204, expresses the shot 
 in the proposed pile. 
 
 QUESTION X. 
 
 To find the shot of the oblong pile abcdef, fig. 3 ; in 
 which BF = 16, and bc = 7. 
 
 SOLUTION. 
 
 The oblong pile proposed, consisting of the square pile 
 ABCD, whose bottom row is 7 shot ; besides 9 arithmetical 
 triangles or progressions, in which the first and last term, as 
 also the number of terms, are known ; it follows, that, 
 if to the contents of the square pile - 140 
 
 we add the sum of the 9th progression - 252 
 
 their total gives the contents required - 392 shot. 
 
 REMARK I. 
 
 The shot in the triangular and the square piles, as also 
 ihe shot in each horizontal course, may at once be ascer- 
 
 VoL. I. 30 tained 
 
226 
 
 ALGEBRA. 
 
 tained by the following table : the vertical column a, con- 
 taifis the s^ot> in the bottom row, horn I to 20 iDclu§ive ; 
 the robmiu b contains the triangular numbers, or nuuiher 
 of each course ; the column c contains the sum of the 
 triangular numbers, that is, the sh>>t contained in a triangular 
 pile, con;mor.ly called pyramidal nunibers ; the column d 
 contains the square of the numbers of the column a, that is, 
 the shot coritaijK d in each square horizontal course ; and the 
 column E contams the sum of these squares or shot in a 
 square pile. 
 
 c 
 
 B 
 
 A 
 
 D 
 
 E 
 
 
 
 
 
 S..U. . i-- 
 
 pyramidal 
 numbers. 
 
 Triangular 
 numbers. 
 
 Natural 
 numbers. 
 
 •Square ol 
 ihe natural 
 numbers. 
 
 ' hese 
 
 tquyre 
 
 nur» berg. 
 
 1 
 
 1 
 
 1- 
 
 1 
 
 1 
 
 4 
 
 3 
 
 2 
 
 4 
 
 5 
 
 , 10 
 
 6 
 
 3 
 
 9 
 
 14 
 
 20 
 
 10 
 
 4 
 
 16 
 
 30 
 
 35 
 
 15 
 
 5 
 
 25 
 
 55 
 
 66 
 
 21 
 
 6 
 
 36 
 
 91 
 
 84 
 
 28 
 
 7 
 
 49 
 
 140 
 
 120 
 
 36 
 
 8 
 
 64 
 
 204 
 
 165 
 
 45 
 
 9 
 
 81 
 
 285 
 
 220 
 
 55 
 
 10 
 
 100 
 
 385 
 
 286 
 
 66 
 
 li 
 
 121 
 
 506 
 
 364 
 
 78 
 
 12 
 
 144 
 
 650 
 
 455 
 
 91 
 
 13 
 
 169 
 
 819 
 
 560 
 
 105 
 
 14 
 
 K;6 
 
 1015 
 
 680 
 
 120 
 
 15 
 
 225 
 
 1240 
 
 816 
 
 136 
 
 16 
 
 256 
 
 1496 
 
 969 
 
 153 
 
 17 
 
 289 
 
 1785 
 
 1140 
 
 171 
 
 18 
 
 324 
 
 2109 
 
 1330 
 
 190 
 
 19 
 
 361 
 
 2470 
 
 1540 
 
 210 
 
 20 
 
 400 
 
 28*/© . 
 
 Thus, the bottom rrw in a triangular pile, consisting of 
 9 shot, the contents will he 165 ; and when of 9 in the square 
 pile, 285. — !n the same manner, the contents either of a 
 scuare or tri?-nj.nilar pile being given, the shot in the bottom 
 row may be easily ascertained. 
 
 The contents of any oblong pile by the preceding table 
 may be also with little trouble ascertained, the less side not 
 exceedir g 20 shot, nor the difference between the less and 
 the greater side SO. Thus, to find the shot in an oblong pile, 
 
 the 
 
PILING OF BALLS. S2T 
 
 the loss side being 15, and the greater 35, we are first to 
 find tlie contents of the sqaare pile by means of which the 
 oblong pile may he conceived to be formed ; that is, we are 
 to find the contents of a square pile, whose bottom row is 
 15 shot V whicii being 1240, we are, secondly, to add these 
 1240 to the product 2400 of the triangular number 120, 
 answering to 15, the number expressing the bottom row of the 
 arithmetical triangle, multiplied by 20, the number of those 
 triangles ; and their sum, being 3G40, expresses the number 
 of shot in the proposed oblong pile. 
 
 REMARK II. 
 
 The following algebraical expressions, deduced from the 
 investigations of the sums of the powers of numbers in 
 arithmetical progression, which are seen upon many gunners' 
 callipers*, serve to compute with ease and expedition the shot 
 or shells in any pile. 
 That serving to compute any triangular } n ■{- 2 x » -r ^ X n 
 
 pil«, is represented by ^ 6 
 
 That serving to compute any square ( n -f~ 1 X 2n -}- I X n 
 
 pile, is represented by ( & 
 
 In each of these, the letter n represents the number in the 
 bottom row ; hence, in a triangular pile, the number in the 
 bottom row being 30 ; then this pile will he 30 + 2 X 30 -^ 1 
 X y = 4960 shot or shells. In a square pile, the number 
 in the bottom row being ali^o 30 ; then this pile will be 
 
 30 + 1 X 60 f 1 X Y = ^-^-^-^ shot or shells 
 . That serving to compute any oblong pile, is represented bj 
 
 2n + 1 4-''¥n X n'i^l X ?i 
 
 — , in which the letter n denotes 
 
 * Callipers ar« large compasses, with bowed shank?, serving to take 
 the diameters of convex and concave bodies. Tiie gunners* callipers 
 cons. St of two thin rules op plates, which are moveible quite round 
 a joint, by the plates folding one over ihe other : the length of each 
 rule or piate is 6 inches, the breadth about 1 mch. It is usual to re- 
 prese.u, on the platen, a variety of scales, tables, proportions, &c. 
 such as are otesme I useful to be known by persons employed about 
 artillery ; but except the meaaurm^ of the caliber of shot and cannon, 
 and the measuring of saliant and re-entering an^^les, none of the 'it ti- 
 des, with which the callipers are usually rilled, are essential to that 
 inetrument. 
 
 the 
 
^28 ALGEBRA. 
 
 the number of courses, and the letter m the number of shot, 
 less one, in the top row : hence, in an oblong pile the num- 
 ber of courses being; 30, and the top row 31 ; this pile will 
 be 60^4- 1 +"90 X 30 -f 1 + V = ^3405 shot or shells. 
 
 ^GEOMETRICAL PROPORTION. 
 
 (geometrical Proportion contemplates the relation of 
 quantities considered as to what part or what multiple one is^ 
 of another, or how often one contains, or is contained in, 
 another.- — Of two quantities compared together, the ^rst is 
 called the Antecedent, and the second the Consequent. Their 
 ratio is the quotient which arises from dividing the one by the 
 other. 
 
 Four Quantities are proportional, when the two couplets 
 have equal ratios, or when the first is the same part or mul- 
 tiple of the second, as the third is of the fourth. Thus, 3, 
 6, 4, 8, and a, ar, 6, br, are geometrical proportionals. 
 
 ar br 
 For I = I = 2, and — = — = r. And they are stated thus, 
 
 a b 
 3 : 6 : : 4 : 8, &c. 
 
 Direct Proportion is when the same relation subsists be- 
 tween the first term and the second, as between the third and 
 the fourth : As in the terms above. But reciprocal, or In- 
 verse Proportion, is when one quantity increases in the same 
 proportion, as another diminishes : As in the'ce, 3, 6, 8, 4 ; 
 and these, a, ar, br, b. 
 
 The Quantities are in geometrical progression, or con- 
 tinuous proportion, when every two terms have always the 
 same ratio, or when the first has the same ratio to the second, 
 as the second to the third, and the third to the fourth, &c. 
 Thus, 2, 4, 8, 16, 32, 64, &c. and a, ar, ar^ , ar^,ar'^, ar^ , 
 &c. are series in geometrical progression. 
 
 The most useful part of geometrical proportion is contained 
 in the following theorems ; which are similar to those in 
 Arithmetical Proportion, using multiplication for addition, &c. 
 
 1. When 
 
GEOMETRICAL PROPORTION. 229 
 
 1. When four quantities are in geometrical proportion, the 
 product of the two extremes is equal to the product of the 
 two means. As in these, 3, 6, 4, 8, where 3X8 = 6X4 = 
 24 ; and in these, a, ur, 6, 6r, where a X 6r = ar X 6 = 
 abr. 
 
 2. When four quantities are in geometrical proportion, 
 the product of the means divided by either of the extremes 
 gives the other extreme. Thus, if 3 : 6 : : 4 : 8, then 
 6X4 6X4 
 
 = 8, and = 3 ; also if a : ar i : b : br, then 
 
 3 8 
 abr abr 
 = 6r, or = a. And this is the foundation of the 
 
 a br 
 
 Rule of Three. 
 
 3. If, any continued geometrical progression, the product 
 of the two extremes, and that of any other two terms, equally 
 ^listant from them, are equal to each other, or equal to the 
 square of the middle term when there is an odd number 
 of them. So in the series 1, 2, 4, 8, 16, 32, 64, &c. it is 1 
 X 64 = 2 X 32 = 4 X 16 = 8 X 8 = 64. 
 
 4. In any continued geometrical series, the last term is 
 equal to the first multiplied by such a power of the ratio as 
 is denoted by 1 less than the number of terms. Thus, in the 
 series 3, 6, 12, 24, 48, 96, 4-c. it is 3 X ^^ = 96. 
 
 6. The sum of any series in geometrical progression, is 
 found by multiplying the last term by the ratio, and dividing 
 the difference of this product and the first term by the dif- 
 ference between 1 and the ratio. Thus, the sum of 3, 6, 
 192 X 2-3 
 
 12, 24, 48, 96, 192, is = 384-3 = 381. And 
 
 2—1 
 the sum of n terms of the series, a, ar, ar^^ ar^, ar'^, &c. to 
 ar"~^ X r — a ar^ — a r" — 1 
 
 ar»~^t is — — • = = a. 
 
 r— 1 r— 1 r — 1 
 
 6. When four quantities, o, «r, 6, br, or 2, 6, 4, 12, are 
 proportional ; then any of the following forms of those quan 
 tities are also proportionl, viz. 
 
 1. Directly, a : ar : : b : 6r ; or 2 : 6 : ; 4 : 12. 
 
 2. Inversely, ar : a : : br : b ; or 6 : 2 : : 12 : 4. 
 
 3. Alternately, a : b :: ar : br ; or 2 : 4 : : 6:12. 
 
 4. Com- 
 
S.MO ALGEBRA. 
 
 4. Compoundedly, a : a-\-ar : :b : b-\-br ; or 2 : 8 : : 4 : 16. 
 
 5. Dividedly, a ; ar — a : : 6 : 6r — 6 ; or 2 : 4 : ; 4 : 8. 
 
 e. Mixed,ar4-a : ar — a : : 6r +6 : 6r— 6 ; or 8 : 4 : : 16 : 8. 
 7. Multiplication, ac : arc : : be : 6rc ; or 2.3 : 6.3 : : 4 : 12. 
 
 a ar 
 S. Division, — : — : : fc : fer ; or 1 : 3 : : 4 : 12. 
 
 c c 
 9. The numbers a, 6, c, d, are in harnaonical proportion, 
 when a : d : : a ^ b : cm d \ or when their reciprocal? 
 1111 
 
 — , — , — , — , are in arithmetical proportion. 
 abed 
 
 EXAMPLES. 
 
 1. Given the first term of a geometric series 1, the ratio 
 2, and the number of terms 12 ; to find the sum of the series ? 
 First, 1 X 2» 1 = 1 X 2048, is the last terra. 
 
 2048 X 2 — 1 4096 - 1 
 
 Then = = 4095, the sum required. 
 
 2 ~ 1 1 
 
 2. Given the first term of a geometrical series i, the ratio 
 i, and the number of terms 8 ; to find the sum of the series ? 
 First, i X {\Y = 1 X yig = 3^^, is the last term. 
 Then,a~^i^ X 1) ^(1- i) = (i-^x-)4.^ =f|f X ^ 
 
 = Iff. the sum required. 
 
 3. Required the sum of 12 terms of the series 1, 3, 9, '27, 
 31, &c. Ans. 266720. 
 
 4. Required the sum of 12 terms of the series 1, \,\, ^V^ 
 3V, &c. Ans. fff IH. 
 
 5. Required the sum of 100 terms of the series J, 2, 4, 8, 
 16, 32, &c. Ans. 1267650600228229401496703205376. 
 
 See more of Geometrical Proportion in the Arithmetic, 
 
 SIMPLE EQUATIONS. 
 
 An Equation in the expression of two equal quantities, with 
 the sign of equality (=) placed between them. Thus, 10— 
 4 = 6 is an equation, denoting the equality of the quantities 
 10-4 and 6. 
 
 Equations 
 
SIMPLE EQUATIONS. 23) 
 
 Equations are either simple or compound.* A Simple 
 Equation, is that which contains only one power of the un- 
 known quantity, without including different powers. Thus, 
 j; — a = 6 -f" c> or ax^ = 6, is a simple equation containing 
 only one power of the unknown quantity x. But x^ — 2a3' 
 aaai 6^ li a compouud one. 
 
 GENERAL RULE. 
 
 Reduction of Equations, is the finding the value of the 
 unknown quantity. And this consists in disengaging that quan- 
 tity from the known ones ; or in ordering the equation so, that 
 the unknown letter or quantity may standalone on one side of 
 the equation, or of the mark^ of equality, without a co-effi- 
 cient ; and all the rest, or the known quantities, on the other 
 side.-T-In general, the unknown quantity is disengaged from 
 the known ones, by performing always the reverse operations. 
 So if the known quantities are connected with it by -{-or ad- 
 dition, they must be subtracted ; if by minus ( -), or subtrac- 
 tion, they must be added ; if by multiplication, we must divide 
 by them ; if by division, we must multiply ; when it is in any 
 power, we must extract the root ; and when in any radical, 
 we must raise it to the power. As in the following particular 
 rules ; which are founded on the general principle of perform- 
 ing equal operations on equal quantities ; in which case it 
 is evident that the results must still be equal, whether by 
 equal additions, or subtractions, or multiplications, or divisions, 
 or roots, or powers. 
 
 PARTICULAR RULE L 
 
 When known quantities are connected with the unknown 
 by ^ or — ; transpose them to the other side of the equation, 
 and change their signs. Which is only adding or subtracting 
 the same quantities on both sides, in order to get all the un- 
 known terms on one side of the equation, and all the known 
 ones on the other side*. 
 
 Thus, 
 
 * Here it is earnestly recommended that the pupil be accustom-, 
 ed, at every line or step »n the reduction of the equations, to 
 name the paiticiilar operation to be performed on the equation 
 in the Ime, in order to pr<'duce the next form or state of the equa- 
 tion, tn applying each of these rules, according as the particular 
 fornos of the equatipn may require ; applying them according to the 
 
 order 
 
332 ALGEBRA. 
 
 Thus, if a:-j-5 = ^ » then transposing 5 gives x = 8 — 3 = 3. 
 And, ifa;'-3-{-7=9; then transposing the 3, and 7, gives 
 
 x—9 + 3-7=6. 
 Also, '\i x—a -\- b = cd : then by transposing a and h, 
 
 it is a: = a — 6 -|- cd. 
 In like manner, if 6x — 6 = 4a: 4- 10, then by transposing 
 6 and 4a:, it is 6a; — 4j: = 10 -|- 6, or a; = IQ. 
 
 RULE II. 
 
 When the unknown term is multiplied by any quantity- ; 
 divide all the terms of the equation by it. 
 
 Thus, if ao; = ab — 4a ; then dividing by a, gives a;=6 — 4. 
 And, if 3:c -|- 6 = 20 ; then first transposing 5 gives 2>x 
 = 16 ; and then by dividing by ^, it is x =6. 
 In like manner, if ax-\-^ab = 4c^ ; then by dividing by a, it 
 4c2 4c2 
 
 is x-{-3b = ; and thfen transposing 36, gives x= — '■ — 3b, 
 
 a a 
 
 RULE m. 
 
 When the unknown term is divided by any quantity ; we 
 must then multiply all the terms of the equation by that divi- 
 sor ; which takes it away. 
 
 X 
 
 Thus, if- = 34-2 : then mult, by 4, gives x = 12+8=20. 
 4 
 
 X 
 
 And, if - = 36 + 2c ^ d : 
 a 
 then by mult, a, it gives x = Sab + 2ac — ad. 
 3x 
 
 Also, if 3 = 5 + 2; 
 
 6 
 Then by transposing 3, it is |j7 = 10. 
 And multiplying by 6, it is 3x = 60. 
 Lastly dividing by 3 gives x = 16f . 
 
 •rder in which they are here placed ; and beginning every Lne with 
 the words Then iy, as in the following specimens of Examples; 
 which two words will always bring -to his recollection, tht,t be is to 
 pronuuncf- what particular operation he is to perform on the last line, 
 in order to gi ve the next j allotting always a single line for each ope- 
 ration, and ranging the equations neat'y just under each other, in the 
 several lines, as they are successively produced. 
 
 RULE 
 
SIMPLE EQUATIONS. 233 
 
 RUU5 IV. 
 
 Whew the unknown quantity is included in any root op 
 surd ; transpose the rest of the terms, if there be any, by 
 Rule I ; then raise earh side to such a power as is denoted by 
 the index of the surd ; viz. square each side when it is the 
 square root ; cube each side when it is the cube root ; &c. 
 which clears that radical. 
 
 Thus, if ^j;— 3 = 4 ; then transposing 3, gives ^x = 7 ; 
 
 And squaring both sides gives x = 49. 
 
 And, if y/2x -1-10 = 8: 
 
 Then by squaring, it becomes 2x -}- \0 = 64 ; 
 And by transposing 10, it is So: = 54 ; 
 Lastly, dividing by 2, gives x = 27. 
 
 Also, if 1/ 3x-f4 -{-3 = 6: 
 
 Then by transposing 3, it is ^ 3a: -{- 4 =: 3 ; 
 And by cubing, it is 3ar + 4 = 27 ; 
 Also, by transposing 4, it is 3v = 23 ; 
 Lastly, dividing by 3, gives a: = 7|. 
 
 RULE V. 
 
 When that side of the equation which contains the un* 
 knovvn quantity is a complete power, or can easily be reduced 
 to one, by rule 1, 2, or 3 : then extract the foot of the said 
 power on both sides of the equation ; that is, extract the 
 square root when it is a square power, or the cube root when 
 it is a cube, &c. 
 
 Thus, if a;2 -(- 8a: + 16 = 36, or (x -f- 4)2 = 36 : 
 
 Then by extracting the roots, it is x -j- 4 = 6 ; 
 
 And by transposing 4, it is a; s= 6 — 4 = 2. 
 Andif3r2_19 = 21-h35. 
 
 Then, by transposing 19, it is 3x^ = 75 ; 
 
 And dividing by 3, gives x'^ =25 ; 
 
 And extracting the root, gives a; = 5. 
 Also, if5a;2— 6 = 24. 
 
 Then transposing 6, gives f a:^ = 30 ; 
 
 And multiplying by 4, gives 3x- = 120 ; 
 
 Then dividing by 3, gives a:^ = 40 ; 
 
 Lastly, extracting the root, gives a; = -^Z 40 =: 6-324555. 
 
 Vol. I. 31 
 
 RULE 
 
S34 ALGEBRA. 
 
 RULE VI. 
 
 When there is any analogy or proportion, it is te be 
 changed into an equation, by multiplying the two extreme 
 terms together, and the two means together, and making the 
 One product equal to the other. 
 Thus, if 2j: : 9 : : 3 : 5. 
 Then, mult, the extremes and means, gives lOx = 27 ; 
 And dividing by 10, gives «.== 2^^. 
 
 And if ^x : a : : 5b : 2c. 
 Then mult, extremes and means gives fca: = bab ; 
 And multiplying by 2, gives 3cx = lOab ; 
 
 lOab 
 
 Lastly, dividing by 3c, gives x = . 
 
 3c 
 Also, if 10— a; : fa; : : 3 : 1. 
 Then mult, extremes and means, gives 10— a: = 2x j 
 And transposing x, gives 10 = 3.r ; 
 Lastly, dividing by 3, gives 3^ = x. 
 
 RULE vn. 
 
 When the same quantity is found on both sides of art 
 equation, with the same sign, either plus or minus, it may be 
 left out of both : and when every term in an equation is either 
 multiplied or divided by the same quantity it may be struck 
 out of them all. 
 
 Thus, if 3a: 4- 2a = 2a -f 6 : 
 
 Then by taking away 2a, it is 3a; = 5. 
 
 And, dividing by 3, it is a; = ^b. 
 Also if there be 4aa; -f 6ab — lac. 
 
 Then striking out or dividing by a, gives 4.r 4' 66 = Ic. 
 
 Then, by transposing 66, it becomes 4j7 = 7c — 66 ; 
 
 And then dividing by 4 gives a; = |c — |6. 
 
 Again, if fa; - | = V — h 
 Then, taking away the |, it becomes fa; = y ; 
 And taking away the 3's, it is 2jr = 10 ; 
 Lastly, dividing by 2 gives j: = 5. 
 
 MISCELLANEOUS EXAMPLES. 
 
 1. Given 7a; - 18 = 4a: -f- 6 ; to find the value of x. 
 First, transposing 18 and 5x gives 3a; = 24 ; 
 Then dividing by 3, gives x == 8. 
 
 2. Givcm 
 
SIMPLE EQUATIONS. 235 
 
 2. Given ?0 — 4j? — 12 = 92— lOx ; to find x. 
 First transposing 20 and 12 and lUa:, gives 6j? = 84 ; 
 Then dividing by 6, gives x = 14. 
 
 3. Let 4«x— 56 = 3rfx -f- 2c be given ; to find x. 
 
 First, by trans. 66 and 3(ix, it is 4ax — 3dar = 56 + 2c : 
 
 56 4- 2c 
 
 Then dividing by 4a •— 3c?, gives x = — ^. 
 
 4a — Ji 
 
 4. Let 5jr2_l2jr: = 9j; -{- 2^2 be given ; to find x. 
 First, by dividing by j?, it is 5-c — 12 = 9 + 2-c j 
 Then transposing 12 and 2x, gives 3j: = 21 ; 
 Lastly, dividing by 3, gives x = 7. 
 
 5. Given 9ax^ — 15a6a:2 = 6ax^ -{- 12a:c2 . to find x. 
 First, dividing by 3ax^ , gives 3jc — 56 = 2a; -h 4 ; 
 Then transposing 56 and 2a;, gives x =- Sb -{- 4, 
 
 ^. Let 1 = 2 be given, to find x. 
 
 3 4 5 
 First, multiplying by 3, gives r — ^x -j- fa: = 6 , 
 Then multiplying by 4, gives x + */a; = 24. 
 Also multiplying by 5, gives 11 x = 120. 
 Lastly, dividing by 17, gives x = lj\, 
 
 x—B X X— 10 
 
 7. Given -] =12 ; to find x. 
 
 3 2 3 
 
 First, mult, by 3, gives jr-5 -f fjr = 36 — :r + 10 
 Then traasposing 5 and x, gives 2x -f- f ^ = 61 ; 
 And multiplying by 2, gives 7a; = 102. 
 Lastly, dividing by 7, gives x = 14|. 
 
 3a; 
 
 8. Let ^ h 7 = 10, be given ; to find x. 
 
 First, transposing 7, gives y^f a? = 3 ; r 
 
 Then squaring the equation, gives f x = 9 ; 
 Then dividing by 3, gives ^x = 3 ; 
 Lastly, multiplying by 4, gives x = 12. 
 
 6a= 
 
 9. Let 2x 4- 2 V«^-f -ra = :;/"-2'q:^i-' be given ; to findx. 
 
 First, mult, by ^a^ -{-a:- , gives 2x ^ a^ -{-x^ ■i-2a2 -\-2x^ 
 = 5a2. 
 
 Then trans. 2a3 and 2x^, gives 2x y/a^ -i-ar^ = Sa^ -.2a:« ; 
 
 Then 
 
Vse 
 
 ALGEBRA. 
 
 Then by squaring, it is 4*^ X a^+x- = 3a2 ^^a-*^* ; 
 
 That is, 4a2j^2 ^ ^j^a = 9a* ~ VZa^x^ + 4a-* ; 
 
 By taking 4x* from both sides, itis 402^^2=9^4 ^ i2a^x^; 
 
 Then transposing 12a2a;3, gives 16a2a;2, = 9a* ; 
 
 Dividing by a^ , gives 16a:2 = 9a2 ; 
 
 And dividing by 16, gives x^ = J^a^ ; 
 
 Lastly extracting the root, gives a: = |a. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Given 2x — 6-4-16 = 21 ; to find or. Ans. jr = 6 
 
 2. Given 9a; — 15 = x 4- 6 : to find x, Ans. x = 2|* 
 
 3. Given 8— 3a;-|-l2=30 — 6a:-f 4 ; to find x. Ans. a; = 7, 
 
 4. Given x + j-^— 4^ = 13 ; to find x. Ans. x = 12. 
 
 5. Given 3a: + \x-\- 2 = 5jc— 4 ; to find x. Ans. a: = 4. 
 
 6. Given 4aa7 + ^a — 2 = aa; — bx ; to find ar. 
 
 6— a 
 
 Ans. X = 
 
 Sa+36 
 
 7. Given ^x-^^x -{- la? = i ; to find x. Ans. a? = f^. 
 
 8. Given ^ 4 -f a: == 4 — ^a: ; to find x. Ans. ar = 2i, 
 
 a72 
 
 9. Given 4a -{-,x = ; to deter, x, Ans x =--2a, 
 
 4a+ar 
 
 10. Given y/4a^ + ar2 = */ 46* + x^ j to find a;. 
 
 6*~-4a* 
 
 Ans. X z= ^ « 
 
 2a2 
 
 4a__ 
 
 11. Given ^ x -{- ^ 2a -{- x =^2^ • ^ ; to find x. 
 
 Ans. X = |a. 
 a a 
 
 12. Given { = 26 ; to find a;. 
 
 1 +2a: 1 — 2a: 
 
 6-« 
 
 Ans. x=i^ . 
 
 6 
 
 13. Given o -f- a: = ^a^-\-x>^ 46a ^ ^.a . to findx. 
 
 Ans. X = ^ «. 
 
 « a 
 
 OF 
 
SIMPLE EQU^TIONTS. J37 
 
 ©F REDUCING DOUBLE, 1 RIPLF. &c. EQUATIONS, COVTAININ© 
 TWO, THREE, OR MO^tE UNKNOWN QUANTITIES. 
 
 PROBLEM I. 
 
 To Exterminate Two Unknown Qnantities ; Or^ to Reduce the 
 Th}0 Simple Equations containing the/n, to a Single one, 
 
 RULE L 
 
 Find the value of one of the unknovvn letters, in terms of 
 the other quantities, in each of the equations, hy the methods 
 already explained. Then put those two values equal to each 
 other for a new equatiorr, with only one unknown quantity in 
 it, whose value is to be found as before. 
 
 JVote. It is evident that we must first begin to find the 
 values of that letter which are easiest to be found in the two 
 proposed equations. 
 
 EXAMPLES. 
 
 1. Given j5^J:2y=uS' *"^°^^^"^^- 
 
 In the 1st equat. transp. 3y and div. by 2, gives x = ; 
 
 2 
 14 + 22/ 
 
 In the 2d transp. 2y and div. by 5, gives x = ; 
 
 5 
 14 + 23^ ll-Sy 
 
 Putting tlifese two values eaual, gives = ; 
 
 5 2 
 
 Then mtilt. by 6 and 2, gives 28 -{- 4i/ = 85 — 15^/ ; 
 Transposing 28 and loy, gives 19t/ = &7 ; 
 And dividing by 19, gives y = 3. 
 And hence x = 4. 
 Or, to do the s^me by finding two values of y, thus : 
 
 17-2a: 
 
 In the 1st equat. tr. 2a; and div. by 3, gives y = ; 
 
 3 
 Sr— 14 
 
 In the 2d tr. Zy and 14, and div. by 2, gives y = — ; 
 
 2 
 6ar— 14 17 -2x 
 
 Putting these two values equal, gives = ; 
 
 2 3 
 
 Mult, by 2 and by 3, gives 16ar-42 = 34-4x ; 
 
 Transp. 
 
2Sa ALGEBRA. 
 
 Transp. 42 and 4a:, gives lOar = 76 ; 
 
 Dividing by 19, gives a: = 4. ' . 
 
 Hence y = 3, as before. 
 
 2. Given JpH^r^J; to find a; and j/. 
 
 Ans. x = a -}- b, and y = la -^ i j*. 
 
 3. Given 3x -\- y =^ 22, and 3i/ -{- a; = 18 ; to find x and y. 
 
 Ans. X = 6i and y = 4. 
 
 4. Given 5 f T i H q, J ; to find ^ and y. 
 
 i 3"^ "T 32/ — y 3 > 
 
 Ans. J? = 6, and y = 3, 
 2a7 31/ 22 3jc 22/ 67 
 
 k. Given {- — . = — , and f* — = — ; to find x andy. 
 
 3 5 5 5 3 15 
 
 Ans. a; = 3, and y = 4. 
 
 6. Given or + ^y = s, and jt^ — 41^2 = c/2 . ^q find a: and y. 
 
 s2 -I- t^a s3 _ d^ 
 
 Ans. J? = , and y =■ . 
 
 25 4s 
 
 7. Givea^F — 2y = d, and x : t/ : : a : 6 ; to find x and y. 
 
 ad bd 
 
 Ans. :r = , and y = . 
 
 a — 26 a — 26 
 
 RULE n. 
 
 Find the value of one of the unknown letters, in only one 
 ef the equations, as in the former rule ; and substitute this 
 value instead of that unknown quantity in the other equation, 
 and there will arise a new equation, with only one unknown 
 quantity, whose value is to be found as before. 
 
 jsfote. It is evident that it is best to begin first with that 
 letter whose value is easiest found in the given equations. 
 
 EXAMPLES. ^ 
 
 1. Given I ?^ _ 1^ Z \l I ; to find x and y. 
 
 This will admit of four ways of solution ; thus : First, 
 
 17-32/ 
 
 In the 1st eq. trans. 3y and div. by 2, gives x == ; 
 
 2 
 85—162/ 
 This val. subs, for x in the 2d, gives, ■ 2y=14 ; 
 
 Mult, by 2, this becomes 8& — 16y — . 41/ = 28 j 
 
 Transp. 
 
SIMPLE EQUATIONS. 23? 
 
 Transp. I6y and 4y and 28, gives 67 = 1% ; 
 And dividing by 19, gives 3 = t/. 
 .17— 3y 
 
 Then x == = 4. 
 
 2 
 
 14-f2y 
 
 2dly, in the 2d trans. 2y and div. by 6, gives x = — ; 
 
 5 
 28-f4^ 
 
 This subst. for x in the 1st, gives {- 3y = 17 ; 
 
 6 
 Mult, by 6, gives 28 + 4y -f ISy = 85 j 
 Transpos. 28. gives 19y = 57 ; 
 And dividing by 1 9, gives y = 3. 
 144-21/ 
 
 Then x = = 4, as before. 
 
 5 
 
 17-2a: 
 
 3dly, in the 1st trans. 2x and div. by 3, gives y = ; 
 
 3 
 34— 4a; 
 
 This subst for y in the 2d, gives, 5a? = 14 ; 
 
 3 
 Multiplying by 3 gives 16a: — M -|- 4* = 42 ; 
 Transposing 34, gives 19a; = 76 ; 
 And dividing by 1 9, gives a: = 4. 
 17— 2x 
 
 Hence y = = 3, as before. 
 
 3 
 
 6a;- 14 
 
 4thly, in the 2d tr. 2y and 14 and div. by 2, gives y = . <• 
 
 2 
 16a; - 42 
 
 This substituted in the 1st, gives 2x -| = 17 ; 
 
 2 
 Multiplying by 2, gives 1 9a;- 42 = 34 ; 
 Transposing 42, gives 19a; = 76 ; 
 And dividing by 19, gives x = 4. 
 oa:- 14 
 
 Hence y = = 3, as before. 
 
 2 
 
 2. Given 2a; + 3y = 29, and 3jr - 2^/ = 11 ; to find x and y, 
 
 Ans. a? = 7, and y =^ 5. 
 
 3. Given l^ly^ '2^; tofinda;andy. 
 
 Ans. X = 8, and y = 6. 
 4. Given 
 
240 ALGEBRA. 
 
 4. Given } ^^ ^ ' ' L_' oq J J ^o find x and y. 
 
 Ans. X = 6, and y = 4. 
 
 X y 
 
 5. Given f- 3y = 21, and h3x = 29 ; to find x and y, 
 
 3 3 
 
 Ans. a; =± 9, and y = 6. 
 
 •^ y ^ — y X 
 
 6. Given 10 = ^ ^ 4, and 1 2 = 
 
 2 3 2 4 
 32/— x 
 ■--! ; to find X and y. Ans. x= S, dmdy = 6. 
 
 7. Given x : y : : 4 : 3, and x^ -^ y^ = SI ; to find x and y. 
 
 Ans. a; = 4, and y = 3> 
 
 RULE TIL 
 
 Let the given equations be so multiplied, or divided, &c. 
 and by such numbers or quantities, as will make the terms 
 which contain one of the unknown quantities the same in 
 both equations ; if they are not the sanie when first pro- 
 posed. 
 
 Then by adding or subtracting the equations, according as 
 the sines may require, there will remain a new equation, with 
 only one unknown quantity, as before. That is, add the two 
 equations, when the sines are unlike, but subtract them when 
 the signs are alike, to cancel that common term. 
 
 JV()ie. To make two unequal terms become equal, as above, 
 multiply each term by the co-efficient of the other. 
 
 EXAMPLES. 
 
 «-«°S2^ + 5 = 16?"°''''^"'°"'^- 
 Here we may either make the two first terms, containing 
 •c, equal, or the two 2d terms, containing y, equal. To make 
 the two first terms equal, we must multiply the 1st equation 
 by 2, and the 2d by 5 ; but to make the two 2d terms equel, 
 we must muhiply the 1st equation by 5, and the 2d by 3 ; as 
 follows* 
 
 l.By, 
 
SIMPLE EQUATIONS. 241 
 
 1. By making the two first terms equal : 
 
 Mult, the 1st equ. Wy 2, gives lOx — 6y = 18 
 And mult the 2d by 5, gives IOjt -f- 2by ="80 
 
 Subtr. the upper from tHe under, gives Sly = 62 
 And dividing by 31, gives y =■ ^. 
 
 9 + 3y 
 
 Hence, from the 1st given equ. x = = 3. 
 
 5 
 
 2. By making the two 2d terms equal : 
 Mult, the 1st equat. by 5, gives 2bx — {by = 4b j 
 And mult, the 2d by 3, gives 6x -{■ 15r/ = 48 ; 
 Adding these two, gives Six = 93 ; 
 And dividing by 31 , gives x = S ; 
 
 bx — 9 
 Hence, from the 1st equ. y = = 2. 
 
 MISCELLANEOUS EXAMPLES. 
 
 ^4-8 y-\-6 
 
 1. Given |-%=21,and — '■ h Bar = 23 ; to find x 
 
 4 3 
 
 and y, Ans. x = 4, and y = S. 
 
 Sx — y Sy -i- X 
 
 2. Given h 10= 13, and \~ b == 12; to find 
 
 4 2 
 
 X and y. Ans. a: = 5, and y = S. 
 
 Sx -{' 4y X 6x — 2?/ y 
 
 3. Given 1 = It), and -|-_-.= i4;to 
 
 5 4 3 6 
 
 find X and y. Ans :c = 8, and y = 4. 
 
 4. Given Sir + 4y = 38, and 4x — Sy = 9 ; to find x and y. 
 
 Ans. X = 6, and y = b, 
 
 PROBLEM XL 
 
 To Exterminate Three or More Unknown Quantities ; Or, to 
 Reduce the Simple Equations, containing them, to a Single 
 one. 
 
 RULE. 
 
 This may be done by any of the three methods in the last 
 problem : viz. 
 
 1. After the manner of the first rule in the last problem, 
 find the value of one of the unknown letters in each of the 
 given equations : next put two of these values equal to each 
 other, and then one of these and a third value equal, and so 
 on for all the values of it ; which gives a new set of equations, 
 
 Vol. 1. 32 with 
 
242 ALGEBRA. 
 
 with which the same process is to be repeated, and so oa til! 
 there is.onl}^ one equation, to he reduced by the rules for a 
 single equation. 
 
 2. Or, as in the 2d rule of the same problem, find the value 
 of one of the unknown quantities in one of the equations 
 only ; then substitute this value instead of it in the other 
 equations ; which gives a new set of equations to be resolved 
 as before, by repeating the operation. 
 
 3. Or, as in the 3d rule, reduce the equations, by multiply- 
 ing or dividing them, so as to make some of the terms to agree : 
 then, by adding or subtracting them, as the signs may require, 
 one of the letters may be exterminated, &c. as before. 
 
 EXAMPLES. 
 
 2/4- ^= 9) 
 1. Given < x -}- 2y + 32- = 16 > ; to find x, y, and z. 
 ■-3^4-4^ = 31^ 
 
 
 1. By the 1st method : 
 Transp. the terms containing t/ and z in each equa. gives 
 X =■ 9 — y — ^, 
 jr= 16 — 2y — 3z, 
 a7 = 2I — 3y— 42-; 
 Then putting the 1st and 2d values equal, and the 2d and 34 
 values equal, give 
 
 9— 2/— 2r= 16 — 2y— 3z, 
 16 — 22/ — 32 = 21 - 3^ — 42 ; 
 lu the I St trans. 9, z, and 2y, gives y = 7 — 2^ ; 
 1b the 2d trans. 16, 32 and 3y, gives y = 6 — z \ 
 Putting these two equal, gives 3 — 2r = 7 — 2^ ; 
 Trans. 5 and 22, gives z =«: 2. 
 Hence ?/= 6 — 2 = 3, and 37 = 9 — ^ — ^ = 4. 
 
 ' 2dly. By the 2d method : 
 
 From the 1st equa. ^ = 9 —y — ^ \ 
 This value of x substit. in the 2d and 3d, gives 
 9 -f y -f- 2z = 16, 
 9 -f 2y 4- 3^ = 21 ; 
 In the 1st trans. 9 and 22, gives y = 7 — 22 ; 
 This substit. in the last, gives 23 — 2- = 21 j 
 Trans, z and 21, gives 2 = 2. 
 Hence again y = 7 — 22 = 3, and jc = 9 — y — 2r = 4. 
 
 3dly. By 
 
SIMPLE EqUATIONfi. 243 
 
 3dly. By the 3d method ; subtracting the 1st equ. from 
 the 2d, and the 2d from the 3d, gives 
 
 y -f 2^ = 7, 
 y -\- 2 = 5; 
 Subtr. the latter from the former, gives 2 = 2. 
 Hence y = 6 — 2r .= 3, and re = 9 -— y — z = 4. 
 
 { X -h3y -{- 2^ = 38) ; to 
 
 i ^ + ^y + i^= 1^') 
 
 Ans. X 
 
 i ^ + i3/ + i^=27) 
 Given { :r-hJ-2/ + 1^ = 20} 5 
 ( ^ + i2/ + i^= J^) 
 
 2. Given < jp -j- 3y 4- 2^ = 38 ) ; to find ar, y, and 2. 
 10 5 
 Ans. X = 4f y = 6, 2 = 
 
 to tind x, y, and ^. 
 Ans. i: = 1,^ = 20, 2" = 60. 
 
 4. Given x — y = 2, x — 2 = 3, and y — 2 = 1; to 
 find or, y, and z. Ans. J7 = 7; y = 6; 2:=, 4. 
 
 ^ 2^ -{- 3t/ 4- 4^ = 34 ^ 
 
 5. Given ? 3x -j- >?/ -f 62- = 46 J ; to find a:, y, and z. 
 
 ( 4x + 5y 4- 6^ = 68 ) 
 
 A COLLECTION OF QUESTIONS PRODUCING SIMPLE 
 EQUATIONS. 
 
 Quest. 1. To find two nambers, such, that their sum shall 
 be 10, and their diflference .6. 
 
 Let X denote the greater number, and y the less*. 
 Then, by the 1st condition j: -f- y — 10, 
 And by the 2d - - x ^— y = 6^ 
 Transp. y in each, gives a: = 10 — y, 
 and jr = 6 -f- y ; 
 Put these two values equal, gives 6 -{- y = 10 — y ; 
 Transpos. 6 and — y, gives - 2y = 4 ; 
 Dividing by 2, gives - - y == 2. 
 
 And hence - - - x =x. 6 -h y = ^' 
 
 * In all these solutions, as many unknown letters are always used 
 as there are unknown numbers to be found, purposely the better to 
 exercise the modes of reducing the equations : avoidinj^ the short 
 ways of notation, which though giving a shorter solution, are for that 
 reason less useful to the pupi!. as affording less exercise in practising 
 the several rules in reducing equations. 
 
 Quest. 2. 
 
244 ALGEBRA. 
 
 Quest. 2. Divide 100/. among a, b, c, so that a may 
 have 20/. more than b, and b 10/. more than c. , 
 
 Let X = a's share, y = b's, and c = c's. 
 Then :x: -^ y -\- z = 100, 
 a: = 2/ 4- ^0, 
 >y ^ 2 ~\- 10. 
 In the 1st substit. y -{- 20 for :r, gives 2^/ -f z + 20 = 100 ; 
 In this substituting z -\- 10 for y, gives 3z -f- 40 = 100 y 
 By transposing 40, gives - '- ' 32 =* GO ; 
 And dividing by 3, gives - - z = 20 
 
 Hence y = z -\- 10 = 30, and :c = ^ + 20 = 60. 
 
 Qbest. 3. A prize of 600/. is to be divided between two 
 persons, so as their shares may be in proportion as 7 to 8 ; < 
 required the share of each. 
 
 Put X and y for the two shares ; then by the question, 
 
 t : S : : j^ : y, or mult, the extremes 
 and the means, 7j/ = 8x, 
 ^ and X -\- y = 600 ; 
 
 Transposing y, gives x =■ 500 — y ; 
 This substituted in the 1st, gives ly ="4000— Sy ; 
 By transposiujg By, it h lay = 4000 ; 
 By dividing by 15, it gives y = 266| ; 
 And hence j: = 600 — j/ = 233i. 
 
 Quest. 4. What number is that whose 4th part e^iceeds 
 its 6th part by 10 ? 
 
 Let X denote the number sought. 
 Then by the question ^x — J-o: = 10 ; 
 By mult, by 4, it becomes x — ijc = 40 ; 
 By mult, by 6 it gives x = 200," the number sought. 
 
 Quest. 5. What fraction is that to the numerator of 
 which if 1 be added, the value will be | ; but if one be added 
 to the denominator, its value will be J- ? 
 
 x 
 Let — denote the fraction. 
 
 y 
 
 . X -{-' 1 -^x 
 
 Then by the quest. = ^, and = i. 
 
 y ^ 2/ + 1 
 
 The 1st mult, by 2 and y, gives 2x -\- 2 =^ y ; 
 The 2d mult, by 3 and ?/ -f 1 is 3j7 == y + 1 ; 
 The upper taken from the under leaves x — 2=1; 
 By trans pos. 2, it gives x = 3. 
 And hence 2/ = 2a: -|- 2 = 8 ; and the fraction is |. 
 
 Quest. 6., 
 
SIMPLE EQUATIONS. 245 
 
 QuBST. 6. A labourer engaged to serve for 30 days on 
 'those conditions : that for every day he worked, he was to 
 receive 20d^ but for every day he played, or was absent, he 
 was to forfeit lOd. Now at the end of the ime he had to 
 receive just 20 shillings, or 240 pence. It is required to find 
 how many days he worked, and how many he was idle ? 
 
 Let jc be the days worked, and y the days idle. 
 Then 20x is the pence earned, and lOy the forfeits ; 
 Hence, by the question - x -\^ y = 30y 
 
 andSOx— 101/ = 240; 
 The 1st mult, by 10, gives Wx -f Wy = 300 ; 
 > These two added give - 30x = 640 ; 
 This div. by 30, gives - or = 18, the days worked ; 
 Hence - 2/ = 30 — x = 12, the days idled. 
 
 Quest. 7. Out of a cask of wine, which had leaked away i, 
 30 gallons tvere drawn ; and then, being gaged, it appeared 
 to be halt full ; how much did it hold ? 
 
 Let it be supposed to have held x gallons. 
 Then it would have leaked ^x gallons, 
 Conseq. there had been taken away ^x -\- 30 gallons. 
 Hence ir = ix + 30 by the question. 
 , Then mult, by 4, gives 2x = :¥ -{- 120 ; 
 And transposing x, gives x = 120 the contents. 
 
 Quest. 8. To divide 20 into two such parts, that 3 times 
 the one part added to 5 times the other may make 76. 
 
 Let X and y denote the two parts. 
 Then by the question ^ - j7 -f- ^ = 20, 
 and 3x 4- 52/ = 76. 
 Mult, the 1st by 3, gives - - 3x -j- 3t/ = 60 ; 
 Subtr. the latter from the former, gives 2y = 16 ; 
 And dividing by 2, gives - - - y z= Q. 
 
 Hence, from the 1st, - x=20 — y = 12. 
 
 Quest. 9. A market woman bought in a certain number of 
 eggs at 2 a penny, and as many more at 3 a. penny, and sold 
 them all out again at the rate of 5 for two-pence, and by so 
 doing; contrary to expectation, found she lost 3d. ; what num- 
 ber of eggs had she ? 
 
 Let X = number of eggs of each sort. 
 Then will |x = cost of the first sort, 
 And ^x = cost of the second sort j 
 
 But 
 
246 ALGEBRA. 
 
 But 5 : 2 : : 2jr (the whole numher of eggs) : ^x ; 
 Hence f x = price of both sorts, at 6 for 2 pence ; 
 Then by the question ^x + i-^ — f^ — 3 ; 
 Mult, by 2, gives - x + ^x — |a7 = 6 ; 
 And mult, by 3, gives 6.r — 2_4^ = jg . 
 Also mult, by 5, gives x = 90, the number of egg» of 
 each sort. 
 
 Quest. 10. Tvro persons, a and b, engage at play. Before 
 they begin, a has 80 guineas, and b has 60. After a certain 
 number of games won and lost between them, a rises with 
 three times as many guineas as b. Query, bow many guineas 
 did a win of b ? 
 
 Let X denote the number of guineas a won. 
 Then a rises with 80 -{- x. 
 And B rises with 60 — x ; / 
 Theref ,by the quest. 80 + ^ = 180 — 3a: ; 
 Transp. 80 and 3jr, gives 4jr = 100 ; 
 And dividing by 4, gives x = 25, the guineas won. 
 
 QUESTIONS FOR PRACTICE. 
 
 1. To determine two numbers such, that their difference 
 may be 4, and the difference of their squares 64. 
 
 Ans. 6 and 10. 
 
 2. To find two numbers with these conditions, viz. that 
 half the first with a 3d part of the second may make 9, and 
 that a 4th part of the first with a fifth part of the second may 
 make 5. Ans. 8 and 15. 
 
 3. To divide the number 20 into two such parts, that a 3d 
 of the one part added to a fifth of the other, may make 6. 
 
 Ans. 15 and 5. 
 
 4. To find three nurnbers such, that the sum of the 1st and 
 2d shall be 7, the sum of the 1st and 3d 8, and the sum of the 
 2d and 3d 9. Ans. 3, 4, 6. 
 
 6. A father, dying, bequeathed his fortune, which was 
 2800L to his son and daughter, in this manner ; that for every 
 half crown the son might have, the daughter was to have a 
 shiUing. What then were their two. shares ? 
 
 Ans. The son 2000/. and the daughter 800/ 
 
 6. Three persons, a, b, c, make a joint contribution, 
 which in the whole amounts t© 400Z. : of which sum b con- 
 
 tributeg 
 
SIMPLE EQjUATIONS. 247 
 
 tributes twice as much as a and 20/. more ; and c as much as 
 k and B together. What sum did each contribute ? 
 
 Ans. A 60/. B 140/. and c 200/. 
 
 7. A person paid a bill of 100/, with half guineas and crowns, 
 using in all 202 pieces ; how many pieces were there of each 
 sort ? Ans. 180 half guineas, and 22 crowns. 
 
 8. Says a to b, if you give me 10 guineas of your money, 
 I shall then have twic^ as much as you will have left ; but 
 says B to A, give me 10 of your guineas, and then I shall have 3 
 times as many as you. How many had each ? 
 
 Ans. A 22, B 26. 
 
 9. A person goes to a tavern with a certain quantity of 
 money in his pocket, wh*^re he spends 2 shillings ; he then 
 borrows as much money as he had left, and going to another 
 tavern, he there spend 2 shillings also ; then borrowing again 
 as much money as was left, he went to a third tavern, 
 where likewise he spent 2 shillings ; and thus repeating the 
 same at a fourth tavern, he then had nothing remaining. What 
 sum had he at ^rst ? Ans. 3s. 9d, 
 
 10. A man with his wife and child dine together at an inn. 
 The landlord charged 1 shilling for the child ; and for the 
 woman he charged as much as for the child and i as much as 
 for the man ; and for the man he charged as much as for the 
 woman and child together. How much was that for each ? 
 
 Ans. The woman 20d. and the man 32c/. 
 
 11. A cask, which held 60 gallons, was filled with a mix- 
 ture of brandy, wine, and cyder, in this manner, viz. the 
 cyder was 6 gallons more than the brandy, and the wine was 
 as much as the cyder and } of the brandy. How much was 
 there of each. Ans. Brandy 15, cyder 21, wine 24. 
 
 12. A general, disposing his army into a square form, finds 
 that he has 284 men more than a perfect square ; but increas- 
 ing the side by 1 man, he then wants 25 men to be a complete 
 square. Then how many men had he under his command ? 
 
 Ans. 24000. 
 
 13. What number is that, to which if 3, 5, and 8, be 
 severally added, the three sums shall be in geometrical pro- 
 gression ? Ans. 1. 
 
 14 The stock of three traders amounted to 860/. the 
 shares of the first and second exceeded that of the third 
 
 by 
 
248 ALGEBRA. 
 
 by 240 ; and the sum of the Sfd and 3d exceeded the 6rst by 
 360. What was the share of each ? 
 
 Ans. The 1st 200, the 2d 300, the 3d 260. 
 
 15. What two numbers are those, which, being iij the ratio 
 of 3 to 4, their product is equal to 12 times thojraum ? 
 
 Ans. 21 and 28. 
 
 16. A certain company at a tavern, when they came to 
 settle their reckoning, found that had there beeil 4 more in 
 company, they might have paid a shilling a-piece less than 
 they did ; but that if there had been 3 fewer in company, they 
 must have paid a shilling a-piece more than they did. What 
 thenvvas the number of persons in company, what each paid, 
 and what was the whole reckoning ? 
 
 Ans. 24 persons, each paid Is, and the whole 
 reckoning 8 guineas. 
 
 17. A jocky has two horses : and also two saddles, the one 
 yalued at 18/ the other at 3/. Now when he sets the better 
 saddle on the 1st horse, and the worse on the 2d, it makes the 
 first horse worth double the 2d : but when he places the bet- 
 ter saddle on the 2d horse, and the worse on the first, it makes 
 the 2d horse worth three times the 1st. What then were 
 the values of the two horses ? Ans. The Ist 61. and the 2di9Z. 
 
 18. What two numbers are as 2 to 3, to each of which if 6 
 te added, the sums will be as 4 to 5 ? Ans. 6 and 9. 
 
 19. What are those two numbers, of which the greater is 
 to the less as their sum is to 20, and as their difference is to 
 10 ? Ans. 15 and 45. 
 
 20. What two numbers are those, whose difference, sum, 
 and product, are to each other, as the three numbers 2, 3, 5 ? 
 
 Ans. 2 and 10. 
 
 21. To find three numbers in arithmetical progression, of 
 which the first is to the third as 5 to 9, and the sum of all three 
 is 63 ? Ans. 16, 21, 27. 
 
 22. It is required to divide the number 24 into two such parts, 
 that the quotient of the greater part divided by the less, may 
 be to the quotient of the less part divided by the greater, as 4 
 to 1. Ans. 16 and 8. 
 
 23. A gentleman being asked the age of his two sons, 
 answered, that if to the sum of their ages 18 be added, 
 the result will be double the age of the elder : but if 6 be 
 
 taken 
 
Q,UADRATIC EQUATIONS. 249 
 
 taken from the difference of their ages, the remainder will 
 be equal to the age of the younger. What then were their 
 ages ? Ans. 30 and 12. 
 
 24. To find four numhers such, that the sum of the 1st, 
 2d, and 3d, shall be 13 ; the sum of the 1st, 2d, and 4th, 15 ; 
 the sum of the 1st, 3d, and 4th, 1& ; and lastly the sum of the 
 2d, 3d, and 4th, 20. Ans. 2, 4, 7, 9. 
 
 26. To divide 48 into 4 such parts, that the 1st increased 
 by 3, the second diminished by 3, the third multiplied by 3, 
 ^nd the 4th divided by 3, may be all equal to each other. 
 
 Ans. 6, 12, 3, 27. 
 
 QUADRATIC EQUATIONS. 
 
 Quadratic Equations are either simple or compound. 
 
 A simple quadratic equation, is that which involves the 
 square of the unknown quantity only. As ax^ = b. And 
 the solution of such quadratics has been already given in 
 simple equations. 
 
 A compound quadratic equation, is that which contains the 
 square of the unknown quantity in one t^rm, and the first 
 power in another term. As ax^ -\- bx = c. 
 
 All compound quadratic equations, after being properly 
 reduced, fall under the three following forms, to which they 
 must always be reduced by preparing them for solution. 
 
 1. x^ -i- ax == b 
 
 2. ^2 — rtJT = 6 
 
 3. x^ — ax = — b. 
 
 The general method of solving quadratic equations, is by 
 what is called completing the square, which is as follows : 
 
 1. Reduce the proposed equation to a proper simple form, 
 as usual, such as the forms above ; namely, by transposing 
 all the terms which contain the unknown quantity to one 
 side of the equation, and the known terms to the other ; 
 placing the square term first, and the single power second ; 
 dividing the equation by the co-efficient of the square or 
 first term, if it has one, and changing the signs of all the 
 terms, when that term happens to be negative, as that term 
 must always be made positive before the solution. Then 
 the proper solution is by completiag the square as follows, 
 viz. 
 
 VdL. I. s^i 2. Complete» 
 
280 ALGEBRA. 
 
 2* Complete the unknown side to a square, in this man- 
 ner, viz. Take half the co elhcient of the second term, and 
 square it ; which square add io both sides of the equation, 
 then that side which contains the unknown quantity will be a 
 complete square. 
 
 3. Then extract the square root on both sides of the 
 equation*, and the value of the unknown quantity will be 
 
 determined, 
 
 * As the square root of any quantity may be either + or — , there- 
 fore all quadratic equations admit of two solutions. Thus, the square 
 root of 4- «* is either -^ n or — n ; fov + n X + « and — n X -" " 
 are each equal to -f. n*. But the square root of — n*, or ^Z -" ^*» 
 is imaginary or impossible, as neither 4* n nor — n, when squared, 
 gives - n2. 
 
 ^ So, i n the first form, x2 + aj7 = ^, where a: •{' la is fou nd «= y/ 
 h + la^, the root may be either + ^6-f- ^*, or -*- ^^ + ia«, 
 since either of them being multiplied by itself produces b + fa^- 
 And this ambiguity is expressed by writing the uncertain or double 
 sign dt before ^yb + ^a* ; thus or =» :t ^b -f la^ — ^a- 
 
 In this form, where a: =« ± ■•^ + k'l^ — ia. the first value of 
 X, viz. X ■=» + y/b + |:02 — ^a. in always affirmative ; for sincSe 
 ifl2 -j- b is greater than la 2, the greater square must neces- 
 sarily have the greater root ; therefore ^ b -h ^a^ will always 
 be greater than ^ja^, or its equal ia ; and consequently -f- 
 ^b + \aS — |a will always be affirmative. 
 
 The second value, viz. 07 *= — <^b + |a* — ^a will always be 
 negative, because it is composed of two nega tive term s. Therefore 
 when x^ -^ ax ==* b, we shall have j: == -f- y/b -J- ^a^ — . |a for the 
 affirmative value of x, and or =* -f- <yb +■ Ja* — ^a for the nega- 
 tive value of X. 
 
 In ths second form , where x =- ±:.s/b -|- ia* + ^a the first 
 value, viz. a: -« + ^^4- i«* + ia is always affirmative, since it 
 is composed of two affirmative terms. But the second value, viz* 
 a: ==" — v^ b + Ja2 -f- |a, will always be negative ; for since 
 b -{- ia^ is greater than ^a^, therefore ^b -j l'. '^ will be greater 
 than v/:iu2, or its equal \a ; and consequently -- y/b + \a^ + ^a is 
 always a negative quantity. 
 
 Therefore, 
 
QUADRATIC EQUATIONS. 251 
 
 determined, making the root of the known side either -f or 
 — , which will give two roots of the equation, or two values 
 •f the unknown quantity. 
 
 KotCy 1. The root of the first side of the equation, is 
 always equal to the root of the first term, with half the 
 co-efficient of the second term joined to it, with its sign, 
 whether -f- or — . 
 
 2. All equations, in which there are two terms including 
 the unknown quantity, and which have the index of the one 
 just double that of the other, are resolved like quadratics, 
 by completing the square, as above. 
 
 Thus, J?* -f ax2 =^ 6, or jr^n _|, ^^.n _.. ^^ ^^ ^^ ^^.2 s=j^ 
 are the same as quadratics, aad the value of the unknown 
 quantity may be determined accordingly. 
 
 Therefore, when x^ -ar = 6, we shall have a: = 4 V'^ + i«^ 
 + \a for the affiimative value of x ; and x = - y/b -f ia^ -f- \a 
 for tbe negative value of ar ; so that in both the first and secnd 
 forms, the unknown quantity has always two values, one of which is 
 positive, and the other nogative. 
 
 But in the third form, where x — ±. ^ ia^ — 6 + ^a, both the 
 values of x will be posHive when |o2 is greater than b. For ihe 
 first value, viz. x =^ •\- »/ \a^ - 6-f"i^ will then be aflSrmative, 
 being composed of two affirmative terms. 
 
 The second value, viz. x =■ — v' i'^^ — 6 -J- |a is affiima- 
 tive also J for since \a^ is greater than \a^ - 6, thecefo-e »/ ,\a^ or 
 ^ is greater than ^ \a'^ - Aj and consequently — v^ :Ju*— 6 + ia 
 ,will always be an afii mative qjantity. So that» when jr* — ax 
 «= -. d,_we shill have x = -f- ^ }a2 — A + ^a, And also x = — 
 */ ia2 «.A -f. ifl, for the vaUes of Xt both positive. 
 
 But m this third form, if 6 be greater .than ^a*, the solution of 
 the proposed question will be impossible. For since the square of 
 any quantity (whether that quantity be affirmative or negative; 
 is always affirmative, the square root of a negative quantity is im- 
 possible, and cannot be assigned. But when h is greater than 
 ^a*, then ^,r,2— A is a negative quantity; and therefore its root 
 ^ i'Z« — ^ is impossible, or imaginary ; consequently, in that case, 
 ^ = i^ ± >/ i'J^— ^» oi* tbe two roots or values of x^ are both im - 
 
 possible, or imaginary qwantities. 
 
 EXAMPLES. 
 
252 ALGEBRA. 
 
 EXAMPLES. 
 
 1. Given x^ -|- 4jr = 60 ; to find x. 
 
 First, by completing the square, x^ -f- 4^ + 4 = 64 ,' 
 
 Then by extracting the root, j? + 2 = ± 8 ; 
 
 Then, transpos 2, gives, j: = 6 or — 10, the two roots. 
 
 2. Given x^ — 6ar4-10 = 66 ; to find x. 
 First trans. 10 gives x^ — 6a: = 65 ; 
 
 Then by complet. the sq. it is x^ — 6j7 -f 9 = 64 ; 
 And by extr. the root, gives j?— 3 = ±: 8 ; 
 Then trans. 3, gives x = 1 1 or - 6. 
 
 3. Given 2x2 -f 8x - 30 = 60 ; to find x. 
 First by transpos, 20, it is 2x3 -f 8x = 90 ; 
 Then div. by 2, gives x^ -}" 4.r = 45 ; 
 
 And by compl. the sq. it is x^ + 4x -|- 4 = 49 ; 
 Then extr. the root, it is x +2 = ± 7 ; 
 And transp. 2, gives x == 5 or — - 9. 
 
 4. Given 3x2 -, 3x + 9 = ^ y to find x. 
 First div. by 3, gives x* — x -}- 3 = 2f ; 
 Then transpos. 3, gives x2 — . x = — f ; 
 And compl. the sq. gives x2 — x -\- \ =■ -^ ] 
 Then extr. the root gives x — ^ = ± | ; 
 And transp. i, gives x = | or ^. 
 
 5. Given ^x^ — ^x + 30^ = 52|, to find x. 
 First by transpos 30^, it is ^x^ — \x = 22i ; 
 Then mult, by 2 gives x^ — |x = 44^ ; 
 
 And by compl. the sq. itis x2 — 2.J7 -j- i = 44A ■ 
 Then extr. the root, gives x — i = ^^ 6| j 
 And transp. ^, gives x = 7 or — 6^. 
 
 6. Given ax^ —^bx = c; to find x. 
 
 b c 
 
 First by div. by a, it is x2 x = — ; 
 
 a a 
 
 b b c b^ 
 
 Then compl. the sq. gives~x2 x -j = ! ; 
 
 a 4a2 a 4a' 
 
 b 4ac + fc3 
 
 And extrac. the root, gives x = ± ^ — • 
 
 2a 4a2 
 
 b 4ac + i2 6 
 
 Then transp. — , gives x = ± y/ 1 . 
 
 2o 4a2 2a 
 
 7. Given x* -.2ax2 = t ; to find x. 
 
 First by compl. the aq. gives x* — Sax* ^ a^ = a2 -j- ^ ; 
 
 And 
 
QUADRATIC EQUATIONS. 263 
 
 And extract, the root, gives z^ — a — ±. ^ofi •\- h ; 
 
 Then transpos. a, gives x^ = ± ^a^ - ^ b -{- a ; 
 
 And extract, the root, gives x=±:^a±^a'-\-b. 
 And thns. by always using similar words at each line, the 
 pupil will resolve the following examples. 
 
 EXAMPLES FOR PRACTICJB. 
 
 1. Given a:2,r- 6x — 7 = 33 ; to find x, Ans. x = 10. 
 
 2. Given x^ -, — 5x — 10 = 14 ; to find a;. Ans. a; = 6. 
 
 3. Given 5x^ -f 4a; — 90 = 114 ; to find x. Ans. a: = 6* 
 
 4. Given |a;2 — ix -{- 2 = 9 ; to find a:. Ans. a: = 4. 
 6. Given 3a:* — 2x* = 40 ; to find x. Ans. a; = 2. 
 
 6. Given ^x — i^ a: = 1^^ ; to find x. Ans. ac = 9. 
 
 7. Given ^x^ -f f a; = f ; to find x. Ans. x = -727766, 
 
 8. Given x^ -\- 4x^ = 12 ; to find x, 
 
 Ans. a: = 3/ 2 = 1-269921. 
 
 ^. Given a;^ + 4a; = a^ -f 2 ; to find x, 
 
 Ans. X = v^ oMhe— 2. 
 
 QUESTIONS PR0DUCIN6 QUADRATIC EQUATIONS. 
 
 1. To find two numbers whose difference is 2, and product 80. 
 Let X and y denote the two required numbers*. 
 
 Then the first condition gives x — y = 2, 
 
 And the second gives xy = 80. 
 
 Then transp. y in the 1st gives x = y -{- 2 ; 
 
 This value of x substitut. in the 2d, is y^ -f- 2i/ = 80 
 
 Then comp. the square gives y- -]- 2y -{- I =^ Ql ; 
 
 And extrac. the root gives y -{- 1 = 9 ; 
 
 And transpos. 1 gives y = 8 ; 
 
 And therefore a; = y 4* 2 = 10. 
 
 ♦ These questions* like those in siniple equations) aire also solved 
 by using as tnany unknown letters, as are the numbers required, for 
 ^e better exercise in reducing equations ; not aiming at the shortest 
 modes of solution, which would not affoi"d so much useful practice. 
 
 2. To 
 
264 ALGEBRA. 
 
 2. To divide the number 14 into two such psirta, that their 
 product may be 48. 
 
 Let X and y denote the two numbers. 
 
 Then the 1st condition gives x -{- y = 14, 
 
 And the 2d gives xy = 4S. 
 
 Then transp^ y in the 1st gives x = 14 — y ; 
 
 This value subst. for x in the 2d, is 14y — y^ = 43 j 
 
 Changing all the signs, to make the square positive, 
 
 gives y^ — 14y = — 48 ; 
 Then compl. the square gives y^ — Hy + 49 = 1 ; 
 And extrac. the root gives y — 7 = ± 1 ; 
 Then transpos. 7, gives ?/ = 8 or 6, the two parts. 
 
 3. Given the sum of two numbers = 9, and the sum of 
 their squares = 46 ; to find those numbers. 
 
 Let X and y denote the two numbers. 
 
 Then by the 1st condition or -{- 2/ = 9. 
 
 And by the 2d x^ + y^ = 45. 
 
 Then transpos. y in the 1 st gives jt = 9 — 1/ » 
 
 This value subst in the 2d gives 81 — '\8y -{- 2y^ = 45 ; 
 
 Then transpos. 81, gives 2y^ — I82/ = — 36 ; 
 
 And dividing by 2 gives y^ — % = — 18 ; 
 
 Then compl. the sq gives y^ •— 9y 4- Y = f ; 
 
 And extrac. the root gives y — f = ± | j 
 
 Then transpos. f gives 1/ = 6 or 3, the two numbers. 
 
 4. What two numbers are those, whose sum, product, and 
 difference of their squares, are all equal to each other ? 
 
 Let X and y denote the two numbers. 
 Then the 1st and 2d expression give x -\- y = xy^ 
 And the 1st and 3d give x -\- y =■ x^ — y^. 
 Then the last equa. div. hy x -\- y^ gives 1 = jr- — y. ; 
 And transpos. 7/, gives y -{- \ =^ x ; 
 This val. substit- in the 1st gives 2?/ 4" 1 = 2/^ + 'iji 
 And transpos. 2y, gives 1 = 2/^ — y ; 
 Then complet. the sq. gives f = 2/^ — 2/ + 4 ;. 
 And extracting the root gives ^ a/ 5 = y — ^y 
 And transposing ^ gives i \/ 6 -f- ^ = ^ ; 
 And therefore x = ^4-l=iv'6 + t« 
 And if these expressions be turned into numbers, by ex- 
 tracting the root of 6, &;c. they give a: = 2 6180 -f-, and 
 2/= 1-6180 -f. 
 
 5. There are four number^ in arithmetical progression, of 
 
 whioK 
 
Q.tJADRATIC EQUATIONS. 26^ 
 
 which the product of the two extremes is 22, and that of the 
 means 40 ; what are the numbers ? 
 
 Let j: = the less extreme, 
 
 and y = the common difference ; 
 Then x^x 4- y, J? + 2y, Ir 4* %> will be the four numbers. 
 Hence by the 1st condition x^ + Sxy = 22, 
 And by the 2d x^ -f- 3xy + 2y^ = 40. 
 Then subtracting the first from the 2d gives 2i/a = 18 ; 
 And dividing by 2 gives 2/^=9; 
 And extracting the root gives i/ = 3. 
 Then substit 3 for y in the 1st, gives x^ -{- 9x = 22 ; 
 And completing the square gives x^ -}- 9x -{■ y = * p ; 
 Then extracting the root gives ^ 4* f = V ; 
 And transposing f gives x = 2 the least number. 
 Hence the four numbers are 2, 5, 8, 11. 
 
 6. To find 3 numbers in geometrical progression, whose sum 
 shall be 7, and the sum of their squares 21. 
 
 Let X, y, and z, denote the three numbers sought. 
 Then by the 1st condition xz = y^, 
 And by the 2d x + y -{- 2r = 7, 
 And by the 3d x^ -{- y^ + z^ = 21. 
 Transposing y in the 2d gives x -{- z= 7 -^ y ; 
 Sq. this equa. gives x^ -\- 2xz -\- z^ -f- = 49 — 14y-{-y^; 
 Substi. 2i/2 for 2x2", gives x^ -\-2y^ -{-z^ = 49—141/4-2/2; 
 Subtr. y^ from each side, leaves x^ -{-y^ -{-z^ = 49— 14y ; 
 Putting the two values of x^ + 1^2 4. 2,2 i ^^__.q__ ... 
 equal to each other, gives 5 ^^ 
 
 Then transposing 21 and 14i/, gives 14i/ = 28 ; 
 And dividing by 14, gives y = 2. 
 Thensubstit. 2 for y in the 1st equa. gives xz = 4, 
 And in the 4th, it gives x + 2 = 5 ; 
 Transposing z in the last, gives x = 5 — z ; 
 This substit. in the next above, gives 5z — z^ = 4 j 
 Changing all the signs, gives z^ — 5z = — 4 ; 
 Then completing the square, gives z^ — 52- -|- y = a j 
 And extracting the root gives 2- — f = -t f ; 
 Then transposing f gives z and x = 4 and 1, the two 
 
 other numbers ; 
 So that the three numbers are 1,2, 4. 
 
 QUESTIONS FOR PRACTICE. 
 
 1. What number is that which added to its square makes 
 42? Ans. 6. 
 
 2, T# 
 
256 ALGEBRA. 
 
 .2. To find two numbers such, that the less miy be to the 
 greater as the greater is to 12, and that the sum of their squares 
 may be 45. Ans. 3 and 6. 
 
 3. What two numbers are those, whose difference is 2, and 
 the difference of their cubes 98 ? Ans. 3 and 5. 
 
 4. What two numbers are those whose sum is 6, and the 
 sum of their cubes 72 ? Ans. 2 and 4. 
 
 5. What two numbers are those, whose product is 20, and 
 the difference of their cubes G 1 ? Ans. 4 and 5. 
 
 6. To divide the number 1 1 into two such parts, that the 
 product of their squares may be 784 Ans. 4 and 7, 
 
 7. To divide the number 5 into two such parts, that the sum 
 of their alternate quotients may be 4^, that is of the two 
 quotients of each part divided by the other^ 
 
 Ans. 1 and 4. 
 
 8. To divide 12 into two such parts, that their product may 
 be equal to 8 times their difference. . Ans. 4 and 8. 
 
 9. To divide the number 10 into two such parts, that the 
 square of 4 times the less part, may be 112 more than the 
 square of 2 times the greater. Ans. 4 and 6. 
 
 10. To find two numbers such, that the sum of their squares 
 may be 89, and their sum multiplied by the greater may pro- 
 duce 104. Ans. 5 and 8. 
 
 11. What number is that, which being divided by the 
 product of its two digits, the quotient is 5i ; but when 9 is 
 subtracted from it, there remains a number having the same 
 digits inverted ? Ans. 32. 
 
 12. To divide 20 into three parts, such that the continual 
 product of all three may be 270, and that the difference of the 
 first and second may be 2 less than the difference of the second 
 and third. Ans. 5, 6, 9. 
 
 13. To find three numbers in arithmetical progression, such 
 that the sum of their squares may be 56, and the sum arising 
 by adding together once the first and 2 times the second and 
 3 times the third, may amount to 28. Ans. 2, 4, 6. 
 
 14. To divide the number 13 into three such parts, that 
 their squares may have equal differences, and that the sum of 
 those squares may be 75. Ans. 1, 5, 7. 
 
 15. To find three numbers having equal differences, so that 
 their sum may be 12, and the sum of their fourth powers 962. 
 
 Ans. 3, 4, 6. 
 
 16. To find three numbers having equal differences, and 
 such that the square of the least added to the product of the 
 two greater may make 28, but the square of the greatest add- 
 ed to the product of the two less may make 44. 
 
 Ans. 2, 4, 6. 
 
 17. Three 
 
CUBIC, &c. EqUATIOJ^S. S5^ 
 
 17. Three merchants, a, b, c, on comparing their gains 
 find, that among them all they have gained 1444/. ; and that 
 b's gain added to the square root of a's made 920/ ; but if 
 added to the square root of c's it made 912. What were 
 their several gains ? Ans. a 400, b 900, c 144. 
 
 18. To find three numbers in arithmetical progression, so 
 that the sum of their squares shall be 93 ; also if the first be 
 multiplied by 3, the second by 4, and the third by 5, the sum 
 of the products may be 66. Ans. 2, 5, 8. 
 
 19. To find four numbers such, that the first may be to the 
 second as the third to the fourth ; and that the first may be 
 to the fourth as 1 to 5 ; also the second to the third as 5 to 9 ; 
 and the sum of the second and fourth may be 20. 
 
 Ans. 3. 5, 9, 15. 
 
 20. To find two numbers such that their product added to 
 their sum may make 47, and their sum taken from the sum 
 df their squares may leave 62. Ans. 5, and 7. 
 
 RESOLUTION OF CUBIC AND HIGHER EQUATIONS. 
 
 A Cubic Equation, or Equation of the 3d degree or power, 
 lis one that contains the third power of the unknown quantity. 
 As x^ —ax^ + fca: = c. 
 
 A Biquadratic^ or Double Quadratic, is an equation that 
 contains the 4th Power of the unknown quantity : 
 As X* —ax^ + bx^ -"CX = d. 
 
 An Equation of the 6th Power or Degree, is one that con- 
 tains the 5th power of the unknown quantity : 
 As x^ ^ax* -h bx^ cx^ -{- dx = e. 
 
 And so on, for all other higher powers. Where it is to 
 be noted, however, that all the powers, or terms in the 
 equation, are supposed to be freed from surds or fractional 
 exponents. 
 
 There are many particular and prolix rules usually given 
 for the solution of some of the above-mentioned powers 
 or equations. But they may be all readily solved by the 
 following easy rule of Double Position, sometimes called 
 Trial-and-error. 
 
 RULE. 
 
 1. Find, by trial, two numbers, as near the true toot as 
 
 ^ou can, and substitute them separately in the given equation, 
 
 instead of the unknown c^uantity ; and find how much the 
 
 V OL. i. ^34 terras 
 
 '< 
 
26« ALGEBRA. 
 
 terms collected together, according to their signs ^^4" or — , 
 differ from the absolute known term of the equation, marking 
 whether these errors are in excess or defect. 
 
 2. Multiply the difference of the two numbers, found or 
 taken by trial, by either of the errors, and divide the pro- 
 duct by the difference of the errors, when they are alike, 
 but by their sum when they are unlike. Or say. As the 
 difference or sum of the errors, is to the difference of the 
 two numbers, so is either error to the correction of its sup- 
 posed number. 
 
 3. Add the quotient, last found, to the number belonging to 
 that error, when its supposed number is too little, but subtract 
 it when too great, and the result will give the true root 
 nearly, 
 
 4. Take this root and the nearest of the two former, or 
 any other that may be found nearer ; and, by proceeding in 
 like manner as above, a root will be had still nearer than 
 before. And so on to any degree of exactness required. 
 
 Note 1. It is best to employ always two assumed numbers 
 that shall differ from each other only by unity in the last 
 figure on the right hand ; because then the difference, or 
 multiplier, is only 1. It is also best to use always the least 
 error in the above operation. 
 
 JVote 2. It will be convenient also to begin with a single 
 figure at first, trying several single figures till there be found 
 the two nearest the truth, the one too little, and the other 
 too great ; and in working with them, find only one more 
 figure. Then substitute this corrected result in the equation, 
 for the unknown letter, and if the result prove too little, 
 substitute also the number next greater for the second sup- 
 position ; but contrarywise, if the former prove too great, 
 then take the next less number for the second supposition : 
 and in working with the second pair of errors, continue the 
 quotient only so far as to have the corrected number to four 
 places of figures. Then repeat the same process again with 
 this last corrected number, and the next greater or less, as 
 the case may require, carrying the third corrected number 
 to eight figures ; because each new operation commonly 
 doubles the number of true figures. And thus proceed to 
 any extent that may be wanted. 
 
 EXAMPLES. 
 
 Ex. 1. To find the root of the cubic equation a?^ 4- x^ -f- 
 X = 100, or the value of x in it. 
 
 H€re 
 
CUBIC, &c. EQUATIONS. 
 
 269 
 
 Here it is soon found that 
 X lies between 4 and 5. As- 
 sume therefore these two num- 
 bers, and the operation will be 
 as follows : 
 1st Sup. 
 
 4 - X 
 
 \^ - x^ 
 
 64 - a;3 
 
 84 
 100 
 
 -16 
 
 - sums - 
 but should be 
 
 - errors - 
 
 + 66 
 
 the sum of which is 71. 
 Then a« 71 : 1 : ; 16 : 
 Hence x = 4*2 nearly. 
 
 Again, suppose 4-2 and 4-3 ; 
 and repeat the work as fol- 
 lows : 
 
 1st Sup. 
 4-2 
 
 17-64 
 74-088 
 
 x^ 
 
 <p3 
 
 2d Sup. 
 
 4-3 
 
 18-49 
 
 79-607 
 
 96-928 
 100 
 
 sums 102-297 
 100 
 
 — 4-072 errors -f 2-297 
 
 the sum of which is 6-369. 
 As 6-369 : 1 : : 2-297 : 0036 
 This taken from - 4-300 
 
 leaves x nearly = 4-264 
 
 Again, suppose 4*264, and 4-266, and work as follows : 
 4-264 - X - 4-266 
 
 18-181696 - -ra - 18-190225 
 
 77-526762 - x^ - 77-581310 
 
 99-972448 
 100 
 
 100-036635 
 100 
 
 —0-027552 - errors - -fO-036535 
 
 the sum of which is -064087. 
 
 Then as -064087 : -001 : : -027562 : 0-0004299 
 To this adding - 4-264 
 
 gives a: very nearly = 4-2644299 
 
 The work of the example above might have been much 
 shortened, by the use of The Table of Powers in the Arith- 
 metic, which would have given two or three figures by in- 
 spection. But the example has been worked out so particu- 
 larly as it is, the better to show the method. 
 
 Ex. 2. To find the root of the equation x'^~~\hx^ -|- 63 a: 
 = 50, or the value of x in it. 
 
 Here it soon appears that x is very little above 1, 
 
 Suppose 
 
260 
 
 ALGEBRA. 
 
 Suppose therefore 1 '0 and 1*1, 
 and work as follows : 
 
 X - M 
 
 63:c 
 same 
 
 69-3 
 -.18-15 
 1-331 
 
 62-481 
 60 
 
 — 1 - errors - 4-2*481 
 3-481 sum of the errors. 
 As 3-481 : 1 : : -1 : '03 correct. 
 1-00 
 
 Hence or "= 1 03 nearly 
 
 Again, suppose the two num- 
 bers 103 and 1-02, &c. as 
 follows : 
 103 > a: - 1»02 
 
 64-89 - 63j: 6?2« 
 -16-9136 — 15x2-15-6060 
 1-092727 x^ l'06120e 
 
 60-069227 sums 49-? 16208 
 60 60 
 
 -f -069227 errors —-284792 
 •284792 
 
 As 
 
 364019: -01 :: -069227: 
 •0019666 
 This taken from 1-03 
 
 leaves x nearly = 1 02804 
 
 Note 3. Every equation has as many roots as it contains 
 dimensions, or as there are units in the index of its highest 
 power. That is, a simple equation has only one value of the 
 root ; hut a quadratic equation has two values or roots, a cubic 
 equation has three roots, a biquadratic equation has four roots, 
 and so on. 
 
 And when one of the roots of an equation has been found 
 by approximation, as above, the rest may be found as follows. 
 Take, for a dividend, the given equation, with the known 
 term transposed, with its sign changed, to the unknown side 
 of the equation ; and for a divisor, take x minus the root just 
 found. Divide the said dividend by the divisor, and the quo- 
 tient will be the equation depressed a degree lower than the 
 given one. 
 
 Find a root of this new equation by approximation, as before, 
 or otherwise, and it will be a second root of the original equa- 
 tion. Then, by means of this root, depress the second equa- 
 tion one degree lower, and from thence find a third root, and 
 so on, till the equation be reduced to a quadratic ; then the 
 two roots of this being found, by the method of completing 
 the square, they will make up the remainder of the roots. 
 Thus in the foregomg equation, having found one ' root to be 
 1-02804, connect it by minus with x for a divisor, and the 
 equation for a dividend, &c. as follows : 
 
 ^^ 1-02804) x^ ~ \bx^ + 63:j? — 60 (x^ — 13-97196x 4- 
 
 48-63627=-0 
 Theo 
 
«UBIC, &c. EQJJATIONS. 2^t' 
 
 'Then the two roots of this quadratic equation, or — — — 
 aja — 13-97196 a: = - 48-63627, by completing the square, 
 are 6-57tti3 and 7-39443, which are also the other two roots 
 «f the given cubic equation. So that all the three roots of 
 that equation, viz. a;^ — 16 x^ -{- 63x = 60. 
 
 *^^ l*c^f?t )and the sum of all the roots is found to be 
 
 A ^'tltliw^^ ^^^^S 6^"^^ *® ^^^ co-efficient of the 
 an d 7-c?t^54J Vg^ ^^^^ ^^ ^^^ equation, which the sum of 
 
 . r /^/^^w^^ I the Toots alwavs ought to be, when they are 
 sum '5 000001 •.. J & J J 
 
 Note 4. It is also a particular advantage of the foregoing 
 rule, that it is not necessary to prepare the equation, as for 
 «ther rules, by reducing it to the usual 6nal form and state 
 of equations. Because the rule may be applied at once to an 
 unreduced equation, though it be ever so much embarrassed 
 by surd and compound quantities. As in the following ex- 
 ample. 
 
 Ex. 3. Let it be required to find the root x of the equation 
 ^ 144a:a-(x3 -[- 20)^ + y/ IdQx^ -(x^-^- 24)2" = 1 H, or 
 the value of x in it. 
 
 By a few trials, it is soon found that the value of x is but 
 little above 7. Suppose therefore first that a; is = 7, and 
 then X = 8. 
 
 First, when x = 7. Second, when a: = 8. 
 
 47-906 - y/ 144^2 _ (j.2 -I- 20)2 - 46-476 
 
 65'384 - v^ 196x2 - (a;2 4- 24)2 . 69-283 
 
 113-290 - the sums of these - 115-759 
 
 114-000 - the true number - 114-000 
 
 — 0-710 - the two errors ^ -f 1-759 
 -fl-759 - 
 
 As 2-469 : 1 : : 0-710 : 0-2 nearly . 
 7-0 
 
 Therefore x = 7-2 nearly 
 
 Suppose again x = 7-2, and then, because it turns out too 
 great suppose x also = 7-1, &c. as follows : 
 
 Supp» 
 
^62- 
 
 ALGEBRA. 
 
 • 
 
 
 Supp. X = 7-2 
 
 
 Supp. X = 
 
 71 
 
 47-990 - 
 
 \/ 144x2 _ (jr2 4. 20)3 
 
 - 47-973 
 
 
 66-402 - 
 
 ^ 196x2 - (x2 + 24)2 
 
 the sums of these 
 the true number 
 
 the two errors 
 
 ::-123: -024 the correcti 
 7-100 add 
 
 - 65-904 
 
 
 ir4-392 - 
 114 000 - 
 
 113-877 
 114-000 
 
 
 -fO'392 - 
 0-123 
 
 — 0-123 
 
 30, 
 
 
 As -515 :'l 
 
 
 Therefore x = 7-124 nearly the root required. 
 
 Noit 5. The same rule also among other more difficult 
 forms of equation, succeeds very well in what are called 
 exponential ones, or those which have an unknown quantity in 
 the exponent of the power ; as in the following example ; 
 
 Ex. 4. To find the value of x in the exponental equation 
 
 a;^= 100. 
 
 For more easily resolving such kinds of equations, it is 
 convenient to take the logarithms of them, and then compute 
 the terms by means of a table of logarithms. Thus, the loga- 
 rithms of the two sides of thie present equation are x X log. 
 of a: = 2 the log. of 100. Then, by a fewt rials, it is soon 
 perceived that the value of x is somewhere between the two 
 numbers 3 and 4, and indeed nearly in the middle between 
 them, but rather nearer the latter than the former. Taking 
 therefore first x = 3-6, and then = 3-6, and virorking with 
 the logarithms, the operation will be as follows : 
 
 First Supp. X = 3-5, 
 Log. of 3'5 = 0-544068 
 then 3-5Xlog. 3-5 = 1-904238 
 the true number 2-000000 
 
 error, too little, — -096762 
 •002689 
 
 Second Supp. x = 3*6. 
 Log. of3-6 = 0-656303 
 then 3-6 X log. 3-6 = 2-002689 
 the true number 2-00000© 
 
 error, too great-I-002689 
 
 -098451 sum of the errors. Then, 
 
 As -098451 : -1 : : -002689 : 0-00273 the correction 
 taken from 3-60000 
 
 leaves - 3-59727 = x nearly. 
 
 On 
 
CUBIC, &c. EQUATIONS. 263 
 
 On trial, this is found to be a very small matter too little. 
 Take therefore again, x = 3-69727, and next = 3-59728, and 
 repeat the operation as follows : 
 
 First, Supp. X = 3-59727. 
 Log. of 3-69727 is 0-655973 
 3-59727 X log. 
 
 of 3-69727 = 1-9999864 
 
 the true number 2-0000000 
 
 error, too little, —0-0000146 
 --0-0000047 
 
 Second, Supp. x = 3-59728 
 Log. of 3-69728 is 0-565974 
 3-59728 X log. 
 
 of 3-59728 = 1-9999953 
 
 the true number 2-0000000 
 
 error, too little, —0-0000047 
 
 0-0000099 diff. of the errors. Then, 
 As -0000099 : -00001 : : -0000047 ; -00000474747* the cor. 
 added to - 3-59728000000 
 
 gives nearly the value of re = 3-59728474747 
 
 Ex. 6. To find the value of a: in the equation sc » + lOa:^ 
 -f- 6a: = 260. Ans. a; = 4-1179867. 
 
 Ex. 6. To find the value of x in the equation x^ — Src = 60. 
 
 Ans.3-S648854. 
 
 ^ ♦ The Author has here followed the general rule in finding' as many 
 additional figures as were known before : viz. The 6 figurcis 3-59728t 
 but as the logarithms here used are to 6 places only, we f.annot de- 
 pend on more than 6 .figures in the answer ; we have no reason, 
 therefore, to suppose any of the figures in -000004747'47, to be 
 coprect. 
 
 The log. of 3-59728 to 15 places is 0-55597 42431 34677 
 the log. of 3«59727 to 15 places is 0-55597 30358 47267 
 which logarithms multiplied by their respective numbers gi ve the fol- 
 lowing products : 
 
 i'S 5?i6 S} •»* '""^ *<• *^ '"* «e"^- 
 
 Therefore, the errors are - - - 49746 564\88 
 
 and - - - 148773 37702 
 
 and the difference of errors - - - 99026 812 14. 
 
 Now since only 6 additional figures are to be obtained, We n my omit 
 the last three figures in these errors ; and state thuff : as diff. t >f errors 
 9902681 : diff. of sup. 1 : : error 4974656 : the correction SI^IT^^. t, which 
 united to 3-59728 gives us the true value of a: « 3-597285023 54- , 
 
 i 
 
 ISx. 7, 
 
264 ALGEBRA. 
 
 Ex. 7. To find the value of x in the equation s:^ -|- 2x^ 
 
 — 23 J? = 70. Ans. x = 5134o7. 
 Ex. 8. To find the value of x in the equation x^ — VI x^ 
 
 4- h^x = 350. Ans. x = 14-95407. 
 
 Ex. 8. To find the value of x in the equation jr* — Sx^ 
 
 — 75a: = 10000. Ans. x ~ 10-2609. 
 Ex. 10. To find the value of x in the equation 2j?'* — 16a: 
 
 4- 40 ar2 — 30j7 = — 1. Ans. x = 1-284724. 
 
 Ex. II. To find the value of x in the equation x^ ■\- 2a;* 
 
 -f 3j73 4- 4a;2 + 6a: = 54321. Ans. x — 8-414455. 
 
 Ex. 12. To find the value of x in the equation x = 
 123456789. Ans. x = 8-6400268. 
 
 Ex. 13. Given 2a: - "Ix^ '\- \\x^ ^ ^x — \\ , to find x. 
 Ex. 14. To find the value of x in the equation 
 
 (3x2 — 2 ^ a: -f 1)« — (x2 — 4x ^ X -f 3 ^x^^ = 66. 
 
 Ans. X = 18-360877. 
 
 To resolve Cubic Equations hy Cardenas Rule, 
 I 
 Though the foregoing general method, by the application 
 of Double Position, be the readiest way, in real practice, of 
 finding the roots in numbers of cubic equations, as well as of 
 all the higher equations universally, we may here add the 
 particular method commonly called Garden's Rule, for re- 
 solving cubic equations, in case any person should choose 
 occasionally to employ that method. 
 
 The form that a cubic equation must necessarily have to 
 be resolved by this rule, is this, viz. z^ -{- az = b, that is, 
 wanting the second term, or the term of the 2d power z^. 
 Therefore, after any cubic equation has been reduced down 
 to its final usual form, x^ + px^ -f- qx = r^ freed from the 
 co-eflicient of its first term, it will then be necessary to take 
 away the 2d term px^ ; which is to be ^one in this manner : 
 Take ^p, or ^ of the coefficient of the second term, and 
 annex it, with the contrary sign, to another unknown letter 
 Zy thus z — i p ; then substitute this for x, the unknown letter 
 in the original equation x^ -|- px'^ -{- qx = r^ and there will 
 result this reduced equation z^ -\- az = b, of the form projser 
 for applying the following, or Garden's rule. C^r take 
 c = ia, and d = ^b, by which the reduced equation takes 
 this form 2 3 -I- 3c /= 2d, 
 
 Thejr^ 
 
CUBIC, &c. EQUATIONS. 26JSI 
 
 Then substitute the values of c and d in this 
 
 form z = ^ci + y (d^ -j- c^) -f ^d^^(^d^ + c^). 
 
 or. = 3/d -f v^(ci^ ^c3)-777^-^^^ 
 
 and the value of the root ., of the reduced equation 2^ + 
 az = b^ will be obtained. Lastly, take x = z — ip, which 
 will give the value of a;, the required root of the original equa- 
 tion x^ 4* px^ + 9^ = »'i first proposed. 
 
 One root of this equation being thus obtained, then de- 
 pressing the original equation one degree lower, after the man- 
 ner described p 260 and 261, the other two roots of that equa- 
 tion will be obtained by means of the resulting quadratic 
 equation. 
 
 JVote. When the co-efficient a, or c, is negative, and c^ is 
 greater thanc?^^ this is called the irreducible case, because 
 then the solution cannot be generally obtained by this rule. 
 
 Ex. To find the roots of the equation x^—6x^ + 10a; = 8. 
 
 First, to take away the 2d term, its co- efficient being— 6, 
 its 3d part is — 2 ; put therefore x = . + 2 ; then 
 x^ = z^ -{- 6.2 ^ i2z -{-8 
 — 6x2 = — 6.2 — 24. ■— 24 
 + lOx = 4- 10. 4- 20 
 
 theref. the sum. .» i^ — 2. + 4 = 8 
 or .3 ♦ — 2. = 4 
 Here then a = — 2, 6 = 4, c = — |, rf = 2. 
 
 Theref. l /d^^{d^ 4 .c3)=3^24-v^(4-^\)=V2-t-v^Vo/-= 
 
 V2 4- V" x/ 3 = 1-57785 
 
 and 3/d-y(^+c3)= 3/2— ^(4-^\)=^24VVV'= 
 3/2— V" ^/^= 0-42265 
 then the sum of these two is the value of . = 2. 
 Hence a: = . 4" 2 e=: 4, one root of a; in the eq. x^ — 6a;2 -^ 
 lOx = 8. 
 To find the two other roots, perform the division, &c. as in 
 p. 261, thus : 
 
 a: — 4) x3 — 6a;3 -|- lOa: — 8 {x^ —2a; + 2 = 0. 
 
 ^3 4a;2 
 
 — 2x2 + lOx 
 
 — 2x2 + 8x 
 
 
 
 2x- 
 
 2x — 
 
 -8 
 -8 
 
 Vot. I. 3i Hence 
 
266 ALGEBRA. 
 
 Hence x^ —2a? =—2, or a;2 — 2a: -f I = ~ 1, and a: — I 
 = ^: v^ - 1 ; ^ = 1 + v^ ~- 1 or = 1 — v^ — 1, the two 
 ether sought. 
 
 Ex. 2. To find the roots of x^ — 9x2 + 28x = 30. 
 
 Ans. a: = 3, or = 3 H- ^ — 1, or = 3 — ^ — 1. 
 Ex. 3. To find the roots of x^ - Tx^ + Hot = 20, 
 
 Ans. a; = 6, or = 1 + ^ — 3, or = 1 - ^ — 3. 
 
 OF SIMPLE INTEREST. 
 
 As the interest of any sum, for any time, is directly pro- 
 portional to the principal sum, and to the time ; therefore the 
 interest of 1 pound, for 1 year being multiplied by any given 
 principal sum, and by the time of its forbearance, in years and 
 parts, will give its interest for that time. That is, if there 
 be put 
 
 r = the rate of interest of 1 pound per annum, 
 p = any principal sum lent, 
 t = the time it is lent for, and 
 
 d = the amount or sum of principal and interest ; then 
 is prt = the interest of the sum /?, for the time t, and conseq, 
 p -\- prt oTp X (1 -{- rt) = a, the amount for that time, i 
 
 From this expression, other theorems can easily be deduced, 
 for finding any of the quantities above mentioned ; whiph theo- 
 rems; collected together, will be as below : 
 1st, a = /) -{- prty the amount. 
 a 
 
 2d, p = , the principal, 
 
 1 + ri 
 a—p 
 
 3d, r = , the rate, 
 
 pt 
 a—p 
 
 4th, < = the time. 
 
 pr 
 For Example. Let it be required to find, in what time 
 any principal sum will double itself, at any rate of simple in- 
 terest. 
 
 In this case, we must use the first theorem, a = p -{- prt, 
 in which the amount a must be made = 2/?, or double the 
 principal, that is, p -|- prt = 2p, or prt = p, or rt = 1 ; 
 
 1 
 and hence t = -. 
 
 r Here, 
 
COMPOUND INTEREST. 267 
 
 Here, r being the interest of IL for 1 ^^ear, it follows, that 
 the doubling at simple interest, is equal to the quotient of 
 any sum dirided by its interest for 1 year. So, if the rate of 
 interest be 6 per cent, than 100 -i- 5 = 20, is the time of 
 doubling at that rate. 
 
 Or the 4th theorem gives at once 
 a-p 2p— /) 2— 1 1 
 
 t = = = s= — , the same as before. 
 
 pr pr r r 
 
 COMPOUND INTEREST. 
 
 Besides the quantities concerned in Simple Interest 
 namely, 
 
 p = the principal sum, 
 
 r = the rate or interest of 11. for 1 year, 
 
 a = the whole amount of the principal and interest, 
 
 t = the time, 
 there is another quantity employed in Compound Interest, 
 viz. the ratio of the rate of interest, which is the amount 
 of 1/. for 1 time of payment, and which here let be denoted 
 by R, viz. 
 
 R = 1 -}- r, the amount of 1/. for 1 time. 
 Then the particular amounts for the several times may be 
 thus computed, viz. As 1/. is to its amount for any time su is 
 any proposed principal sum, to its amount for the same time ; 
 that is, as 
 
 11. : R : : p : pR, the 1st year's amount, 
 
 11. : R : : pR : pR^ , the 2d year's amount, 
 
 1/. : R : : pR^ ; pR^, the 3d year's amount, 
 
 and so on. 
 Therefore, in general, pR* = a is the amount for the i year, 
 or t time of payment. Whence the following general theoi» 
 reras are deduced : 
 
 1st, a = joRt, the amount, 
 a 
 
 2d, /? = — the principal, 
 
 a 
 3d, R = ^ — , the ratio, 
 
 P 
 log. of a — log. of p 
 
 4th, t = , the time. 
 
 lo£. of R 
 
 From 
 
^6d 
 
 ALGEBRA. 
 
 From which, any one of the quantities may be found, wheji 
 the rest are given. 
 
 As to the whole interest, it is found by barely subtracting 
 the principal p from the amount a. 
 
 Example. Suppose it be required to find, in how many 
 years any principal sum will double itself, at any proposed 
 rate of compound interest. 
 
 In this case the 4th theorem must be employed, making 
 (0 = 2p ♦ and then it is, 
 
 ' log. a — log./) log. 2p — log. /) log. 2 
 
 log. R log. R log. R 
 
 So, if the rate of interest be 5 per cent, per annum ; then 
 ^R = 1 -f -06 = 1 -05 ; and hence 
 log. 2 -301030 
 
 t — -' — = = 14-2067 nearly ; 
 
 log. 106 -021189 
 that is, any sum doubles itself in 14a years nearly, at the rate 
 of 6 per cent, per annum compound interest. 
 
 Hence, and from the like question in Simple Interest, 
 above given, are deduced the times in which any sum doubles 
 itself, at several rates of interest, both simple and com- 
 pound ; viz. 
 
 At-, 
 
 per cent, per annum 
 interest, 1/. or any 
 
 other sum, will 
 double itself in the 
 
 following years. 
 
 " At Simp. Int. 
 
 At Comp. Int. 
 
 in 50 
 
 in 35-0028 
 
 40 
 
 28-0701 
 
 33i 
 
 23-4498 
 
 284 
 
 20-1488 
 
 25 «1 
 
 1 7-6730 «5 
 
 22f S 
 
 15-7473 S 
 
 20 ? 
 
 14-2067?' 
 
 16| 
 
 11-8967 
 
 14f 
 
 10-2448 
 
 12i 
 
 9-0065 
 
 IH 
 
 8-0432 
 
 L 10 
 
 1 7 2725 
 
 The 
 
eOMPOTJND INTEREST. 
 
 f69 
 
 Th€ following Table will very much facilitate calculations 
 •f compouad interest on any sum, for any number ot years, 
 at various rates of interest. 
 
 
 The Amounts of 
 
 \l. in an) 
 
 r Number of Year« 
 
 . 
 
 YTi: 
 
 3 
 
 H 
 
 4 
 
 H 
 
 5 
 
 6 
 
 1 
 
 1-0:^00 
 
 1-0350 
 
 1 ()4i-0 
 
 1-04.^0 
 
 1 0500 
 
 1 0600 
 
 2 
 
 1-0609 
 
 J -0712 
 
 10816 
 
 1-0920 
 
 1102o 
 
 1 1^36 
 
 3 
 
 10927 
 
 1-1087 
 
 11249 
 
 1.1412 
 
 1-1576 
 
 M910 
 
 4 
 
 11255 
 
 11475 
 
 1-1699 
 
 11925 
 
 i-2155 
 
 1-2625 
 
 5 
 
 1 1593 
 
 1-1877 
 
 1-2167 
 
 1-2462 
 
 1-2763 
 
 1-3382 
 
 6 
 
 11941 
 
 1-2293 
 
 1-2653 
 
 1 -302^ 
 
 1-3401 
 
 1-4185 
 
 7 
 
 1 2:^99 
 
 1-2723 
 
 1-3159 
 
 1 36U9 
 
 1-4071 
 
 1-5036 
 
 8 
 
 1-2668 
 
 1 3168 
 
 1-3686 
 
 , 4221 
 
 1-4775 
 
 1-5939 
 
 9 
 
 1 3048 
 
 1-3629 
 
 1-4233 
 
 1-4861 
 
 1-5513 
 
 1 -6895 
 
 10 
 
 1-3439 
 
 1-4106 
 
 1-4802 
 
 1-5530 
 
 1-628^ 
 
 1-7909 
 
 11 
 
 1-3842 
 
 1-4600 
 
 1-5895 
 
 1-6229 
 
 1-710;': 
 
 1-8983 
 
 12 
 
 14258 
 
 1-6111 
 
 1 6010 
 
 1-6959 
 
 1-7950 
 
 2-0122 
 
 13 
 
 1-4685 
 
 1-5640 
 
 1-66^1 
 
 1 7722 
 
 1 -8866 
 
 2 1329 
 
 14 
 
 1-6126 
 
 1-6187 
 
 1-7317 
 
 1-8519 
 
 19799 
 
 2-2609 
 
 15 
 
 1-5580 
 
 1-6753 
 
 1-8009 
 
 1-9353 
 
 2-0789 
 
 2-3966 
 
 16 
 
 1-6047 
 
 1-7340 
 
 1-8730 
 
 2-0224 
 
 2-1829 
 
 2-6404 
 
 17 
 
 1-6528 
 
 1-7947 
 
 1-9479 
 
 2-1134 
 
 2-2920 
 
 2-6928 
 
 18 
 
 1-7024 
 
 1-8575 
 
 2-0258 
 
 2-2085 
 
 2-4066 
 
 2-8543 
 
 19 
 
 1-7535 
 
 1-9225 
 
 2-1068 
 
 2-3079 
 
 2-5270 
 
 3 -02^6 
 
 20 
 
 1 1-8061 
 
 1-9898 
 
 2-1911 
 
 2-4117 
 
 2-6533 
 
 ,3-2071 
 
 The use of this Table, which contains all the powers, r*, 
 to the 20th power, or the amounts of 1/. is chiefly to cal- 
 culate the interest, or the amount of any principal sum, for 
 any time, not more than 20 years. 
 
 For example, let it be required to find, to how much 523/. 
 will amount in 15 years, at the rate of 5 per cent, per annum 
 /compound interest. 
 
 In the table, on the line 15, and in the column 5 per cent, 
 is the amount of 1/. viz. - - 2-0789 
 this multiplied by the priDcipai - 523 
 
 gives the amount - - - 1087-2647 
 
 or - - - 1087/. 5s. S^d, 
 
 and therefore the interest is - 664/. 5s. 3\d. 
 
 Note 1. When the rate of interest is to be determined to 
 
 any other time than a year ; as suppose to i a year, or ^ a 
 
 year, &c. \ the rules are still the same ; but then t will 
 
 express 
 
27^. ALGEBRA. = 
 
 express that time, and r must be taken the amount for that 
 time also. 
 
 JsToie 2. When the compound interest, or amount, of any 
 sura is required for the parts of a jear ; it may be deter- 
 mined in the following manner : 
 
 \st. For any time which is some aliquot part of a year : — 
 Find the amount of 1/. for 1 year, as before ; then that root 
 of it which is denoted by the aliquot part, will be the amount 
 of IL This amount being multiplied by the principal sum, 
 will produce the amount of the given sum as required. 
 
 2{/, When the time is not an aliquot part of a year : — 
 Reduce the time into days, and take the 366tii root of the 
 amount of \l. for 1 year, which will give the amount of the 
 same for 1 day. Then raise this amount to that power whose 
 index is equal to the number of days, and it will he the amount 
 for that time. Which amount being multiplied by the princi- 
 pal sum, will produce the amount of that sum as before.— And 
 in these calculations, th^ operation by logarithms will be 
 Tery useful. 
 
 OF ANNUITIES. 
 
 ANNUITY is a term used for any periodical income, aris- 
 ing from money lent, or from houses, lands, salaries, pen- 
 sions. &c. payable from time to time, but mostly l?y annual 
 payments. 
 
 Annuities are divided into those that are in Possession, and 
 those in Reversion : the former meaning such as have com- 
 menced ; and the latter such as will not begin till some par- 
 ticular event has happened, or till after some certain time has 
 elapsed. 
 
 When an annuity is forborn for some years, or the pay- 
 ments not made for that time, the annuity is said to be in 
 Arrears. 
 
 An annuity may also be for a certain number of years ; 
 or it may be without any limit, and then it is called a Per- 
 petuity. 
 
 The Amount of an annuity, forborn for any number of 
 years, is the sum arising from the addition of all the annui- 
 ties for that number of years, together with the interest due 
 upon each after it bqcwnes d.ue. 
 
 The 
 
ANNUITIES. i'71 
 
 The Present Worth nr Vahie of an annuity, is the price or 
 sum wliich ought to .e given for it, supposing it to be bought 
 off, or paid all at once. 
 
 Let a = the annuity, pension, or yearly rent ; 
 n = the number of years forborn, or lent for ; 
 R = the aiTiount of ll. for 1 year ; 
 fn = the amount of the annuity ; 
 V = its value, or its present worth. 
 Now, 1 being the present value of the sum r, by propor- 
 tion the present value of any other sum a, is thus found : 
 
 a 
 as R : 1 : : a : — the present value of a due 1 year Iienc3. 
 
 In hke mannner — is the present value of a due 2 years 
 
 R2 
 
 ct a a a a 
 
 hence ; for r : 1 : : — : — So also — , — , — , &c. will 
 
 R R2 r3 r.4 r5 
 
 be the present values of a, due at the end of 3, 4, 6, &c. 
 years respectively. Consequently the sum of all these, or 
 a a a a 1111 
 
 — + -- + — + — 4-&C. = (— + — + — +-&c.)X a, 
 
 R r2 r3 r4 R R2 r3 ^4 
 
 continued to n terms, will be the present value of all the n 
 years' annuities. And the value of the perpetuity, is the sum 
 of the series to infinity. 
 
 But this series, it is evident, is a geometrical progressi(Mi, 
 1 
 having — both for its first term and common ratio, and the 
 
 ft 
 number of its terms n ; therefore the sum u of all the terms 
 ©r the present value of all the annual payments, will be 
 11 1 
 
 R R R" r" — 1 a 
 •» = X a, or = X — . 
 
 1 R — 1 R" 
 
 1 — 
 
 R 
 
 When the annuity is a perpetuity ; n being infinite, r" ib 
 
 1 
 also idfinite, and therefore the quantity — becomes = 0. 
 
 R" 
 
 a 1 
 
 therefore X — also = ; consequently the expression 
 
 R- I R» 
 
 becOHie^ 
 
27g ALGEBRA. 
 
 a 
 becomes barely v = ; that is, any annuity divided by 
 
 R-l 
 
 tbe interest of \l for 1 year, gives the value of the perpetuity. 
 So, if the rate of interest be 6 per cent, 
 
 Then lOOa -r- 5 = 20rt is the value of the perpetuity at 
 £ per cent : Also lOOa -f- 4 = 25a is the value of the per- 
 petuity at 4 per cent : And 100a -J- 3 = 33^a is the value of 
 the perpetuity at 3 per cent : and so oa. 
 
 Again, because the anriount of 1/. in n years, is r", its 
 
 increase in that time will be r'* — 1 ; but its interest for one 
 
 iingle year, or the annuity answering to that increase, is 
 
 R — 1 ; therefore as r — 1 is to r" — 1, so is a to m ; that 
 
 R" — 1 
 
 is, m = X a. Hence, the several cases relating ta 
 
 R — 1 
 Annuities in Arrear, will be resolved by the following equa- 
 feons : 
 
 R»-~l 
 
 ■m = '— X a = -VR^ I 
 
 R 
 
 1 
 
 
 
 
 
 
 
 
 R"- 
 
 1 
 
 X— : 
 
 R» 
 
 m 
 5 
 
 R« 
 
 
 
 
 
 
 R 
 
 1 
 
 
 R 
 
 1 
 
 Xm 
 
 R 
 
 
 
 1 
 
 X VK*; 
 
 
 
 R«-- 
 
 1 
 
 R° 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 mR — w 
 
 4- 
 
 tb 
 
 log. m — log. "& 
 
 n = ■ 
 
 log. R 
 log. m — log. t 
 
 Log. R = 
 
 1 1 
 
 RP R" R — - 1 
 
 In this last theorem, r denotes the present value. of aa 
 annuity in reversion, after/) years, or not commencing till 
 after the first p years, being found by taking the difference 
 r" — ■ 1 a rP — 1 a 
 
 between the two values X — and —7 X — , for 
 
 R — 1 r" r — 1 rP 
 n years and p years. 
 
 But the amount and present vaUie of any annuity for any 
 number of years, up to 21, will be most readily found by the 
 tno following tables. tablk 
 
ANNUITIES. 
 
 27a 
 
 TABLE I. 
 The Amount of an Annuity of ll. at Compound Interest 
 
 YrvjatSperc. 
 
 3) pe: < .. 4 !>er r 
 
 4iperr.| 
 
 5 rei r. 
 
 6 p' re 
 
 1 
 
 10000 
 
 10000 
 
 10000 
 
 1-0000 
 
 1 (000 
 
 1 OCOC 
 
 2 
 
 2 0300 
 
 2 0350 
 
 20400 
 
 2 04.0 
 
 2 0500 
 
 2 Or. 00 
 
 3 
 
 3 0909 
 
 3 3062 
 
 3 1216 
 
 3- 1370 
 
 3 35:;5 
 
 3 1836 
 
 4 
 
 4 18.36 
 
 4 2l4y 
 
 4 :>465 
 
 4 4782 
 
 43101 
 
 4 3746 
 
 5 
 
 5 3091 
 
 5-3625 
 
 5-4»6.. 
 
 54707 
 
 5-5-ot 
 
 56371 
 
 6 
 
 6-4684 
 
 6 5502 
 
 6 6330 
 
 6 7169 
 
 6 8019 
 
 6 9.53 
 
 7 
 
 7-66 -'5 
 
 7-77i'4 
 
 7 8983 
 
 8-0192 
 
 8 1420 
 
 8 3938 
 
 8 
 
 8-8923 
 
 9 0517 
 
 9 142 
 
 9 3800 
 
 9-5191 
 
 9 8975 
 
 9 
 
 10 1591 
 
 10 3685 
 
 105828 
 
 108021 
 
 11-0266 
 
 li 4913 
 
 10 
 
 11 4639 
 
 n 73i4 
 
 12 006! 
 
 12-2882 
 
 12-5779 
 
 13 1808 
 
 U 
 
 12 8078 
 
 13 1420 
 
 13 48 4 
 
 13 8412 
 
 14 2068 
 
 149716 
 
 12 
 
 14 1920 
 
 14 60^<0 
 
 15 0258 
 
 15 4640 
 
 159171 
 
 16 8699 
 
 13 
 
 156178 
 
 \6 1130 
 
 ^ 6 6268 
 
 17 :5y9 
 
 17 7130 
 
 18 8821 
 
 14 
 
 17*0863 
 
 17 6770 
 
 18 2919 
 
 189321 
 
 19 5986 
 
 21 0.51 
 
 15 
 
 18 5989 
 
 19 2957 
 
 20 3236 
 
 ';0 7841 
 
 21 5786 
 
 23 2760 
 
 16 
 
 20-1569 
 
 20 9710 
 
 I 8245 
 
 2.7193 
 
 23 6o75 
 
 25 67^5 
 
 17 
 
 217616 
 
 227050 
 
 23 6975 
 
 ^4 7417 
 
 25-8404 
 
 28-2129 
 
 18 
 
 234141 
 
 24 4997 
 
 25-6454 
 
 26-8551 
 
 28- 1324 
 
 30 9057 
 
 19 
 
 25 1169 
 
 26 357 
 
 27-6712 
 
 290636 
 
 SO 5390 
 
 337600 
 
 20 
 
 268704 
 
 58-2797 
 
 29-7781 
 
 31 3714 
 
 33 0660 
 
 36 7856 
 
 21 
 
 28-6/65 
 
 30 2695 
 
 31 9692 
 
 33-7831 
 
 S5 7193 
 
 39 99 ?7 
 
 TABLE II. The present value of an Annuity of 1^ 
 
 ^K 
 
 at 3perc 
 
 33Pvrc. 
 
 4!ier c. 
 
 4^ per c 
 
 5 jjcrc. 
 
 6 perc. 
 
 1 
 
 9709 
 
 0-9662 
 
 9615 
 
 9569 
 
 9524 
 
 0-9434 
 
 2 
 
 1 9135 
 
 18997 
 
 18861 
 
 1-8727 
 
 1 8594 
 
 1-8334 
 
 3 
 
 2 8286 
 
 28016 
 
 2 7751 
 
 2-7490 
 
 27233 
 
 2-6730 
 
 4 
 
 3-7171 
 
 36731 
 
 3 6299 
 
 3 5875 
 
 35460 
 
 3 4651 
 
 5 
 
 4 5797 
 
 45151 
 
 4-4518 
 
 4 5900 
 
 4-3295 
 
 4-2124 
 
 6 
 
 5 4172 
 
 53286 
 
 5 2421 
 
 51579 
 
 50757 
 
 49173 
 
 7 
 
 62303 
 
 6 1*45 
 
 600-0 
 
 5-8927 
 
 5 7864 
 
 55824 
 
 8 
 
 7 0197 
 
 6-8740 
 
 67327 
 
 6 5959 
 
 6 4632 
 
 6-^^098 
 
 9 
 
 7 7861 
 
 7 6077 
 
 74353 
 
 7-2688 
 
 7 1078 
 
 68017 
 
 10 
 
 8 5302 
 
 8-3166 
 
 8 1109 
 
 7 9127 
 
 7-7217 
 
 7 3601 
 
 U 
 
 9 2526 
 
 , 9 0116 
 
 8 7605 
 
 8 5289 
 
 8-3054 
 
 7 8869 
 
 12 
 
 9-9540 
 
 9 6633 
 
 9 3851 
 
 9 1186 
 
 88633 
 
 83838 
 
 13 
 
 10 6^>50 
 
 10-3027 
 
 9 9857 
 
 9 6829 
 
 9 3936 
 
 885/7 
 
 14 
 
 11 296 i 
 
 10 9205 
 
 10 5631 
 
 10 2228 
 
 9 8986 
 
 9 2950 
 
 15 
 
 1 1-9379 
 
 ,115174 
 
 111184 
 
 10 7396 
 
 '0 3797 
 
 9 712^ 
 
 16 
 
 12 5611 
 
 12 0941 
 
 116523 
 
 11 2340 
 
 10 8378 
 
 10 10 59 
 
 17 
 
 13 1661 
 
 126513 
 
 12 1657 
 
 117072 
 
 11 2741 
 
 10 4773 
 
 18 
 
 13 7535 
 
 13 1897 
 
 126593 
 
 12-1600 
 
 11-6896 
 
 10 8276 
 
 19 
 
 14 3238 
 
 13 7098 
 
 13 1339 
 
 12*5933 
 
 12 0853 
 
 11 1581 
 
 20 
 
 14 8775 
 
 l4-oi24 
 
 135903 
 
 13 0079 
 
 12-4622 
 
 114699 
 
 21 
 
 154150 
 
 14 6980 
 
 140292 
 
 13 4047 
 
 ' 12-8212 
 
 11 7641 
 
 Vol. L 
 
 36 
 
 n 
 
274 ALGEBRA. 
 
 To find the Amount of^ any annuity forborn a certain number 
 of years. 
 
 Take out the amount of \L from the first table, for the 
 proposed rate and time ; then multiply it hy the given annuity; 
 and the product will be the amount, for the same number of 
 years, and rate of interest. — And the converse to find the rate 
 or time. y 
 
 Exam. To find how much an annuity of 60Z. will amount to 
 in 20 years, at 3^ per cent, compound interest. 
 
 On the line of 20 years, and in the column of 3i per cent, 
 stands 28*2797, which is the amount of an annoity of \l. for 
 ihe 20 years. Then 28-2797 X 50 gives 1413'986/. = 
 iW13/. 19s. 8d. for the answer required. 
 
 To find the present Value of any annuity for any number of 
 years. — Proceed here by the 2d table, in the same manner as 
 above for the 1st table, and the present worth required will 
 be found. 
 
 Exam. 1. To find the present value of an annuity of 501, 
 which is to continue 20 years, at 3^ per cent. — By the table, 
 the present value of IZ, for the given rate and time, is 
 14-2124 ; therefore 14-2124 X 50 = 710-6,2/. (yr 710Z. 126-. 4d, 
 is the present value required. 
 
 Exam, 2. To find the present value of an annuity of 20/. 
 to commence 10 years hence, and then to continue for 1 1 
 years longer, or to terminate 21 years hence, at 4 per cent, 
 interest. — in such cases as this, we have to find the difierence 
 between the present values of two equal annuities, for the 
 two given times ; which therefore will be done by subtract- 
 ing the tabular value of the one period from that of the 
 other, and then multiplying by the given annuity. Thus, 
 tabular value for 21 years 14*0292 
 ditto for - - 10 years 8-1109 
 
 the difference 5-9183 
 multiplied by 20 
 
 gives - 118-366/. 
 
 or - - 118/. 7s. 3^d. the answer. 
 
 END (IF THE ALGEBRA. 
 
GEOMETRY. 
 
 DEFINITIONS. 
 
 1. -A. POINT is that wh^ch has positioa, 
 but no magnitude, nor dimensions ; neither 
 length, breadth, nor thickness. 
 
 2. A Line is length, without breadth or 
 thickness. 
 
 3. A Surface or Superficies, is an extension 
 or a figure, of two dimensions, length and 
 breadth ; but without thickness. 
 
 4. A Body or Solid, is a figure of three di- 
 mensions, namely, leagth, breadth, and depth, 
 or thickness. 
 
 6. Lines are either Right, or Curved, or 
 Mixed of these two. 
 
 6. A Right Line, or Straight Line, lies all 
 in the same direction, between its extremities ; 
 and is the shortest distance between two points. 
 
 When a line is mentioned simply, it means 
 a Right Line. 
 
 7. A Curve continually changes its direction 
 between its extreme points. 
 
 8. Lines are either Parallel, Oblique, Per- 
 pendicular, or Tangential. 
 
 9. Parallel Lines are always at the same per- 
 pendicular distance ; and they never meet though 
 ever so far produced. 
 
 10. ObUque lines change their distance, and 
 would meet, if produced on the side of the 
 least distance. 
 
 11. One line is Perpendicular to another, 
 when it inclines not more on the one side 
 
 than 
 
 (f7 
 
 Jrd7 
 
L/ 
 
 276 GEOMETRY, 
 
 than the other, or when the angles on both 
 sides of it are equal. 
 
 12. A line or circle is Tangential, or a 
 Tangent to a circle, or other curve, when it 
 touches it, without cutting, when both are pro- 
 duced. 
 
 13. An Angle is the inclination or opening 
 of two lines, having different directions, and 
 meeting in a point. 
 
 14. Angles are Rigkt or Oblique, Acute or 
 ©btuse. 
 
 15. A Right Angle is that which is made by 
 one line perpendicular to another. Or wheii 
 the angles on each side are equal to one an- 
 other, they are right angles. 
 
 16. An Oblique Angle is that which is 
 made by two oblique lines ; and is either less 
 or greater than a right angle. 
 
 17. An Acute Angle is less than a right 
 angle. 
 
 18. An Obtuse Angle is greater than a right 
 angle. 
 
 19. Superficies are either Plane or Curved. 
 
 20. A Plane Superficies, or a Plane, is that with which 
 a right line may, every way coincide. Or, if the line touch 
 the plane in two points, it will touch it in every point. But, 
 if not, it is curved. 
 
 21. Plane figures are bounded either by right lines or 
 curves. 
 
 22. Plane figures that are bounded by right lines have 
 names according to the number of their sides, or of their 
 angles ; for they have as many sides as angles ; the least 
 number being three. 
 
 23. A figure of three sides and angles is called a Triangle. 
 And it receives particular denominations from the relations 
 of its sides and angles. 
 
 24. An Equilateral Triangle is that whose 
 three sides are all equal. 
 
 A 
 
 26. An Isosceles Triangle is that which has / \ 
 
 two sides equal. /y^ 
 
DEFINITIONS. 
 
 277 
 
 26. A Scalene Triangle is that whose three 
 sides are all enequal. 
 
 27. A Right-angled Triangle is that which 
 has one ritcht-angle 
 
 28. Other triangles are Oblique-angled, and 
 are either Obtuse or Acute 
 
 29. An Obtuse-angled Triangle has one ob- 
 tuse angle. 
 
 30. An Acute-angled Triangle has all its three 
 angles acute. 
 
 31. A figure of Four sides and angles is called 
 a Quadrangle, or a Quadrilateral. 
 
 32. A Parallelogram is a quadrilateral which 
 has both its pairs of opposite sides parallel. 
 And it takes the following particular names, 
 viz. Rectangle, Square, Rhombus, Rhomboid. 
 
 33. A Rectangle is a parallelogram having a 
 right angle. 
 
 34. A Square is an equilateral rectangle ; 
 having its length and breadth equal. 
 
 35. A Rhomboid is an oblique-angled paral- 
 lelogram. 
 
 36. A Rhombus is an equilateral rhomboid ; 
 having all its sides equal, but its angles ob- 
 lique. 
 
 37. A Trapezium is a quadrilateral which 
 hath not its opposite sides parallel. 
 
 38. A Trapezoid has only one pair of oppo- 
 site sides parallel. i V 
 
 39. A Diagonal is a line joining any two op- 
 
 posite angles of a quadrilateral. /N. I 
 
 40^ Plane figures that have more than four sides, are, in 
 
 general, called Polygons : and they receive other particular 
 
 names, according to the number of their sides or angles 
 
 ^ Thus, * 
 
 41. A Pentagon is a polygon of five sides : a Hexagon, of 
 six sides ; a Heptagon, seven ; an Octagon, eight ; a Nona- 
 gon, nine; a Decagon, ten ; an Undecagoa, eleven ; and a Do- 
 decagon, twelve sides. 
 
 42. A 
 
27S GEOMETRY. 
 
 42. A Regular Polygon has all its sides and all its angles 
 equal. — If they are not both equal, the polygon is Irregular. 
 
 43. An Equilateral Triangle is also a Regular Figure of 
 three sides, and the Square is one of four ; the fornjer being 
 also called a i rigon, and the latter a Tetragon. 
 
 44. Any figure is equilateral, when all its sides are equal : 
 and it is equiangular T\hen all its angles are equal. When 
 both these are equal, it is a regular figure. 
 
 46. A Circle is a plain figure bounded by 
 a curve line, called the Circnuiference, which 
 is every where equidistant from a certain point 
 within, called its Centre. 
 
 The circumference itself is often called a 
 circle, and also the Periphery. 
 
 46. The Kadius of a circle is a line drawn 
 from the centre to the circumference. 
 
 47. The Diameter of a circle is a line drawn 
 through the centre, and terminating at the cir- 
 cumference on both sides. 
 
 48. An Arc of a circle is any part of the 
 circumference. 
 
 49. A Chord is a right line joining the ex- 
 tremities of an arc. 
 
 50. A Segment is any part of a circle bound- 
 ed by an arc and its chord. 
 
 61. A Semicircle is half the circle, or a seg- 
 ment cut off by a diameter. 
 
 The half circumference is sometimes called 
 the semicircle. 
 
 62. A Sector is any part of a circle which 
 is bounded by an arc, and two^adii drawn to its 
 extremities. 
 
 63. A Quadrant, or Quarter gf a circle, is a 
 sector having a quarter of the circumference 
 for its arc, and its two radii are perpendicular 
 to each other. A quarter of the circumference 
 is sometimes called a Quadrant. 
 
 64. Tl^,e 
 
DEFINITIONS. 
 
 27S 
 
 54. The Height or Altitude of a 6gure is 
 a perpendicular let fall from ah angle, or its 
 vertex, to the opposite side, called the base. 
 
 55. In a right-augled triangle, the side op- 
 posite the right angle is called the Hypothe- 
 nuse ; and the other two sides are called the 
 Legs, and sometimes the Base and Perpendi- 
 cular. 
 
 56. When an angle is denoted by three 
 letters, of which one stands at the angular 
 point, and the other two on the two sides, 
 that which stands at the angular pomt is read 
 in the middle. Thus the angle contained 
 by the lines BA and AD is called the angle 
 BAD or DAB. 
 
 57. The circumference of every circle is 
 supposed to be divided into 360 equal parts, 
 called Degrees : and each degree into 60 Mi- 
 nutes, each minute into 60 Seconds, and so on. 
 Hence a seniicircle contains 1 80 degrees, and 
 a quadrant 90 degrees. 
 
 58. The Measure of an angle, is an arc of 
 any circle contained between the two lines 
 which form that angle, the angular point being 
 the centre ; and it is estimated by the num- 
 ber of degrees contained in that arc. 
 
 59. Lines, or chords, are said to be Equi- 
 distant from the centre of a circle, when per- 
 pendiculars drawn to them from the centre 
 are equal. 
 
 60. And the right line on which the Great- 
 er Perpendicular falls, is said to be farther 
 from the centre. 
 
 61. An Angle In a segment is that which 
 is contained by two lines, drawn from any 
 point in \he arc of the segment, to the two 
 extremities of that arc. 
 
 62. An Angle On a segment, or an arc, is that which is 
 contained by two lines, drawn from any point in the opposite 
 or supplemental part of the circumference, to the extremi- 
 ties of the arc, and containing the arc between them. 
 
 63. An Angle at the circumference, ifi that 
 whose angular point is any wh^re in the cir- 
 cumference And an angle at the centre, is 
 that whose angular point is at the centre. 
 
 64. A 
 
280 
 
 GEOMETRY. 
 
 64. A right-lined figure is Inscribed in a 
 circle, or the circle Circumscribes it, when 
 all the angular points of the figure are in the 
 circumference of the circle. 
 
 65. A right-lined figure Circumscribes a 
 circle, or the circle is Inscribed in it, when all 
 the sides of the figure touch the circumference 
 of the circle. 
 
 66. One right-lined figure is Inscribed in 
 another, or the latter Circumscribes the for- 
 mer, when all the angular points of the for- 
 mer are placed in the sides of the latter. 
 
 67. A Secant is a line that cuts a circle, 
 lying partly within, and partly without it. 
 
 68. Two triangles, or other right-lined figures, are said to 
 be mutually equilateral, when all the sides of the one are 
 equal to the corresponding sides of the other, each to each : 
 and they are said to be mutually equiangular, when the angles 
 •f the one are respectively equal to those of the Other. 
 
 69. Identical figures, are such as are both mutually equ^" 
 lateral and equiangular ; or that have all the sides and all the 
 angles of the one, respectively equal to all the sides and all the 
 angles of the other, each to each ; so that if the one figure 
 were applied to, or laid upon the other, all the sides of the one 
 would exactly fall upon and cover all the sides of the other ; 
 the two becoming as it were but one and the same figure. 
 
 70. Similar figures, are those that have all the angles of 
 the one equal to all the angles of the other, each to each, and 
 the sides about the equal angles proportional, 
 
 71. The Perimeter of a figure, is the sum of all its sidei 
 taken together. 
 
 72. A Proposition, is something which is either proposed 
 •to be done, or to be demonstrated, and is either a problem or 
 
 a theorem. 
 
 73. A Problem, is something proposed to be done. 
 
 74. A Theorem, is something proposed to be demonstrated. 
 
 75. A Lemma, is something which is premised, or demon- 
 strated, in order to render what follows more easy^ 
 
 76. A CoroUory, is a consequent truth, gained immediately 
 from some preceding truth or demonstration. 
 
 77. A Schohum, is a remark or observation made upcm 
 something going before it. 
 
 AXIOMS. 
 
[ 281 1 
 AXIOMS. 
 
 1. Things which are equal to the same thing are equal 
 to each other. 
 
 2. When equals are added to equals, the wholes are equal. 
 
 3. When equals are taken from equals, the remainders are 
 equal. 
 
 4. When equals are added to unequals, the wholes are 
 unequal. 
 
 5. When equals are taken from unequals, the remainders 
 are unequal. ' 
 
 6. Things which are double of the same thing, or equal 
 things, are equal to each other. 
 
 7. Things which are halves of the same thing, are equal. 
 
 8. Every whole is equal to all its parts taken together. 
 
 9. Things which coincide, or fill the same space, are iden* 
 tical, or mutually equal in all their parts. 
 
 10. All right angles are equal to one another. 
 
 11. Angles that have equal measures, or arcs, are equal. 
 
 THEOREM ] 
 
 If two Triangles have Two Sides and the Included Angle 
 in the one, equal to Two Sides and the Included Angle 
 in the other, the Triangles will be Identical, or equal in all 
 respects. 
 
 In the two triangles abc, def, if 
 the side ac be equal to the side df, 
 and the side bc equal to the side ef, 
 and the angle c equal to the angle f ; 
 then will the two triangles be iden- 
 tical, or equal in all respects. A B D E 
 
 For conceive the triangle abc to be applied to, or placed 
 on, the triangle dbf, in such a manner that the point c may 
 
 Vol. L 37 coincide 
 
GEOMETRY. 
 
 coincide with the point f, and the side ac with the side df, 
 which is equal to it. 
 
 Then, since the angle f is equal to the angle c (hy hyp), 
 the side bc will fall on the side ef. Also, because ac is 
 equal to df, and bc equal to ef (by hyp), the point a will 
 coincide with the point d, and the point b with the pt-int e ; 
 consequently the side ab will coincide with the side de. 
 Therefore the two triangles are identical, arid have all th ir 
 other corresponding parts equal (ax. 9), namely, the ^id;' ab 
 equal to the side de, the angle a to the angle d, and the 
 angle b to the angle e. q. e. d, 
 
 THEOREM n. 
 
 When Two Triangles have Two Angleis and the included 
 Side in the one. equal to '3'wo Anjj;les and the included Side 
 in the other, the Triangles are Identical^ or have their olher 
 sides and angle equal. 
 
 Let the two triangles abc, def, q F 
 
 have the apgle a equal to the angle 
 D, the angle b equal to the angle e, 
 and the side ab equal to the side de ; 
 then these two triangles will be 
 identical. 
 
 For, conceive the triangle abc to be placed on the trfangle 
 DEF. in such nnanner that the side ab may fall exactly on the 
 equal side de. Then, since the angle'^A is equal to the angle^ 
 D (by hyp.), the side ac n«nst hU on the side df ; and, in 
 like manner, because the angle b is equal to the angle e the 
 side bc must fall on the side ef. Thus tbetlree sides of the 
 triangle abc will be exai tly placed on the three sides of the 
 triangle def : consequently the two triangles are identical 
 (ax. 9), having the other two sides ac, bc, equal to ti»e two 
 df, ef, and the remaining angle c equal to the remaining 
 angl^ F. ci. E. D. 
 
 THEOREM HI. 
 
 In an Isos^celes triangle, the Angles at the Base are equal. 
 Or, if a Triangle have Two Sides equal, their Opposite 
 Angles will also be equal. 
 
 if the triangle abc have the side ac equal 
 to the side bc : then will the angle b be 
 equal to the angle a. 
 
 For, conceive the angle c to be bisected, 
 or divided into two equal parts by the line 
 CD, making the angle acd. equal to the 
 angle bc©. 
 
 ^ Then, 
 
THT:0REMS. 283 
 
 Then, the two tr!3n2:les acd, bcd, have two sides and 
 the <: ."t ur;cf] ^ru:U' of the one, equal to two sides and the 
 cop.tainefl a^^5ie of the other, viz. the side ac equal to bc, the 
 BM^Jo A.D <^quU to BCD, and the side cd common ; there- 
 fore thcf^; tvso triangles are indentical, or equal in all re- 
 Sj)!- /s (th. 1) ; and consequently the angle a equal to the 
 
 a-VJ. B. Q. E. I). 
 
 Corol. 1, Honce the line which hisects the verticle angle of 
 aa isosceles triani^lc, bisects the base, and is also perpendicu- 
 lar to it. 
 
 Corol. 2-. Hence too it appears, that every equilateral tri- 
 angle, is also equian^uiar, or has all its angles equal. 
 
 THEOREM IV. 
 
 WtfEN a Tf iangK^ has Two of its x\ngles ^equal, the Sides 
 Opposite to them are also equal. 
 
 If the triargle abc, have the angle a C 
 
 eq'ial to th^ angle b. it will also have the side /K 
 
 AC oqijal to the side bc / I \ 
 
 For, conceiye tb'ii side ab to be bisected / I \ 
 
 in li»e point d, making ad equal to db ; / t \ 
 
 and j.nn DC, dividing the whvol.'^ triangle into A. D B 
 the two triangles acd, bcd. Also c orrceive 
 the t'iangle acd to be turned over upon the 
 triangle bcd, so that ad maj f :U on bd. 
 
 Then, because the line ad is equal to the line db ^by hyp.), 
 the point a coincides with the point b, and the point d with the 
 point D. Also, because the angle a is equal to the an5;;le b by 
 (hyp.), the hne ac will fall on the line bc, and the extremity c 
 of the side ac will coincide with the extremity c of the side bc, 
 becanse dc is common to both ; coiisequently the side ac is 
 equal to bc q. e. d*. 
 
 Corol. Hence every eiijuiangular triangle is also equila- 
 teral. 
 
 THEOREM V. 
 
 When Two Triangles have all the Three Sides in the one, 
 equal to all the Three Sides in the other, the Triangles are 
 Identical, or have also their Three Angles equal, each to each. 
 
 Let the two triangles, abc, abd, 
 have their three sides respectively 
 equal, viz. the side a« equal to ab, 
 AC to AD, and bc to bd ; then shall 
 the two triangles be identical, or have 
 their angles equal, viz. those angles 
 
 * I lis tlemon-iti-ation of Theoiem iv, does not appear to me to be 
 conclusive. Editoii. • 
 
 that 
 
^84 GEOMETR\. 
 
 that are opposite to the equal sides ; C 
 
 namely, the angle bag to the angle ^--^^"^^Nv 
 
 BAD, the angle abc to the angle abd, a <^— -__J__\-d 
 
 and the angle c to the angle d. ^;^~— \~/^ 
 
 For, conceive the two triangles to ^^""^-j/ 
 
 be joined together by their longest D 
 
 equal sides, and draw the line cd. 
 
 Then, in the triangle acd, because the side ac is equal 
 to AD (by hyp.), the angle acd is equal to the angle adc 
 (th. 3). In like manner, in the triangle bcd. the angle bcd is 
 equal to the angle bdc, because the side bc is equal to bd. 
 Hence then, the angle acd being equal to the angle, a©c and 
 the angle bcd to the angle bdc, by equal additions the sum of 
 the two angles acd, bcd, is equal to the sum of the two adc, 
 BDC, (ax. 2), that is, the whole angle acb equal to the whole 
 angle adb. 
 
 Since then, the two sides ac, cb, are equal to the two sides 
 AD, db, each to each, (by hyp.), and their contained angles 
 ACB, adb, also equal, the two triangles abc, and abd, are iden- 
 tical (th. 1), and have the other angles equal, viz. the angle" 
 BAc to the angle bad, and the angle abc to the angle abd. 
 
 Q. E. D. 
 
 THEOREM VI. 
 
 When one Lme meets another, the Angles which it makes 
 on the Same Side of the other,^ are together ^qual to Two 
 Right Angles. 
 
 Let the line ab meet the line cd : then 
 TPill the two angles abc, abd, taken together, 
 be equal to two right angles. 
 
 For, first, when the tw® angles abc, abd . 
 are equal to each other, they are both of 
 them right angles (def. 15.) 
 
 E A 
 
 B D 
 
 But when the angles are unequal, suppose be drawn per- 
 pendicular to CD. Then, since the two angles ebc, ebd, are 
 right angles (def. 15), and the angle ebd, is equal to the two 
 angles eba, abd, together (ax. 8), the three angles, ebc, eba, 
 and abd, are equal to two right Angles. 
 
 But the two angles ebc, eba, are together equal to the 
 angle abc (ax. 8). Consequently the two angles abc, abd, are 
 also equal to two right angles, q. e. d. 
 
 Corol. 1. Hence also, conversely, if the two angles abc, 
 abd, on both sides of the line ab, make up together two 
 right angles, then cb and bd form one continued right line cd. 
 
 Corol, 
 
THEOREMS. 286 
 
 Corol. 2. Hence, all the angles which can be made, at 
 any point b, by any number of lines, on the same side of 
 the right line cd, are, when taken all together, equal to two 
 right angles. 
 
 Corol. 3. And, as all the angles that can be made on the 
 other side of the line cd are also equal to two right angles ; 
 therefore ail the angles that can be made quite round a point 
 B, by any number of lines, are equal to four right angles. 
 
 Corol. 4. Hence also the whole circumfer- 
 ence of a circle, being the sum of the mea- 
 sures of all the angles that can be made about 
 the centre f (def. 67), is the measure of four 
 right angles. Consequently ,^ a semicircle, or 
 180 degrees, is the measure of two right 
 angles : and a quadrant, or 90 degrees, the measure of one 
 right angle. 
 
 THEOREM Vn. 
 
 When two Lines Intersect each other, the Opposite Angles 
 are equal. 
 
 Let the two lines ab, cd, intersect in 
 the point e ; then will the angle aec be 
 equal to the angle bed, and the angle 
 AED equal to the angle ceb. 
 
 For since the Hne ce meets the line 
 AB, the two angles aec, bec, taken to- 
 gether, are equal to two right angles (th. 6). 
 
 In like manner, the line be, meeting the line cd, makes 
 the two angles bec, bed, equal to two right angles. 
 
 Therefore the sum of the two angles aec, bep, is equal 
 to the sum of the two bec, bed (ax. 1). 
 
 And if the angle bec, which is common, be taken away 
 from both these, the remaining angle aec will be equal to 
 the remaining angle bed (ax. 3). 
 
 And in like manner it may be shown, that the angle aed 
 is equal to the opposite angle bec. 
 
 THEOREM Vm. 
 
 When One Side of a Triangle is produced, the Outward 
 Angle is Greater than either of the two Inwajrd Opposite 
 Angles. 
 
 Let 
 
286 
 
 GEOMETRY. 
 
 Let ABC be a triangle, having the 
 side AB produced to d ; then will the 
 outward angle cbd be greater than 
 either of the inward opposite angles a 
 or c. 
 
 For, conceive the side 'bc to be bi- 
 sected in the point e, and drnw the line 
 AE, producing it till ef be equal to ae ; 
 and join BF. - 
 
 Then, since the two triangles aec, bef, have the side 
 ae = the side ef, and the side ce = the side be (by suppos.) 
 and the inrJuded or opposite angles at e also equal (tb. 7), 
 therefore those t;Vo triangles are equal in all respects 
 (th. 1), and have the angle c«= tht corresponding angle ebf. 
 Bat the angle cbd is greater tha» the angle kbf ; consequent- 
 ly the said outward angle cbd is also great^T than the angle c. 
 
 In like manner, if cb be prod-jced to o. and ab be bisected, 
 it Diaj be shown that the outward angle abg, or its equal cbd, 
 is greater than the other. angle a. 
 
 THEOREM IX. 
 
 DB 
 
 ') 
 
 The Greater Side, of every Triangle, is opposite to the 
 Greater Angle ; and the Greater Angle opposite to the Great- 
 er Side. 
 
 Let ABC be a triangle, having the side 
 ab greater than the side ac ; then will 
 the angle acb. opposite the greater side 
 AB, !>e greater than the angle b, opposite 
 the less side ac^ 
 
 For, on the greater side ab, take the 
 part ar equal to thS less side ac, and join cd. Then, sin(ie 
 BCD is a triangle, the outward angle adc is greater than the 
 inward oppoj^ite angle b (th. 8). But the angle acd is equal 
 to .tbe said outward angle adc, because ad is equal to ac (th. 3). 
 Consequently the asigle' acd also is greater than the angle b. 
 And siace the angle acd is only a part of acb, much more 
 must the whole angle acb be greater than the angle b. q. e. d. 
 
 Again, conversely, if the angle c be greater than the angle 
 B, then will the side ab, opposite the former, be greater than 
 the side ac, opposite the latter. 
 
 For, if ab be not greater than ac, it must be either equal 
 to it, or less than it. But it cannot be equal, for then 
 
THEOREMS. 287 
 
 the angle c would be equal to the angle b (th. 3) which it is 
 not, by the supposition. Neither can it be less, for then the 
 auji^le c would be le!?s than the angle b, by the former part of 
 this; ; which is also contrary to the supposition. The side ab, 
 then, being neiilier equal to ac, nor less than it, must neces- 
 sarily be greater, q. e. d. 
 
 THEOREM X. 
 
 The Sum of any Two Sides of a Triangle is Greater than 
 the Third Side. ^ 
 
 Let ABC be a triano;le ; then will the 
 sum of any two of its sides be greater than 
 the third side, as for instances, ac + cb 
 greater than as. 
 
 For, pi'oduce ac till cp be equal to 
 CB, or AD equal to the sum of the two 
 AC + CB ; and join bd : — Then, because ^ ~~B 
 
 CD is equal to cb (by «onstr.), the angle d is equal to the angle 
 CBD (th. 3). But the angle abd is greater than the angle cbd, 
 consequently it must also be greater than the angle d. And, 
 since the greater side of any triangle is opposite to the' greater 
 angle (th. 9), the side ad (of the triangle abd) is greater than 
 the side ab. " But ad. is equal to ac and cd, or ac and cb, 
 taken together (by constr.) ; therefore ac -{- cb is also great- 
 er than AB. Q. E. d. 
 
 CoroL The shortest distance between two points, is a single 
 right line drawn from the one point to the other. 
 
 THEOREM XI. 
 
 The Difference of any Two Sides of a Triangle, is Less than 
 the Third Side. 
 
 Let ABC be a triangle ; then will the 
 difference of any two sides, as ab — ac, 
 be less than the third side bc. 
 
 For, produce the less side ac to d, till 
 ad be equal to the greater side ab, so 
 that CD may be the difference of the two 
 sides AB — AC ; and join bd. Then, be- 
 cause AD is equal to ab (by constr.), the opposite angles o 
 and ABD are equal (th. 3). But the angle cbd is less than the 
 angle abd, and consequently also less than the equal angle d. 
 And since the greater side of any triangle is opposite to the 
 
 greater 
 
2*8 GEOMETRY. 
 
 greater angle (th. 9), the side cd (of the triangle bcd) is less 
 than the side bc. q.. e. d. 
 
 THEOREM Xn. 
 
 When a Line Intersects two Parallel Lines, it makes the 
 Alternate Angles Equal to each other. 
 
 Let the line ef cut the two parallel 
 lines AB, CD ; then will the angle aef be 
 equal to the alternate angle efd. 
 
 For if they are not equal, one of them 
 must be greater than the other ; let it be 
 EFD for instance which is the greater, if 
 possible ; and conceive the line fb to be 
 drawn ; cutting off the part or angle efb equal to the angle 
 AEF ; and meeting the line ab in the point b. 
 
 Then, since the outward angle aef, of the triangle bef, is 
 greater than the inward opposite angle liFB (th. 8) ; and since 
 these two angles also are equal (by the constr.) it follows, that 
 those angles are both equal and unequal at the same time : 
 which is impossible. Therefore the angle efd is not unequal 
 to the alternate angle aef, that is, they are equal to each 
 other. Q. e. d. 
 
 Corol. Right lines which are perpendicular to one, of tw« 
 parallel lines, are also perpendicular to the other. 
 
 THEOREM Xin. 
 
 When a line, cutting Two other Lines, makes the Al- 
 ternate Angles Equal to each other, those two Lines are Pa- 
 rallel. 
 
 Let the line ef, cutting the two lines 
 ab, CD, make the alternate angles aef, 
 »fe, equal to each other ; then will ab 
 be parallel to cd. 
 
 For if they be not parallel, let some - ^ _ 
 
 •ther line, as fg, be parallel to ab. ^ /lE" '""---P 
 Then, because of these parallels, the ^ 
 
 angle aef is equal to the alternate angle efg {th. 12, But 
 \he angle aef is equal to the angle efd (by hyp.). There- 
 fore the angle efd is equal to the angle efg (ax. 1) ; that is, 
 a part is equal to the whole, which is impossible. Therefore 
 BO line but cd can be parallel to ab. q. e. d. 
 
 Corol. Those Hues which are perpendicular to the same 
 line, are parallel to each other. 
 
 THEOREM 
 
THEOREMS, 28^ 
 
 THEOREM XrV. 
 
 When a Line cuts two Parallel Lines, the Outward Angle 
 is Equal to the Inward Opposite one, on the Same Side ; and 
 the two inward Angles, on the Same Side, equal to two Right 
 Angles. 
 
 Let the line ef cut the two parallel 
 lines AB, CD ; then will thei)utward angle 
 EGB be equal to the inward opposite an- 
 gle GHD, on the same side of the line ef; 
 and the two inward angles bgh, ghd, 
 taken together, will be equal to two 
 right angles. 
 
 For, since the two lines ab, cd, are 
 parallel, the angle agh is equal to the alternate angle ghd,. 
 (th. 12). But the angle agh is equal to the opposite angle 
 EGB (th. 7). Therefore the angle egb is also equal to the 
 angle ghd (ax. 1)-. q. e. d. 
 
 Again, because the two adjacent angles egb, bgh, are to- 
 gether equal to two right angles (th. 6) ; of which the an- 
 . gle egb has been shown to be equal to the angle ghd ; there- 
 fore the two angles bgh, ghd, taken together, are also equal 
 to two right angles. 
 
 Corol. 1. And, conversely, if one line meeting two other 
 lines, make the angles on the same side of it equal, those 
 two lines are parallels. 
 
 Corol. 2. If a line, cutting two other lines, make the sum 
 ©f the two inward angles, on the same side, less than two 
 right angles, those two lines will not be parallel, but will m^et 
 each other when produced. 
 
 THEOREM XV. 
 
 Those Lines which are Parallel to the Same Line, are Pa- 
 rallel to each other. 
 
 Let the Lines ab, cd, be each of -h 
 
 them parallel to the line ef ; then shall J^ 1 ^g 
 
 the lines ab, cd, be parallel to each ^ | -p. 
 
 other. ^ Hi 
 
 For, let the line gi be perpendicular £ X F 
 
 to ef. Then will this line be also per- 
 pendicular to both the hnes ab, cd (corol th. 12), and con- 
 sequently the two lines ab, cd, are parallels (corol. th. 13). 
 
 Q. E. D. 
 
 Vol. L 38 THEOKBM 
 
290 GEOMETRY. 
 
 THEOREM XVL 
 
 When one Side of a triangle is produced, the Outward 
 Angle is equal to both the inward Opposite Angles taken to- 
 gether. 
 
 Let the side, ab, of the triangle C J| 
 ABC, be produced to d ; then will the /\ / 
 outward angle cbd be equal to the / \ / 
 sum of the two inward opposite an- / \'/ 
 gles A and c. ^ g 5 
 
 For, conceive be to be drawn pa- 
 rallel to the side ac of the triangle. 
 
 Then bc, meeting the two parallels ac, be, 'makes the alter- 
 nate angles c and cbe equal (th. 12). And ad, cutting the 
 same two parallels ac, be, makes the inward and outward 
 angles on the same side, a and ebd, equal to each other (th. 
 14). Therefore, by equal additions, the sum of the two an- 
 gles A and c, is equal to the sum of the two cbe and ebd, that 
 is, to the whole angle cbd (by ax. 2). «i. e. d. 
 
 THEOREM XVII. 
 
 In any Triangle, the sum of all the Three Angles is equal t© 
 Two Right Angles. 
 
 C 
 
 A. 
 
 Let ABC be any plane triangle ; then 
 the sum of the three angles a -f" b "h c 
 is equal to two right angles. 
 
 For, let the side ab be produced to d. 
 
 Then the outward angle cbd is equal ^ ■ — ^ *v 
 
 to the sum of the two inward opposite 
 angles a -f c (th. 16). To each of these equals add the 
 inward angle b, then will the sum of the three inward an- 
 gles A -f B -f c be equal to the^um of the two adjacent angles 
 ABC -{- CBD (ax. 2). But the sum of these two last adjacent 
 angles is equal to two right angles (th. 6). Therefore also 
 the sum of the three angles of the triangle a -f b -|- c ia 
 equal to two right angles (ax. 1). q. e. d. 
 
 Carol. 1 . If two angles in one triangle, be equal to two 
 angles in another triangle, the third angles will also be equal 
 (ax. 3), and the two triangles equiangular. 
 
 Cbrol. 2. If one angle in one triangle be equal to one 
 angle in another, the sums of the remaining angles will also 
 be equal (ax. 3.) 
 
THEOREMS. 
 
 251 
 
 Corol. 3. If one angle of a triangle be richt, the sum of 
 the other two will also be equal to a right ani^le, and each 
 of them singly will be acute, or less than a right angle. 
 
 Corol. 4. The two least angles of every triangle are acute, 
 or each less than a right angle. 
 
 l^HEOREM XVin. 
 
 In any Quadrangle, the sum of all the Four Inward Angles, 
 is equal to Four Right Angles. 
 
 Let ABCD be a quadrangle ; then the 
 sum of the four inward angles, a + » 4* 
 c -j- D is equal to four ri^ht angles. 
 
 Let the diagonal ac be drawn, dividing 
 the quadrangle into two triangles, ABC, ADC. 
 Then, because the sum of the three angles 
 of each of these triangles is equal to two A B 
 
 right angles (th. 17) ; it follows, that the sum of all the 
 angles of both triangles, which make up the four angles of 
 the quadrangle, must be equal to four right angles (ax. 2). 
 
 Q. E. D. 
 
 Corol. 1. Hence, if three of the angles be right ones, the 
 fourth will also be a right angle. 
 
 Corol. 2. And, if the sum of two of the four angles be 
 equal to two right angles, the sum of the remaining two will 
 also be equal to two right angles. 
 
 THEOREM XIX. 
 
 In any figure whatever, the Sum of all the Inward Angles, 
 taken together, is equal to Twice as many Right Angles, 
 wanting four, as the Figure has Sides 
 
 Let ABCDE be any figure ; then the 
 sum of all its inward angles, a -f b + 
 c -}- D -|- E, is equal to twice as many 
 right angles, wanting four, as the figure 
 has sides. 
 
 For, from any point p, within it, draw 
 lines PA, PB, PC, kc. to all the angles, 
 dividing the polygon into as many tri- 
 angles as it has sides. Now the sum of the three angles of 
 each of these triaugles, is equal to two right angles (th. 17) ; 
 therefore the sum of the angles of all the triangles is equal 
 to Twice as miny right angles as the figure has sides. But 
 the sum of .all the angles about the point p, which are so 
 
 manj 
 
29f 
 
 GEOMETRY. 
 
 many of the angles of the triangles, but no part of the in- 
 ward angles of the polygon, is equal to four right angles 
 (cord. 3, th 6). and must be deducted out of the former 
 sum. Hence it follows that the sum of all the inward angles 
 of the polygon alone, A-j-B-f'C-l-D-f'E, is equal to twice 
 as many right angles as the figure has sides, wanting the 
 said four right angles. Q. e. d, 
 
 THEOREM XX. 
 
 When every Side of any Figure is produced out, the 
 Sum of all the Outward Angles thereby made, is equal to 
 Four Right Angles. 
 
 Let A, B, c, &c. be the outward 
 angles of any polygon, made by pro- 
 ducing all the sides ; then will the sum 
 A -f B + c + D + E, of all those outward 
 angles, be equal to four right angles. 
 
 For every one of these outward angles, 
 together with its adjacent inward angle, 
 make up two right angles, as a -f a equal 
 to two right angles, being the two angles 
 made by one line meeting another (th. 6). And there 
 being as many outward, or inward angles, as the figure has 
 sides : therefore the sum of all the inward and outward 
 angles, is equal to twice as many right angles as the figure 
 has sides. But the sum of all the inward angles, with four 
 right angles, is equal to twice as many right angles as the 
 figure has sides (th. 19). Therefore the sum of all the in- 
 ward and all the outward angles, is equal to the sum of all 
 the inward angles and four right angles (by ax. 1). From 
 each of these take away all the inward angles, and there 
 remain all the outward angles equal to four right angles 
 (by ax. 3). 
 
 * 
 
 THEOREM XXI. 
 
 A Perpendicular is the Shortest Line that can be drawn 
 from a Given Point to an Indefinite Line.' And, of any- 
 other Lines drawn from the same Point, those that are Nearest 
 he Perpendicular, are Less than those More Remote. 
 
 If AB, AC, ao, &c. be lines drawn from 
 the given point a, to the indefinite line de, 
 of which AB is perpendicular. Then shall 
 the perpendicular ab be less than ac, and 
 AC, less than ad, &;c. 
 
 For, the angle b being a right one, the 
 
THEOREMS. 293 
 
 angle c is acute (by cor. 3, th. 17), and therefore less than 
 the angle b. But the less angle of a triangle is subtended by 
 the less side (th. 9). Therefore the side ab is less than the 
 side AC. 
 
 Again, the angle acb being acute, as before, the adjacent 
 angle acd will be obtuse (by th. 6) ; consequently the angl« 
 » is acute (corol. 3, th. 17), and therefore is less than the angle 
 €. And since the less side is opposite to the less angle, there- 
 fore the side ac is less than the side ad. - q. e. d 
 
 Corol. A perpendicular is the least distant of a given point 
 ^om a line. 
 
 THEOREM XXII. 
 
 The Opposite Sides and Angles of any Parallelogram are 
 equal to each other ; and the Diagonal divides it into two 
 Equal Triangles. 
 
 Let abcd be a parallelogram, of which ^^ P 
 
 the diagonal is bd ; then will its opposite 
 sides and angles be equal to each other, 
 and the diagonal bd will divide it into two 
 equal parts, or triangles. 
 
 For, since the sides ae and do are pa- 
 rallel, as also the sides ad and bc (detin. 
 32), and the line bd meets them ; therefore the alternate 
 angles are equal (th. 12), namely, the angle aed to the angle 
 €db, and the angle adb to the angle cbd. Hence the two 
 triangles, having two angles in the one equal to two angles 
 ia the other, have also their third angles equal (cor: 1, th. 17), 
 namely, the angle a equal to the angle c, which are two of the 
 opposite angles of the parallelogram. 
 
 Also, if to the equal angles abd, cdb, be added the equal 
 angles cbd, adb, the wholes will be equal (ax. 2), namely, the 
 whole angle abc to the whole ado, which are the other two 
 opposite angles of the parallelogram. q.' e. d. 
 
 Again, since the two triangles are mutually equiangular and 
 have a side in each equal, viz. the common side bd ; there- 
 fore the two triangles are identical (th. 2), or equal in all re- 
 spects, namely, the side ab equal to the opposite side dc, and 
 ad equal to the opposite side bc, and the whole triangle abd 
 equal to the whole triangle BCB. q. e. d, 
 
 Corol. 
 
294 GEOMETRY. 
 
 Corol. 1. Hence, if one angle of a parallelogrskm be a right 
 angle, all the other three will also be li^-ht angles, and the 
 parallelogram a rectangle. 
 
 Corol. 2. Hence also, the sam of any two adjacent angles of 
 a parallelogram is equal to two right angles. 
 
 THEOREM XXIII. 
 
 Every Quadrilateral, whose Opposite Sides are equal, is a 
 Parallelogram, or has its Opposite Sides Parallel. 
 Let ABCD be a quadrangle having the 
 opposite sides equal, namely, the side ab 
 equal to dc, and ad equal to eg ; then 
 shall these equal sides be also parallel, 
 and the figure a parallelogram. 
 
 For, let the diagonal bd be drawn. 
 Then, the triangles, abd, cbd, being 
 mutually equilateral (by hyp.), they are 
 also mutually equiangular (th. 6), or have their correspond- 
 ing angles equal ; consequently the opposite sides are parallel 
 (th. 13) ; viz. the side ab parallel to do, and ad parallel to bc, 
 and the figure is a parallelogram. q. e d. 
 
 THEOREM XXIV. 
 
 Those Lines which join the Corresponding Extremes of 
 two Equal and Parallel Lines, are themselves Equal and 
 Parallel. ^^ 
 
 Let ab, DC, be two equal and parallel lines ; then will the 
 lines AD, bc, which join their extremes, be also equal and pa- 
 rallel. [See the fig. above.] 
 
 For, draw the diagonal bd. Then, because ab and dc are 
 parallel (bv hyp), the angle abd is equal to the alternate 
 angle bdc (th 12), Hence then, the two triangles having two 
 sides and the contained angles equal, viz. the side ab equal to 
 the side dc, and the side bd common, and the contained 
 angle abd equal to the contained angle bdc, they have the 
 remaining sides and angles also respectively equal (th. 1) ; 
 consequently ad is equal to bc, and also parallel to it (th. 12). 
 
 Q. E. D. 
 THEOREM XXV. 
 
 Parallelograms, as also Triangles, standing on the Same 
 Base, and between the Same Parallels, are equal to each 
 «ther. 
 
 Let 
 
THEOREMS, 
 
 id's 
 
 Let ABCD, ABEF, be two parallelo- 
 grams, and ABC, ABF, two triangles, 
 stalling on the same base ab, and be- 
 tween the same parallels ab, de : then 
 will the parallelogram ABCD,be equal to 
 the parallelogram ABEF, and the triangle 
 ABC equal to the triangle abf. 
 
 For, since the line de cuts the two 
 parallels ap, be, and the two ad, bc, it makes the angle e 
 equal to the angle afd, and the angle d equal to the angle bce 
 (th. 14) ; the two triangles adf, bce, are therefore equiangu- 
 lar (cor. 1, th. 17) ; and having the two corresponding sides, 
 ad, bc, equal (th. 22), being opposite sides of a parallelogram, 
 these two triangles are identical, or equal in all respects 
 (th. 2). If each of these equal triangles then be taken from 
 the whole space abed, there will remain the parallelogram 
 ABEF in the one case, equal to the parallelogram abcd in the 
 other (by ax. 3). 
 
 Also the triangles abc, abf, on the same base ab, and be- 
 tween the same parallels, are equal, being the halves of thfe 
 said equal parallelograms (th. 22). q. E d. 
 
 Carol. 1. Parallelograms, or triangles, having the same base 
 and altitude, are equal. For the altitude is the same as the 
 perpendicular or distance between the two parallels, which is 
 every where equal, by the definition of parallels. 
 
 Carol. 2. Parallelograms, or triangles, having equal bases 
 and altitudes, are equal. For, if the one figure be applied 
 with its base on the other, the bases will coincide or be the 
 same because they are equal : and so the two figures, having 
 the same base and altitude, are equal. 
 
 THEOREM XXVI. 
 
 3) C E 
 
 If a Parallelogram and a Triangle stand on the Same Base, 
 and between the Same Parallels, the Parallelogram will be 
 Double the Triangle, or the Triangle Half the Parallelogram. 
 
 Let abcd be a parallelogram, and abe, a 
 triangle, on the same base ab, and between 
 the same parallels, ab, de ; then will the pa- 
 rallelogram ABCD be double the triangle abe, 
 or the triangle half the parallelogram. 
 
 For, draw the diagonal ac of the parallelo- 
 gram, dividing it into two equal parts (th. 22). 
 Tken because the triangles abc, abe, on the 
 
 same 
 
296 GEOMETRY. 
 
 same base, and betfveen the same parallels, are equal (th. 25) ; 
 and because the one triangle abc is half the parallelogram 
 ABCD (th. 22), the other equal triangle abe is also equal to half 
 the same parallelogram abcd. q. e. d. 
 
 Corol. 1. A triangle is equal to half a parallelogram of the 
 same base and altitude, because the aHitude is the perpendi- 
 cular distance between the parallels, which is every where 
 equal, by the definition of parallels. 
 
 Corol. 2. If the base of a parallelogram be half that of a 
 triangle, of the same altitude, or the base of the triande be 
 double that of the parallelogram, the two figures will be equal 
 to each other. 
 
 THEOREM XXVn. 
 
 Rectangles that are contained by Equal Lines, are Equal 
 to each other. 
 
 Let BD, FH, be two rectangles, having J) C' H G 
 
 the sides ab, bc, equal to the sides ef, fg, 
 each to each ; then will the rectangle bd 
 be equal to the rectangle fh. 
 
 / 
 
 For, draw the two diagonals ac, eg, di- j^ j^ j^ ]gi 
 Tiding the two parallelograms each into 
 two equal parts. Then the two triangles 
 ABC, efg, are equal to each other (th. 1), because they have 
 the two sides ab, bc, and the contained angle, b equal to the 
 two sides ef, fg, and the contained angle p (by hyp). But 
 these equal triangles are the halves, of the respective rectan- 
 angles. And because the halves, or the triangles, are equal, 
 the wholes, or the rectangles, db, hf, are also equal (by ax. 6). 
 
 Q. E. i>. 
 
 Carol. The squares on equal lines are also equal ; for 
 every square is a species of rectangle. 
 
 THEOREM XXVUI. 
 
 The Complements of the Parallelograms, which are about 
 the Diagonal of any Parallelogram, are equal to each other. 
 
 Let AC be a parallelogram, bd a dia- 
 gonal, eif parallel to ab or dc, and gih 
 parallel to ad or bc, making ai, ic com- 
 plements to the parallelograms eg, hf, 
 which are about the diagonal db : then ^ jj 
 
 will the complement ai be equal to the 
 complement IC. 
 
 For, 
 
THEOREMS. 
 
 297 
 
 For, since the diagonal db biserts the three parallelograms 
 AC, EG, HF, (th. 22) ; therefore, the whole triangle dab 
 being equal to the whole triangle dcb, and the parts dei, 
 IHB, respectively equal to the parts dgi, ifb, the remaining 
 parts Ai, ic, must also be equal (by ax. 3). q. e. d. 
 
 THEOREM XXIX. 
 
 A Trapezoid, or Trapezium having two 'Sides Parallel, 
 is equal to Half a Parallelogram, whose Base is the Sum of 
 those two Sides, and its Altitude the Perpendicular Distance 
 between them. 
 
 D C H F 
 
 A. GB E 
 
 Let ABCD be the trapezoid, having its 
 two sides ab, do, parallel ; and in ab 
 produced take be equal to do, so that 
 AE may be the sum of the two parallel 
 sides ; produce dc also, and let ef, gc, 
 bh, be all three parallel to ad. Then is 
 AF a parallelogram of the same altitude with the trapezoid 
 ABCD, having its base ae equal to the sum of the parallel sides 
 of the trapezoid ; and it is to be proved that the trapezoid 
 ABCD is equal to half the parallelogram af. 
 
 Now, since triangles, or parallelograms, of equal bases 
 and altitude, are equal (corol. 2, th. 25), the parallelogram 
 DG is equal to the parallelogram he, and the triangle cgb 
 equal to the triangle chb ; consequently the line bc bisects, 
 or equally divides, the parallelogram af, and abcd is the 
 half of it. (i. E. D. 
 
 THEOREM XXX. 
 
 The Sum of .all the Rectangles contained under one 
 Whole Line, and the several Parts of another Line, any way 
 divided, is Equal to the Rectangle contained under the Two 
 Whole Lines. 
 
 Let' AD be the one line, and ab the 
 other, divided into the parts ae, ef, ^ G H C 
 fb ; then will the rectangle contained by 
 ad and ab, be equal to the sum of the 
 
 rectangles of ab and ae, and ad and ef, 
 
 and ad and fb : thus expressed, ad . ab A. E P IB 
 
 = AD . AE -j- AD . ef + AD . FB. 
 
 For, make the rectangle ac of the two whole lines ad, 
 AB ; and draw eg, fh, perpendicular to ab, or parallel to 
 AD, to which they are equal (th. 22). Then the whole 
 rectangle ac is made up of all the other rectangles ag, 
 
 Vol. L 39 eh, 
 
GEOMETRY. 
 
 i> one 
 
 E FB 
 
 I r B 
 
 EH, Fc. But these rectangles are contain- 
 ed by AD and ae, eg and ef, fh and fb ; 
 which are equal to the rectangles of ad 
 and AE, AD and ef, ad and fb. because 
 AD is equal to ea<^h of the two eg; fh. 
 Therefore the rectangle ad . ab is' equal to the sunn of all 
 the other rectangles ad . ae, ad . ef, ad . fb. q,. e. d. 
 
 Carol. If a right line be divided into any two parts ; the 
 square on the whole line, is equal to both the rectangles of 
 the whole line and each of the parts. 
 
 THEOREM XXXI. 
 
 The Square of the Sura of two Lines is greater than the 
 Sum of their Squares, by Twice the Rectangle of the said 
 Lines. Or, the Square of a whole Line, is equal to the 
 Squares of its two Parts, together with Twice the Rectangle 
 of those Parts. ~ 
 
 Let the line ab be the sum of any two 
 lines AC, cb : then will the square of ab 
 be equal to the squares of ac, cb, together 
 with twice the rectangle of ac . cb. That 
 
 is, AB2 = AC2 -J-.CB2 -f 2aC . CB. 
 
 For, let ABDE be the square on the sum a C B 
 
 or whole line ab, and acfg the square 
 on the part ac Produce cf and gf to the other side at n 
 and i. 
 
 From the lines ch, gi, which are equal, being each 
 equal to the sides of the square ab or bd (th. 22), take the 
 parts cf, gf, which are also equal, being the sides of the* 
 square af, and there remains fh equal to fi, which are 
 also equal to dh, di, being the opposite sides of a parallelo- 
 gram. Hence the figure hi is equilateral : and it has all 
 its angles right ones (coroL 1, th. 22) ; it is therefore a 
 square on the line fi, or the square of its equal cb. Also 
 the figures ef, fb, are equal to two rectangles under ac 
 and cb, because gf is equal to ac, and fh or fi equal 
 to cB. But the whole square ad is made up of the four 
 figures, viz. the two squares af, fd, and the two equal rect- 
 angles ef, fb That is, the square of ab is equal to the 
 squares of ac, cb, together with twice the rectangle of ac, 
 
 CB. ^ Q. E. D, 
 
 Corol. Hence, if a line be divided into two equal parts ; 
 the square of the whole line, will be equal to four times the ^ 
 squa^re of half the liae. 
 
 THEOREM 
 
 
 
 y 
 
 
THEOREMS. 
 
 299 
 
 a 
 
 F 
 
 E 
 
 
 
 EC 
 C 
 
 D 
 
 
 A. B 
 
 
 K. r 
 
 THEOREM XXXII. 
 
 The Square of the Difference of two Lines, ie less than 
 the Sum of their Squares, by Twice the Rectangle of the 
 said Lines. 
 
 Let AC, Bc, be any two li|res, and ab 
 their difference : then will the square of 
 AB be less than the squares of ac, bc, by 
 twice the rectangle of ac and bc. 'Or, 
 
 AB^ = AC^ -{- BC2 -» 2 AC . BC. 
 
 For let ABDE be the square on the dif- 
 ference AB, and ACFG the square on the 
 line AC. Produce ed to h ; also produce 
 DB and Hc, and draw ki, making bi the square of the other 
 line BC. 
 
 Now it is visible that the square ad is less than the twcr 
 squares af, bi, by the two rectangles ef, di. But gf is equal 
 to the one line ac, and ge, or fh is equal to the other line bo; 
 consequently the rectani^le ef, contained under eg and gf, is 
 equal to the rectangle of ac and bc. 
 
 Again, fh being equal to ci or bc or dh, by adding the 
 common part hc, the whole hi will He equal to the whole fc, 
 or equal to ac ; and consequently the figure i>i is equal to the 
 rectangle contamed by ac and bc. 
 
 Hence the two figures ef, di, are two rectangles of the 
 two lines ac, bc ; and consequently the square of ab is less 
 than the squares of ac, bc, by twice the rectangle ac . bc 
 
 ft. E. D. 
 
 THEOHEM XXXni. 
 
 The Rectangle under the Sum and Diff<p»rence of tw© 
 Lines, is equal to the Difference of the Squares of those 
 Lines. 
 
 Let ab, ac, be any two unequal lines ; E K D 
 
 then will the difference of the squares of ~ 
 
 ab, AC, be equal to a rectangle under G- 
 their sum and difference, "^fhat is. 
 
 F 
 
 B 
 
 H 
 
 AB2 — AC2 = AB + AC . AB — AC. Ji. C 
 
 For, let ABDE be the square of ab, and 
 ACFG the square of ac. Produce db 
 till BH be equal to ac ; draw hi parallel 
 to AB or ED. and produce fc both ways to 
 I and K. 
 
 Then the difference of the two squares ad, af, is evi- 
 dently 
 
sou 
 
 GEOMETRY. 
 
 dently the two rectangles ef, kb. But the rectangles ef, bi, 
 are equal, being contained under equal lines ; for ek and bh 
 are each equal to ac, and ge is equal to ce, being each equal 
 to the difference between ab and ac, or their equals ae and 
 AG. Therefore the two ef, kb, are equal to the two kb, bi, 
 or to the whole kh ; and consequently kh is equal to the dif- 
 ference of the squares ad, af. But kh is a rectangle contE^in- 
 ed by dh, or the sum of ab and ac, and by kd, or the differ- 
 ence of AB and AC Therefore the difference of the squares 
 of AB, AC, is equal to the rectangle under their sum and dif- 
 ference. €t. E. D. 
 
 THEOREM XXXIV. 
 
 In any Right-angled Triangle, the square of the Hypothe- 
 nuse, is equal to the Sum of the Squares of the other two 
 Sides. 
 
 Let ABC be a right-angled triangle, 
 having the right angle c ; then will the 
 square of the hypothenuse ab, be equal 
 to the sum of the squares of the other 
 two sides ac, cb. Or ab^ = ac^ -f- 
 
 BC^. 
 
 For, on ab describe the square ae, 
 anl on ac, cb, the squares ag, bh ; 
 then draw ck parallel to ad or be ; and 
 join Ai, BF, CD, ce. D K El 
 
 Now, because the line ac meets the two cg, cb, so as to 
 make two right angles, these two form one straight line gb 
 (corol. 1, th. 6). And because the angle fac is equal to the 
 angle dab, being ^ch a right angle, or the angle of a square ; 
 to each of these equals add the common angle bac, so will 
 the whole angle or sum fab, be equal to the whole angle or 
 sum cad. But the line fa is equal to the line ac, and the 
 line ab to the line ad, being sides of the same square ; so 
 that the two sides fa, ab, and their included angle fab, are 
 equal to the two sidos ca, ad, and the contained angle cad, 
 each to each ; therefore the whole triangle afb is equal to the 
 whole triangle acd (th. 1). 
 
 But the square ag is double the triangle afb, on the same 
 base fa, and between the same parallels fa, gb (th. 26) ; in 
 like manner, the parallelogram ak is double the triangle acd, 
 on the same base ad, and between the same parallels ad, ck. 
 And since the doubles of equal things, are equal (by ax. 6) ; 
 therefore the square ag is equal to the parallelogram ak. 
 
THEOREMS. 301 
 
 In like manner, the other square bh is proVed equal to the 
 other parallelogram bk. Consequently the two squares ag 
 and BH together, are equal to the two parallelograms ak and 
 BK together, or to the whole square ae. That is, the sum of 
 the two squares on the two le§fe sides, is equal to the square 
 on the greatest side. q. e. d. 
 
 Corol. 1. Hence the square of either of the two less sides, 
 is equal to the difference of the squares of the hypothenuse 
 and the other side (ax. 3) ; or, equal to the rectangle contain- 
 ed by the sum and diflference of the said hypothenuse and 
 other side (th. 33). 
 
 Corol. 2. Hence also, if two right-angled triangles have 
 two sides of the one equal to two corresponding sides of the 
 other ; their third sides will also be equal, and the triangles 
 identical. 
 
 THEOREM XXXV. 
 
 In any Triangle, the Difference of the Squares of the 
 two Sides, is Equal to the Difference of the Squares of the 
 Segments of the Base, or of the two Lines, or Distances, 
 included between the Extremes of the Base and the perpen- 
 dicular. 
 
 Let ABC be any triangle, having 
 CD perpendicular to ab ; then will 
 the difference of the squares of ac, 
 Bc, be equal to the difference of ^ / l 
 
 the squares of ad, bd ; that is, ac ^ y I a 
 
 — Bc2 = ad2 -bd2 a B 3) a. D B 
 
 For, since ac^ is equal to ad^ -f- cd^ > ,, , . 
 
 and Bc2 is equal to bd^ -j- cd^. J ^ ^ * / ' 
 Theref. the difference between ac^ and bc^, 
 is equal to the difference between ad^ -f- cd^ 
 and bd2 -f- cd^, 
 or equal to the difference between ad^ and bd^, 
 by taking away the common square cd^ q. e. d 
 
 Corol. The rectangle of the sum and difference of the 
 two sides of any triangle, is equal to the rectangle of the 
 sum and difference of the distances between the perpendi- 
 cular and the two extremes of the base, or equal to the rect- 
 angle of the base and the difference or sum of the segments, 
 according as the perpendicular falls within or without the tri- 
 angle. 
 
 That 
 
30^ 
 
 GEOMETRY. 
 
 That is, AC -f- bc . ac — bc = ad -{- bd . ad — - bd 
 
 Or, AC -{- BC . AC — BC = AB . AD — BD in the 2d figure- 
 And AC -f- BC . AC — BC = AB . Ao -|" BD in the 1st figure. 
 
 THEOREM XXXVI. 
 
 In any Obtuse-angled Triangle, the Square of the Side sub- 
 tending the Obtuse Angle, is Greater than the Sum of the 
 Squares of the other two Sides, by Twice the Rectangle of 
 the Base and the Distance of the Perpendicular from the Ob- 
 tuse Angle. 
 
 Let ABC be a triangle, obtuse angled at b, and cd perpen ~ 
 dicular to ab ; then will the square of ac be greater than the 
 - squares of ab, bc, by twice the rectangle of ab, bd. That is, 
 Ac2 = AE^ -\- Bc2 -{- 2ab . BD. See the 1st fig. above or be- 
 low. 
 
 For, since the square of the whole line ad is equal to the 
 squares of the parts ab, bd, with twice the rectangle of the 
 same parts ab, wd (th. 31) ; if to each of these equals there 
 be added the square of cd, then the squares of ad, cd, will 
 be equal to the squares of ab, bd, cd, with twice the rectan- 
 gle of ab. bd (by ax. 2). 
 
 But the squares of ad, cd, are equal to the square of ac ; 
 and the squares of bd, cd, equal to the square of bc (th. 34) ; 
 therefore the square of ac is equal to the squares of ab, bc> 
 together with twice the rectangle of ab, bd. ^. e. jk 
 
 THEOREM XXXVII. 
 
 In any Triangle, the Square of the Side subtending an 
 Acute Angle, is Less than the Squares of the Base and the 
 other Side, by Twice the Rectangle of the Base and the Dis- 
 tance of the Perpendicular from the Acute Angle. 
 
 Let ABC be a triangle, having 
 the angle a acute, and cd perpen- 
 dicular to AB ; then will the square 
 of BC, be less than the squares of 
 AB, AC, by twice the rectangle of 
 AB, AD. That is, Bc^ = AB^ -h 
 
 AC2 2aB . AD. 
 
 A. B D ,A. 2) J3 
 
 For, 
 
THEOREMS. 
 
 SOS 
 
 For, in fig. 1, ac^ is = bc^ -f- ab^ -f- 2ab . bd (th. 36). 
 To eaoii of these equals add the square of ab, 
 then is ab^ -f ac^ = bc'^ -f 2ab-^ -f ^ab . ed (ax. 2), 
 or = Bc3 4. 2ab . AD (th. 30). 
 
 ^. E. D. 
 
 Again, in fig. 2. ac^ is = ad^ + dc^ (th. 34). 
 And ab^ = ads -{- db^ -f- 2ad . db (th. 31). 
 Theref.AB2 -j- ac2 = bd^ -{- dc^ -f Sad^ 4- 2ad . bb (ax. 2), 
 . or = Bc3 4- 2ad 2 4 2ad . db (th. 34), 
 or = BC2 4 2aB . AD (th. 30). 'q. E. J3. 
 
 THEOREM XXXVIII. 
 
 In any Triangle, the Double of the Square of a Line drawn 
 from the Vertex to the Middle of the Base, together with 
 Double the Square of the half Base, is Equal to the Sum of the 
 Squares of the other Two Sides. 
 
 Let ABC be a triangle, and cd the lirie 
 drawn from the vertex to the middle of 
 the base ab, dividing it into two equal 
 parts ad, db ; then will the sum of the 
 squares of ac, cb, be equal to twice the 
 sum of the squares of c^, bd ; or ac^ 4 
 •gb^ = 2cd2 4 2db3. 
 
 For, let CE be perpendicular to the 
 
 A DEB 
 
 base ab. Then, 
 since (by th 36) ac^ exceeds the sum of the two squares ad^ 
 
 and cd3 (or bd^ and cd^) by the double rectangle 2ad . de 
 (or ^bd , de) ; and since (by th. 37) bc^ is less than the same 
 0Mm by the said double rectangfe ; it is manifest that both ac* 
 and Bc^ together must be equal to that sum twice taken ; the 
 excess on the one part making up the defect on the other. 
 
 Q. E. 15. 
 
 THEOREM XXXIX. 
 
 In an Isosceles Triangle, the Square of a Linfe drawn fro» 
 the Vertex to any Point in the Base, together with the Rectan- 
 gle of the Segments of the Base, is equal to the Square of one 
 of the Equal Sides of the Triang e. 
 
 Let ABC be the isosceles triangle, and 
 CD a line drawn from the vertex to any 
 point D in the base : then will the square 
 of AC, be equal to the square of cd, to- 
 gether with the rectangle of ac and db. 
 
 That, is Aca = cd= 4- ad . db. 
 
 For, AD:B 
 
304 GEOMETRY. 
 
 For, let cE bisect the vertical ano;le ; then will it also 
 bisect the base ab perpendicularly making ae = eb (cor. 1, 
 th. 3). 
 
 But, in the triangle acd, obtuse angled at d, the square 
 
 AC2 is = CD2 _|- AD^-f- 2aD . DE (th. 36), 
 
 or = cd2 + ad . AD 4- 2de (th. 30), 
 
 Ot = CD2 -f- ad . AE 4" DE, 
 or = CD^ -{- ad . BE -f DE, 
 or = CD^ -f- ad . DB, A 
 
 ft. E. D. AD£ 33 
 THEOREM XL. 
 
 In any Parallelogram, the two Diagonals Bisect each other ; 
 and the Sum of their Squares is equal to the Sum of the 
 Squares of all the Four sides of the Parallelogram. 
 
 Let ABCD be a parallelogram, whose 
 diagonals intersect each other in e : then 
 will AE be equal to ec, and be to ed ; and 
 the sum of the squares of ac, bd, 
 will be equal to the sum of the squares 
 •f AB, Bc, CD, DA. That is, 
 
 AE = EC, and BE = ED, 
 
 and Ac2 4- bd2 = ab^ -|- bc^ -f cd^ -}- da^. 
 For, the triangles aeb, dec, are equiangular, because they 
 have the opposite angles at e equal (th. 7), and the two lines 
 AC, BD, meeting the parallels ab, dc, make the angle bae equal 
 to the angle dce, and the angle abe equal to the angle cde, 
 and the side ab equal to the side dc (th. 22) ; therefore these 
 two triangles are identical, and have their corresponding sides 
 equal (th. 2), viz. ae = ec, and be = ed. 
 
 Again, since ac is bisected in e, the sum of the squares ad^ 
 -f Dc2 = 2ae2 -f 2de3 (th. 38). 
 
 In like manner, ab^ -j- bc^ = 2ae2 -f- 2be2 or 2de2 , 
 
 Theref. ab^ -f bc2 -j- cd^ -f da3 = 4ae2 + 4de2 (ax. 2). 
 
 But, because the square of a whole line is equal to 4 times 
 the square of half the line (cor. th. 31), that is, ac^ = 4ae2, 
 and bd2 = 4de2. 
 
 Theref. ab^ -f- bc^ -{- cd^ + t)a^ = ac^ -f- bd^ (ax. 1). 
 
 Q. E, d. 
 THEOREM 
 
THEOREMS. 306 
 
 THEOREM XLI. 
 
 ip a Line, drawn through or from the Centre of a Circle, 
 Bisect a Chord, it will be Perpendicular to it ; or if it be 
 Perpendicular to the Chord, it will Bisect both the Chord 
 and the arc of the Chord. 
 
 Let AB be any chord in a circle, and cd 
 a line drawn from the centre c to the 
 chord. Then, if the chord be bisected in 
 the point d, cd will be perpendicular to 
 
 AB. 
 
 For, draw the two radii ca, cb. Then, 
 the two triangles acd, bcd, having ca 
 equal to cb (def. 45), and cd comnaon, also 
 AD equal to db (by hyp.) ; they have all the three sides of 
 the one, equal to all the three sides of the other, and so have 
 their angles also equal (th 5). Hence then, the angle adc 
 being eqiial to the angle bdc, these angles are right angles, 
 and the line cd is perpendicular to ab (def. II). 
 
 Again, if cd be perpendicular to ab, then will the chord 
 AB be bisected at the point d, or have ad equal to db ; and 
 the arc aeb bisected in the point e, or have ae equal eb. 
 
 For, having drawn ca, cb, as before. Then, in the tri- 
 angle ABC, because the side ca is equal to the side cb, their 
 opposite angles a and b are also equal (th. 3) Hence then, 
 in the two triangles acd, bcd, the angle a is equal to the 
 angle b, and the angles at d are equal (def U) ; therefore 
 their third angles are also equal (corol. 1, th 17). And 
 having the side cd common, they have also the side ad equal 
 to the side db (th. 2). 
 
 Also, since the angle ace is equal to the angle bce, the 
 arc AE, which measures the former (def. 57), is equal to the 
 arc BE, which measures the latter, since equal angles must 
 have equal measures. 
 
 Corol. Hence a lirfe bisecting any chord at right angles, 
 passes through the centre of the circle. 
 
 THEOREM XLIL 
 
 If More than Two Equal Lines can be drawn from *oy 
 Point within a Circle to the Circumference, that Point will be 
 the centre. 
 
 Vor.. k 4e Ipet 
 
306 
 
 GEOMKTRY. 
 
 Let ABC be a circle, and d a point 
 within it : then if three lines, da, bb, 
 DC, drawafrom the point d to the cir- 
 cumference, be equal to each other, 
 the point d wil! be the centre. 
 
 For, draw the chords ab. bc, which 
 let be hisected in the point e, f, and 
 
 join DE, DF. 
 
 Then, the two triangles, dae, dbe, have the side da equal 
 to the side db by supposition, and the side ae equal to the 
 side eb by hypothesis, also the side de common : therefore 
 these two triangles are identical, and have the angles at b 
 equal to each other (th. 6) ; consequently de is perpendicu- 
 lar to the n»iddle of the chord ab (def 11), and therefore 
 passes through the centre of the circle (corol. th. 41). 
 
 In like manner, it may be shown that df passes through 
 the centre. Consequently the point d is the centre of the 
 circle, and the three equal lines, da, db, dc, are radii. 
 
 q. E. D. 
 
 THEOREM XUir. 
 
 If two Circles touch one another Internally, the Centres of 
 the Circles, and the Point of Contact will be all in the Same 
 Right Line. 
 
 L*»t the two circles abc, ade, touch 
 one another internally in the point a ; 
 then will the point a and the centres of 
 those circles be all in the same right 
 line. 
 
 For, let F be the centre of the circle 
 ABC, through which' draw the diameter 
 AFC. Then, if the centre of the other 
 circle can be out of this line ac, let it bp 
 supposed in some other point as g ; through which draw the 
 line FG cutting the two circles in b and d. 
 
 Now, in tjhe triangle afg, the sum of the two sides, fg, 
 GA, is greater than the third side af (th JO), or greater than 
 its equal radius fb. From each of these take away the 
 common part fg, and the remainder ga will be greater 
 than the remainder gb. But the point g being supposed 
 the centre of the inner circle, its two radii, ga, gd, are 
 equal to each other ; consequently gd will also be greater 
 than gb. But ade being the inner circle, gd is necessarily 
 
 less 
 
THEOREMS. 
 
 307 
 
 less than gb. So that od is both greater and less than gb ; 
 which is absurd. Consequently the centre g cannot be out of 
 the line ajc. q. e. d. 
 
 I'HEOREM XLIV. 
 
 If two Circles Touch one another Externally, the Centres 
 of the Circles and the Point of Contact will be all in the Same 
 Right Line. 
 
 Let the two circles abc, ade, touch one 
 another externally at the point a ; then will 
 the point of contact a and the centres of the 
 two circlejj be all in the same right line. 
 
 For, let F be the centre of the circle abc, 
 through which draw the diameter afc, and 
 produce it to the other circle at e Then, if 
 the centre of the other circle ade can be out 
 of the lineFE, letit, if possible, be supposed 
 in some other point as g ; and draw the lines 
 AG, FB, DG, cutting the two circles in b and d. 
 
 Then, in the triani^le afg, the sum of the two sides af, 
 AG, is greater than the third .«ide fg (th. 10) But, f and g 
 being the centres of the two circles, the two radii ga, gd, 
 are equal, as are also the two radii af, fb. Hence the sum 
 of ga, af, is equal to the sura of gd, bp ; and therefore this 
 latter sum also, gd, bf, i.^ greater than gf, which is absurd. 
 Consequently the centre o cannot be out of the line ef. 
 
 Q. E. D, 
 
 THEOREM XLV. 
 
 Any Chords in a Circle, which are Equally Distant from the 
 Centre, are Equal to each other ; or if they be Equal to each 
 other, they will be Equally Distant from the Centre. 
 
 Let AB, CD, be any two chords at equal 
 distances from the centre g : then will 
 these two chords ab, cd, be equaltoeach 
 other. 
 
 For, draw the two radii ga, gc, and 
 the two perpendiculars ge. gf, which are 
 the equal distances from the centre g, 
 Then, the two right angled trianj^les. 
 gae, gcf, having the side ga equal the side gc, and the side' 
 GE equal the side gf, and the angle at e equal to the an^-Ie 
 
 ''at 
 
308 
 
 GEOMETRY. 
 
 at F, therefore the two triangles gae, 
 GCF, are identical (cor. 2, th. 34), and 
 have the hne ae, equal the line cf. 
 But AB is the double of ae, and cd is 
 the double of of (th. 41) ; therefore 
 ^B is equal to cd (by ax. 6). ft. e. d. 
 
 Again, if the chord ab be equal to 
 the chord cd : then will their distances from the centre, ge^ 
 GF, also be equal to each other. 
 
 For, since ab is equal cd by supposition, the half ae is 
 equal the half cf. Also the radii ga, gc, being equal, as well 
 as the right angles e and f, therefore the third sides are equal 
 (cor. JjJ, th. 34), or the distance ge equal the distance gf. 
 
 ft. e. d. 
 
 THEOREM XLVl. 
 
 A Line Perpendicular to the Extremity of a Radius, is a Tan- 
 gent to the Circle. 
 
 Let the line adb be perpendicular to the 
 radius cd of a circle ; then shall ab touch 
 the circle in the point D only. 
 
 For, from any other point e in the line 
 ab, draw cfe to the centre, cutting the 
 circle in f. 
 
 Then, because the angle i), of the trian- 
 gle CDE, is a right angle, the angle at e is acute (th. 17, cor. 3), 
 and consequently less than the angle d. But the greater 
 side is always opposite to the greater angle (th. 9) ; therefore 
 the side ce is greater than the side cd, or greater than its 
 equal cf. Hence the point e is without the circle : and the 
 same for every other point in the line ab. Consequently the 
 whole line is without the circle, and meets it in the point 9 
 only. 
 
 THEORE 
 
THEOREMS. 309 
 
 THEOREM XLVn. 
 
 When a Line is a Tangent to a Circle, a Radiufs drawn t© 
 the Point of Contact is Perpendicular to the Tangent. 
 
 Let the line ab touch the circumference of a circle at the 
 point D ; then will the radius cd be perpendicular to the 
 tangent ab. [See the last figure.] 
 
 For, the line ab being wholly without the circumference 
 except at the point d, every other line, as ce drawn from 
 the centre c to the line ab, mu«t pass out of the circle to 
 arrive at this line. The line cd is therefore the shortest that 
 can be drawn from the point c to the line ab, and consequent- 
 ly (th. 21) it is perpendicular to that line. 
 
 Corol. Hence, conversely, a line ^rawn perpendicular to a 
 tangent, at the point of contact, passes through the centre o£ 
 the circle. 
 
 THEOREM XLVin. 
 
 The Angle formed by a Tangent and Chord is Measured by 
 Half the Arc of that Chord. 
 
 Let AB be a tangent to a circle, and cd 
 a chord drawn from the point of contact c ; 
 then is the angle bcd measured by half the 
 arc CFD, and the angle acd measured by 
 half the arc cgd. 
 
 For, draw the radius ec to the point of 
 contact, and the radius ef perpendicular to 
 the chord at h. 
 
 Then the radius ef, being perpendicular to the chord cd, 
 bisects the arc cfd (th. 4 J). Therefore cf is half the arc 
 
 CFB. 
 
 In the triangle ceh, the angle h being a right one, the sum 
 of the two remaining angles e and c is equal to a right angle 
 (corol. 3, th. 17), which is equal to thfe angle bce, becau!>e 
 the radius ce is perpendicular to the tangent From each of 
 these equals take away the common part or angle g, and there 
 remains the angle e equal to the angle bcd. But the angle 
 E is measured by the arc cf (def 67), which is the half of cfd; 
 therefore the equal angle bcd must also have the same mea- 
 sure, namely, half the arc cfd of the chord cd. 
 
 Again, 
 
310 GEOMETRY. 
 
 Again, tbe line oef, being perpendicular 
 to the chord cd, bisects the arc cgd, (th. 
 41). Therefore cg is half the arc cgd. 
 Now, since the line ce, meeting fg, makes 
 the sum of the two angles at e equal to 
 two right angles (th. 6), and the line cd 
 makes with ab the sum of the two angles 
 at c equal to two right angles ; if from 
 these two equal sums there be taken away the parts or angles 
 CEH and BCH which have been proved equal, there remains 
 the angle ceg equal to the angle ach. But the former of 
 these, ceg, being an angle at the centre, is measured by the 
 arc CG (def. 67) ; consequently the equal angle acd must 
 also have the same measure cg, which is half the arc cgd of 
 the chord cd. q. ,e. d. 
 
 Corol. 1. The sum of two right angles is measured by 
 half the circumference. For the two angles. bcd, acd, which 
 make up two rijsht angles, are measured by the arcs, cf, cg, 
 which make up half the circumference, fg being a diameter. 
 
 Corol. 2. Hence also -one right angle must have for its 
 measure a quarter of the circumference, or 90 degrees. 
 
 THEOREM XUX. 
 
 An Angle at the Circumference of a Circle, is measured by 
 Half the Arc that subtends it. 
 
 Let BAC be an angle at the circumference ; 
 it has for its measure, hall the arc bc which 
 subtends it. 
 
 For, suppose the tangent de passing 
 through the point of contact a. Then, the 
 angle dac being measured by half the arc 
 ABC, and the angle dab by half the arc ab 
 (th. 48y ; it follows, by equal subtraction that the difference, 
 or angle BAC, must be measured by half the arc bc, which it 
 stands upon. Q. £. d. 
 
 THEOREM 
 
THEOREMS. 
 
 311 
 
 THEOREM L. 
 
 All Angles in the Same Segment of a Circle, or Standing on 
 the Same Arc, are equal to each other. 
 
 Let and d be two angles in the same 
 segment acdb, or, which is the same thing, 
 St mdingon the supplemental arc aeb ; then 
 will the angle c be equal to the angle d. 
 
 For each of these angles is measured by 
 half the arc aeb ; and thus, having equal 
 measures, they are equal to each other (ax. 
 U). 
 
 THEOREM LI. 
 
 An Angle at the Centre of a Circle is Double the Angle at the 
 Circumference, when both stand on the Same Arc. 
 
 Let c be an angle at the centre c, and 
 D an angle at the circumference, both stand- 
 ing on the same arc or same chord ab : then 
 will the angle c be double of the angle d, 
 or the angle d equal to half the angle g. 
 
 For, the angle at the centre c is measur- 
 ed by the whole arc aeb (def. 57), and the 
 angle at the circumference d is measured by half the same 
 arc aeb (th. 49) ; therefore the angle d is only half the angle 
 c, or the angle c double the angle d. 
 
 THEOREM UI. 
 
 An Angle in a Semicircle, is a Right Angle. 
 
 If ABC or ADC be a semicircle ; then 
 any angle d in that semicircle, is a right 
 angle. 
 
 For, the angle d, at the circumference, 
 is measured by half the arc abc (th. 49), 
 that is, by a quadrant of the circumference. 
 But a quadrant is the measure of a right 
 an^le (corol. 4. th. 6 ; or corol. 2, th. 48). 
 angle d is a right angle. 
 
 O 
 
 B 
 Therefore the 
 
 THEOREM 
 
312 GEOMETRY. 
 
 THEOUfiM UII.^ 
 
 The Angle formed by a Tangent to a Circle, and a Cheri 
 drawn from the Point of Contact, is Equal to the Angle in the 
 Alternate Segment. 
 
 If AB be a tangent, and ac a chord, and , 
 
 D any anj^e in the alternate segment adc ; 
 then will the angle d be equal to the angle 
 BAG made by the tangent and thord, of the 
 arc AEc. 
 
 For the angle d, at the circumference, 
 is measured by lialf the arc aec (th. 49) ; 
 and the angle bac, made by the tangent and chord, is als» 
 measured by the same half ar« aec (th. 48) ; therefore these 
 two angles are equal (ax. 11). 
 
 THEOREM LIV. 
 
 The Sum of any Two Opposite Angles of a Quadrangle In- 
 scribed in a Circle, is Equal to Two Right Angles. 
 
 Let ABCD be any quadrilateral inscribed 
 in a circle ; then shall the sum of the two 
 opposite angles a aud c, or b and d, be 
 equal to two right angles. 
 
 For the angle a is measured by half the 
 arc ©CB, which it stands on, and the angle' 
 c by half the arc dab (th. 49) ; therefore 
 the sum of the two angles a and c is measured by half the 
 sum of these two arcs, that is, by half the circumference. 
 But half the circumference is the measure of two right an- 
 gles (corol. 4, th. 6) ; therefore the sum of the two opposite 
 angles a and c is equal to two right angles. In hk^ manner 
 it is shown, that the sum of the other two opposite angles, b 
 and B, is equal tp two right angles. Q. e. d. 
 
 THEOREM LV. 
 
 If any Side of a Quadrangle, Inscribed in a Circle, be Pro- 
 duced out, the Outward Angle will be Equal to the InwardI 
 Opposite Angle. 
 
 If the side ab, of the quadrilateral abcd, 
 inscribed in a circle, be produced to e ; the 
 outward angle DAE.will be equal to the 
 inward opposite angle c. 
 
 For, 
 
THEOREMS. ' 313 
 
 For, the sum of the two adjacent angles dae and dab is 
 equal to two right angles (th. 6) ; and the sum of the two 
 opposite angles c and dab is also equal to two right angles 
 (th. S4) ; therefore the former sum, of the two angles dae 
 and dab, is equal to the latter sum, of the two c and da» 
 (ax. 1). From each of these equals taking away the common 
 angle dab, there remains the angle dae equal the angle c. 
 
 ^. B. D. 
 
 THEOREM LVL 
 
 Any Two Parallel Chords Intercept Equal Arcs. 
 
 Let the two chords ab, cd, be parallel : 
 then will the arcs ac, bd, be equal j or 
 
 AC = BD. 
 
 For» draw the line bc. Then, because 
 the Hues ab, gd, are parallel, the alternate 
 angles b and c are equal (th. 12). But the 
 angle at the circumference b, is measured by half the arc 
 AC (th. 49) ; and the other equal angle at the circumference 
 c is measured by half the aire bd : therefore the halves of the 
 arcs AC, BD, and consequently the arcs themselves, are also 
 „ equal. <l. E. D. 
 
 THEOREM LVn. 
 
 When a Tangent and Chord are Parallel to each other, they 
 Intercept Equal Arcs. 
 
 Let the tangent Ape be parallel to the 
 ' jchord DF ; then are the arcs bd, bf, equal ; 
 that is, BD = BF. 
 
 For, draw the chord bd. Then, be- 
 cause the Unes ab, df, are parallel, the al- 
 ternate angles d and b are equal (th. 12). 
 But the angle b, formed by a tangent and chord, is measured 
 by half the arc bd (th. 48) ; and the other angle at the circum- 
 ference D is measured by half the arc bf (th. 49) ; there- 
 fore the arcs bd, bf, are equal. q. e. d. 
 
 Vol. I. 41 THEQRBM 
 
314 GEOMETRY. 
 
 THEOREM LVra. 
 
 The Angle formed, Within a Circle, by the Intersection of two 
 Chords, is Measured by Half the Sum of the Two Inter- 
 cepted Arcs. 
 
 Let the two chords ab, cd, intersect at 
 the point e : then the angle aec, or deb, is 
 measured by half the sum of two arcs ac, 
 
 DB. 
 
 For, draw the chord af parallel to cd. 
 Then, because the lines af, cd, are parallel, 
 and ab cuts them, the angles on the same 
 side A and deb are equal (th. 14). But the angle at the cir- 
 cumference a is measured by half the arc bf, or of the sum of 
 FD and DB (th. 49) ; therefore the angle e is also measured by 
 half the sum of fd and db. 
 
 Again, because the chords af, cd, are parallel, the arcs ac 
 fd, are equal (th. 56) ; therefore the sum of the two arcs ac, 
 DB is equal tq the sum of the two f D, db ; and consequently 
 the angle e, which is measured by half the latter sum, is also 
 measured by half the former. r. e. d. 
 
 THEOREM LIX. 
 
 The Angle formed. Without a Circle, by two Secants, is 
 Measured by Half the Diflference of the Intercepted 
 
 Arcs. 
 
 Let the angle e be formed by two se- 
 cants eab and ecd ; this angle is measured 
 by half the difference of the two arcs 
 AC, DB, intercepted by the two secants. 
 
 Draw the chord af parallel to cd* Then, 
 because the lines af, cd, are parallel, and 
 AB cuts them, the angles on the same side a 
 and BED are equal (th. 14). But the angle a, at the circum- 
 ference, is measured by half the arc bf (th. 49), or of the 
 difference of df and db : therefore the equal angle e is als» 
 measured by half the difference of df, db. 
 
 Again, because the chords af, cd, are parallel, the arcs 
 AC, FD, are equal (th. 56) ; therefore the difference of the 
 
 two 
 
THEOREMS. 316 
 
 two arcs ac, db, is equal to the difference of the two df, db. 
 Consequently the angle e, which is measured hy half the latter 
 difference, is also measured by half the former. ^, e. d« 
 
 THEOREM LX. 
 
 The Angle formed by Two Tangents, is Measured by Half 
 the Difference of the two Intercepted Arcs. 
 
 Let eb, ed, be two tangents to a circle 
 at the points a, c ; then the angle e is mea- 
 sured by half the difference of the two arcs, 
 
 ^FA, CGA. 
 
 For, draw the chord af parallel to ed. 
 Then, because the lines af, ed, are parallel, 
 and EB meets them, the angles on the same 
 side A and e are equal (th. 14). But the 
 angle a, formed by the chord af and the tangent ab, is mea- 
 sured by half the arc af (th. 48) ; therefore the equal angle 
 E is als© measured by half the same arc af, or half the diffe- 
 rence of the arcs cfa and cf, or cga (th. 67). 
 
 Corol. In like manner it is proved, that 
 the angle e formed by a tangent ecd, and 
 a secant eab, is measured by half the dif- 
 ference of the two intercepted arcs ca and 
 cfb. 
 
 l-HEOREM LXL 
 
 When two Lines, meeting a Circle each in two Points, Cut one 
 another, either Within it or Without it ; the Rectangle of 
 the Parts of the one, is Equal to the Rectangle of the 
 Parts of the other ; the Parts of each being measured from 
 the point of meeting to the two intersections with the cir- 
 cumference. 
 
 l«ET 
 
316 
 
 GEOMETRY. 
 
 Let the two lines, ab, cd, meet each 
 other in e ; then the rectangle of ae, eb, 
 will be equal to the rectangle of ce, ed. Or, 
 
 AE . EB = CE . ED. 
 
 For, through the point e draw the diame- 
 ter FG ; also, from the centre h draw the 
 radius dh, and draw hi perpendicular to 
 
 CD. 
 
 Then, since deh is a triangle, and the 
 perp. HI bisects the chord cd (th. 41), the 
 line ce is equal to the difference of the 
 segments di, ei, the sum of them being 
 DE. Also, because h is the centre of the 
 circle and the radii dh, fh, gh, are all equal, the line eg is 
 equal to the sum of the sides dh, he ; and ef is equal to their 
 diflference. 
 
 But the rectangle of the sum and difference of the two sides 
 of a triangle, is equal to the rectangle of the sum and dif- 
 ference of the segments of the base (th. 36) ; therefore the 
 rectangle of fe, eg, is equal to the rectangle of ce, ed. In 
 like manner it is proved, that the same rectangle of fe, eg, 
 is equal to the rectangle of ae, eb. Consequently the rectan- 
 gle of AE, EB, is also equal to the rectangle of ce, ed (ax. 1). 
 
 ■'■■'''■ Q. E. D. 
 
 Corol. 1. When one of the lines in the 
 second case, as de, by revolving about the 
 point E, comes into the position of the tan- 
 gent EC or ED, the two points c and d run- 
 ning into one ; then the rectangle of ce, ed, 
 becomes the square of ce, because ce and 
 DE are then equal. Consequently the rect- 
 angle of the parts of the secant, ae, eb, is 
 equal to the square of the tangent ce^ . 
 
 Corol. 2. Hence both the tangents ec, ef, drawn from the 
 same point e, are equal; since the square of each is equal to 
 the same rectangle or quantity ae . EB. 
 
 THEOREM LXn. 
 
 In Equiangular Triangles, the Rectangles of the Correspond- 
 ' ing or Like Sides, taken alternately, are equal. 
 
 Let 
 
THEOREMS. 
 
 3Vi 
 
 Let ABC, DEP, be two equiangular 
 triangles, having the angle a = the 
 angle d, the angle b = the angle e, 
 and the angle c = the angle f ; also 
 the like sides ab, de, and ac, df, being 
 those opposite the equal angles : then 
 will the rectangle of ab, df, be equal 
 to the rectangle of ac, de. 
 
 In BA produced take ag equal to dp ; and through the 
 three points b, c, g, conceive a circle ^bcch to be described, 
 meeting ca produced at h, and join gh. 
 
 Then the angle g is-equal to the angle c on the same arc 
 BH, and the angle h equal to the angle b on the same arc cg 
 (th. 60) ; also the opposite angles at a are equal (th. 7) : 
 therefore the triangle agh is equiangular to the triangle 
 acb, and consequently to the triangle dfe also. But the 
 two like sides AG, df, are also equal by supposition; conse- 
 quently the two triangles agh, dfe, are identical (th. 2), 
 having the two sides ag, ah, equal to the two df, de, each to 
 each. 
 
 But the rectangle ga . ab is equal to the rectangle ha . ac 
 (th. 61) : consequently the rectangle df . ab is equal the 
 rectangle de . ac q. e. d. 
 
 THEOREM LXUL 
 
 The Rectangle of the two* Sides of any Triangle, is Equal to 
 the Rectangle of the Perpendicular on the third Side and 
 the Diameter of the Circumscribing Circle. 
 
 Let CD be the perpendicular, and ce 
 the diameter of the circle about the tri- 
 angle ABC ; then the rectangle ca . cb is 
 = the rectangle cd . ce. 
 
 For, join be : then in the two trian- 
 gles acd, ecb, the angles a and e are 
 
 equal, standing on the same arc bc (th. 60) : also the right an- 
 gle D is equal to the angle b, which is also a right angle, being 
 in a semicircle (th. 52) : therefore these two triangles have also 
 their third angles equal, and are equiangular. Hence, ac, ce, 
 and CD, cb, being like sides, subtending the equal angles, the 
 rectangle ac . cb, of the first and last of them, is equal to the 
 rectangle ce .cd, of the other two (th. 62). 
 
 THEOREM 
 
318 
 
 GEOMETRY. 
 
 THEOREM LXIV. 
 
 The Square of a line bisecting any Angle of a Triangle, 
 together with the Rectangle of the Two Segments of the 
 opposite Side, is Equal to the Rectangle of the two other 
 Sides including the Bisected Angle. ♦ 
 
 Let CD bisect the angle the c of triangle 
 ABC ; then the square cd^ -f- the rectangle 
 AD . DB is = the rectangle ac . cb. 
 
 For, let CD be produced to meet the cir- 
 cumscribing circle at e, and join ae. 
 
 Then the two triangles ace, bcd, are 
 equiangular : for the angles at c are equal 
 by supposition, and the angles b and e are equal, standing on 
 the same arc ac (th. 50) ; consequently the third angles 
 at A and© are equal (corol. 1, th 17) : also ac, cd, and 
 CE, CB, are like or corresponding sides, being opposite to 
 equal angles ; therefore the rectangle ac . cb is = the 
 rectangle cd . ce (th 62). But the latter rectangle cd . ce 
 is = cd2 -f the rectangle cd . de (th 30) ; therefore also 
 the former rectangle ac , cb is also = cd^ -f- cd . de, or equal 
 to cd2 -j- ad . db, since cd . de is = ad . db (th. 61). . 
 
 ^. E. D. 
 
 THEOREM LXV. 
 
 The Rectangle of the two Diagonals of any Quadrangle In- 
 scribed in a Circle, is equal to the sum of the two Rect- 
 angles of the Opposite Sides. 
 
 Let abcd be any quadrilateral inscribed 
 in a circle, and ac, bd, its two diagonals : 
 then the rectangle ac . bd is = the rectan- 
 gle ab . DC -}- the rectangle AD . bc. 
 
 For, let ce be drawn, making the angle 
 BCE equal to the angle dca. Then the two 
 triangles acd, bce, are equiangular ; for the angles a and 
 B are equal, standing on the same arc dc ; and the angles 
 dca, bce, are equal by supposition ; consequently the third 
 angles adc, bec are also equal ; also, ac, bc, and ad, be, are 
 like or corresponding sides ; being opposite to the equal an- 
 gles : therefore the rectangle ac . be is = the rectangle ad . bc 
 (th. 62V 
 
 Again, 
 
THEOREMS. 319 
 
 Again, the two triangles abc, dec, are equiangular : for 
 the angles bag, bdc, are equal, standing on the same arc bc ; 
 and the angle dce is equal to the angle bca, hy adding the 
 common angle ace to the two equal angles dca, bce ; therefore 
 the third angles e and abc are also equal : but ac, dc, and ab, 
 DE, are the like sides : therefore the rectangle ac . de is «« 
 the rectangle ab . dc (th. 62). 
 
 Hence, by equal additions, the sum of the rectangles ac . 
 BE + AC . DE is = AD , BC + ab . DC But the formcr sum 
 of the rectangles ac . be + ac . de is = the rectangle ac . 
 BD (th. 30) : therefore the same rectangle ac . bd is equal to 
 the latter sum, the rect. ad . bc 4- the rect. ab . dc (ax. 1). 
 
 Q. E. D. 
 
 OF RATIOS AND PROPORTIONS. 
 
 DEFINITIONS. 
 
 Def. 76. Ratio is the proportion or relation which one 
 magnitude bears to another magnitude of the same kind with 
 respect to quantity. 
 
 Note. The measure, or quantity, of a ratio, is conceived, 
 by considering what part or parts the leading quantity, called 
 the Antecedent, is of the other, called the Consequent ; or 
 what part or parts the number expressing the quantity of the 
 former, is of the number denoting in like manner the latter. 
 So, the ratio of a quantity expressed by the number 2, to a 
 like quantity expressed by the number 6, is denoted by 6 
 divided by 2, or f or 3 : the number 2 being 3 times con- 
 tained in 6, or the third part of it. In like manner, the ratio 
 of the quantity 3 to 6, is measured by f or 2 ; the ratio of 4 
 to 6 is I or \\ ; that of 6 to 4 is f or | ; &c. 
 
 77. Proportion is an equality of ratios. Thus, 
 
 78. Three quantities are said to be Proportional, when the 
 ratio of the first to the second is equal to the ratio of the second 
 to the third. As of the three quantities a (2), b (4), c (8), 
 where |. = | == 2, both the same ratio. 
 
 79. Four quantities are said to be Proportional, when the 
 ratio of the first to the second, is the same as the ratio of the 
 third to the fourth. As of the four, a (2), b (4), c (6), d (10), 
 where 4 — . u — 2, both the same ratio. 
 
320 6E0METRY. 
 
 Note, To denote that four quantities, a, b, c, d, are pro- 
 portional, they are usually stated or placed thus, a : b : : c : d ; 
 and read thus, a is to b as c is to d. But when three quanti- 
 ties are proportional, the middle one is repeated, and they are 
 written thus, a : b : : b : c. 
 
 80. Of three proportional quantities, the middle one is said 
 to be a Mean Proportional between the other two ; and the 
 last, a Third Proportional to the first and second. 
 
 81. Of four proportional quantities, the last is said to be a 
 Fourth Proportional to the other three, taken in order. 
 
 82. Quantities are said to be Continually Proportional, or 
 in Continued Proportion, when the ratio is the same between 
 every two adjacent terms, viz. when the first is to the second, 
 as the second to the third, as the third to the fourth, as the 
 fourth to the fifth, and so on, all in the same common ratio. 
 
 As in the quantities 1, 2, 4, 8, 16, &c. ; where the common 
 ratio is equal to 2. ... 
 
 - 83. Of any number of quantities, a, b, c, d, the ratio of the 
 first. A, to the last d, is said to be Compounded of the ratios 
 of the first to the second, of the second to the third, and so on 
 to the last. 
 
 84. Inverse ratio is, when the antecedent is made the con- 
 sequent, and the consequent the antecedent. — Thus, if 1 : 2 : : 
 3:6; then inversely, 2 : 1 : : 6 : 3. 
 
 85. Alternate proportion is, when antecedent is compared 
 with antecedent, and consequent with consequent, — As, if 
 1 : 2 : : 3 : 6 ; then, by alternation, or permutation, it will be 
 1 : 3 : : 2 : 6. 
 
 86. Compounded ratio is, when the sum of the antecedent 
 and consequent is compared, either with the consequent, or 
 with the antecedent. — Thus, if 1 : 2 : : 3 : 6, then, by composi- 
 tion, 1 + 2 : 1 : : 3 + 6 : 3, and 1 4- 2 : 2 : : 3 -f 6 : 6. 
 
 87. Divided ratio, is when the difference of the antecedent 
 and consequent is compared, either with the antecedent or 
 with the consequent. — Thus, if 1 : 2 : : 3 : 6, then, by division, 
 2-1:1 :: 0-3: 3, and 2-1 : 2 : : 6-3 : 6. 
 
 Note. The term Divided, or Division, here means subtract- 
 ing, or parting ; being used in the sense opposed to com- 
 pounding, or adding^ in def. 86. 
 
 THEOREM 
 
THEOREMS. 321 
 
 THEOliEM LXVI. 
 
 Equimultiples of any two Quantities have the same Ratio as 
 the Quantities themselves. 
 Let a and b be any two quantities, and tma, mB, any 
 equimultiples of them, m being any number whatever ; thea 
 will mk and mB have the same ratio as a and b, or a : b : : 
 mA : mB. 
 
 mB B 
 For — = — , the same ratio. 
 
 WIA A 
 
 Corol. Hence, like parts of quantities have the same ratio 
 as the wholes ; because the wholes are equimultiples of the 
 like parts, or a and b are like parts of mA and mB. 
 
 THEOREM LXVII. 
 
 If Four Quantities, of the Same Kind, be Proportionals ; 
 they will be in Proportion by Alternation or Permutation, 
 or the Antecedents will have the Same Ratio as the Con- 
 sequents. 
 Let a : b : : mA : mB ; then will a : mA : : b : mB. 
 
 mA mB 
 
 For — = m, and — = m, both the same ratio. 
 
 A B 
 
 THEOREM LXVm. 
 
 If Four Quantities be Proportional ; they will be in Pro- 
 portion by Inverlion, or Inversely. 
 Let a : b : : mA : mB ; then will b : a : : ms : mA. 
 
 mA A 
 For — = — , both the same ratio. 
 mB B 
 
 THEOREM LXIX. 
 
 If Four Quantities be Proportional ; they will be ia Pro- 
 portion by Composition and Division. 
 Let a : b : : mA : mB ; 
 Then will b ±: a : a : : mB zt mk : mA, 
 and B ± A : B : : mB it mA : mB. 
 mA A mB B 
 
 For, = ; and = . 
 
 mB db mA b ± a mB rb mA ft ± a 
 
 V*r.. L 42 CoroL 
 
32£ GEOMETRY. 
 
 Corol. It appears ffom hence, that the Sum of the Greatest 
 and Least of four proportional quautities, of the same kind, 
 
 exceeds the Sum of the Two Means. For, since 
 
 A : A -|- B : : mA : mA -{- mB, where a is the least, and 
 wiA -j- »«B the greatest ; then m -f 1 • a + wb, the sum of 
 the greatest and least exceeds y/i + 1 . a + b the sum of 
 the two means. 
 
 THEOREM LXX. 
 
 If, of Four Proportional Quantities, there betaken any Equi- 
 multiples whatever of the two Antecedents, and any Equi- 
 multiples whatever of the two Consequents ; the quantities 
 resulting will still be proportional. 
 
 Let a ; b : : mA : mB ; also, let pA and pmA be any 
 equimultiples of the two antecedents, and ^b and ^mB any 
 
 equimultiples of the two consequents ; then will 
 
 jpA : 9B : : pmA : ^mB. 
 qniB qB 
 
 For = — , both the same ratio. 
 
 pmA pA 
 
 THEOREM LXXI. 
 
 If there be Four Proportional Quantities, and the two Conse- 
 quents be either Augmented or Diminished by Qjaantities 
 that have the Same Ratio as the respective Antecedents ; 
 the Results and the Antecedents will still be Propor- 
 tionals. 
 
 Let a : b : : mA : jiiB, and ua and nmA any two quan- 
 tities having the same ratio as the two antecedents ; then will 
 A : B ± nA : : mA : mB ± nmA. 
 mB ■±L nmA b ±: nA 
 
 For — = , both the same ratio. 
 
 mA A 
 
 THEOREM LXXn. 
 
 If any Number of Quantities be Proportional, then any 
 one of the Antecedents will be to its Consequent, as the 
 Sum of all the Antecedents is to the Sum of all the Con- 
 sequents. 
 Let A : b : : mA : mB : : ka : wb,&c. ; then will 
 
 A : B : : a -|- mA -j- »a : : b -f mB + «»> &c. 
 
 B -f- »"B -f- flB b 
 
 X For " = — , the same ratio. 
 
 a + mA -{- riA A 
 
 THEOREM 
 
THEOREMS. 3^3 
 
 THEOREM LXXin. 
 
 U a Whole Magnitude be to a Whole, as a Part taken from 
 the first, is to a l^art taken from the other ; then the Re- 
 mainder will be to the Remainder, as the whole to the 
 whole. 
 
 m m 
 
 Let A : B : : — a : — b ; 
 
 then will a:b::a a:b 1 
 
 n n 
 
 -B 
 
 B 
 
 For ° = — , both the same ratio. 
 
 m 
 A--A A 
 
 THEOREM LXXIV. 
 
 If any Qjuantities be Proportional ; their Squares, or Cubes, 
 or any Like Powers, or Roots, of them, will also be Pro- 
 portional. 
 
 Let A ; b : : mA : wiB ; then will a" : b" : : m^A" : m" b", 
 m"B" b" 
 
 For = — , both the same ratio. 
 
 w"a" a" ^ 
 
 THEOREM LXXV. 
 
 If there be two Sets of Proportionals ; then the Products or 
 Rectangles of the Corresponding Terms will also be Pro- 
 portional. 
 
 Let a : b : : fOA : wb, 
 and c : D : : nc : no ; 
 then will ac : bd : : mnAC : mriBD. 
 
 mriBD bd 
 For = — , both the same ratio. 
 
 TnnAC AC 
 
 THEOREM LXXVI. 
 
 If Four Quantities be Proportional ; the Rectanj^le or Product 
 of the two Extremes, will be Equal to the Rectangle or 
 Product of the two Means. And the converse. 
 Let a : b : : mA : niB ; 
 then is A X mB = b X mA = mAB, as is evident. 
 
 THEOREM 
 
324 GEOMETRY. 
 
 THEOREM LXXVII. 
 
 If Three Quantities be Continued Proportionals ; the Rect- 
 angle or Product of the tv.o Extremes, will be Equal to the 
 Square of the Mean. And the converse. 
 Let a, mA, tw'^a be three proportionals, 
 or A : mA : : mA : m-A ; 
 then is A X m^A = m^AS, as is evident. 
 
 THEOREM LXXVUI, 
 
 If any Number of Quantities be Continued Proportionals j 
 the Ratio of the First to the Third, will be duplicate or the 
 Square of the Ratio of the First and Second ; and the Ratio 
 of the First and Fourth will be triplicate or the cube of 
 that of the First and Second ; and so on. 
 Let A, mA, m^A, m^A, &c. be proportionals ; 
 mA m^A m^A 
 
 that is — = m ; but = rn^ ; and = m^ ; &c. 
 
 THEOREM LXXIX. 
 
 Triangles, and also Parallelograms, having equal Altitudes, 
 ' are to each other as their Bases. 
 
 Let the two triangles ado, def, have I C K. 
 
 the same altitude, or between the same 
 parallels ae, cf ; then is the surface of 
 the triangle adc, to the surface of the 
 triangle def, as the base ad is to the / U -X fi-ix i s^ 
 
 base DE. Or, ad : de : : the triangle 
 ADC : the triangle def. 
 
 For, let the base ad be to the base de, as any one num- 
 ber m (2), to any other number n (3) ; and divide the respec- 
 tive bases into those parts, ab, bd, dg, gh, he, all equal 
 to one another ; and from the points of division draw the 
 lines Bc, fg, fh, to the vertices c and f. Then will 
 these lines divide the triangles adc, def, into the same num- 
 ber of parts as their bases, each equal to the triangle 
 ABC, because those triangular parts have equal bases and 
 altitude (corol. '2, th. 26) ; namely, the triangle abc, equal 
 to each of the triangles bdc, dfg, gfh, hfe. ^o that 
 the triangle adc, is to the triangle dfe, as the number of 
 
 parts 
 
THEOREMS. 326 
 
 parts m (S) of the former, to the numher n (3) of the latter, 
 that is, as the base ad to the base de (def. 79) 
 
 In like manner, the parallelogram adki is to the parallelo- 
 gram DEFK, a« the base ad is to the base de ; each of these 
 having the same ratio as the number of their parts, m to n. 
 
 Q. E. D»- 
 
 THEOREM LXXX. 
 
 Triangles, and also Parallelograms, having Equal Bases, arc 
 to each other as their Altitudes. 
 
 Let ABC, BEF, be two triangles 
 having the equal basses ab, be, and 
 whose altitudes are the perpendicu- 
 lars CG, FH ; then will the triangle 
 ABC : the triangle bef : ; cg : fh. 
 
 For, let BK be perpendicular to ab, 
 and equal to cg ; in which let there 
 be taken bl=fh; drawing AK and AL. 
 
 Then, triangles of equal base and heights being equal 
 (corol. 2, th. 25), the triangle abk is = abc, and the triangle 
 ABL = BEF But considering now abk, abl, as two triangles 
 on the bases bk, bl, and having the same altitude ab, these 
 will be as their bases (th. 79), namely the triangle abk : the 
 triangle abl : : bk : bl. 
 
 But the triangle abk = abc, and the triangle abl = bef, 
 
 also bk = CG, and bl = fh. 
 Theref. the triangle abc : triangle bef : : cg : fh. 
 
 And since parallelograms are the doubles of these triangles, 
 having the same bases and altitudes, they will likewise have 
 to each other the same ratio as their altitudes. q. e. d. 
 
 Corol. Since, by this theorem, triangles and parallelogram?, 
 when their bases are equal, are to each other as their ahi- 
 tudes ; and by the foregoing one, when their altitiKJes are 
 equal, they are to each other as their bases ; therefore uni- 
 versally, when neither are equal, they are to ezth other in 
 the compound ratio,' or as the rectangle or product of their 
 bases and altitudes. 
 
 THEOREM 
 
326 GEOMETRY. 
 
 THEOREM LXXXI. 
 
 If Four Lines be Proportional ; the Rectangle of the Ex- 
 tremes will be equal to the Rectangle of the Means. 
 And, conversely, if the Rectangle of the Extremes, of four 
 Lines, be equal to the Rectangle of the Means, the Four 
 Lines, taken alternately, will be Proportional. 
 
 
 Q 
 B 
 
 A 
 
 P D 
 
 R 
 
 Let the four lines, a, b, c, d, be A 
 
 proportionals, or a : b : : c : d ; B 
 
 then will the rectangle of a and d be S. 
 
 equal to the rectangle of b and c ; 
 or the rectangle a . d = b . c. 
 
 For, let the four lines be placed 
 with their four extremities meeting 
 in a common point, forming at that 
 
 point four right angles ; and draw lines parallel to them to 
 complete the rectangles p, q, r, where p is the rectangle of a 
 and D, Q the rectangle of b and c, and r the rectangle of b 
 and D. 
 
 Then the rectangles p and r, being between the same 
 parallels, are to each other as their bases a and b (th, 79) ; 
 and the rectangles q and r, being between the same pa- 
 rallels, are to each other as their bases c and d. But the 
 ratio ef a to b, is the same as the ratio of c to d by hypo- 
 thesis ; therefore the ratio of p to r, is the same as the'ratio of 
 Q to R ; and consequently the rectangles p and ^ are equal. 
 
 Q. E. D. 
 
 Again, if the rectangle of a and d, be equal to the 
 rectangle of b and c ; these lines will be proportional, or 
 A : B : : c ; D. 
 
 For, the rectangles being placed the same as before : then, 
 because parallelograms between the same parallels are to one 
 another as their bases, the rectangle p : r : : a : b, and 
 q : R : : c ; d. But as p and q are equal, by supposition, 
 they have the same ratio to r, that is, the ratio of a to b is 
 equal to the ratio of c to d, or a : b : : c : d. ct. e. d. 
 
 Corol. 1. When the two means, namely, the second and 
 third terms, are equal, their rectangle becomes a square of 
 the second term, which supplies the place of both the second 
 and third. And hence it follows, that when three lines are 
 proportionals, the rectangle of the two extremes is equal to 
 
 the 
 
THEOREMS. 327 
 
 I the square of the mean ; and, conversely, if the rectangle of 
 
 R the extremes be equal to the square of the mean, the three 
 
 f' lines are proportionals. 
 
 Corol. 2. Since it appears, bj the rules of proportion in 
 Arithmetic and Algebra, that when four quantities are propor- 
 tional, the product of the extremes ia equal to the product of 
 the two means ; and, by this theorem, the rectangle of the 
 extremes is equal to the rectangle of the two means ; it fol- 
 lovvs, that the area or space of a rectangle is represented or 
 expressed by the product of its length and breadth multiplied 
 together. And, in general, a rectangle in geometry is simi- 
 lar to the product of the measures of its two dimensions of 
 length and breadth, or base and height. Also, a square is si- 
 milar to, or represented by, the measure of its side multiplied 
 by itself So that, what is shown of such products, is to be 
 understood of the squares and rectangles. 
 
 Corol. 3. Since the same reasoning, ^s in this theorem, 
 holds for any parallelograms whatever, as well as for the rect- 
 angles, the same property belongs to all kinds of parallelO'- 
 grams, having equal angles, and also to triangles, which are 
 the halves of parallelograms ; namely, that if the sides about 
 the equal angles of parallelograms or triangles, be reciprocally 
 proportional, the parallelograms or triangles will be equal ; 
 and, conversely, if the parallelograms or triangles be equal, 
 their sides about the equal angles will be reciprocally propor- 
 tionali 
 
 Coi-ol. 4. Parallelograms, or triangles, having an angle in 
 each equal, are in proportion to each other as the rectangles 
 of the sides which are about these equal angles. 
 
 THEOREM LXXXIL 
 
 If a Line be drawn in a Triangle Parallel to one of its sides, 
 it will cut the other Sides Proportionally. 
 
 Let de be parallel to the side bc of the 
 triangle abc ; then will ad : db : : ae ; eg. 
 
 For draw be and cb. Then the tri- 
 angles DBE, DOE, are equal to each other, 
 because they have the same base de, and 
 are between the same parallels de, bc 
 (th 25). But the two triangles ade, bde, 
 on the bases ad, db, have the same alti- 
 
 tilde ; 
 
328 GEOMETRY. 
 
 tude ; and the two triangles a©e, cde, on 
 the bases ae, ec, have also the same alti- 
 tude ; and because triangles of the same 
 altitude are to each other as their bases, 
 therefore 
 
 the triangle ade : bde : : ad : db, 
 
 and triangle ade : cde : : ae : ec. 
 
 But BDE is = CDE ; and equals must have to equals the same 
 ratio ; therefore ad : db : : ae : ec. q. e. d. 
 
 Corol. Hence, also, the whole lines ab, ac, are proportional 
 to their corresponding proportional segments (corol. th. 66). 
 
 VIZ. AB : AC : : ad : ae, 
 and AB : AC : : bd : ce. 
 
 THEOREM LXXXm. 
 
 A Line which Bisects any Angle of a Triangle, divides the 
 opposite Side into Two Segments, which are Proportional 
 to the two other Adjacent Sides. 
 
 E 
 
 Let the angle acb, of the triangle abc, be 
 bisected by the line en, making the angle r >, 
 
 equal to the angle s : then will the seg- ^ / 
 
 ment ad be to the segment db. as the side 
 AC is to the side cb. Or, ad : db : : ac : cb. 
 
 For, let be be parallel to cd, meeting 
 AC produced at e. Then, because the line bc cuts the two 
 parallels cd, be, it makes the angle cbe equal to thte alternate 
 angle s (th. 12), and therefore also equal to the angle r, which 
 is equal to s by the supposition. Again, because the line ae - 
 cuts the two parallels dc, be, it makes the angle e equal to the 
 angle r on the same side of it (th. 14). Hence, in the triangle 
 bce, the angles b and e, being each equal to the angle r, are 
 equal to each other, and consequently their opposite sides cb, 
 ce, are also equal (th. 3). 
 
 But now, in the triangle abe, the lime cd, being draw^ 
 parallel to the side be, cots the two other sides ab, ae propor- 
 tionally (th. 82), making ad to db, as is ac to ce or to its 
 equal cb. a. e d. 
 
 THEOREM 
 
THEOREMS. 
 
 329 
 
 THEOREM LXXXIV. 
 
 Equiangular Triangles are Similar, or ha?e their Like Sides 
 Proportional. 
 
 Let ABC, DEF, be two equiangular tri- q 
 
 angles, having the angle a equal to the 
 angle d, the angle b to the angle e, and 
 consequently the angle c to the angle f ; 
 then will AE : AC : : de : df. 
 
 For, make dg = ab, and dh = ac, 
 and join gh. Then the two triangles 
 ABC, DGH, having the two sides ab, ac, 
 equal to the two do, dh, and the con- 
 tained angles a and d also equal, are iden- 
 tical, or equal in all respects (th. l),name- 
 ly the angles b and c are equal to the 
 
 angles g and h. But the angles b and c are equal to the angles 
 E and F by the hypothesis ; therefore also the angles g and h 
 are equal to the angles e and f (ax I), and consequently the 
 line GH is parallel to the side ef (cor. 1, th. 14). 
 
 Hence then, in the triangle def, the line gh, being parallel 
 to the side ef, divides the two other sides proportionally, 
 making dg : dh : : de : df (cor. th. 82). But dg and «h are 
 equal to ab and ac ; therefore also ab : ac : : de : df. 
 
 q,. E. D. 
 
 THEOREM LXXXV. 
 
 Triangles which have their Sides Proportional, are Equi- 
 angular.^ 
 
 In the two triangles abc, def, if 
 ab : : de : : AC : df : : bc : ef ; the two 
 triangles will have their corresponding 
 angles equal. 
 
 For, if the triangle abc be not equian- 
 gular with the triangle def, suppose some 
 other triangle, as deg, to be equiangular 
 with ABC. But this is impossible : for if 
 the two triangles abc, deg, were equian- 
 gular, their sides would be proportional 
 (th. 84). So that, ab being to de as ac 
 to dg, and ab to np as eg to eg, it follows that dg and e» 
 Wmg fourth proportionals to the same three quantities 
 
 Von. I. • 43 a9 
 
33U 
 
 GEOMETRY. 
 
 as well as the two df, ef, the former dg, eg, would be equal 
 to the latter, df, ef. Thus then, the two trianjj^les, def, deg, 
 having t^ieir three sides equal, would be identical (th. 6) : 
 which is absurd, since their angles are unequal. 
 
 THEOREM LXXXVL 
 
 Triangles, which have an Angle in the one Equal to an Angle 
 in the other, and the Sitles about these angles Proportional, 
 are Equiangular. 
 
 Let ABC, DEF, be two triangles, having 
 the angle a — the angle d, and the sides 
 AB, AC, proportional to the sides de, df : 
 then will the triangle abc be equiangular 
 with the triangle def. 
 
 For, make dg = ab, and dh = ac, 
 and join gh. 
 
 Then, the two triangles abc, dgh, hav- 
 ing two sides equal, and the contamed 
 angles a, and d equal, are identical and 
 equiangular (th. 1), having the angles g 
 and H equal to the angles b and c. But, since the sides dg, 
 DH, are proportional to the sides de, df, the line gh is parallel 
 to ef (th. 82) ; hence the angles e and f are equal to the an- 
 gles G and H (th. 14), and consequently to their equals e and c. 
 
 G E 
 
 THEOllEM LXXXVIL 
 
 in a Right- Angled Triangle, a Perpendicular from the Right 
 Angle, is a Mean Proportional between the Segments of 
 the Hypothenuse ; and each of the Sides, about the Right 
 Angle, is a Mean Proportional between the Hypothenuse 
 and the adjacent segment, 
 
 ^,. .G 
 
 Let abc be a right-angled triangle, and 
 CD a perpendicular from the right angle 
 c to the hypothenuse ab ; then will 
 
 CD be a mean proportional between ad and db ; 
 
 AC a mean proportional between ab and ad ; 
 
 bo a mean proportional betwen ab and bd. 
 
 Or, ad : CD : : CD : db ; and ab : bc : : bc : bd ; and 
 
 ab : AC : : ac. 
 
 For. 
 
THEOREMS. 
 
 331 
 
 For, the two trian?;les abc, adc, having the right angijes 
 at c an(J d equal, and the angle a common, have their third 
 angles equal, and are equiangular (corol. I, th. 17). In like 
 manner, the two triangles abc, bdc, having the ri^ht angles 
 at c and d equal, and the angle b common, have the third an- 
 gles equal, and are equiangular. 
 
 Hence then, all the three triangles abc, adc, bdc, 
 l?eing equiangular will have their like sides proportional 
 (th. 84;. 
 
 VIZ. AD ; CD : 
 and AB : AC : 
 and AB : bc : 
 
 : CD : DB ; 
 ; AC : AD ; 
 : BC : BD. 
 
 Corol, Because the angle in a semicircle is a right angle 
 (th. 52) ; it follows, that if, from any point c in the periphery 
 of the semicircle, a perpendicular be drawn to the diameter 
 AB ; and the two chords ca, cb, be drawn to the extremities of 
 the diameter : then are ac, bc, cd, the mean proportionals as 
 in this theorem, or (by th. 77), cd^ = ad . db ; ac^ = ab . ad ; 
 
 and BC2 = AB . BD. 
 
 THEOREM LXXXVIIl. 
 
 Equiangular or Similar Triangles, are to each other as the 
 Squares of their Like Sides. 
 
 Let ABC, DEF, be two equi- 
 angular triangles, ab and de 
 being two like sides ; then will 
 the triangle abc be to the tri- 
 angle DEF, as the square of ab 
 is to the square of de, or as 
 
 AB^ to DE^. 
 
 For, let AL and dn bc the 
 
 y 
 
 B D 
 
 K 
 
 squares on ab and de ; also draw their diagonals bk, ev, and 
 the perpendiculars cg, fh, of the two triangles. 
 
 Then, since equiangular triangles have their like sides 
 proportional (th. 84), in the two equiangular triangles abc, 
 DEF, the side ac : df : : ab : de.; and in the two acg, 
 DFH, the side ac : df : : cg : fh ; therefore, by equaUty 
 CG : FH : : ab : de, or cg : ab : : fh : de. 
 
 But because triangles on equal bases are to each other as 
 their altitudes, the triangles abc, abk, on the same base 
 AB, are to each other, as their altitudes cg, ak, or ab ; 
 
 and 
 
332 OEOMETR?. 
 
 and the triangles dkf, dem, on the same base de, are as theic 
 altitudes fh, dm, or de. 
 
 that is. triangle abc : triangle abk : : cg : ab, 
 and triangle def : triangle dem : : fh : de. 
 
 But it has been shown that cg : ab : : fh : de ; 
 
 theref. of equality A abc : A abk : : A def : A dem, 
 
 er alternately, as A abc : A def : : A abk : A dem. 
 
 But the squares al, dn, being the double of tiie triangleu 
 ABH, DEM, have the same ratio with them ; 
 
 therefore the A abc : A def : : square al : square dn. 
 
 ft. E. D. 
 THEOREM LXXXIX. 
 
 All Similar Figures are to each other, as the Squares of their 
 Like Sides. 
 
 Let abcde, fghik, be 
 any two similar figures, the 
 like sides? being ab, fg, and 
 bc, gh, and so on in the same 
 order : then will the figure 
 ABtDE be to the figure fghik, 
 as the square of ab to the 
 square of fg, or as ab» to rc^. 
 
 For, draw be, bd, gk, gi, dividing the figures into asi 
 equal number of triangles, by lines from two equal angles 
 b and G. 
 
 The two figures being similar (by suppos.), they, are equi- 
 angular, and have their like sides proportional (def 70). 
 
 Then, since the angle a is = the angle f, and the sidee 
 AB, ae, proportional to the sides fg, fk, the triangles 
 ABE, fgk, are equiangular (th. 86). In like manner, the 
 two triangles bcd, ghi, having the angle c = the angle h, 
 and the sides bc, cd, proportional to the sides gh, hi, are 
 also equiangular Also, if from the equal angles aed, fki, 
 there be taken the equal angles aeb, fkg, there will remain 
 the equals bed, gki ; and it from the equal angles cdc, 
 hik, be taken away the equals cdb, hig, there will remain 
 the equals bde, gik ; so that the two triangles bde, gik, 
 having two angles equal, are also equiangular. Hence each 
 triangle of the one figure, is equiangular with each corres- 
 ponding triangle of the other. 
 
 But equiangular triangles are similar, and are proportional 
 to the squares of their hke sides (th. 88). 
 
 Therefore 
 
THEOREM!?. 
 
 333 
 
 Therefore the A abe : A 
 and A bcd : A 
 and A bde : A 
 
 FGK : : ab2 : fg«, 
 GHi : : Bc2 ; GH^ ; 
 GiK : : de2 : ik^. 
 
 But as the two polygons are similar, their hke sides are pro- 
 portional, and consequently their squares also proportional ; 
 so that all the ratios, ab^ to fg2, and bc3 to gh2 and de2 to 
 ik2, are equal among themselves, and consequently the cor- 
 responding triangles also, abe to fgk, and bcd to ghi, and 
 BDE to GIK, have all the same ratio, viz. that of ab^ to fg^ : 
 and hence all the antecedents, or the figure abode, have to 
 all the consequents, or the figure fghik, still the same ratio, 
 viz. that of ab2 to fg2 (th. 72;. Q. e. c 
 
 THEOREM XC. 
 
 "Jimilar Figures Inscribed ia Circles, have their Like Sides, 
 and also their Whole Perimeters, in the Same Ratio as the 
 Diameters of the Circles in which they are Inscribed. 
 
 Let abode, fghik, 
 be two similar figures, 
 inscribed in the circles 
 whose diameters are al 
 and FM : then will each 
 side ab, bc, &.c. of the 
 one figure be to the like 
 side GF, GH, &-C. of the 
 other figure, or the whole perimeter ab -{- bc -}- &c. of the 
 one figure, to the whole perimeter fg -}- gh -f- kc. of the 
 other figure, as the diameter al to the diameter fm. 
 
 For, draw the two corresponding diagonals, ac, fh, as 
 also the lines bl, gm. Then, since the polygons are similar, 
 they are equiangular, and their like sides have the same ratio 
 (def. 70) ; therefore the two triangles abc, fgh, have the 
 angle b = the angle g, and the sides ab, bc, proportional 
 to the two sides fg, gh, consequently these two triangles 
 are equiangular (th. 86), and have the angle acb = fhg. 
 But the angle acb = alb, standing on the same arc ab ; 
 and the angle fhg = fmg, standing on the same arc fg ; 
 therefore the angle alb = fmg (th. 1). And since the 
 angle abl = fgm, being both right angles, because in a 
 semicircle ; therefore the two triangles, abl, fgm, having 
 two angles equal, are equiangular ; and consequently their 
 ^ like 
 
334 ©EOMETRY. 
 
 like sides are proportional (th. 84) ; hence ab : fg : : the 
 diameter al : the diaraeter fm. 
 
 in like manner, each side bc, gd, &c. has to each side 
 GH, HI, &c. the same ratio of al to fm : and consequently 
 th^ sums of them are still in the same, ratio ; viz. ab -|- b<^ 4* 
 CD, &c. : FG -f- GH -|- HI, &c. : : the diam. al : the diam. 
 
 FM (th. 72). ft. E. D. 
 
 THEOREM XCI. 
 
 Similar Figures Inscribed in Circles, are to each other as 
 the Squares of the Diameters of those Circles. 
 
 Let abcde, fghik, 
 be two similar figures in- 
 scribed in the circles 
 whose diameters are al 
 and fm ; then the surface 
 of the polygon abode 
 will be to the surface of 
 the polygon fghik, as al^ to fm^ . ^ 
 
 For, the figures being similar, are to each other as the 
 squares of their like sides, ab^ to fg^ (th. 88). But. by 
 the last theorem, the sides ab, fg, are as the diameters al, 
 FM ; and therefore the squares of the sides ab^ to rc^ , as the 
 squares of the diameters al2 to fm^ (th. 74). Consequently 
 the polygons abcde, fghik, are also to each other as the 
 squares of the diameters al^ to fm^ (ax. 1). Q. e d.> 
 
 THEOREM XCH. 
 
 The Circumferences of all Circles are to each other- as their 
 Diameters. 
 
 Let D, d, denote the diameters of two circles, and c, c, 
 their circumferences ; 
 
 then will d : </ : : c : c, or d : c : : d : c. 
 
 For, (by theor. 90,) sinjilar polygons inscribed in circles 
 have their perimeters, in the same ratio as the diameters of 
 those circles. 
 
 Now, as this property belongs to all polygons, whatever 
 the number of the sides may be ; conceive the number of the 
 sides to b«: indefinitely great, and the length of each inde- 
 finitely small, till they coincide with the circumference of 
 
 the 
 
THEOREMS. 335 
 
 the circle, and be equal to it, indefinitely neap. Thee tbe 
 perimeter of the poly^^on of an infinite number of sides, is 
 the same thing as the circumference of the circle. Hence it 
 appears that the circumferences of the circles, being the same 
 as the perimeters of such polygons, are to each other in the 
 same ratio as the diameters of the circles. q. e. d. 
 
 THEOREM XCni. 
 
 The Areas or Spaces of Circles, are to each other as the 
 Squares of their Diameters, or of their Radii. 
 
 Let a, a, denote the areas or spaces of two circles, and d, 
 d their diameters ; then a : a : : d^ : d^ . 
 
 For (by theorem 91) similar polygons inscribed in circles 
 are to each other as the squares of the diameters of the cir- 
 cles. 
 
 Hence, conceiving the number of the sides of the polygons 
 to be increased more and more, or the length of the sides to 
 become less and less, the polygon approaches nearer and near- 
 er to the circle, till at length, by an infinite approach, coin- 
 cide, and become in effect equal ; and then it follows that the 
 spvices of the circles, which are the same as of the polygons, 
 will be to each other as the squares of the diameters of the 
 circles. q. e. b. 
 
 Coral, The spaces of circles are also to each other as the 
 squares of the circumferences ; since the circumferences are 
 in the same ratio as the diameters (by theorem 92). 
 
 THEOREM XCIV. 
 
 The Area of any Circle, is Equal to the Rectangle of Half 
 its Circumference and half its Diameter. 
 
 Conceive a regular polygon to be in- ^ — ^ 
 
 scribed in the circle : and radii drawn to /f\ \ /\\ 
 
 all the angular points, dividing it into as // ^ \:/ \\ 
 
 many equal triangles as the polygon has r — ~Q^v ^ 
 
 sides, one of which abc, of which the Vv / :\ // 
 
 altitude is the perpendicular cd from the \/ \ \/ / 
 
 centre to the base ae. j^T^SP^ 
 
 Then the triangle abc, being equal to 
 a rectangle of half the base and equal altitude (th. 26, cor. 2), 
 is equal to the rectangle of the half base ad and the altitude co ; 
 
336 GEOMETRY. 
 
 consequently the whole polygon, or all 
 the triangles added together which com- 
 pose it, is equal to the rectangle of the 
 common altitude cd. and the halves of all 
 the sides, or the half perimeter of the 
 polygon. 
 
 Now, conceive the number of sides of the polygon to be 
 indefinitely increased ; then will its perimeter coincide with 
 the circumference of the circle, and consequently the altitude 
 CD will become equal to the radius, and the whole polygon 
 equal to the circle. Consequently the space of the circle, or 
 of the polygon in that state, is equal to the rectangle of the 
 radius and half the circumference. q. e. b. 
 
 OF PLANES AND SOLIDS. 
 
 DEFINITIONS. 
 
 Def. 88. The common Section of two Planes, is the line 
 in which they meet, to cut each other. 
 
 89. A Line is Perpendicular to a Plane, when it is per- 
 pendicular to every line in that plane which meets it. 
 
 90. One Plane is Perpendicular to Another, when every 
 line of the one, which is perpendicular to the line of their 
 common section, is perpendicular to the other. 
 
 91. The inclination of one Plane to another, or the angle 
 they form between them, is the angle contained by two 
 lines drawn from any point in the common section, and at 
 right angles to the same, one of these lines in each plane. 
 
 92. Parallel Planes, are such as being produced ever so 
 far both ways, will never meet, or which are every where at 
 an equal perpendicular distance. 
 
 93. A Solid Angle, is that which is made by three or more 
 plane angles, meeting each otlier in the same point. 
 
 94. Similar 
 
DEFINITIONS. 
 
 337 
 
 \ N 
 
 1 
 
 ^ ^ 
 
 94. Similar Solids, contained by plane figures, are such as 
 have all their solid angles equal, each to each, and are bound- 
 ed by the same number of similar planes, alike placed. 
 
 95. A Prism, is a solid whose ends are parallel, equal, and 
 like plane figures, and its sides, connecting those ends, are 
 parallelograms. 
 
 96. A Prism takes particular names according to the figure 
 of its base or ends, whether triangular, square, rectangular, 
 pentagonal, hexagonal, &.c^ 
 
 97. A Right or Upright Prism, is that which has the 
 planes of the sides perpendicular to the planes of the ends 
 or base. 
 
 98. A Parallelopiped, or Parallelopipedon, is 
 a prism bounded by six parallelograms, every 
 opposite two of which are equal, alike, and pa- 
 rallel. 
 
 99. A Rectangular Parallelopipedon, is that whose bound- 
 ing planes are all rectangles, which are perpendicular to each 
 other. 
 
 100. A Cube, is a square prism, being bound- 
 ed by six equal square sides or faces, and are 
 perpendicular to each other. 
 
 101. A Cylinder, is a round prism, having cir- 
 cles for its ends ; and is conceived to be formed 
 by the rotation of a right line about the circum- 
 ferences of two equal and parallel circles, always 
 parallel to the axis. 
 
 102. The Axis of a Cylinder, is the right line 
 joining the centres of the two parallel circles, about which 
 the figure is described. 
 
 103. A Pyramid, is a solid, whose base is any 
 right-lined plane figure, and its sides triangles, 
 having all their vertices meeting together in a 
 point above the base, called the Vertex of the 
 pyramid. 
 
 104. A pyramid, like the prism, takes particular names 
 from the figure of the base. 
 
 105. A Cone, is a round pyramid, having a cir- A 
 cular base, and is conceived to be generated by / \ 
 the rotation of a right line about the circumference L--\ 
 of a circle, one end of which is fixed at a point CJ' 
 above the plane of that circle. 
 
 J^ 
 
 
 7 
 
 
 
 
 A 
 
 
 ^ 
 
 Vol. I. 
 
 44 
 
 19S. The 
 
338 GEOMETRY. 
 
 106. The Axis of a cone, is the right line, joining the 
 vertex, or fixed pQint, and the centre of ilie circle about 
 which the figure is described. 
 
 107. Similar Cones and Cylinders, are such as have their 
 altitudes and the diameters of their bases proportional. 
 
 108. A Sphere, is a solid bounded by one curve surface, 
 which is every where equally distant from a certain point 
 within, called the Centre. It is conceived to be generated 
 by the rotation of a semicircle about its diameter, which re- 
 mains fixed. 
 
 109. The Axis of a Sphere, is the right line about which 
 the semicircle revolves ; and the centre is the same as that 
 of the revolving semicircle. 
 
 110. The Diameter of a Sphere, is any right line passing 
 through the centre, and terminated both ways by the surface. 
 
 111. The Altitude of a Sohd, is the perpendicular drawn 
 from the vertex to the opposite side or base. 
 
 THEOREM XCV. 
 
 A Perpendicular is the Shortest Line which can be draws 
 from any Point to a Plane. 
 
 Let ab be perpendicular to the plane . 
 
 DE ; then any other line, as ac, drawn \ 
 
 from the same point a to the plane, will \ 
 
 be longer than the line ab. ^^^—a^.,,.^^ 
 
 In the plane draw the line bc, joining dC B — ^CyE 
 the points B, c. 
 
 Then, because the line ab is perpendi- 
 cular to the plane de, the angle b is a right angle (def. 89), 
 and consequently greater than the angle c ; therefor^ the 
 line.AB opposite to the less angle, is less than any other line 
 AC, opposite the greater angle (th. 21). q, e- d. 
 
 THEOREM XCVI. 
 
 A Perpendicular Measures the Distance of any Point from a 
 Plane. 
 
 The distance of one point from another is measured by a 
 right line joining them, because this is the shortest line which 
 can be drawn from one point to another. So, also, the 
 distance from a point to a line, is measured by a perpendi- 
 cular, because this line is the shortest which can be drawn 
 
 from 
 
THEOREMS. 
 
 339 
 
 from the point to the line. In like manner, the distance from 
 a point to a plane, must be measured by a perpendicular 
 drawn from that point to the plane, because this is the short- 
 est line which can be drawn from the point to the plane. 
 
 THEOREM XCVII. 
 
 The common Section of Two Planes, is a Right Line. 
 
 .\ 
 
 A' 
 
 \ 
 
 1 
 
 Let acbda, aebfa, be two planes 
 tfuttiogeach pther, and a, b, two points 
 in which the two planes meet : drawing 
 the line ab, this line will be the common 
 intersection of the two planes. 
 
 For because the right line ab touches 
 the two planes in the points a and b, it 
 touches them in all other points (def. 
 20) : this hue is therefore common to the two planes. That 
 is, the common intersection of the two planes is a right line. 
 
 Q. E. D. 
 
 D 
 
 THEOREM XCVm. 
 
 B 
 
 If a Line be Perpendicular to two other Lines, at their Com- 
 mon Point of Meeting ; it will be Perpendicular to the 
 Plane of those Lines. 
 Let the line ab make right angles 
 
 with the lines ac, ad ; then will it be 
 
 perpendicular to the plane cde which 
 
 passes through these lines. 
 
 If the line ab were not perpendicular 
 
 to the plane cde, another plane misjht 
 
 pass through the point a, to which the 
 
 line ab would be perpendicular. But 
 
 this is impossible ; for, since the angles bag, bad, are right 
 
 angles, this other plane must pass through the points c, d. 
 
 Hence, this plane passing through the two points a, c, of the 
 
 line AC, and through the two points a, d, of the line ad, it 
 
 will pass through both these two lines, and therefore be the 
 
 same plane with the former. q. e. d. 
 
 CS) 
 
 THEOREM 
 
340 
 
 eEOMETRY. 
 
 THEOREM XCIX. 
 
 Tf Two Lines be Perpendicular to the Same Plane, they will 
 be Parallel to each other. 
 
 Let the two lines ab, cd. be both per- 
 pendicular to the same plane ebdf ; then 
 will AB be parallel to cd. 
 
 For, join b, d, by the line bd in the 
 plane. Then, because the lines ab, cd, 
 are perpendicular to the plain ef, they 
 are both perpendicular to the line bd (def. 89) in that plane ; 
 and consequently they are parallel to each other (corol. th. 
 13). Q. e: d*. 
 
 Corol. If two lines be parallel, and if one of them be 
 perpendicular to any plane, the other will also be perpendicu* 
 far to the same plane. 
 
 THEOREM C, 
 
 If Two planes Cut each other at Right Angles, and a Line be 
 drawn in one of the Planes Perpendicular to their Common 
 Intersection, it will be Perpendicular to the other Plane. 
 
 A^ 
 
 ^, 
 
 Let the two planes acbd, aebf, cut 
 each other at right angles ; and the line 
 CG be perpendicular to their common 
 section ab ; then will cg be also perpen- 
 dicular to the other plane aebf. 
 
 For, draw eg perpendicular to ab. -q 
 
 Then because the two lines gc, ge, are 
 
 perpendicular to the common intersection ab, the angle cge 
 is the angle of incUnation of the two planes (def. 91). But 
 since the two phnes cut each other perpendicularly, the 
 angle of inclination cge is a right angle. And since the line 
 CG is perpendicular to the two lines ga, ge, in the plane aebf> 
 it is therefore perpendicular to that plane (th. 98). 
 
 Q. e. d. 
 
 * This demonstratimi of Theorem, xcix. does not appear to me' to 
 fee conclude.. EDI roR. . THEOREM 
 
THEOREMS. 341 
 
 THEOREM CI. 
 
 If one Plane Meet another Plane, it will make angles with 
 that other Plane, which are together equal to two Right 
 Angles. 
 
 Let the plane acbc meet the plane aebf ; these planes 
 make with each other two angles whose sum is equal to two 
 right angles. 
 
 For, through any point Oj in the common section ab, dravr 
 CD, EF, perpendicular to ab. Then, the line cg makes with 
 EF two angles together equal to two right angles. But these 
 two angles are (by def. 91) the angles of inclination of the two 
 planes. Therefore the two planes make angles with each 
 other, which are together equal to two right angles. 
 
 Corol. In like manner it may be demonstrated, that planes 
 which intersect, have their vertical or opposite angles equal ; 
 also, that parallel planes have their alternate angles equal ; 
 and so on, as in parallel lines. 
 
 THEOREM CU. 
 
 If Two Planes be Parallel to each other ; a Line which is 
 Perpendicular to one of the Planes, will also be Perpendi- 
 cular to the other. 
 
 Let the two planes cd, ef, be parallel, 
 
 and let the line ab be perpendicular to the ^ — -^ 
 
 plane cd : then shall it also be perpendicu- ^f^ — 1 )^ 
 
 lar to the other plane ef. ^J — r 
 
 For, from any point g, in the plane ef, 
 draw GH perpendicular to the plane cd, and 
 draw . 
 
 raw ah, bg, ^ ^^ 
 
 Then, because ba, gh, are both perpendi- ^\A. — H/^ 
 
 cular to the plane cd, the angles a and h are 
 
 both right angles. And because the planes cd, ef, are 
 parallel, the perpendiculars ba, gh, are equal (def 92). 
 Hence it follows that the lines bg, ah, are parallel (^Aef, 9)* 
 And the line ab being perpendicular to the line ah, is also per- 
 pendicular to the parallel line bg (cor. th. 12). 
 
 In like manner it is proved, that the line ab is perpendi- 
 cular to all other lines whick can be drawn from the point b 
 
 in 
 
342 
 
 GEOMETRY. 
 
 in the plane ef. Therefore the line ab is perpendicular to 
 the whole plane ef (def. 92). ^ * ^. e. d. 
 
 THEOREM cm. 
 
 If Two Lines be Parallel to a Third Line, though not in the 
 same Plane with it ; they will be Parallel to each other. 
 
 F 
 
 .^^Tx 
 
 Let the lines ab, cd, be each of them 
 parallel to the third line ef, though not in 
 the same plane with it ; then will ab be pa- "Q 
 
 rallel to cd. 
 
 For, from any point g in the line ef, let gh, 
 Gi, be each perpendicular to ef, in the planes H 
 
 eb, ED, of the proposed parallels. . 
 
 Then, since the line ef is perpendicular /E\ 
 
 to the two lines gh, gi, it is perpendicular ^ 
 to the plane ghi of those lines (th 98). And because ef is 
 perpendicular to the plane ghi, its parallel ab is also perpen- 
 dicular to that plane (cor. th. 99.) For the same reason, the 
 line CD is perpendicular to the same plane ghi. Hence, be- 
 cause the two hnes ab, cd, are perpendicular to the same plane 
 these two lines are parallel ^th. 99). ^. e. d. 
 
 THEOREM CIV. 
 
 If Two Lines, that meet each other, be Parallel to Two other 
 Lines that meet each other, though not in the same Plane 
 with them ; the Angles contained by those Lines will be 
 equal. 
 
 Let the two lines ab, bc, be parallel to 
 the two lines de, ef ; then will the angle abc 
 be equal to the angle def. 
 
 For, make the lines ab, bc, de, ef, all 
 equal to each other, and join ac, df, ad, be, 
 of. . 
 
 Then, the lines ad, be, joining the equal 
 and parallel lines ab, de, are equal and paral- j^ 
 lei (th. 24). For the same reason, cf, be, 
 are equal and parallel. Therefore ad, cf, are equal and pa- 
 rallel (th. 15) ; and consequently also ac, DF(th. 24). Hence 
 the two triangles abc, def, having all their sides equal, 
 ) each 
 
 kKl 
 
THEOREMS. 
 
 343 
 
 each to each, have their angles also equal, and consequently 
 the angle abc = the angle def. q. e. d. 
 
 THEOREM CV. 
 
 The Sections made by a Plane cutting two other Parallel 
 Planes, are also Parallel to each other. 
 
 Let the two parallel planes ab, cd, be 
 out by the third plane ekhg, in the lines 
 EF, GH : these two sections ef, gh, will 
 be parallel. 
 
 Suppose FG, FH, be drawn parallel to 
 Bach other in the plane eehg ; also let ei, 
 FK, be perpendicular to the plane cd ; and 
 let ig, kh, be joined. 
 
 Then eg, fh, being parallels, and ei, fk, hemg both per- 
 pendicular to the plane cd, are also parallel to each other 
 (th. 99) ; consequently the angle hfk is equal to the angle 
 GEi (th. 104). But the angle fkh is also equal to the angle 
 EiG, being both right angles ; therefore the two triangles are 
 equiangular (cor. 1, th. 17) ; and the sides fk, ei, being the 
 equal distances between the parallel planes (def. 92), it fol- 
 lows that the sides fh, eg, are also equal- (th. 2) But these 
 two hnes are parallel (by suppos.), as well as equal ; conse- 
 quently the two lines ef, gh, joining those equal parallels, are 
 also parallel (th. 24). Q. e. d. 
 
 THEOREM CVI. 
 
 If any Prism be cut by a Plane Parallel to its Base, the Sec- 
 tion will be equal and Like to the Base. 
 
 Let AG be any prism, and il a plane pa- 
 rallel to the base ac ; then will the plane il 
 be equal and like to the base ac, or the two 
 planes will have all their sides and all their 
 angles equal. 
 
 For the two planes ac, il, being parallel, 
 by hypothesis ; and two parallel planes, cut 
 by a third plane, having parallel sections ^ J3 
 
 (th. 105) ; therefore ik is parallel to ab, and 
 KL to Bc, and im to cd, and iiM to ad. But ai and bk are 
 parallels (by def. 95) consequently ak is a parallelogram ; 
 and the opposite sides ab, ik, are equal (th. 22). In like 
 
 manner, 
 
 3 
 
 b: g 
 
344 
 
 GEOMETRY. 
 
 IrL 
 
 .1 
 
 G 
 
 K 
 
 B 
 
 manner, it is shown that kl is = bc, and lm 
 = CD, and iM = AD, or the two planes ac, il, 
 are mutually equilateral. But these two 
 planes, having their corresponding sides pa- 
 rallel, have the angles contained by them also 
 equal (th. 104), namely, the angle a = the 
 angle i, the angle b = the angle k, the angle 
 c = the angle l, and the angle d = the angle 
 M. So that the two planes ac, il, have all 
 their corresponding sides and angles equal, or they are equal 
 and like. > q. e. d. 
 
 THEOREM CVII. 
 
 If a Cylinder be cut by a Plane Parallel to its Base, the Sec- 
 tion will be a Ci^-cle, Equal to the Base. 
 
 Let af be a cylinder, and ghi any 
 section parallel to the base abc ; then will 
 QHi, be a circle equal to abc 
 
 For, let the planes ke, kf pass through 
 the axis of the cylinder mk, and meet the 
 section ghi in the three points h, i, l ; 
 and join the points as in the figure* 
 
 Then, since kl, cl, are parallel (by def. 
 101) ; and the plane ki meeting the two 
 parallel planes abc, ghi, makes the two sectiens kc, li, pa- 
 rallel (th. 105) ; the figure klic is therefore a parallelogram, 
 and consequently has the opposite sides li, kc, equal, where 
 Kc is a radius of the circular base. 
 
 In like manner it is shown that lh is equal to the radius 
 KB ; and that any other lines, drawn from the point l to the 
 circumference of the section ghi, are all equal to radii of the 
 base ; consequently ghi is a circle, and equal to abc. q. e. d. 
 
 THEOREM CVIII. 
 
 All Prisms and Cylinders, of equal Bases and Altitudes, are 
 Equal to each other. 
 
 Let AC, DF, be two 
 prisms, and a cylinder, 
 ©n equal bases ab, de, 
 and having equal alti- 
 tudes B«, FF ; then will 
 the solids ac, df, be 
 equal. 
 
 Foxy let PQ, Rs, be' 
 > T, . any 
 
 H 
 
 / / 
 
 / ) 
 
THEOREMS. 
 
 345 
 
 any two sections parallel to the bases, and equidistant from 
 them. Then, by the last two theorems, the section pq is 
 equal to the base ab, and the section rs equal to the base 
 DE. But the bases ab, de, are equal, by the hypothesis ; 
 therefore the sections pq, rs, are equal also. In like manner, 
 it may be shown, that any other corresponding sections are 
 equal to one another. 
 
 bince then every section in the prism ac, is equal to its 
 corresponding section in the prism or cylinder df, the prisms 
 and cylinder themselves, which are composed of an equal 
 number or all those equal sections, must also be equal, q.. e. d. 
 
 Carol. Every prism, or cylinder, is equal to a rectangular 
 parallelopipedon, of an equal base and altitude. 
 
 THEOREM CIX. 
 
 Rectangular Parallelopipedons, of Equal Altitudes, are to 
 each other as their Bases. 
 
 \ ^ 
 
 V ^ 
 
 ^ \ 
 
 D 
 I 
 
 K 
 
 B 
 
 \ ^ 
 
 \ ^ 
 
 \ \ 
 
 V ^ 
 
 a' 
 
 
 
 
 Let AC, EG, be two fectan- 
 gular parallelopipedons, having Q R S ^C T "V G 
 
 the equal altitudes ad, eh ; 
 then will the solid ac be to the 
 solid E6, as the base ab is to 
 the base ef. 
 
 For, let the proportion of the 
 
 base AB to the base ef, be that ^^ I^ M K E P 
 
 of any one number m (3) to any 
 
 other number n (2). And conceive ab to be divided into m 
 equal parts, or rectangles, ai, lk, mb, (by dividing an into 
 that number of equal parts, and drawing il, km, parallel 
 to bn). And let ef be divided, in like manner, into n equal 
 parts, or rectangles, eo, pf : all of these parts of both bases 
 being mutually equal among themselves. And through the 
 lines of division let the plane sections lr, ms, fv, pass parallel 
 
 to AQ, ET. 
 
 Then, the parallelopipedons ar, ls, mc, ev, pg, are all 
 equal, having equal bases and altitudes. Therefore the soHd 
 AC is to the solid eg, as the number of parts in the former, 
 to the number of equal parts in the latter ; or as the number 
 of parts in ab to the number of equal parts in ef, that is, as 
 the base ab to the base ef. q. e. d. 
 
 Corol. From this theorem, and the corollary to the last, it 
 
 appears, that all prisms and cylinders of equal altitudes, are 
 
 Vo*. 1. 45 t« 
 
346 
 
 <IEOMBTRY. 
 
 to each other as their bases ; every prism and cylinder being 
 equal to a rectangular pirallelopipedon of an equal base and 
 altitude. 
 
 tHEOREM ex. 
 
 Rectaiigular Parallelopipedons, of Equal Bases, are to each 
 other as their Altitudes. 
 
 Let ab, cd, be two rectan- 
 gular parallelopipedons, stand- 
 ing on the equal bases ae, cf ; 
 then will the solid ab be to the* 
 solid CD, as the altitude eb is to 
 the altitude fb. 
 
 For, let AG be a rectangular 
 parallelopipedon on the base ae, 
 and its aUitude eg equal to the altitude fd of the solid cd. 
 
 Then ag and cd are equal, being piisnos of equal bases 
 and altitudes. But if hb, hg, be considered as bases, the 
 solids ab, AG, of equal altitude ah, will be to each other 
 as those bases hb, hg. But these bases hb, hg, being 
 parallelograms of equal altitude he, are to each other as 
 .their bases eb, eg ; therefore the two prisms ab, ag, are 
 to each other as the Hues eb, eg. But ag is equal to 
 CD, and eg equal to fd ; consequently the prisms ac, cd, 
 are to each other as their altitudes eb, fd ; that is, - - - 
 AB : CD : : eb : fd. q. e. d. 
 
 CoroL 1. Ftom this theorem, and the corollary to theorem 
 1G8, it appears, that all prisms and cylinders, of equal bases, 
 are to one another as their altitudes. 
 
 Carol. 2. Because, by corollary 1, prisms and cylinders are 
 as their altitudes, when their bases are equal. And. by the 
 corollary to the last theorem, they are as their bases, when 
 their altitudes are equal. Therefore, universally, when nei- 
 ther are equal, they are to one another as the product of their 
 bases and altitudes. And hence also these products are the 
 proper numeral measures of their quantities or magnitudes. 
 
 THEOREM CXI. 
 
 Similar Prisms and Cylinders are to each other, as the 
 Cubes of their Altitudes, or of any other Like Linear Di- 
 mensions. 
 Let abcd, efgh, be two similar prisms ; then will the^ 
 
 prism CD be to the prism gh, as ab^ to ef* or ad^ to eh 3. 
 
 For 
 
THEOREMS. 
 
 347 
 
 O 
 
 For the solids are to each other as 
 the product of their bases and alti- ^ . 
 tudes (th. 1 10, cor. 2), that is, as N 
 AC . AD to EG . EH. But the bases, H^''^'*^ 
 
 being similar planes, are to each [S^ / 
 
 other as the squares of their like 
 sides, that is, ac to eg as ab^ to ^j^ 
 
 ef2 , therefore the solid cd is to the _ 
 
 solid GH, as ab3 . ad to ef^ . eh. B IF 
 
 B^it ED and FH, being similar planes, have their like sides 
 
 proportional, that is, ab : ef : : ad : eh, 
 
 or ab2 : ef2 : : ad^ : eh^ : therefore ab^ : ad : ef^ : eh : : ab ' : ef ' , 
 or : : ad^ : eh^ ;, oonseq. the solid cd : solid gh : : ab^ ; 
 |sf3 : : ad3 : eh'. Q* e. d.^ 
 
 ^ 
 
 & 
 
 THEOREM CXIL 
 
 In any Pyramid, a Section Parallel to the Base is similar to 
 the Base ; aud these two planes are to each other as the 
 Squares of their Distances from th^ Vertex. 
 
 Let abcd be a pyramid, and efg a sec- 
 tion parallel to the base bcd, also aih a 
 'line perpendicular to the two planes at h 
 and I : then will bd, eg, be two similar 
 planes, and the plane bd will be to the 
 plane eg, as ah^ to ai^. 
 
 For, join ch, fi. Then, because a plane 
 cutting two parallel planes, makes parallel 
 sections 'th. 105), therefore the plane abc, 
 meeting the two parallel planes bd, eg, makes the sections 
 bc, ef, parallel : In like manner, the plane acd makes the 
 sections cd, fg, parallel. Again, because two pair of paral- 
 lel lines make equal angles (th. 104), the two ef, fg, which 
 are parallel to bc, cd, make the angle efg equal the angle 
 BCD. And in Hke manner it, is shown, that each angle in the 
 plane eg is equal to each angle in the plane bd, aud conse- 
 quently those two planes are equiangular. 
 
 Again, the three lines ab, ac, ad, making with the 
 parallels bc, ef, and cd, fg. equal angles (th. 14), and 
 the angles at a being common, the two triangles abc, aef, 
 are equiangular, as also the two triangles acd, afg, and 
 have therefore their like sides proportional, namely, - - - 
 
348 
 
 GEOMETRY. 
 
 AC : AF : : BC ; EF : : CD : fg. And ia 
 like manner it may be shown, that all the 
 lines in the plane fg, are proportional to all 
 the corresponding lines in the base bd. 
 Hence these iwo planes, having their an- 
 gles equal, and their sides proportional, are 
 similar, by def. 68. 
 
 But, similar planes being to each other as the squares of 
 their like sides, the plane bd : eg : : bc^ : ef^, or : : ac^: 
 AF^, by what is shown above. Also, the two triangles 
 AHC, AiF, having the angles h and i right ones (th. 98), 
 and the angle a common, are equiangular, and have there- 
 fore their like sides proportional, namely, ac : af : : ah : ai, 
 or Ac3 : af2 : : ah^ : ai^. Consequently the two planes 
 BD, EG, which are as the former squares ac^, af^, will 
 
 be also as the latter squares ah^, ai^, that is, 
 
 :bd : EG : : ah^ : ai^. q,. e. d. 
 
 THEOREM CXIII. 
 
 In a Cone, any Section Parallel to the Base is a Circle ; and 
 
 this Section is to the Base, as the Squares of their Distances 
 
 from the Vertex. 
 
 Let ABCD be a cone, and ghi a section 
 parallel to the base bcd ; then will ghi be 
 a circle, and bcd, ghi, will be to each other, 
 as the squares of their distances from the 
 vertex. 
 
 For, draw alf perpendicular to the two 
 parallel planes ; and let the planes ace, 
 ADE, pass through the axis of the cone 
 ake, meeting the section in the- three points 
 
 H, I,-K. 
 
 Then, since the section ghi is parallel to the base bcd, and 
 the planes ck, dk, meet them, hk is parallel to ce, and ik to 
 de (th 105). And because the triangles formed by these 
 lines are equiangular, kh : ec : : ak : ae : : ki : ed. But 
 EC is equal to ed, being radii of the same cir< le ; therefore 
 Ki is also equal to kh. And the same may be shown of any 
 other lines drawn from the point k to the perimeter of the 
 ■section ghi, which is therefore a circle (def. 45). 
 
 Again, by similar triangle?, al : af : : ak : ae or 
 Ki : ED, hence al^ : af^ : : ki^ : fd^ ; but ki^ : ed^ 
 circle ghi : circle bcd (th. 93) j therefore al^ : af^ 
 circle ghi : circle bgd. q e. d. 
 
 THEOKEM 
 
THEOREMS. 
 
 349 
 
 THEOREM CXIV. 
 
 All Pyramids, and Cones, of Equal Basos and Altitudes, are 
 Equal to one another. 
 
 Let ABC, DEF, be 
 any pyramids and 
 cone, of equal bases 
 Bc, EF, and equal al- 
 titudCvS AG, DH ; then 
 will the pyramids 
 and cone abc and 
 DEF, be equal. 
 
 For, parallel to the 
 bases and at equal distances an, do, from the vertices, sup- 
 pose the planes ik, lm, to be drawn. 
 
 Then, by the two preceding theorems, - 
 
 Do2 : dh2 : : lm : ef, and 
 AN^ : ag2 : : IK : bc. 
 
 But since an2, ag^^ are equal to do^, dh^, 
 
 therefore ik : bc : : lm : ek. But bc is equal to ef, 
 by hypothesis ; therefore ik is also equal to lm. 
 
 In like manner it is shown, that any^ other sections, at equal 
 distance from the vertex, are equal to'each other. 
 
 Since then, every section in the cone, is equal to the cor- 
 responding section in the pyramids, and the heights are equal, 
 the solids abc, def, composed of all those sections, must be 
 #qual also. q. e. d. 
 
 THEOREM CXV. 
 
 Every Pyramid is the Third Part of a Prism of the Same Base 
 and Altitude. 
 
 Let abcdef be a prism, and bdef a A 
 
 pyramid, on the same triangular base 
 DEF : then will the pyramid, bdef be a 
 third part of the prism abcdef. 
 
 E 
 
 / 
 
 %: 
 
 For, in the planes of the three sides of 
 the prism, draw the diagonals bf, bd, cd. 
 Then the two planes bdp, bcd, divide the 
 whole prism into the three pyramids bdef, dabc, dbcf, which 
 are proved to be all equal to one another, as follows. 
 
 Since the opposite ends of the prij^m are equal to each other, 
 the pyramid whose base is abc and vertex d, is equal to the 
 
 pyramitl 
 
350 
 
 GEOMETRY. 
 
 pyramid whose base is def and vertex b 
 (th. 114), being pyramids of equal base and 
 altitude. 
 
 But the latter pyramid, whose base is^ 
 DEF and vertex b, is the same solid ias the 
 pyramid whose base is bef and vertex d, 
 and this is equal to the third pyramid 
 whose base is b< f and vertex d, being py- 
 ramids of the same altitude and equal bases 
 bef, bcf. , . 
 
 Consequently all the three pyramids, which compose the 
 •prism, are equal to each other, and each pyramid is the third 
 part of the prism, or the prism is tripje of the pyramid. 
 
 Q. E. rj. 
 
 Hence also, every pyramid, whatever its figure may be, is 
 the third part of a prism of the same base and altitude ; since 
 the base of the prism, whatever be its figure, may be divided 
 into triangles, and the whole solid into triangular prisms and 
 pyramids. 
 
 Corol. Any cone is the third part of a cylinder, or of a 
 prism, of equal base and altitude ; since it has been proved 
 that a cylinder is equal to a prism, and a cone equal to a pyra- 
 mid, of equal base and altitude. 
 
 Scholium. Whatever has been demonstrated of the propor- 
 tionality of prisms, or cylindere, holds equally true of pyra- 
 mids, or cones ; the former bemg always triple the latter ; viz. 
 that similar pyramids or cones are as the cubes of their like 
 linear sides, or diameters, or ailtiludes, &,c. And the same 
 for all similar solids, whatever, viz. that they are in proportion^ 
 to each other, as the cubes of their like linear dimensions, 
 since they are composed of pyramids every way similar. 
 
 THEOREM CXVI. 
 
 If a Sphere be cut by a Plane, the Section will be a Circle. 
 
 Let the sphere aebf be cut by the 
 plane adb ; then will the section adb be 
 a circle. 
 
 Draw the chord ab, or diameter of 
 the section ; perpendicular to which, or 
 to the section adb, draw the axis of the 
 sphere ecgf, through the centre c, 
 which will bisect the chord ab in the 
 point G (th. 41). Also, join ca, cb ; 
 
 and 
 
THEOREMS. 
 
 351 
 
 and draw cd, «d, to any point d in the perimeter of the sec- 
 tion AOB. 
 
 Then, because cg is perpendicular to the plane a»b, it is 
 perpendicular both to ga and gd (def. 90). So that cga, cgd, 
 are two risfht-angled triangles, having the perpendicular cc 
 cornmon. and the two hypothenuses ca, cd, equal, being both 
 radii of the sphere ; therefore the third sides ga, gd, are also 
 equal (cor. 2, th. 34). In like manner it is shown, that any 
 other line, drawn from the centre g to the circumference of 
 the section adb, is equal to ga or gb ; consequently that sec- 
 tion is a circle. 
 
 Corol. The section through the centre, is a circle having 
 the same centre and diameter as the sphere, and is called a 
 great circle of the sphere ; the other plane sections being \iU 
 tie circles. 
 
 THEOREM CXVII. 
 
 Every Sphere is Two-Thirds of its Circumscribing Cylinder 
 
 ^IC 
 
 "^ 
 
 /^\ 
 
 ^.^ 
 
 K 
 
 y 
 
 B 
 
 Let abcd be a cylinder, circumscrib- a 
 
 ing the sphere efgh ; then will the 
 sphere efgh be two-thirds of the cylin- 
 der ABt.D. 
 
 For, let the plane ac be a section of 
 the sphere and cylinder through the 
 
 centre i. Join ai, bi. Also, let fih be ^ 
 
 parallel to ad or bc, and eio and kl pa- D H 
 
 rallel to ab or dc, the base of the cy- 
 linder ; the latter line kl meeting bi in m, and the circular 
 section of the sphere in n. 
 
 Then, if the whole plane hfbc be conceived to reVolve 
 about the line hf as an axis, the square fg will describe a 
 cylinder ag, and the quadrant ifg will describe a hemisphere 
 EFG, and the triangle ifb will describe a cone iab. Also, in 
 the rotation, the three lines or parts kl, kn, km, as radii, will 
 describe corresponding circular sections of those solids, name • 
 ly, KL a section of the cylinder, KJf a section of the sphere, 
 and KM a section of the cone. 
 
 Now, FB being equal to fi or ig, and kl parallel to fb, 
 then by similar triangles ik is equal to km (th. 82). And since, 
 in the right-anglv-d triangle ikn, in^ is equal to ik^ -f. kx^ 
 (th. 34) ; and because kl is equal to the radius le or in, and 
 
352 
 
 GEOMETRY. 
 
 A 
 O 
 
 XQ- 
 
 
 D 
 
 KM = IK, therefore ki.^ is equal to km^ 
 -f KN^, or the square of the longest ra- 
 dius, of the said circular sections, is 
 equal to the sum of the squares ol the 
 two others. And because circles are to 
 each other as the squares of their diam- 
 eters, or of their radii, therefore the 
 circle described by kl is equal to both 
 the circles des< iliied by km and kn ; or 
 the section of the cylinder, is equal to both the corresponding 
 sections of the sphere and cone. And as this is always the 
 case in every parallel position of kl, it follows, that the cylin- 
 der EB, which is composted of all the former s^ections, is equal 
 to the hemisphere efg and cone iab, which are composed of 
 of all the latter sections. 
 
 But the cone iab is a third part of the cylinder fb (cor. 2, 
 th. 116) ; consequentl) the heoiisphere efg is equal to the 
 remaining two-thirds ; or the whole sphere efgh equal to two- 
 thirds ot the whole cylinder abcd. q. e. d 
 
 Corol. 1. A cone, hemisphere, and cylinder of the same 
 base and altitude, are to each other as the numbers 1, 2, 3. 
 
 Corol. 2. All spheres are to each other as the cubes of their 
 diameters ; all these being like parts of their circumscribing 
 cylinders. 
 
 Corol. 3. From the foregoing demonstration it also appears^ 
 that the spherical zone or frustrum egnp, is equal to the diffe- 
 rence between the cylinder eglo and the cone imq. all of the 
 same common height ik. And that the spherical segment pfn, 
 is equal to the difference between the cylinder ablo and the 
 conic frustrum a^imb, all of the same common altitude fk. 
 
 PROBLEMS. 
 
[353} 
 
 PR0BLJ2M&. 
 PROBLEM I. 
 
 To Bisect a Line ab 
 
 that is, to divide it iato two Equal 
 Parts. 
 
 From the two centres a and b, with 
 any equal radii, describe arcs of circles, 
 intersecting each other in c and d ; and 
 draw the line cd, which will bisect the 
 given line ab in the point e. 
 
 For, draw the radii ao, bc, ad, bd. 
 Then, because all these four radii are 
 equal, and the side cd common, the two 
 triangles acd, bcd, are mutually equilateral : consequently 
 they are also mutually equiangular (th. 5), and have the angle 
 ACE equal to the angle bce. 
 
 Hence, the two triangles ace, bce, having the two sides 
 AC, CE, equal to the two sides bc, ce, and their contained an- 
 gles equal, are identical (th. 1), and therefore have the side 
 ae equal to eb. q. e. d. 
 
 PROBLEM II. 
 To Bisect an Angle bac. 
 
 From the centre a, with any radius, de- 
 scribe an arc, cutting oflf the equal lines 
 ad, ae ; and from the two centres d, e, 
 with the same radius, describe arcs inter- 
 secting in F ; then draw af, which will bi- 
 sect the angle a as required. 
 
 For, join df, ef. Then the two tri- 
 angles adf, aef, having the two sides 
 AD, DF, equal to the two ae, ef (being equal radii), and the 
 side AF common, they are mutually equilateral ; consequently 
 they are also mutually equiangular (th. 3), and have the angle 
 BAF equal to the angle caf. 
 
 ScholiuiYi. In the sacne manner is an arc of a circle, bi- 
 sected. 
 
 Vot, I.- 4« 
 
 PROBt^M 
 
354 
 
 GEOMETHY. 
 
 PROBLEM in. 
 
 EB 
 Then the two 
 
 At a Given Point c, in a Line ab, to Erect a Perpendicular. 
 
 From the given point c, with any radius, 
 cut off any equal parts cd, ce, of the given 
 line ; and, from the two centres d and e, 
 with any one radius, describe arcs intersect- 
 ing in F ; then join cf, which will be per- 
 pendicular as required. 
 
 Tor, draw the two equal radii df, ef. 
 triangles CDF, cef, having the two sides cd, df, equal to 
 the two CE, EF, and cf common, are mutually equilateral ; 
 consequently they are also mutually equiangular (th. 5), and 
 have the two adjacent angles at c equal to each other ; there- 
 fore the line cf is perpendicular to ab (def. 11). 
 
 Otherwise. 
 When the Given Point c is near the End of the line. 
 
 From any point d, assumed above the 
 line, as a centre, through the given point 
 c describe a circle, cutting the given Hne 
 at E ; and through e and the centre d, 
 draw the diameter edf ; then join cf, 
 which will be the perpendicular required. 
 
 For the angle at c, being an angle in a semicircle, is a 
 right angle, and therefore the line cf is a perpendicular 
 (by def 15). 
 
 PROBLEM lY. 
 
 From a Given point a to let fall a Perpendicular on a given 
 Line bc. 
 
 From the given point a as a centre, with 
 any convenient radius, describe an arc, cut- 
 ting the given line at the two points d and 
 £ ; and from the two centres d, e, with 
 any radius, describe two arcs, intersecting 
 at f ; then draw agf, which will be perpen- 
 dicular to bc as required- 
 
 For, draw the equal radii ad, ae, and 
 DF, EF. Then the two triangles adf, aef, having the two 
 sides ad, df, equal to the two ae, ef, and af common, are 
 
 mutually 
 
PROBLEMS. 356 
 
 mutually equilateral ; consequently they are also mutually 
 equiangular (th. 5), and have the ani^le dag equal the angle 
 EAG. Hence then, the two triangles adg, aeg, having the 
 two sides ad, ag, equal to the two ae, ag, ahd their included 
 angles equal, are therefore equiangular (th 1), and have the 
 angles at g equal ; consequently ag is perpendicular to bc 
 (def. 11). 
 
 Otherwise. 
 
 When the Given Point is nearly Opposite the end of the 
 Line. 
 
 From any point d, in the given line 
 Be, as a centre, descrihe the arc of a 
 circle through the given point a, cutting 
 BC in B ; and from the centre e, with the 
 radius ea, describe another arc, cutting 
 the former in f ; then draw agf, which 
 will be perpendicular to bc as required. F 
 
 For, draw the equal radii da, df, and ea, ef. Then the 
 two triangles dae, dfe, will be mutually equilateral ; conse- 
 quently they are also mutually equiangular (th. 5), and have 
 the angles at d equal. Hence, the two triangles dag, dfg, 
 having the two sides da, dg, equal to the two df, dg, and the 
 included angles at d equal, have also the angles at g equal 
 (th. 1) ; consequently those angles at g are right angles, and. 
 the line ag is perpendicular to dg. 
 
 PROBLEM V^ 
 
 At a Given Point a, in a Line ab, to make an Angle Equal 
 to a Given Angle c. 
 
 From the centres a and c, with any one JEi 
 
 radius, describe the arcs de, fg. Then, 
 with radius de, and centre f, describe an 
 
 arc, cutting fg in g. Through g draw C D 
 
 the line ag, and it will form the angle re- ^^ 
 
 quired. ' ^,^^ \ 
 
 For, conceive the equal lines or radii, A FB 
 
 de, fg, to be drawn. Then the two triangles cde, afg, bemo" 
 mutually equilateral, are mutually equiangular (th. 5), and 
 have the angle at a equal to the angle c. 
 
 PROBLEM 
 
V 
 
 366 ' GEOMETRY. 
 
 PROBLEM VI. 
 
 Through a Given Point a, to draw a Line Parallel to a Given 
 Line bc. 
 
 From the given point a draw a line ad EA 
 
 to any point in the given line bc Then 
 draw the hne eaf making the angle at a 
 equal to the angle at d (by prob. 5) ; so jg~" jS^ 
 
 shall EF be parallel to bc as required. 
 
 For. the angle d being equal to the alternate angle a, th^ 
 lines BC, EF, are parallel, by th. 13. 
 
 PROBLEM VII. 
 
 To Divide a Line ab into any proposed Number of Equal 
 Parts. 
 
 m 
 
 Draw any other line ac, forming any 
 iangle with the given line ab ; on which 
 set off as many of any equal parts, ad, de, 
 EF, Fc, as the line ab is to be divided into. 
 Join bc ; parallel to which draw the other A. I H & B 
 
 lines FG, EH, Di : then these will divide 
 ab in the manner as required. — For those parallel lines divide 
 l)oth the sides ab, ac, proportionally, by th. 82. 
 
 PROBLEM \1IL 
 
 To find a Third Proportional to Two given Lines ab, ac,v 
 
 PtACE the two given lines ab, ac, 
 
 forming any angle at a ; and in ab take A — B 
 
 ?ilso AD equal to ac Join bc, and A -C 
 
 draw DE parallel to it ; so will ae be -. C 
 
 the third proportional sought. . ""^^ 
 
 For, because of the parallels bc, de, 
 
 the two lines ab, ac, are cut propor- ^ " 
 
 tionally (th. 82) ; 90 that ab : ac : : ad or ac : ae ; there-- 
 fbre AE is the third proportional to ab, ac. 
 
 PROBLEM IXt 
 
 To find a Fourth Proportional to three Lines ab, ac, ad. 
 
 Place two of the given lines ab, ac, making any angle 
 at A ; also place ad on ab. Join bc j and parallel to it draw 
 
 de : 
 
PROBLEMS. 
 
 367 
 
 Dfi : SO shall AE be the fourth propor- 
 tional as required. 
 
 For, because of the parallels bc, de, 
 the two sides ab, ac, are cut propor- 
 tionally (th. 82) ; so that . 
 AB : AC : : AD : ae. 
 
 A — 
 
 A 
 
 B 
 
 A 
 
 D^ 
 
 
 ^r\ 
 
 A. 
 
 " I> B 
 
 PROBLEM X. 
 
 To find a Mean Proportional between Two Lines ab, bc. 
 
 Place ab, bc, joined in one straight p^ g 
 
 line AC : on which as a diameter, des- 
 cribe the semicircle adc ; to meet which 
 erect the perpendicular bd ; and it will 
 be the mean proportional sought, be- 
 tween AB and Bc (by cor. th. 87). 
 
 D— C 
 
 OB 
 
 PROBLEM XI. 
 To find the Centre of a Circle. 
 
 Draw any chord ab ; and bisect it per- 
 pendicularly with the line cd, which will 
 be a diameter (th. 41, cor.). Therefore, 
 &D bisected into o, will give the centre, 
 as required. 
 
 PROfiLEA^ XII. 
 
 To describe the Circamference of a Circle through Three 
 Given Points a, b, c. 
 
 From the paiddle point b draw chords 
 BA, BC to the two other points, and bi- 
 sect these chords perpendicularly by 
 lines meeting in o, which will be the 
 centre. Then from the centre o, at the 
 distance of any one of the points, as oa, 
 describe a circle, and it will pass through 
 the two other points b, g, as required. 
 
 For, the two right-angled triangles oad, obd, havino- the 
 sidfis AD, DB, equal (by constr.), and od common with the 
 included right angles at d equal, have their third sides oa, 
 OB, also equal (th. J). And in like manner it is shown, that 
 oc, is equal to ob or oa. So that all the three oa, ob, oc, 
 ^eing equal, will be radii of the same circle. 
 
 PROBLEM 
 
358 
 
 GEOMETRY. 
 
 PROBLEM XIII. 
 
 To draw a Tangent to a Circle, through a Giren Point i. 
 
 When the given point a is in the cir- 
 cumference of the circle : Join a and the 
 centre o ; perpendicular to which draw 
 BAG, and it will be the tangent, by th. 46. 
 
 But when the given point a is out of 
 the circle : draw ao to the centre o ; 
 on which as a diameter describe a semi- 
 circle, cutting the given circumference 
 in D ; through which draw badc, which 
 will be the tangent as required. 
 
 For, join do. Then the angle ado, 
 in a semicircle, is a right angle, and 
 consequently ad is perpendicular to the 
 radius, do, or is a tangent to the circle 
 (th. 4tJ) 
 
 PROBLEM XIV. • 
 
 On a Given Line b to describe a Segment of a Circle, to 
 Contain a Given Angle c. 
 
 At the ends of the given line make 
 angles DAB, dba, each equal to the 
 given angle c. Then draw ae, be, 
 perpendicular to ad, bd ; and with the 
 centre e and radius ea or eb, describe 
 a circle ; so shall Afb be the segment 
 required, as an angle f made in it will 
 be equal to the given angle c. 
 
 For, the two lines ad, bd, being per- 
 pendicular to the radii ea, eb (by constr.), are tangents to the 
 circle (th. 46) ; and the angle a or b, which is equal to the 
 given angle c by construction, is equal to the angle f in the 
 alternate segment afb (th. 63). 
 
 PROBLEM XV. 
 
 To Cat ofl' a Segment from a Circle, that shall Contsan a 
 Given Angle c. 
 
 Draw any tangent ab to the given 
 circle ; and a chord ad to make the 
 angle dab equal to the given angle c ; 
 then DEA will be the segment required, 
 an angle e made in it being equal to 
 the given angle e. 
 
PROBLEMS. 
 
 35» 
 
 For tbe angle a, made by the tans;ent and chord, which is 
 equal to the given angle c by construction, is also equal to any 
 ti^e E in the alternate segment (th. 53). 
 
 PROBLEM XVI. 
 To make an Equilateral Triangle on a Giren Line ab. 
 
 From the centres a and b, with the 
 distance ab, describe arcs, intersecting 
 in c. Draw ac, bc, and abc will be the 
 equilateral triangle 
 
 For the equal radii ac, bc, are, each 
 of them, equal to ab. 
 
 PROBLEM XVn. 
 
 To make a Triangle with Three Given Lines ab, ac, bc. 
 
 With the centre a, and distance ac, 
 describe an arc. With the centre b, 
 and distance bc, describe another arc, 
 cutti;)}i; the former in c Draw ac, bc, 
 and ABC will be the triangle required. 
 
 For the radii, or sides of the triangle, 
 AC, BC, are equal to the given lines ac, 
 BC, by construction. 
 
 PROBLEM XVIIL 
 
 To make a Square on a Given Line ab. 
 
 Raise ad, bc, each perpendicular and 
 equal to ab ; and join dc ; s'o Shall abcd 
 be the square sought. 
 
 For all the three sides ab, ad, bc, are 
 equal, by the construction, and dc is equal 
 an«) parallel to ab (by th. 24) ; so that all j 
 
 the four sides are equal, and the opposite 
 ones are parallel. Again, the angle a or b, of the parallelo- 
 gram, being a right angle, the angles are all right ones (cor. 
 1, th. 22). Henre, then, the figure, having all its sides equal, 
 sind all its angles right, is a square (def. 34). 
 
 B 
 
 PROBLEM 
 
36d 
 
 GEOMETRY. 
 
 PROBLEM XrX. 
 
 To make a Rectangle, or a Parallelogram, of a Given Lengtk 
 and Breadth, ab, bc. 
 
 Erect ad, bc, perpendicular to ab, and I> C 
 
 each equal to bc ; then join dc, and it is 
 done. 
 
 The demonstration is the same as the last j 
 
 prohlem. B- 
 
 And in the same manner is descrihed any oblique parallel- 
 ogram, only drawing ad and bc to make the given oblique an 
 gle with ab, instead of perpendicular to it. 
 
 B 
 
 PROBLExM XX. 
 
 To Inscribe a Circle in a Given Triangle abc. 
 Bisect any two angles a and b, with 
 the two lines ad, bd. From the inter- 
 section D, which will be the centre of 
 the circle, draw the perpendiculars de, 
 DF, dg, and they will be the radii of the 
 circle required. 
 
 For, since the angle dae is equal to 
 the angle dag, and the angles at e, g, 
 
 right angles (by constr.), the two triangles ade, adg, are equi- 
 angular ; and, having also the side ad common, they are iden- 
 tical, and have the sides de, dg, equal (th. 2). In like man- 
 ner it is shown, that df is equal to de or dg. 
 
 Therefore, if with the centre d, and distance de, a circle 
 be described, it will pass through all the three points e, f, g, 
 in which points also it will toucuj the three sides of the triangle 
 (th. 46), because the radii de, df, dg, are perpendicular t© 
 them. 
 
 PROBLEM XXL 
 
 To Describe a Circle about a Given Triangle abc 
 
 Bisect any two sides with two of the 
 perpendiculars de, df, dg, and d will be 
 the centre. 
 
 For, join DA, DB, dc. Then the two 
 right-angled triangles dae, dbe, have the 
 two sides de, ea, equal to the two de, 
 eb, and the included angles at e equal : 
 those two triangles are therefore identical 
 
 (th. 
 
PROBLEMS. 
 
 361 
 
 (th. 1), and have the side da equal to db. In like manner 
 it is shown, thaf dc is also equal to da or db. So that all 
 the three da, db, dc, being equal, they are radii of a circle 
 passing through a, b, and c. 
 
 PROBLEM XXn. 
 
 To Inscribe an Equilateral Triangle in a Given Circle- 
 
 Through the centre c draw any oTa- 
 meter ab. From the point b as a centr<|| 
 with the radius bc of the given circle, 
 describe an arc dce. Join ad, ae, de, 
 and ADE is the equilateral triangle sought. 
 
 For, join db, do, eb, eg. Then dcb 
 is an equilateral triangle, having each 
 side equal to the radius of the given circle. 
 In like manner, bce is an equilateral triangle. But the angle 
 ADE is equal to the angle abe or cbe, standing on the same 
 arc AE ; also the angle aed is equal to the angle cbd, on, the 
 same arc ad ; hence the triangle dae has two of its angles, 
 ADE, aed, equal to the angles of an equilateral triangle, and 
 ther^ore the third angle at a is also equal to the same ; so 
 that triangle is equiangular, and therefore equilateral. 
 
 PROBLEM xxnr. 
 
 To Inscribe a Square in a Given Circle. 
 
 Deaw two diameters ac, bd, crossing 
 at right angles in the centre ». Then 
 join the four extremities a, b, c, d, with 
 right lines, and these will form the in- 
 scribed square abcd. 
 
 For, the four right-angled triangles 
 aeb, beg, ged, »ea, are identical, be- 
 cause they have the sides ea, eb, eg, ed, 
 all equal, being radii of the circle, and 
 the four included angles at e all equal, 
 being right angles, by the construction. Therefore all their 
 third sides ab, bg, cd, da, are equal to one another, and the 
 figure ABCD is equilateral. Also, all its four angles, a, b, c, 
 D, are right ones, being angles in a semicircle. Consequently 
 the figure is a square. 
 
 VOB. h 
 
 47 
 
 PROBLPM 
 
f 
 
 ^ 
 
 \^ 
 
 V 
 
 362 GEOMETRY. 
 
 PROBLEM XXIV. 
 To Describe a Square about a Given Circle. 
 
 Draw two diameters ac, bd, crossing 
 at right angles in the centre e. Then ^ J^y^^ ? 
 through their four extremities draw fg, 
 iH, parallel to ac, and fi, gh, parallel to 
 BD, and they will form the square fghi. 
 
 For, the opposite sides of parallelo- 
 grams being equal, fg and ih are each r — ^^ j^ g." 
 equal to the diam||ter ac, and fi and gh 
 each equal to the diameter bd ; so that 
 the figure is equilateral. Again, because the opposite angles 
 of parallelograms are equal, all the four angles f, g, h, i, are 
 right angles, being equal to the opposite angles at e. So that 
 the figure fghi, having its sides equal, and its angles right 
 ones, is a square, and its sides touch the circle at the four 
 points a, b, c, d, being perpendicular to the radii drawn to 
 those, points. 
 
 I>ROBLEM XXV. 
 To Inscribe a Circle in a Given Square. 
 
 Bisect the two sides fg, fi, in the points a and b (last fig.). 
 Then through these two points draw ac parallel to fg or ih, 
 and BD parallel to fi or gh. Then the point of intersection 
 E will be the centre, and the four lines ea, eb, ec, ed, radii 
 of the inscribed circle. 
 
 For, because the four parallelograms eb*, eg, eh, ei, have 
 their opposite sides and angles equal, therefore all the four 
 lines ea, eb, ec, ed, are equal, being each equal to half a 
 side of the square. So that a circle described from the centre 
 E, with the distance ea, will pass through all the points a, b, 
 c, d, and will be inscribed in the square, or will touch its four 
 sides in those points, because the angles there are right ones. 
 
 PROBLEM XXVL 
 
 To Describe a Circle about a Given Square, 
 (see fig. Prob. xxiii.) 
 
 Draw the diagonals ac, bd, and their intersection e will be 
 the centre. 
 
 For the diagonals of a square bisect each other (th. 40), 
 making ea, eb, ec, ed, all equal, and consequently these 
 are radii of a circle passing through the four points a, b, c, d. 
 
 PROBLEM 
 
PROBLEMS. 
 
 36d 
 
 Then, adf 
 
 PROBLEM XXVII. 
 To Gut a Given Line in Extreme and Mean Ratio. 
 
 Let ab be the giren line to be divided 
 
 in extreme and mean ratio, that is, so as 
 
 that the whole line may be to the greater 
 
 part, as the greater part is to the less part. 
 Draw Bc perpendicular to ab, and equal 
 to half AB. Join ac ; and with centre c 
 and distance ce, describe the circle bd ; 
 then with centre a and distance ad, de- 
 scribe the arc de ; so shall ab be divided 
 in E in extreme and mean ratio, or so that 
 AB : AE,; : ae ; eb. 
 
 For, produce ac to the circumference at 
 being a secant, and ab a tangent, because b is a right angle ; 
 therefore the rectangle af . ad is equal to ab^ (cor. 1, th. 61) ; 
 consequently the means and extremes of these are proportion- 
 al (th. 77), viz. ab : af or ad -f- df : : ad : ab. But ae is 
 equal to ad by construction, and ab = 2bc = df ; therefore, 
 ab : ae -I- ab : : ae : ab ; and by division, 
 AB :. AE : : ak : eb. 
 
 PROBLEM XXVIIL 
 
 To Inscribe an Isosceles Triangle in a Given Circle, that shall 
 have each of the Angles at the Base Double the Angle at 
 the Vertex. 
 
 Draw any diameter ab of the given 
 circle ; and divide the radius cb, in the 
 point d, in extreme and mean ratio, by 
 the last problem. From the point b 
 apply the chords be, bf, each equal to 
 the greater part cd. Then join ae, af, 
 EF ; and aef will be the triangle requir- 
 ed. 
 
 For, the chords be, bf, being equal, their arcs are equal ; 
 therefore the supplemental arcs and chords ae, af, are also 
 equal ; consequently the triangle aef is isosceles, and has 
 the angle e equal to the angle f ; also the angles at g are 
 right angles. 
 
 Draw OF and df. Then, bc : cd : : cd : bd, or 
 EC : bf : : bf : bd by constr. And ba :' bf : : bf : bg 
 (by tb. 87). But bc = 4^ba ; therefore bg = Ibd = gd ; 
 therefore the two triangles gbf, gdf, are identical (th. 1), 
 
 and 
 
364 
 
 €frEOMETRY. 
 
 and each equiangular to abf and agf (th. 87). Therefofe 
 their doubles, bfd, afe, are isosceles and equiangular, as 
 well as the triangle bcf ; having the two sides bc, cf, equal, 
 and the angle b common with the triangle bfd. But cb 
 is = DF or BF ; therefore the angle c = the angle dfc (th. 
 4) ; consequently the angle bdp, which is equal to the sum of 
 these two equal angles (th. 16), is double of one of them c ; or 
 the equal angle b or cfb double the angle c. So that cbf ifi 
 an isosceles triangle, having each of its two equal angles 
 double of the third angle c. Consequently the triangle aef 
 (which it has been shown is equiangular to the triangle cbf) 
 has also each of its angles at the base double the angle a at 
 the vertex. 
 
 PROBLEM XXIX. 
 
 To Inscribe a Regular Pentagon in a Given Circle. 
 
 Inscribe the isosceles triangle abc 
 having each of the angles abc, acb, 
 double the angle bac (prob. 28). Then 
 bisect the two arcs adb, afc, in the 
 points D, E ; and draw the chords ad, 
 »b, ae, EC, so shall adbce be the in- 
 scribed equilateral pentagon required. 
 
 For, beeause equal angles stand on 
 equal arcs, and double angles on double arcs, also the angles 
 ABC, ACB, being each double the angle bac, therefore the arcs 
 ADB, AEc, subtending the two former angles, each one double 
 the arcs bc subtending the latter. And since the two former 
 arcs are bisected in d and e, it follows that all the five arcs 
 AD, db, bc, ce, ea, are equal to each other, and consequently 
 the chords also which subtend them, or the five sides of the 
 pentagon, are all equal. 
 
 Note In the construction, the points d and e are most easily 
 found, by applying bd and ce each equal to bc. 
 
 PROBLEM XXX. 
 
 To Inscribe a Regular Hexagon in a Circle. 
 
 Apply the radius ao of the given circle 
 as a chord, ab, bc, cd, &c. quite round the 
 circumference, and it will complete the 
 regular hexagon abcdef. 
 
 For, draw the radii ao, bo, co, do, eo, 
 yo, completing six equal triangles ; of 
 which any one, as abo, being equilateral 
 
PROBLEMS. 365 
 
 by constr.)it8 three angles are all equal (cor. 2, th. 3), and 
 any one of them, as aub, is one-third of the whole, or of two 
 right angles (th. 17), or one-sixth of four right angles. But 
 the whole circumference is the measure of four right angles 
 (cor 4, th. 6). Therefore the arc ab is one-sixth of the cir- 
 cumference of the circle, and consequently its chord ab one 
 aide of an equilateral hexagon inscribed in the circle. And 
 the same of the other chords. 
 
 Corol, The side of a regular hexagon is equal to the radius 
 of the circumscribing circle, or to the chord of one-sixth 
 part of the circumference. 
 
 PROBLEM XXXI. 
 
 To describe a Regular Pentagon or Hexagon about a Circle. 
 
 In the given circle inscribe a regular A^ 
 
 polygon of the same name or number ~ 
 
 of sides, as abode, by one of the forego- 
 ing problems. Then through all its an- 
 gular points draw tangents (by prob. 13) 
 and these will form the circumscribing 
 polygon required. 
 
 For, all the chords, or sides of the 
 inscribing figure, ab, bc, &c. being equal, and all the radii 
 OA, oB, &c. being equal, all the vertical angles about the point 
 o are equal. But the angles oef, oaf, oag, obo, made by 
 the tangents and radii, are right angles ; therefore oef + oaf 
 = two right angles, and oag + obg = two right angles ; 
 consequently, also, aoe -f afe = two right angles, and aob 
 -|- AGB = two right angles (cor. 2, th. 18). Hence, then, the 
 angles aoe -\- afe being = aob -f agb, of which aob is = 
 AGE ; consequently the remaining angles f and g are also 
 equal. In the same manner it is shown, that all the angles f, 
 G, H, I, K, are equal. 
 
 Again, the tangents from the same point fe, fa, are equal, 
 as also the tangents ag, gb (cor. 2, th. 61) ; and the angles 
 F atid G of the isosceles triangles afe, agb, are equal, as well 
 as their opposite sides ae, ab ; consequently those two trian- 
 gles are identical (th. 1), and have their other sides ef, fa, ag, 
 GB, all equal, and fg equal to the double of any one of them. 
 In like manner it is shown, that all the other sides gh, hi, ik, 
 KF, are equal to fg, or double of the tangents gb, bh, &c. 
 
 Hence, then, the circumscribed figure is both equilateral 
 and equiangular, which was to be shown. 
 
 Corol, 
 
366 . 
 
 GEOMETRY. 
 
 Corol The inscribed circle touches the middles of the sides 
 of the polygon. 
 
 PROBLEM XXXII. 
 
 To inscribe a circle in a Regular Polygon. 
 
 Bisect any two sides of the polygon 
 by the perpendiculars go, fo, and their 
 intersection o will be the centre of the 
 inscribed circle, and o'g or of will be 
 the radius. 
 
 For the perpendiculars to the tangents 
 
 AF, AG, pass through the centre (cor. th. 
 
 47) ; and the inscribed circle touches C D 
 
 the middle points f, g, by the last corollary. Also, the two 
 sides AG, Ao, of the right-angled triangle aog, being equal 
 to the two sides af, ao, of the right-angled triangle aof, the 
 third sides of, og, will also be equal (cor. th. 45). Therefore 
 the circle described with the centre o and radius og, will pass 
 through F, and will touch the sides in the points g and f. And 
 the same for all the other sides of the figure. 
 
 PROBLEM XXXni. 
 
 To Describe a Circle about a Regular Polygon. 
 
 Bisect any two of the angles, c and d, 
 with the lines co, do ; then their inter- 
 section o will be the centre of the. cir- 
 cumscribing circle : and oc, or od, will 
 be the radius, 
 
 For, draw ob, oa, oe, &c. to the 
 angular points of the given polygon. Then 
 the triangle ocd is isosceles, having the angles at c, and d 
 equal, being the halves of the equal angles of the polygon 
 BCD, CDE ; therefore their opposite sides co, do, are equal 
 (th. 4). But the two triangles ocd, ocb, having the two sides 
 oc, CD, equal to the two oc, cb, and the included angles ocd, 
 ocb also equal, will be identical (th. l),and have their third 
 sides bo, od, equal. In like manner it is shown, that all the 
 lines OA, ob, oc, od, oe, are equal. Consequently a circle 
 described with the centre o and radius oa, will pass through 
 all the other angular points, b, c, d, &c. and will circumscribe 
 the polygon. 
 
 PROBLEM 
 
PROBLEMS. 
 
 367 
 
 PROBLEM XXXIV. 
 
 To make a Square Equal to the Sum of two or more Giveo 
 
 Squares. 
 
 Let ab and ac be the sides of two 
 given squares. Draw two indefinite 
 lines AP, AQ, at right angles to each 
 other ; in which place the sides ab, 
 AC, of the given squares ; join bc ; 
 then a square described on bc will be 
 
 equal to the sum of the two squares 
 
 described on ab and ac (th. 34). P B D A 
 
 In the same manner, a square may be made equal to the 
 sum of the three or more given squares. For, if ab, ac, ad, 
 be taken as the sides of the given squares, then, making ae = 
 Bc, AD = AD, and drawing de, it is evident that the square on 
 fiE will be equal to the sum of the three squares on ab, ac, ad. 
 And so on for more squares. 
 
 PROBLEM XXXV. 
 
 To make a Square Equal to the Diflference of two Given 
 Squares. 
 
 ^ 
 
 Let ab and ac, taken in the same 
 straight line, be equal to the sides of the 
 two given squares. — From the centre a, 
 
 with the distance ab, describe a circle ; 
 
 and make cd perpendicular to ab, meet- A C B 
 
 ing the circumference in d : so shall a square described on cd 
 be equal to ad^ — ac^ , or ab^ — ac^ , as required (cor. 1 , th. 34). 
 
 PROB LEM XXXVL 
 
 To make a Triangle Equal to a Given Quadrangle abcd. 
 
 Draw the diagonal ac, and parallel 
 to it de, meeting ba produced at e, and 
 join ce ; then will the triangle ceb be 
 equal to the given quadrilateral abcd. 
 
 For, the two triangles ace, acd, be- 
 ing on the same base ac, and between 
 the same parallels ac, de, are equal (th. 25) ; therefore, if 
 ABC be added to each, it will make bce equal to abcd (ax. 2). 
 
 PROBLEM 
 
368 
 
 GEOMETRY. 
 
 PROBLEM XXXVH. 
 To make a Triangle Equal to a Given Pentagon abcde. 
 
 Draw da and db, and also ef, cg, 
 parallel to them, meeting ab produced 
 at F and c ; then draw df and dg ; so 
 shall the triangle dfg be equal to the 
 given pentagon abcde. 
 
 For the triangle dfa = dea, and 
 the triangle dgb = dcb (th. 25) ; 
 therefore, by adding dab to the equals, 
 
 the sums are equal (ax. 2), that is, dab + daf + dbg = dab 
 -f dab + DBC, or the triangle dfg = to the pentagon abcde. 
 
 PROBLEM XXXVUL 
 To make a Rectangle Equal to a Given Triangle abc. 
 
 Bisect the base ab in d ; then raise de 
 and bf perpendicular to ab, and meeting cf 
 parallel to ab, at e and f : so shall df be 
 the rectangle equal to the given triangle 
 ABC (by cor. 2, th. 26). 
 
 PROBLEM XXXIX. 
 To make a Square Equal to a Giv^n Rectangle abcd. 
 
 D 
 
 JF 
 
 Produce one side ab, till be be 
 equal to the other side bc. On ae 
 as a diameter describe a circle, meet- G-^. 
 
 ing BC produced at f : then will bf 
 be the side of the square bfgh, 
 
 equal to the given rectangle bd, as , 
 
 required ; as appears by cor. th. 87, A Jl BE 
 
 and th. 77. 
 
 C\ 
 
 APPLICATION 
 
[ 369 1 
 
 APPLICATION OF ALGEBRA 
 
 GEOMETRY. 
 
 Wk 
 
 HEN it is proposed to resolve a geometrical problena- 
 algebraically, or by algebra, it is proper, in the first place, 
 to draw a figure that shall represent the several parts or con- 
 ditions of the problem, and to suppose that figure to be the 
 true one. Then, having considered attentively the nature of 
 the problem, the figure is next to be prepared for a solution, 
 if necessary, by producing or drawing such lines in it as ap- 
 pear most conducive to that end. 'J'his done, the usual sym- 
 bols or letters, for known and unknown quantities, are em- 
 ployed to denote the several parts of the figure, both the 
 known and unknown parts, or as many of them as necessary, 
 as also such unknown Une or lines as may be easiest found, 
 whether required or not. Then proceed to the operation, 
 by observing the relations that the several parts of the figure 
 have to each other ; from which, and the proper theorems ia 
 the foregoing elements of geometry, make out as many equa- 
 tions independent of each other, as there are unknown quan- 
 tities employed in them : the resolution of which equations, in 
 the same manner as in arithmetical problems, will determine 
 the unknown quantities, and resolve the problem proposed. 
 
 As no general rule can be given for drawing the lines, and 
 selecting the fittest quantities to substitute for, so as always 
 to bring out the most simple conclusion, because different 
 problems require different modes of solution ; the best way to 
 gain experience, is to try the solution of the same problem 
 in different ways, and then apply that which succeeds best, 
 to other cases of the same kind, when they afterwards occur. 
 The following particular directions, however, may be of some 
 
 l5^, In preparing the figure, by drawing lines, let them be 
 either parallel or perpendicular to other lines in the figure, 
 or so as to form sirinilar triangles. And if an angle be given, 
 it will be proper to let the perpendicular be opposite to that 
 angle, and to fall from one end of a given Une, if possible. 
 
 Vol, I. • 48 H. 
 
370 APPLICATION OF ALGEBRA 
 
 9,d, In selecting the quantities proper to substitute for, 
 those are to be chosen, whether requiied or not. tvbich he 
 nearest the known or given parts of the iigure, and by nie-ans 
 of which the next adjacent parts may be expressed by addi- 
 tion and subtraction only^ without using surds. 
 
 3c/, When two lines or quantities are alike related to other 
 part*? of the figure or problem, the best way is, not to make 
 use of either of them separately, but to substitute for their 
 sum, or difference, or rectangle, or the sum of their alternate 
 quotients, or for some line or lines, in the figure, to which 
 they have both the same relation. 
 
 4th, When the area, or the perimeter, of a figure, is §iven» 
 or such parts of it as have only a remote relation to the parts 
 required : it is sometimes of use to assume another figure 
 similar to the proposed one, having one side equal to unity, or 
 some other known quantity. For hence the other parts of the 
 figure may be found, by the known proportiong of the like 
 sides, or parts, and so an equation be obtained. For exam- 
 ples, take the following problems. 
 
 PROBLEM I. 
 
 In a Right-angled Triangle^ having given the Base (3)yand the 
 Sum of the Hypothenuse and Perpemiicular (9) ; to find 
 both these two Sides. 
 
 Let ABC represent the proposed triangle, 
 right-angled at b. Put the base ab = 3 = 6, 
 and the sum ac -f bc of the hypothenuse 
 and perpendicular = 9 = 5 ; also, let x de- 
 note the hypothenuse ac, and y the perpen- . 
 dicular ec. 
 
 Then by the question - - - ^. + y = », 
 and by theorems 34, - - - x-=y--\-b^. 
 By transpos. y in the 1st equ. gives x = s — y, 
 This value of x substi. in the 2d, 
 
 gives ---'---- 52 — ^sy -\-y^ = 2/^ + b^\ 
 Takingaway 2/^ on both sides leaves s2 — 2sy = b^. 
 By transpos. ^sy and 6*, gives s^ — 6^ = est/, 
 
 52—63 
 
 Aiid dividing by 2s, gives - - ~ ^ = 4 = bc. 
 
 2s 
 Hence x = s — y 6 = ac. 
 
 N. B. In this solution, and the following ones, the nota- 
 liop is made by using as many unknown letters, x and y, as 
 
 there 
 
TO GEOMETRY. 371 
 
 there are unknown sides of the triangle, a separate letter for 
 each ; in preference to using only one unknown letter for one 
 side, and expressing the other unknown side in terms of that 
 letter and the given sum or difference of the sides ; though 
 this latter way would render the solution shorter ; because the 
 former way gives occasion for more and better practice in re- 
 ducing equations, which is the very end and reason for which 
 these problems are given at all. 
 
 PROBLEM II. 
 
 In a Right-angled Triangle, having given the Hypothenuse (5) ; 
 and the sum of the Base and Perpendicular (7) ; to find both 
 these two Sides, 
 
 Let ABC (see last %.) represent the proposed triangle, 
 right-angled at b. Put the given hypothenuse ac = 5 = a, and 
 the sum ab -f- bc of the base and perpendicular = 7 = s; also 
 let X denote the base ab, and y the perpendicular bc. 
 Then by the question - - - x -{- y = s 
 
 and by theorem 34 - - - x^-\-y^=-a^ 
 By transpos. y in the 1st, gives x = s — y 
 By substitu. this valu. for a;, gives s^ — 2sy -f ^y^ = a^ 
 By transposing s^ , gives - - 2^2 — 2vy = a^ — 52 
 By dividing by 2, gives - - y^ — sy = ia^ — ±s3 
 By completing the square, gives y"^ — sy •\- \s^ = la^ — ^52 
 
 By extracting the root, gives y — is = ^ la^ — \s^ 
 
 By transposing \s, gives - - y = \s ±. ^ la^ ^is^ = 
 
 4 and 3, the values oi x and y, 
 
 PROBLEM III. 
 
 In a Rectangle, having given the Diagonal (10), and the Peri- 
 meter, or Sum of ail the Four Sides (28) ; to fmd each of the 
 Sides severally. 
 
 Let abcd be the proposed rectangle ,; DC 
 
 and put the diagonal ac = 10 = c?. and '" 
 
 half the perimeter ab -j- bc oir ad -|- dc 
 = 14 = a ; also put one side ab = a-, 
 
 and the other side bc = y. Hence,, by ^ jj 
 
 right-angled 
 
372 
 
 APPLICATION Ot^ ALGEBRA 
 
 right-angled triangles, - - - - a;2 _|> ^2 = ^a 
 
 And by the question, - -- - x -\-y =a 
 
 Then by transposing y in the 2d, gives x =a — y 
 
 Thisvahiesubstitutedinthe lst,givesa? — 2ay-\-2y^ = d- 
 
 Transposing a^ , gives - - 2y^ — 2ay = r/2 — a^ 
 
 And dividing by 2, gives - y^ — oy = ^d^ — ia^ 
 
 By completing the square, it is i/^ — ay -{- \a^ = ic?2 __ 1^2 
 
 And extracting the root, gives y — i^ = ^ ^d^ — ^a^ 
 
 la±y/\d^^\a^ = 8 
 
 and transposing ia, gives y 
 
 or 6. the values of x and y. 
 
 PROBLEM IV. 
 
 Having given the Base and Perpendicular of any Triangle ^ 
 to find the Side of a Square Inscribed in the Same. 
 L^T ABC represent the given triangle, f^ 
 
 and EFGH its inscribed square. Put the 
 base AB = b, the perpendicular cd = a, 
 and the side of the square gf or gh = Gr, 
 
 Di =1 x; then will ci = cd — di = 
 
 Then, because the like lines in the 
 sinoilar triangles abc, gfc, are propor- 
 tional (by theor. 84, Geom.), ab 
 
 H DEB 
 
 GF : ci, that 
 is b : a : : X : a — x. Hence, by multiplying extremes and 
 means, ab — fex = ax, and transposing bx, gives ab = ax 
 
 ab 
 
 -}- bx ; then dividing by a -\- b, gives x = = gf 
 
 a -f 6 
 ©r GH the side of the inscribed square : which therefore is of 
 the same magnitude, whatever the species or the angles of 
 the triangles may be. 
 
 PROBLEM V. 
 
 In an Equilateral Triangle^ havirig given the lengths of the three 
 Perpendiculars, drawn from a certain Point within, on the . 
 ihree Sides : to determine the Sides. 
 Let ABC represent the equilateral tri- 
 angle, and BE, DF, DG, the given per- 
 pendiculars {ron\ the point d. Draw the 
 lines DA, DB, DC, to the three angular 
 points ; and let fall the perpendicular ch 
 on the base ab. Put the three given per- 
 pendiculars DE s= a, DF = 6, DG = C, 
 
 and put Of = AH or bh, half the side of 
 
TO GEOMETRY. 373 
 
 the equilateral triangle. Then is ac or bc = 2r, and by 
 
 right angled triangles the perpendicular ch = ^ ac^ — ah^ 
 
 = ^ 4x2 _ x3 = y Sx^ = or ^ 3. 
 
 Now, since the area or space of a rectangle, is expressed 
 by the product of the base and height (cor. 2, th. 81 Geom.), 
 and that a triangle is equal to half a rectangle of equal base 
 and hei^hl (cor. 1 , th 26), it follows that, 
 
 the whole triangle abc is = ^ab XcH=xXxy/3 = x^^3, 
 the triangle abd = Iab X dg = a; X c = cjt, 
 the triangle bcd = Ibc X de = a; X a = ax, 
 the triangle acd = ^ac X df = a; X 6 = 6x. 
 
 But the three last triangles make up, or are equal to, the 
 whole former, or great triangle ; 
 
 that is, x^^ 3 = ax -{- bx -\- ex ; hence, dividing by rr, gives 
 X ^ 3 = a 4- ^ +^> and dividing by ^3, gives 
 '\ia+b-i-cl 
 X =^ ^, half the side of the triangle sought. 
 
 Also, since the whole perpendicular ch is = ar y^ 3, it is 
 therefore = a -f 6 -j- c. That is, the whole perpendicular 
 CH, is just equal to the sum of all the three smaller perpen- 
 diculars DE 4* DP + DG taken together, wherever the point d 
 is situated. 
 
 PROBLEM VI. 
 
 In a Right-angled Triangle, having given the Base (3), and 
 the Difference between the Hypothenuse and Perpendicular 
 (1) ; to find both these two Sides. 
 
 PROBLEM Vn. 
 
 In a Right-angled Triangle, having given the Hypothenuse 
 rs), and the Difference between the Base and Perpendicular 
 (1) ; to determine both these two Sides. 
 
 PROBLEM VUL 
 
 • Having given the Area, or Measure of the Space, of a 
 Rectangle, inscribed in a given Triapgle j to determine the 
 Sides of the Rectangle. 
 
 PROBLEM 
 
:m APPLICATION OF ALGEBRA 
 
 PROBLEM IX. 
 
 In a Triangle, having given the Ratio of the two Sides^ 
 together with both the Segments of the Base, raade by a Per- 
 pendicular from the Vertical Angle j to determine the Sides of 
 the Triangle. 
 
 PROBLEM X. 
 
 In a Triangle, having given the Base, the Sam of the other 
 two Sides, and the Length of a Line drawn from the Vertical 
 Angle to the Middle of the Base ; to find the Sides of the Tri- 
 angle. 
 
 PROBLEM XI. 
 
 In a Triangle, having given the two Sides about the Vertical 
 Angle, with the Line bisecting that Angle, and terminating in 
 the Base : to find the Base. 
 
 PROBLEM XU. 
 
 To determine a Right-angled Triangle; having given the 
 Lengths of two Lines drawn from the acute angles, to the 
 Middle of the opposite Sides. 
 
 PROBLEM XIII. 
 
 To determine a Right- Angled Triangle ; having given the 
 Perimeter, and the Radius of its Inscribed Circle, 
 
 PROBLEM XIV. 
 
 To determine a Triangle ; having given the Base the Per- 
 pendicular, and the Ratio of the two Sides. 
 
 PROBLEM XV. 
 
 To determine a Right-angled Triangle ; having given the 
 Hypothenuse, and the Side of the Inscribed Square. 
 
 PROBLEM 
 
W GEOMETRY. 37i 
 
 iPBOBUEM 3CVI. 
 
 To determine the Radii of ihtee Equal Circles, described 
 in a given Circle, to touch each other and also the Circumfer- 
 ence of the given Circle. 
 
 PROBLEM XVn. 
 
 In a Right-angled Triangle, having given the Perimeter or 
 Sum of all the Sides, and the Perpendicular let fall from the 
 Rii^ht Angle on the Hypdthenuse ; to determine the 1 rianglC; 
 that is, its Sides. 
 
 PROBLEM XVIIL 
 
 To determine a Right-angled Triangle ; liaving given the 
 Hypothenuse, and the Difference of two lines drawn from the 
 two acute angles to the Centre of the Inscribed Circle. 
 
 PROBLEM XIX. 
 
 To determine a Triangle ; having given the Base, the Per 
 pendicular, and the Difference of the two other Sides. 
 
 PROBLEM XX, 
 
 To determine a Triangle ; having given the Base, the Per- 
 pendicular, and the Rectangle or Product of the two Sides. 
 
 PROBLEM XXI. 
 
 To determine a Triangle ; having given the Lengths ©f 
 three Lines drawn from the three Angles, to the Middle of the 
 opposite Sides. 
 
 PROBLEM XXII. 
 
 In a Triangle, having given all the three Side* ; to find the 
 Radius of the Inscribed Circle. 
 
 PfiQBLfiM 
 
37« APPLICATION OF ALGEBRA, ko. 
 
 PROBLEM XXIIL 
 
 To determine a Right-angled Triangle ; having given the 
 Side of the Inscribed Square, and the Radius of the Inscribed 
 Circle. 
 
 PROBLEM XXiy. 
 
 To determine a Triangle, and the Radius of the Inscribed 
 Circle ; having given the Lengths of three Lines drawn from 
 the three Angles, to the Centre of that Circle. 
 
 PROBLEM XXV. 
 
 To determine a Right-angled Triangle ; having given the 
 Hypothenuse, and the Radius of the Inscribed Circle. 
 
 PROBLEM XXVI. 
 
 To determine a Triangle ; having given the Base, the Line 
 bisecting the Vertical Angle, said the Diameter, of the Cir- 
 cumscribing Circle. 
 
 PLANE 
 
[ 377 ] 
 
 PLANE TRIGONOMETRY. 
 
 DEFINITIONS. 
 
 1. JT LANE TRIGONOMETRY treats of the rela- 
 tions and calculations of the sides and angles of plane tri- 
 angles. 
 
 2. The circumference of every circle (as before observed 
 in Geom. Def 67) is supposed to be divided into 360 equal 
 parts, called Degrees ; also each degree into 60 Minutes, 
 each minute into 60 Seconds, and so on. Hence a semicircle 
 contains 180 degrees, and a quadrant 90 degrees. 
 
 3. The measure of an angle (Def ^8, Geom.) is an arc of 
 any circle contained between the two lines which form that 
 angle, the angular point being the centre ; and it is estimated 
 by the number of degrees contained in that arc. 
 
 Hence, a right angle, being measured by a quadrant, or 
 quarter of the circle, is an angle of 90 degrees ; and the 
 sum of the three angles of every triangle, or two right angles, 
 is equal to 180 degrees. Therefore, in a right-angled triangle, 
 taking one of the acute angles from 90 degrees, leaves the 
 other acute angle ; and the sum of two angles, in any trian- 
 gle, taken from 180 degrees, leaves the third angle : or one 
 angle being taken from U'O degrees, leaves the stim of the 
 other two angles. 
 
 4. Degrees are marked at the top of ihe figure with a small **, 
 minute with ', seconds with ", and so on. Thus 57 •" 30' 12", 
 denote 67 degrees 30 minutes and 12 seconds. 
 
 l^A^ 
 
 5. The Complement of an arc, is 
 what it wants of a quadrant or OO**, 
 Thus, if AD be a quadrant, then bd is 
 the complement of the arc ab ; and, 
 reciprocally, ab is the complement of 
 BD. So that, if AB be an arc of 50°, 
 then its complement bd will be 40°. 
 
 6. The Supplement of an are, is 
 what it wants of a semicircle, or 
 180°. Thus, if ade be a semicircle, then bde is the supple- 
 ment of the arc ab ; and, reciprocally, ab is the supplement of 
 the arc bde. So that, if ab be an arc of 60°, then its supple- 
 ment bde will be 130°. 
 
 Vo*. I. 
 
 49 
 
 7. The 
 
378 PLANE TRIGONOMETRY. 
 
 7. The Sine, or Right Sine, of an arc, is the line tJrawn 
 from one extremity of the arc, perpendicular to the diameter 
 which passes through the other extremity. Thus, bf is the 
 sine of the arc ab, or of the supplemental arc bde. Hence 
 the sine (bf) is half the chord (bg) of the double arc (bag). 
 
 8. The Versed Sine of an arc, in the part of the diameter in- 
 tercepted between the arc and its sine. So, af is the versed 
 sine of the arc ab, and ef the versed sine of the arc edb. 
 
 9. The Tangent of an arc, is a line touching the circle ia 
 one extremity of that arc, continued from thence to meet a 
 line drawn from the centre through the other extremity ; 
 which last line is called the Secant of the same arc. Thus, 
 AH is the tangent, and ch the secant of the arc ab. Also, 
 EI is the tangent, and ci the secant, of the supplemental arc 
 BDE. And this latter tangent and secant are equal to the for- 
 mer, but are accounted negative, as being drawn in an oppo- 
 site or contrary direction to the former. 
 
 10. The Cosine, Cotangent, and Cosecant, of an arc, 
 are the sine, tangent, and secant of the complement of that 
 arc, the Co being only a contraction of the word comple- 
 raent. Thus, the arcs ab, bd, being the complements of 
 each other, the sine, tangent, or secant of the one of these, 
 is the cosine, cotangent, or cosecant of the other. So, bf, 
 the sine of ab, is the cosine of bd ; and bk, the sine of 
 BD, is the cosine of ab : in like manner ah, the tangent 
 of ab, is the cotangent of bd ; and dl, the tangent of db, 
 is the cotangent of ab : also, ch, the secant of ab, is the 
 cosecant of bd ; and cl, the secant of bd, is the cosecant of 
 
 Corol. Hence several remarkable properties easily follow 
 from these definitions ; as, 
 
 \st^ That an arc and its supplement have the same sine, 
 tangent, and secant ; but the two latter, the tangent and 
 secant are accounted negative when the arc is greater than a 
 quadrant or 90 degrees. 
 
 2rf, When the arg is 0, or nothing, the sine and tangent 
 are nothing, but the secant is then the radius ca, the least it 
 can be. As the arc increases from 0, the sines, tangents, 
 and secants, all proceed increasing, till the arc becomes a 
 whole quadrant ad, and then the sine is the greatest it can be, 
 
 beinji; 
 
DEFINITIONS^ 
 
 379 
 
 being the radius cd of the circle : and both the tangent and 
 secant are infinite. 
 
 3d, Of an arc ab, the versed sine af, and cosine bk, or cf, 
 together make up the radius .ca of the circle.— The radius 
 CA, the tangent ah, and the secant ch, form a right-angled tri- 
 angle CAH. So also do the radius, sine, knd cosine, form ano- 
 ther right angled-triangle cbf or cbk. As alsathe radius, co- 
 taiiorent, and cosecant, another right-angled triangle cdl. And 
 all these right-angled triangles are similar to each other. 
 
 11. The sine, tangent, or 
 secant of an angle, is the sine, 
 tangent, or secant of the arc 
 by which the angle is measur- 
 ed, or of the degrees, &c. in 
 the same arc or angle. 
 
 12. The method of con- 
 structing the scales of chords, 
 sines, tangents, and secants, 
 usu:\lly engraven on instru- 
 ments, for practice, is exhi- 
 bited io the annexed figure. 
 
 13. A Trigonometrical Ca- 
 non, is a table showing the 
 length of the sine, tangent, and 
 secant, to every degree and 2 
 minute of the quadrant, with ^ 
 respect to the radius, which 
 is expressed by unity or I, 
 with an}' number of cyphers. 
 The logarithms of these sines, 
 tangents and secants, are also 
 ranged in the tables; and these 
 are most commonly used, as they perform the calculations by 
 only addition and subtraction, instead of the multiplication and 
 division by the natural sines, &c. according to the nature of 
 logarithms. Such a table of log. sines and tangents, as well as 
 the logs, of common numbers, are placed at the end of the se- 
 cond volume, and the description and use of them are as fol- 
 low ; viz. of the sines and tangents ; and the other table, of 
 common logs, has been alreadv explained. 
 
 Des- 
 
38i PLANE TRIGONOMETRY. 
 
 Description of the Table of Log. Sines and Tangents. 
 
 In the first column of the table are contained all the arcs, 
 or angles, for every minute in the quadrant, viz from 1' to ' 
 45**, descending from top to bottom by the left-hand side, and 
 then returning back by the right-hand side, ascending from 
 bottom to top, from 45° to 90° ; the degrees being set at top 
 or bottom, and the minutes in the column. Then the sines, 
 cosines, tangents, cotangents, of the degrees and minutes, are 
 placed on the same lines with them, and in the annexed 
 columns, according to their several respective names or titles, 
 which are at the top of the columns for the degrees at the top, 
 but at the bottom of columns for the degrees at the bottom of 
 the leaves. The secants and cosecants are omitted in this 
 table, because they are so easily found from the sines and co- 
 sines ; for, of every arc or angle, the sine and cosecant to- 
 gether tnake up 20 or double the radius, and the cosine and 
 secant together make up the same 20 also. Therefore, if a 
 secant is wanted, we have only to subtract the cosine from 20 ; 
 or, to find the cosecant, take the sine from 20. And the best 
 way to perform these subtractions, because it may be done at 
 sight, is to begin at the left hand, and take every figure from 
 
 9, but the last or right hand figure from 10, prefixing 1, for 
 
 10, before the first figure of the remainder. 
 
 PROBLEM I. 
 
 To compute the Natural Sine and Cosine of a Given Arc. 
 
 This problem is resolved after various ways. One of these 
 is as follows, viz. by means of the ratio between the diameter 
 and circumference of a circle, together with the known seriei 
 for the sine and cosine, hereafter demonstrated. Thus, the 
 semicircumference of the circle, whose radius is 1, being 
 3.141592663589793 &,c. the proportion will therefore be, 
 
 as the number of degrees or minutes in the semicircle, 
 
 is to the degrees or minutes in proposed arc, 
 
 so is 3. 14159265 &c to the length of the said arc. 
 
 This length of the arc being denoted by the letter a ; aaA 
 
 ite 
 
 \ 
 
fROBLEMS. 381 
 
 its s\ne and cosine by 5 and c ; then will these two be express- 
 ed by the two follow ing series, viz. 
 
 s = a 1 h&c. 
 
 2.3 2.3.4.5 2.3.4.5.6.7 
 a^ a^ a 
 
 6 120 5040 
 
 G=; 1 { h&c. 
 
 2 2.3.4 2.3.4.5.6 
 a3 a* a^ 
 
 = 1 H h&c. 
 
 2 24 720 
 
 Exam. 1. If it be required to find the sine and cosine of 
 #ne minute. Thes the number of minutes in ISO** being 
 10800, it will be first, as 10800 : 1 : : 3 1415^266 kc. : 
 •000290888208665 = the length of an arc of one minute. 
 Therefore, in this case, 
 
 a = 0002908882 
 and ia» = -000000000004 &c. 
 the diflf. is s = -0002908882 the sine of 1 minute. 
 Also, from 1 . 
 
 take itt2 = 00000000423079 &c. 
 leave c = -9999999577 the cosine of 1 minute. 
 
 Exam. 2. For the sine and cosine of 5 degrees. 
 Here, as 180<> : 5° : : 3-14159265 &c. : -08726646 = a the 
 length of 5 degrees. Hence a = -08726646 
 »«ia3 = — -00011076 
 4._i_a5 = '-00000004 
 
 these collected give s = -08715574 the sine of S<*. 
 
 And, for the cosine, 1 = 1- 
 
 __ ±a2 =^ — -00380771 
 -f._Ca4= -00000241 
 
 these collected give c = -99(519470 the cosine of 5**. 
 
 After the same manner, the sine and cosine of any other 
 arc mny be computed. But the greater the arc is the slower 
 the <*eries will converge, iti which case a greater number of 
 terms must be taken, to bring out the conclusion to the same 
 degree of exactness. 
 
 Or. 
 
382 FLANE TRIGONOMETRY. 
 
 Or, having found the sine, the cosine will be found from 
 it, by the pro perty of th e right-angled triangle cbf, viz. the 
 cosine of = y' cb^— bf^, or c = ^ 1— a-s. 
 
 There are also other methods of constructing the canon 
 of sines and cosines, which for brevity's sake, are here 
 ooutted. 
 
 PROBLEM II. 
 
 To compute the Tangents md Secants. 
 
 The sines and cosines being known, or found by the 
 foregoing problem ; the tangents and secants will be easily 
 found, from the principle of similar triangles, in the foUow'- 
 ing manner : 
 
 In the first figure, where, of the arc ab, bf is the sine, 
 CF or BK the cosine, ah the tangent, gh the secant, dl 
 the cotangent, and cl the cosecant, the radius being ca or 
 CB or CD ; the three similar triangles cf b, cah, cdl, give the 
 following proportions : 
 
 \st, CF : FB : : GA : AH ; whence the tangent is known, 
 being a fourth proportional to the cosine, sine, and radius. 
 
 ^dy cF : cB : : ga : ch ; whence the secant is known, 
 being a third proportional to the cosine and radius. 
 
 orf, BF : Fc : : CD : DL ; whence the cotangent is known, 
 being a fourth proportional to the sine, cosine, and radius. 
 
 4(h, BF : BG : : cd : cl : whence the cosecant is known, 
 being a third proportional to the sine and radius. 
 
 As for the log. sines, tangents, and secants, in the tables, 
 they are only the logarithms of the natural sines, tangents, 
 and secants, calculated as above. 
 
 HAVING given an idea of the calculation and use of sines, 
 tangents, and secants, we may now proceed to resolve the 
 several cases of Trigonometry ; previous to which, however, 
 it may be proper to add a few preparatory notes and observa* 
 tions, as below. 
 
 - JVote 1. There are usually three methods of resolving tri- 
 angles, or the cases of trigonometry ; namely, Geometrical 
 Construction, Arithmetical Computation, and Instrumental 
 Operation. 
 
 In the first Method^ The triangle is constructed, by making 
 the parts of the given magnitudes, namely, the sides from a 
 scale of equal parts, and the angles from a scale of chords, 
 
• THEOREM I. 333 
 
 or by ?ome other instrument. Then mpasuring the unknown 
 p?irts by the same scales or instruments, the solution will be 
 obtained near the trutlf. 
 
 In the Second Method, Having stated the term§ of the pro- 
 portion according to the proper rule or theorem, resolve it 
 like any other proportion, in which a fourth term is to be 
 found from three given terms, by multiplying the second 
 and third together, and dividing the product by the first, in 
 working with the natural numbers ; or, in working with the 
 loga ritbjD s. add the logs, of the second and third terms to- 
 gether, and from the sum take the log. of the first terra ; 
 then the natural number answering to the remainder in the 
 fourth t**rm sought. 
 
 In the Third Method. Or Instrumentally, as suppose by the 
 log. lines on ^ one side of the common two-foot scales; Ex- 
 tend the Compasses from the first term, to the second or 
 third, which happens to be of the same kind with it ; then 
 that extent will . reach from the other term to the fourth 
 term, as required, taking both extents towards the" same end 
 of the scale. 
 
 JVote 2. Every triangle has six parts, viz. three sides and 
 three angles. And in every triangle, or case in trigonometry, 
 there must be given three of these parts, to find the other 
 three. Also, of the three parts that are given, one of them 
 at least must be a side ; because with the same angles,^ the 
 sides may be greater or less in any proportion. 
 
 JVote 3. All the cases in trigonometry, may be comprised in 
 three varieties only ; viz. 
 
 1st, When a side anc> its opposite angTc are given. 
 
 2rf, When two sides and the contained angle are given. 
 
 3d, When the three sides are given. 
 
 For there cannot possibly be more than these three varieties 
 of cases ; for each of which it will therefore be proper to 
 give a separate theorem, as follows : 
 
 THEOREM I. 
 
 When a Side and its Opposite Angle are two of the Given Parts, 
 
 Then the unknown parts will be found by this theorem ; 
 viz. The sides of the triangle have the same proportion to 
 each other, as the sines of their opposite angles have. 
 That is, As any one side. 
 
 Is to the sine of its opposite angle ; 
 
 So is any other side, 
 
 To the sine of its opposite angle. 
 
 Bemonstr. 
 
384 PLANE TRIGONOMETRY. 
 
 Detnonsir. For, let abc be the pro- C 
 
 po?ed triangle, having ab the greatest 
 side, and bc the least. Take ad = 
 Bc. considerina: it as a radius ; and let 
 fall the perpendiculars de, cf, which j^ j^ J J 
 will evidently be the ^ines of the an- 
 gles A and B, to the radius ad or bc« 
 
 Now the triangles ade, acf, are equiangular ; they therefore 
 have their like sides proportional, namely, ac : cf : : a» 
 or BC : DE ; that is, the side ac is to the sine of its oppos^ite 
 angle b, as the side bc is to the sine of its )pposite angle a. 
 
 JVote 1. In practice, to find an angle, begin the proportion 
 with a side opposite to a given angle. And to find a side, 
 begin with an angle opposite to a given side. 
 
 JVo^e 2. An angle found by thi^ rule is ambiguous, or nn- 
 eertain whether it be acute or obtuse, unless it be a right 
 angle, or unless its magnitude be such as to prevent the ambi- 
 guity ; because the sine answers to two angles, which are 
 supplements to each other ; and accordingly the geometrical 
 construction forms two triangles with the same parts that are 
 given, as in the example below ; and when there is no re- 
 striction or limitation included in the question, either of them 
 may be taken. The number of degrees in the table, answer- 
 ing to the sine, is the acute angle ; but if the angle be obtuse, 
 subtract those degrees from 180«>, and the remainder will be 
 the obtuse angle. When a given angle is obtuse, or a right one, 
 there can be no ambiguity ; for then neither of the other 
 angles can be obtuse, and the geometrical construction will 
 form only one triangle. 
 
 EXAMPLE!. 
 
 In the plane triangle abc, 
 
 { ab 346 yards 
 Given ? bc 232 yards 
 
 (Z A 3'/'' 20' 
 Required the other parts. 
 
 I. Geometrically. 
 
 Draw an indefinite line ; on which set off ab = 34i 
 from some convenient scale of equal parts. — Make the angle 
 A = 37^1. — With a radius of 232, taken from the Scme 
 scale of equal parts, and centre b, cross ac in the two 
 points c, G. — Lastly, join bc, bc, and the figure is con- 
 structed. 
 
THEOREM I. 386 
 
 structed, which gives two triangles, and shows that the case 
 is ambiguous. 
 
 Then, the sides ac measured by the scale of equal parts, 
 and the angles b and c measured by the line of chords, or 
 other instrument, will be found to be nearly as below ; viz. 
 
 AC 174 Z. B 27*' Z. c 115«>i. 
 
 or 374A or 78 a or 64^. 
 
 2. Arithmetically. 
 
 
 First, to find the angles at 
 
 c. 
 
 
 As side bc 232 
 
 - 
 
 log. 2-365488 
 
 To sin. op. 2L a 37*' 20' - 
 
 - 
 
 9-782796 
 
 So side AB 345 
 
 - 
 
 2-537819 
 
 3osin. op. Z.C US*' 36' or 640 
 
 24' 9-955127 
 
 add Z. A 37 20 
 
 37 
 
 20 
 
 the sum 152 66 or 
 
 101 
 
 44 
 
 taken from 180 00 
 
 180 
 
 00 
 
 leaves Z b 27 04 or 
 
 78 
 
 16 
 
 Then, to find the side ac. 
 
 
 
 As sine Z. a 37*^ 20' 
 
 _ 
 
 log. 9-782796 
 
 To op. side bc 232 
 
 
 2-365488 
 
 sosio. ., j-0*' : 
 
 
 9-658037 
 9-990829 
 
 To op. side ac 174-07 
 
 
 2-240729 
 
 or 374-56 
 
 
 2-573521 
 
 3. Instrumentally. 
 
 
 In the first proportion. — Extend the compasses from 232 
 to 345 on the Hne of numbers • then that extent will reach, 
 on the sines, from 37**i to 64*>^, the angle c. 
 
 In the second proportion. — Extend the compasses from 
 37°i to 27<' or 78<'i, on the sines ; then that extent will 
 reach, on the line of numbers, from 232 to 174 or 374^, 
 the two values of the side ac 
 
 EXAMPLE II. 
 
 In the plane triangle abc, 
 
 C AB 366 poles ( ^c ^^** ^" 
 
 Given? Z.A 57o 12' Ans. ( ac 154-33 
 
 ( AB 24 46 ( BC 309-86 
 
 Required the other parts. 
 Vol. I. 59 E3CAMPLfi 
 
38e PLANE TRIGONOMETRY. 
 
 EXAMPLE III. 
 
 In the plane triangle abg, 
 
 5Z.B 64« 35^ 
 or 116 26 
 Z.C 57 57 
 or 7 7 
 AB 112 6 feet. 
 or 16-47 feet. 
 
 THEOREM n. 
 
 When two Sides and their Contained Jingle are given. 
 
 First add the two given sides together, to get their sum, 
 and subtract them, to get their difference. Next subtract the 
 given angle from 180**, or two right angles, and the remainder 
 will be the sum of the two other angles ; thon divide that by 
 2, which will give the half sum of the said unknown angles. 
 Then say, 
 
 As the sum of the two given sides, 
 Is to the difference of the same ^ides ; 
 So is.-the tang, of half the sum of their op. angles, 
 To the tang of half the diff, of the same angles. 
 Then add the half difference of the Jungles, so found, to 
 their half sum, and it will give the greater angle, and sub- 
 tracting the same will leave the less angle ; because the half 
 sum of any two quantities, increased by their half difference, 
 gives the greater, and diminished by it gives the l(s» 
 
 Then all the angles being now known, the unknown side 
 will be found by the former theorem. 
 
 Note. Instead of the tangent of the half sum of the un- 
 known angles, in the third term of the proportion, may be 
 used the cotangent of half the given angle, which is the same 
 thing. 
 
 Demonst. Let abc be the proposed 
 tri^iigie, having the two given sides 
 
 AC, Bc including the given angle c. 
 With the centre c, and radius ca, 
 tiie less of these two sides, describe 
 a semicircle, meeting the other side 
 Br produced in d, e, and the un- 
 known side AB in a, g. Join ae, 
 
 AD, CG, and draw df parallel to ae. ^ 
 
 Then be is the sum of the two given sides ac, cb, or of 
 EC, cb ; and bo is the difference of the same two giv.en sides 
 
 / AC, 
 
THEOREM II. 387 
 
 AC, BC, or of CD, CB. Also, the external angle ace, is equal 
 to the given snna of the two internal angles cab, cba ; but the 
 ar»i»;le ade, at the circumference, is equal to half the angle 
 ACE dt the centre : therefore the same angle ade is equal to 
 h?lf the, given sum of the angles cab, cba. Also, the exter- 
 Dol angle AGC, of the triangle bcg, is equal to the sum of the 
 tT7'> internal angles «cb, gbc, or the angle gcb is equal to the 
 ' diiV lonce of the two angles agc, gbc ; but the angle cab is 
 equal to the said angle agc, these being opposite to the equal 
 sides AC, CG ; and the angle dab, at the circumference, is 
 equal to half the angle dcg at the centre ; therefore the angle 
 dab is equal to half the difference of the two angles cab. cba ; 
 of which it has been shown that ade or cda is the half sum. 
 Now the angle dae, in a semicircle, is a right angle, or ae 
 is perpendicular to ad ; and df, parallel to ae, is also per- 
 pendicular to AD : consequently ae is the tangent of cda the 
 half sum, and df the tangent of dab th£ half difference of 
 the angles, to the same radius ad by the definition of a tan- 
 gent. ' but the tangents ae, df, being parallel, it will be as 
 BE : bd : : AE : DF ; that is, as the sum of the sides is to the 
 difference of the sides, so is the tangent of -half the su-u of 
 the opposite angles, to the tangent of half their difference. 
 
 EXAMPLE I. 
 
 In the plane triangle abc, 
 C ab 346 yards 
 Given ? ag 17 i07 yards 
 ( Z.A 37» 20' 
 Required the other parts. 
 
 1. Geometrically. 
 
 Draw AB = 345 from a scale of equal parts. Make the 
 angle a = 37*» 20'. Set off ac = 174 by the scale of equal 
 parts. Join bc, and it is done. 
 
 ' Then the other parts being measured, they are found to bc 
 nearly as follows ; viz. the side bc 232 yards, the angle b 27**, 
 and the angle c 115*»|^. 
 
 2. Arithmetically. 
 
 The side ab 345 From 180*> 00' 
 
 the side ac 174-07 take/. a 37 20 
 
 their sum 519-07 sum of c and b 142 40 
 their differ. 170-93 half sum of do. 71 20 
 
 As 
 
388 PLANE TRIGONOMETRY. 
 
 As sum of sides ab, ac, - - 619 07 
 
 log. 2-716226 
 
 To diff. of sides ab, ac, - - 170-93 
 
 2-232818 
 
 So tang, half sum Zs c and B 71'' 20' 
 
 - 10-471298 
 
 To tang, half diff. ^s c and b 44 16 
 
 9-988890 
 
 these added give Z. c 116 36 
 
 
 and subtr. give z. b 27 4 
 
 
 Then, by the former theorem, 
 
 
 As sin. Z. c- 1 16° 36' or 64« 24' 
 
 log. 9-956126 
 
 To its op. side ab 346 
 
 2-637819 
 
 Bo sin. of Z. A 370 20' . . - 
 
 9-782796 
 
 To its op. side bc 232 ' - 
 
 2-365489 
 
 3. Instrumentally. 
 
 In the fifst proportioTi. — Extend the compasses from 619 
 to 171, on the Une of numbers ; then that extent will reach, 
 on the tangents, from 71** J- (the contrary way, because the 
 tangents are set back again from 46**) a little beyond 46, 
 which being set so far back from 4:-, falls upon 44**^, the 
 fourth term. 
 
 In the second proportion Extend from 64<*i to 37'>i on 
 
 the sines ; then that extent will reach on the numbers, from 
 346 to 232, the fourth term sought. 
 
 EXAMPLE It 
 lu the plane triangle abc, 
 Given < 
 Required the other parts. 
 
 EXAMPLE IIL 
 
 In the plane triangle abg, 
 
 ( AC 120 yards 
 Given { bc 112 yards Ans. { Z.a 67*' 28 
 
 (Zc 570 67' (zb 64 35 
 
 Required the other parts. 
 
 THEOREM III. 
 
 When the TJiree Sides of a Triangle are given. 
 
 First, let fall a perpendicular from the g^reatest angle on the 
 opposite sine, or base, dividing it into two segments, and the 
 whole triangle into t\»o right-angled triangles : then the pro- 
 portion will be, 
 
 As 
 
 le piane iriangie abc, 
 
 i ab 366 poles ( bc 309 86 
 
 t { AC 154-33 Ans. { Zb 24*' 45' 
 
 (Za57°12' (Zc98 3 
 
 ( AB 112-6 
 
 , ?Za 67*' 28 
 (Zb 64 35 
 
THEOREM Hi. 389 
 
 As the base, or sums of the segments. 
 
 Is to the sum of the other two sides ; 
 
 So is the diflference of those sides, 
 
 To the diflf. of the segments of the base. 
 Then take half this diflference of the segments, and add it to 
 the half sum, or the half base, for the greater segment ; and 
 subtract the same for the less segment. 
 
 Hence in each of the two right-angled triangles, there 
 will be known two sides, and the right angle opposite to one 
 of them ; consequently the other angles will be found by the 
 iirst theorem. / 
 
 Demonstr. By theor, 35, Geom. the rectangle of the sum 
 and difference of the two sides, is equal to the rectangle of 
 the sum and diflference of the two segments. Therefore, by 
 forming the sides of these rectangles into a proportion by theor. 
 70, Geometry, it will appear that the sums and dififerences are 
 proportional as in this theorem. 
 
 EXAMPLE I. 
 
 In the plane triangle abc, 
 Given r^!ff^y^^^« 
 
 the sides > ^^ ^^|.Q^ ^^^^^^^^ 
 
 P B 
 To find the angles. 
 
 I. Geometrically. 
 
 Draw the base ab = 345 by a scale of equal parts. Witli 
 radius 232, and centre a, describe an arc ; and with radius 
 174, and centre b, describe another arc, cutting the former in 
 c. Join AC, Bc, and it is done. 
 
 Then, by measuring the angles, they will be found to be 
 nearly as follows, viz. 
 
 Za 27«>, Zb 37oi, and ^c 115o|. 
 
 2.. Arithmetically, 
 
 Having let fall the perpendicular cp, it will be, 
 As the base ab : ac -f- bc : : ac — bc : ap — bp, 
 that is, as 345 : 406-07 : : 67-93 : 68-18 = ap — bp. 
 
 its half is - 3409 
 
 the half base is 172-50 
 
 the sum of these is 206-59 = ap 
 
 and their diflf. is 138-41 = bp 
 
 Then, 
 
390 PLANE TRIGONOMETRY. 
 
 Then, in the triangle apc, right-angled at p, 
 
 As the side ac - • 
 
 . 232 
 
 log. 2-365488 
 
 To sin. op. ^p • • 
 
 9()*> 
 
 - 10-oooboo 
 
 So is the side ap - - 
 
 ■ 206-59 - 
 
 2-316109 
 
 Tosin. op. Zacp - • 
 
 . 62^tyQ' 
 
 - 9-949621 
 
 Which taken from - 
 
 90 00 
 
 
 leaves the ^^ a 27 04 
 Again, in the triangle bpc, right-angled at p, 
 
 As the side bc - 
 
 - 174-07 
 
 - log. 2-240724 
 
 To sin. op. ^p - 
 
 90° 
 
 - - 10-000000 
 
 So is side bp - 
 
 . 138-41 
 
 - - 2-441168 
 
 To sin. op ^^Bcp - 
 
 - 5S^° 40' 
 
 - - 9-900444 
 
 which tauten from - 
 
 90 00 
 
 
 leaves the Zb 37 20 
 
 Also, the Zacp 62° 66' 
 
 added to Zbcp 62 40 
 
 gives the whole Zacb 115 36 
 
 So that all the three angles are as follow, viz. 
 the Za 27° 4'; the ^b 37*^ 20' ; the Zc 116^36'. 
 
 3. Instrumentally* 
 
 In the first proportion. — Extend the compasses from 345- , ' J 
 to 406, on the line of numbers ; then that extent will reach, • ] 
 on the same line, from 68 to 68 2 nearly, which is the diffe- 
 rence of the segments of the base. 
 
 In the second proportion.— ^Extend from 232 to 206i. on the 
 line of numbers ; then that extent will reach, on the sines, 
 from 90° to eS*'. 
 
 In the third proportion. — Extend from 174 to 138^ ; then 
 that extent will reach from 90** to 52*^1 on the sines. 
 
 EXAMPLE II. 
 
 • "\ 
 
 In the plane 
 
 triangle abc. 
 
 
 Given i^-'f.^Pf' 
 *u« ^A . < AC 164 33 
 
 To find the angles. 
 
 (Za67o j2' 
 Ans. {Zb24 46 
 (Zc98 3 
 
 EKAMPL5 
 
 ! 
 i 
 
 - 
 
 
 1 
 

 THEOREM IV. 
 
 
 
 
 
 EXAMPLE m. 
 
 
 
 
 
 In the plane 
 
 triangle abc 
 
 f 
 
 
 
 .ni AB 
 
 120 
 
 ( 
 
 ZA 
 
 57«» 
 
 28' 
 
 
 112-6 
 
 Ans. ( 
 
 ZB 
 
 57 
 
 57 
 
 f BC 
 
 112 
 
 ( 
 
 Zc 
 
 64 
 
 35 
 
 find the angles. 
 
 
 
 
 
 301 
 
 The three foregoing theorems include all the cases of plane 
 triangles, both right-angled and oblique. But there are other 
 theorems suited to some particular forms of triangles, which 
 are sometimes more expedilious in their use than the general 
 ones ; one of which, as the case for which it serves so fre- 
 quentljr occurs, may be here taken, as follows : 
 
 THEOREM IV. 
 
 When a Triangle is Right-angled ; any of the unknown parts 
 may be found by the following proportions : viz. 
 
 As radius 
 
 Is to either leg of the triangle ; 
 
 So is tang, of its adjacent angle, 
 
 To its opposite leg ; 
 
 And so is secant of the same angle, 
 
 To the hypothenuse. 
 
 Demonstr. ab being the given leg, in the 
 right-angled triangle abc ; with the centre 
 A, and any assumed radius ad, describe an 
 arc DE, and draw df perpendicular to ab,. 
 or parallel to bc. Then it is evident, from 
 the definitions, that df is the tangent, and 
 af the secant of the arc de, or of the an- 
 gle A which is measured by that arc, to the radius ad. Then, 
 because of the parallels bc, df it will be, - - - - as 
 ■ AD : AB : : DF : bc and : : af : ac, which is the same as 
 the theorem is in words. 
 
 Aote, The radius is equal, either to the sine of 90*', or the 
 tangent of 45*^ ; and is expressed by 1, in a table of natural 
 sines, or by 10 in the log. sines. 
 
 EXAMPLE I. 
 
 In the right-angled triangle abc, 
 
 Given J ^^ 530^7' 43" ( ^^ ^"^ ^^ ^"^ ^^' 
 
 i. Geometrically. 
 
392 
 
 PLANE TRIGONOMETRY. 
 
 1. Geometrically. 
 
 Make ab = 162 equal parts, and the angle a = 53^ 7' 48" ; 
 then raise the perpendicular bc, meeting ac in c. So shall 
 AC measure 270, and bc 216. 
 
 
 2 
 
 Arithmetically. 
 
 
 
 As radius 
 
 - 
 
 . 
 
 log. 
 
 10-000000 
 
 To leg AB 
 
 - 
 
 162 
 
 
 2-209515 
 
 So tang. Z. A 
 
 - 
 
 53° r 48" 
 
 - 
 
 10- 124937 
 
 To leg BC 
 
 - 
 
 216 
 
 . 
 
 2-334452 
 
 So secant Z. a 
 
 -■ 
 
 630 r 48" 
 
 - 
 
 10-221848 
 
 To hyp. AC 
 
 3. 
 
 270 
 Instrument ally. 
 
 
 2-431363 
 
 Extend the compasses from 45*» to 53' j, on the tangents. 
 Then that extent will reach from 162 to 216 on the line of 
 numbers. 
 
 EXAMPLE II. 
 
 In the right-angled triangle abc, 
 ^. ^ the leg AB 180 . ^ ac 392-0146 
 
 ^*^^" I the Za 62° 40' ^^^' I bc 348-2464 
 
 To find the other two sides. 
 
 JSTote. There is sometimes given another method for right- 
 angled triangles, which is this : 
 
 ABC being such a triangle, make one 
 leg AB radius ; that is, with centre a, 
 and distance ab, describe an arc bf. 
 Then it is evident that the other leg bc 
 represents the tangent, and the hypo- 
 thenuse ac the secant, of the arc bf, or 
 «f the angle a. 
 
 In like manner, if the leg bc be made 
 radius : then the other leg ab will re- 
 present the tangent, and the hypothenuse ac the secant, of 
 the arc bg or angle c. 
 
 But if the hypothenuse be made radius ; then each leg 
 will represent the sine of its opposite angle ; namely, the leg 
 ab the sine of the arc ae or angle c, and the leg bc the sine 
 ©f the arc cd or angle a. 
 
 Then the general rule for all these cases is this, namely, 
 that the sides of the triangle bear to each other the same pro- 
 portion as the parts which they represent. 
 
 And this is called, Making every side radius. 
 
 JVote 
 
OF HEIGHTS AND DISTANCES. 38^3 
 
 Note 2. When there are given two sides of a right-angled 
 triangle, to find the third side ; this is to be found by the 
 property of the squares of the sides in theorem 34, Geom. 
 Tiz. that the square of the hypothenuse, or longest side, is 
 equal to both the squares of the two other sides together. 
 Therefore, to find the longest side, add the squares of the 
 two shorter sides together, and extract the square root of 
 that sum ; but to find one of the shorter sides, subtract the 
 one square from the other, and extract the root of the re- 
 mainder. 
 
 OF HEIGHTS AND DISTANCES, &c. 
 
 BY the mensuration aad protraction of lines and angles, 
 are determined the lengths, heights, depths, and distances of, 
 bodies or objects. 
 
 Accessible lines are measured by applying to them some 
 certain measure a number of times, as an inch, or foot, or 
 yard. But inaccessible lines must be measured by taking 
 angles, or by such-like method, drawn from the principles of 
 geometry. 
 
 When instruments are used for taking the magnitude of 
 Ihe angles in degrees, the lines are then calculated by trigo- 
 ndmetry : in the other methods, the lines are calculated from 
 the principle of similar triangles, or some other geometrical 
 property, without regard to the measure of the angles. 
 
 Angles of elevation, or of depression, are usually taken 
 cither with a theodolite, or with a quadrant, divided into 
 degrees and minutes, and furnished with a plummet suspend- 
 ed from the centre, and two open sights fixed on one of the 
 radii, or else with telescopic sights. 
 
 To take an Angle of Altitude and Depression with the (Quadrant, 
 
 Let A be any object, as the sun, 
 moon, or a star, or the top of a 
 tower, or hill, or pther eminence : 
 and let it be required to find the 
 measure of the angle, abc, which a 
 line drawn from the object makes 
 above the horizontal line bc. 
 
 Place the centre of the quadrant 
 in the angular point, and move it 
 
 V«L. I. 51 r»uni 
 
394 ; OF HEIGHTS 
 
 found there as a centre, till with one eye at b, the other 
 being shut, you perceive the object a throngb the sights ; 
 then will tlie arc gh of the quadrant, cut off by the plumb- 
 line BH, be the measure of the angle abc as required. 
 
 The angle abc of depression of ^j . ^ ^ . 
 
 any object a, below the horizontal 
 line Bc, is taken in the same man- 
 
 ^ 
 
 ner ; except that here the eye is ap- 
 plied to the centre, and the measure ^ 
 of the angle is the arc gh, on the 
 other side of the plumb-line. 
 
 
 \^ 
 
 The following examples are to be constructed and calcu- 
 lated by the foregoing methods, treated of in Trigonometry. 
 
 EXAMPLE I. 
 
 Having measured a distance of 200 feet, in a direct ho- 
 rizontal line, from the bottom of a steeple, the angle of 
 elevation of its top, taken at that distance, was found to be 
 47° 30' ; from hence it is required to find the height of the 
 steeple.^ 
 
 Construction, 
 
 , Draw an indefinite line ; on which set off ac = 200 equal 
 parts for the measured distance. Erect the indefinite per- 
 pendicular AB ; and draw cb so as to make the angle c = 
 4T* 30*, the angle of elevation ; and it is done. Then ab, 
 measured on the scale of equal parts, is nearly 218^. 
 
 Calculation. 
 
 As radius - - 30*000000 
 
 To AC 200 - - 2-30H)30 / 
 
 So tang. Z c 47° 30' 10 037948 / 
 
 To AB 218-26 required 2-338978 
 
 EXAMPLE IL 
 
 What was the perpendicular height of a cloud, or of a 
 balloon, when its angles of elevation wore 35*^ and 64°, as 
 taken by two observers, at the same tim§, both on the same 
 si/le of it, and in the same vertical plane ; the distance be- 
 tween them being half a mile or 880 yards. And what wa» 
 its distance from the said two observers ? 
 
 Construction i 
 
AND DISTANCES. 
 
 396 
 
 Construction. 
 
 Draw an indefinite ground line, on which set off the given 
 distance ab = 880 : then a and b are the places of the ob- 
 servers. Make the angle a = 35*>, and the angle b = h4^ ; 
 then the intersection of the lines at c will be the place of the 
 balloon : whence the perpendicular cd, being let fall, will be 
 its perpendicular height. Then by measurement are found 
 the distances and height nearly as follow, viz. ac 1631, eg 
 1041, Dfc 936. 
 
 • Calculation. 
 
 First, froro^ b 64» 
 take Z A 35 
 lea>ves ^ acb 29 
 
 ^ / 
 y^ / ' 
 
 Then in the triangle abc, 
 As sin. /, acb 29** 
 
 To op. side ab 880 
 So sin. Z A 33^ - 
 
 To op. side bc 1041-125 - 
 
 B 
 
 I) 
 
 As sin. ^ acb 29<* - 
 
 To op. side ab 880 - 
 So sin. Z b 116® or' 64*^ 
 
 To op. side AC 1631-442 - 
 
 And in the triangle bcd. 
 
 As sin. Z D 90° - 
 
 To op. side bc 1041-125 - 
 So sin. Z b 64° 
 
 To op. side cd 935-757 - 
 
 ^'685571 
 ^•944483 
 9-758591 
 3017503 
 
 9-685571 
 2-944^83 
 9-953660 
 3-212572 
 
 10-000000 
 3 017503 
 9-953660 
 2-971163 
 
 EXAMPLE III. 
 
 Having to find the height of an obelisk standing oij, the top 
 of a declivity, I first measured from its bottom^ distance 
 of 40 feet, and there found the angle, formed by the oblique 
 plane and a Hne imagined to go to the top of the obelisk, 41°; 
 but after measuring on in the same direction 60 feet farther', 
 the like angle was only 23» 45'. What then was the height 
 «f the obelisk ? 
 
 * Construction , 
 
39fi 
 
 «F HEIGHTS 
 
 Construction. 
 
 Draw an inde6nite line for the sloping plane or declitity, 
 in which Assume any point a for the bottom of the obelisk, 
 from which set off the distance ac = 40, and again cd = 60 
 equal parts. Then make the angle c = 41°, and the angle p 
 = 23° 45' ; and the point b where the two lines meet will 
 be the top of the obelisk. Therefore ab, joined, will be its 
 height. 
 
 Calculation ' * r^l^ 
 
 From the Zc 41° 00' 
 take the Z » 23 45 
 leaves theZ DBG 17 15 
 
 DBC, 
 IC, 
 
 121-4 
 41-'^ 
 
 69^ 
 
 42 
 
 
 T 
 
 
 
 Then in the triangle ] 
 
 As sin Z HBc 17^ 15' 
 To op. side DC 60 - 
 So sin. Z D 23 45 
 To op. side CB 81-488 
 
 D 
 
 88 
 188 
 30' 
 24i - . 
 
 1 
 
 9-472086 
 i-778151 
 9-605032 
 l-9110qx 
 
 And in the triangle ab 
 As sum of sides cb, ca 
 To diflf. of sides cb, ca 
 So tang, half sum Zs a, b 
 To tang, halfdifif Zs a, b 
 
 2-084533 
 
 1-617923 
 
 10-427262 
 
 9-960652 
 
 the diff. of these is Z. cba 27 
 
 '^\ 
 
 
 Lastly, as sin.ZcB a 27° b% 
 To op side ca 40 
 So sin. Z c - 41° 0' 
 To op. side AB 57-623 
 
 - " - 
 
 9-658284 
 1-602060 
 9-816943 
 1-760719 
 
 EXAMPLE IV. 
 
 Wanting to know the distance between two inaccessible 
 trees, or other objects, from the top of a tower 120 feet 
 high, which lay in the same right line with the two objects, 
 I took the angles formed by the perpendicular wall and hnes 
 conceived to be drawn from the top of the tower to the bot- 
 tom of each tree, and found them to be 33° and 64°. What 
 then may be the distance between the two objects^ 
 
 Conztfuction. 
 
AND DISTANCES. 
 
 397 
 
 Construction. 
 
 Draw the indefinite ground line 
 BD, and perpendicular to it ba = 
 120 equal parts. Then draw the 
 two lines ac, ad, making the two 
 angles bag, bad, equal to the given 
 angles SS^ and 64^^. So shall c 
 and D be the places of the two ob- 
 jects. 
 
 Calculation. 
 
 'm:^- 
 
 \, 
 
 First, in the right- angled triangle abc. 
 
 As radius 10-OOOOOi 
 
 ToAB - 120 - - - 2079181 
 
 So tang. Z BAG 33«> - , - - 9-812517 
 
 TOBC 
 
 77-929 
 
 1-891698 
 
 Then in the right-angled triangle abd, 
 
 As radius ----- 10 000000 
 
 ToAB - - - 120 - - 2-079181 
 
 So tang. Z BAD 64^ - - 10-321504 
 
 Toed - - 251-585 - - 2-400685 
 From which take bo 77-929 
 leaves the dist. cd 1 $'3-656 as required. 
 
 EXAMPLE V. 
 
 Being on the side of a river, and wanting to know the dis- 
 tance to a house which was seen on the other side, 1 measur- 
 ed 200 yards in a strait line by the side of the river ; and 
 then, at each end of this line of distance, took the horizontal 
 angle formed between the house and the other end of the line ; 
 which angles were, the one of them 68^ 2', and the other 
 73<> 15'. What then were the distances from each end to the 
 house ? 
 
 Construction. 
 
 Draw the line ab = 200 equal parts. Then draw ac so as 
 to make the angle a = 68° 2', and bc to make the angle b = 
 730 15'. So shall the point c be the place of the house re- 
 quired. 
 
 Calculation, 
 
398 OF HEIGHTS 
 
 Calculation. 
 
 To the given ^ a 68** 2' 
 add the given ^ b 73 15 
 then their sum 141 17 
 
 being taken from 1 80 
 
 leaves the third ^q 38 43 
 
 A B , 
 
 Hence, As sin. ^ c 38«» 43' - 9-796206 - 
 
 To op. side AB 200 - 2-301030 
 
 So sin. Z A 68<» 2' - 9-967268 
 
 To op. side Bc 296 64, - 2-472092 
 
 And, As sin. Z c 3?** 43' - - 9-796206 
 
 To op. side AB 200 - 2-301030 
 
 So sin Z B 73P 15' - 9-981171 
 
 To op. side AC 306-19 - 2-485996 
 
 Exam., vi. From the edge of a ditch, of 36 feet wide, sur- 
 rounding a fort, having taken the angle of elevation of the top 
 of the wall, it was found to be 62<* 40' : required the height 
 of the wall, and the length of a ladder to reach from my sta- 
 tion to the top of it ? . ( height of wall 69-64, 
 ^°^- gladder, 78-4 feet. 
 
 Exam. vif. Required the length of a shoar, which being to 
 strut 11 feet from the upright of a building, will support a 
 jamb 23 feet 10 inches from the ground ? 
 
 Ans. 26 feet 3 inches. 
 
 Exam. viii. A ladder, 40 feet long, can be so planted, that 
 it shall reach a window 33 feet from the ground, on one side 
 of the street ; and by turning it over, without moving the foot 
 out of its place, it will do the same by a window 21 feet high, 
 on the other side : required the breadth of the street ? 
 
 Ans. 5b-649 feet. 
 
 Exam. ix. A maypole, whose top was broken off by a blast 
 of wind, struck the ground at 15 feet distance from the foot- of 
 the pole : what was the height of the whole maypole, suppos- 
 ing the broken piece to measure 39 feet in length ? 
 
 Ans. 75 feet. 
 
 Exam. x. At 170 feet distance from the bottom of a tower, 
 the angle of its elevation was found to be 62° 3' : required 
 the altitude of the tower ? Ans. 221-55 feet. 
 
 Exam. xi. From the top of a tower, by the sea-side, of 143 
 feet high, it was observed that the angle of depression of a 
 ship's bottom, then at anchor, measured 35*^ ; what then was 
 the ship's distance from the bottom of the wall '^ 
 
 Ans. 204-22 feet. 
 Exam. 
 
AND DISTANCES. 3Sa 
 
 Exam. xii. What is the perpendicular height of a hill ? 
 its angle of elevation, taken at the bottom of it, being 46**, 
 and 200 yards farther ofif, on a level with the bottom, the angle 
 was SI** ? . Ans. 286-28 yards. 
 
 Exam. xiii. Wanting to know the height of an inacces- 
 sible tower ; at the least distance from it, on the same hori- 
 zontal plane, I took its angle of elevation equal to 58" ; 
 then going 300 feet directly from it, found the angle there to 
 be only 32** ; required its height^, and my distance from it at 
 the iirst station ? . 5 height 307-53 
 
 ^"^- ^ distance 192-15 
 
 Exam. xiv. Being on a horizontal plane, and wanting to 
 know the height of a tower placed on the top of an inac- 
 cessible hill ; I took the angle of elevation of the top of the 
 hill 40**, and of the top of the tower 51° ; then measuring in 
 a line directly from it to the distance of 200 feet farther, I 
 found the angle to the top of the tower to be 23** 45'. W^hat 
 then is the height of the tower ? Ans. 9333148 feet 
 
 Exam. xv. From a window near the bottom of a house, 
 which seemed te be on a level with the bottom of a steeple, 
 I took the angle of elevation of the top of the steeple equal 
 40** : then from another wmdow, 18 feet directly above the 
 former, the like angle was 37« 30' : what then is the height 
 and distance of the steeple ? * ^ height 21044 
 
 ^^^' ^distance 250-79 
 
 Exam. xvi. Wanting to know the height of, and my dis- 
 tance from, an object on the other side of a river, which 
 seemed to be on a level with the place where I stood, close 
 by the side of the river ; and not having room to measure 
 backward, on the satne plane, because of the immediate rise 
 of the bank, I placed a mark where I stood, and measured 
 in a direction from the object, up the ascending ground to 
 the distance of 264 feet, where it was evident that I was 
 above the level of the top of the object ; there the angles 
 of depression were found to be, viz. of the mark left at the 
 river's side 42*>, of the bottom of the object, 27*>, and of its 
 top 19*. Required then the height of the object, and the 
 distance of the mark from its bottom ? 
 
 ^ ^height 67-26 
 ^"^' ^distance 150-50 
 
 Exam. xvir. If the height of the mountain called the 
 Peak of Teneriffe be 2^ miles, as it is nearly, and the angle 
 
 take* 
 
400 OF HEIGHTS 
 
 taken at the top of it, as formed between a plumb-line and a 
 line conceived to to touch the earth in the horizon, or farthest 
 visible point, be 87® 58' ; it is required from these to deter- 
 mine the magnitude of the whole earth, and the utmost dis- 
 tance that can be seen on its surface from the top of the moun- 
 tain, supposing the form of the earth to be perfectly round ? 
 
 . i dist. 140-876 > ., 
 A°^- Idiam. IQ-Sel"^'^^'' 
 
 Exam, xviii. Two ships of war, intending to cannonade 
 a fort, are, by the shallowness of the water, kept so far 
 from it, that they suspect their guns cannot reach it with 
 effect. In order therefore to measure the distance, they se- 
 parate from 6ach other a quarter of a mile,' or 440 yards ; 
 then each ship observes and measures the angle which the 
 other ship and the fort subtends, which angles are 83*" 45 
 and 85° 15'. What then is the distance between each ship 
 and the fort? . J 2292-26 yards. 
 
 ^^^' I 2298-05. 
 
 Exam. xix. Being on the side of a river, and wanting to 
 know the distance to a house which was seen at a distance on 
 the other side ; I measured out for a base 400 yards in a 
 right line by the side of the river, and found that the" two 
 angles, one at each end of this line, subtended by the other 
 end and the house, were 68«* 2' and 7?><* 15. What then was 
 the distance between each station and the house ? 
 
 . i 593-08 yards. 
 ^"^' I 612-38 
 
 Exam. xx. Wanting to know the breadth of a river, I 
 measured a base of 500 yards in a straight line close by one 
 side of it ; and at each end of this hne I found the angles 
 subtended by the other end and a tree close to the bank on 
 the other side of the river, to be 53<' and 79^ 12'. What then 
 was the perpendicular breadth of the river ? 
 
 Ans. 529-48 yards. 
 
 Exam. xxi. Wanting to know the extent of a piece of 
 water, or distance between two headlands; I measured from 
 each of them to a certain point inland, and found the two dis- 
 tances to be 735 yards and 840 yards ; also the horizontal 
 angle subtended between these two lines was S5<^ 40'. What 
 then was the distance required ? Ans. 741 2 yards. 
 
 Exam. xxii. A point of land was observed, by a ship at 
 sea, to bear east-by-south ; and after sailing north-east 12 
 miles, it was found to bear south-east-by-east. It is required 
 
 t9 
 
AND DISTANCES. 401 
 
 t« determine the place of that headland, and the ship's dis- 
 tance from it at the last observation ? Ans. 260728 miles. 
 
 Exam, xxiii. Wanting to know the distance between a 
 house and a mill, which were seen at a distance on the other 
 side of a river, I measured a base line along the side where 
 I was, of 600 yards, and at each end of it took the angles 
 subtended by the othe end and the house and mill which 
 were as follow, viz at one end the angles were 58*^ 20' and 
 95° 20', and at the other end the like angles were 53^ 30' and 
 98<* 45'. What then was the distance between the house and 
 Boill ? Ans. 959-5866 yard«. 
 
 Exam. xxiv. Wanting to know my distance from an in- 
 accessible object 0, on the other side of a river ; and having 
 no instrument for taking angles, but only a chain or cord for 
 measuring distances ; from each of two stations, a and b, which 
 were taken at 500 yards asunder, I measured in a direct line 
 from^;he objectO 100 yards, viz. ac and bd each equal to 100 
 yards ; also the diagonal ad measured 550 yards, and the dia- 
 gonal Bc 560. What then was the distance of the object 
 from each station a and b ? . i ao 536-25 
 
 ^^^' } BO 500-09 
 
 Exam. xxv. In a garrison besieged are three remarkable 
 objects, A, B, c, the distances of which from each other are 
 discovered by means of a map of the place, and are as fol- 
 low, viz. AB 2661, AC 530, bc 327i yards. Now, having to 
 erect a battery against it, at a certain spot without the place, 
 and being desirous to know whether the distances from the 
 three objects be such, as that they may from thence be bat- 
 tered with effect, I took, with an instrument, the horizontal 
 angles subtended by these objects from my station s, and 
 found them to be as follow, viz. the angle asb 13** 30', and 
 the angle bsc 29° 50' ; required the three distances, sa, sb, sc ; 
 the object b being situated nearest to me, and between the 
 two others A and c ? C sa 757-14 
 
 Ans. ? SB 537-10 
 I sc 655-30 
 Exam. xxvr. Required the same as in the last example, 
 when the object b is the farthest from my station, but still 
 seen between the two others as to angular position, and those 
 angles being thus, the angle asb 33*^ 45', and bsc 22° 30', 
 also the tliree distances, as 600, ac 80U, bc 400 yards ? 
 
 C sa 709J- 
 Ans. ? SB 1042f 
 f sc 934 
 Vol. F. B2 MENSI7R4TION 
 
[ 402 J 
 
 MENSURATION OF PLANES. 
 
 THE Area of any plane figure, is the measure of the gpa<ie 
 contained within its extremes or bounds ; without any regard 
 to thickness. 
 
 This area, or the content of the plane figure, is estimated 
 by the number of little squares that may be contained in it ; 
 the side of those little measuring squares being an inrh, a 
 foot, a yard, or any other fixed quantity. And hence tbe area 
 or content is said to be so many square inches, or square feet, 
 or square yards, fitc. 
 
 Thus, if the figure to be measured be 
 the rectangle abcd, and the little square 
 E, whose side is one inch, be the mea- 
 suring unit proposed : then as often as 
 the said little square is contained in the 
 rectangle, so many square inches the 
 rectangle is said to contain which in 
 the present c&se is 12. 
 
 D 
 
 4 
 
 C 
 
 
 
 
 
 
 3 
 
 
 
 
 
 
 
 
 
 
 B 
 
 PROBLEM I. 
 
 To find the Area of any Parallelogram ; •whether it heeLSquart^ 
 a Rectangle, a Rhombus, or a Rhomboid. 
 
 ^ ■ d ;, 
 
 Multiply the length by the perpendicular breadth, or 
 height^ and the product will be the area*. 
 
 EXAMPLES. 
 
 » Tbe truth of this rule is proved in the Geom. theor. 81, cor. 2. 
 
 The same is othcwise proved thu- •. L^-.i tf.e foregoing rectangle 
 be the figure proposed : «nd let the 1 .ngth and bieadth be divided h^to 
 several parts each equal to the linear measuring unit, being; here 4 for 
 the leng'h, and o for the breadth ; and kt the opposite points of divi- 
 sion be connected by right lines — Then it is evident that these lines 
 divide the rectangle into y niimber of little squares, ench equal to the 
 square measuring umt e j and further, that the number of tbes^e little 
 squares, or ihe aiea of tbe figure, is equal to the number of linear 
 measuring units in the lengvh, repealed as often as there are hnear 
 
 measuring 
 
MENSURATION OF PLANES. 403 
 
 EXAMPLES. 
 
 Ex. 1 To and the area of a parallelogram, the length be- 
 ing 12-26 and height 8-5. 
 
 12-25 length 
 8-5 breadth 
 
 6126 
 9800 
 
 104-125 area, 
 
 Ex. 2. To find the area of a square, whose side is 35-25 
 chains. Ans. 124 acres, 1 rood, 1 perch. 
 
 Ex. 3. To find the area of a rectangular board, whose 
 length is 12^ feet, and breadth 9 in€he8. Ans. 9| feet. 
 
 Ex. 4. To find the content of a piece of land, in form of 
 a rhombus, its length being 6-20 chains, and perpendicular 
 height 5-45 Ans. 3 acres, 1 rood, 20 pp.rches. 
 
 Ex. 6. To find the number of square yards of painting in 
 a rhomboid, whose length is 37 feet, and breadth 5 fettt 3 
 inches. Ans. 21 /^square yards 
 
 PROBLEM II. 
 
 Tojind the Area of a Triangle. 
 
 Rule I. Multiply the base by the perpendicular height, and 
 take half the product for the area*. Or, multiply the one of 
 these dimensions by half the other. 
 
 measuring units in the breadth, or height ; that is, equal to the. 
 length drawn into the height ; which here Is 4 X 3 or 12- 
 
 And it is proved, (Geom. theop. 25, cor. 2) that any oblique 
 parallelogram is equal to a rectangle, of equal length and perpendi> 
 cula> breadth. I'herefbre the rule is general for all parallelograms 
 whatever. 
 
 ♦ The truth of this rule is evident, he< use any triangle is the 
 half of a parallelogram of equal base and altitude, by Geom. theor. 
 26. 
 
 examples; 
 
404 MENSURATION 
 
 EXAMPLES. 
 
 Ex. 1. To find the area of a triangle, whose base is 626 ji 
 and perpendicular height 520 hnks ? ^ 
 
 Here 625 X 260 = 162500 square links, 
 
 or equal 1 acre, 2 roods, 20 perches, the answer. 
 Ex. 2. How many square yards contains the triangle, whose 
 base is 40, and perpendicular 30 feet ? 
 
 Ans. 66| square yards. 
 Ex. 3. To find the number of square yards in a triangle, 
 whose base is 49 feet, and height 25^ feet ? 
 
 An^. 684f, or 68-7361. 
 Ex. 4. To find the area of a triangle, whose base is 18 feet 
 4 inches, and height 11 feet 10 inches ? 
 
 Ans. 108 feet, 5| inches^. 
 
 Rule 11 When two sides and their contained angle are 
 
 given : Multiply the two given sides together, and take half 
 
 their product. Then say, as radius is to the sine of the given 
 
 angle, so is that half product, to the area of the triangle 
 
 Or, multiply that half product by the natural sine of the said 
 angle, for the area*. 
 
 Ex. I. What is the area of a triangle, whose two sides are 
 39 and 40, and their contained angle 28*^ 57' ? 
 
 By Natural Numbers . By Logarithms. 
 First 4 X 40 X 30 = 600, \ 
 
 then 1 : 600 : : -484046 sin. 28^ 57' log 9-684887 
 
 600 2 778151 
 
 Answer 290-4276 the area answering 2-463038 
 
 Ex. 2. How many square yards contains the triangle, of 
 which one angle is 45°, and its containing sides 25 and ^MA, 
 fe^t 1 Ans. 20-86947. 
 
 • For, let AB, AC, be the two given sides, 
 including the given angle A. N'^w ^ ab X 
 CP is the area, by the first rule, CP being the 
 perpendicular. But, by trigonometry, as sin. 
 ^ p. or radius : AC : : sin. / A : CP, which is 
 therefore «= AC X sin. Z A, tak.ng radius= 1. 
 Therefore the area ij aB^cp is = ^ AB X ac 
 X sin. Z A, to radius 1 ; or as radius: sin. jt, 
 ^4. : : ^ AB ;<AC : the area. 
 
 P B 
 
 RULE III. 
 
6F PLANES. 
 
 4t5 
 
 Rule III. When the three sides are given : Add all the 
 three sides together, and take half that sum. Next, subtract 
 each side severally from the said half sum, obtaining three 
 remainders. Then multipl}' the said half sum and those three 
 remainders all together, and extract the square root of the 
 last product, for the area of the triangle*. 
 
 * For let ABC be the given 
 trinng-le. D aw the parallels 
 AE. BD, meeting the two sides 
 AC, CB, produced, in D and e, 
 an', making CD = cb, and 
 CE =«« CA. Also draw CFo bi- 
 sect n^ DB and AE perpendicu. 
 laily in F and g ; and fhi pa- 
 ranel to the side ab, meeting 
 AC in H, and ae produced in i. 
 Lastly, with centre h, and radi- 
 us hf, describe a circle meeting 
 
 AC produced in k ; which will pass through o, because c is a right 
 angle, and tfirough i, becausci by means of the parallels, Ai = fb ==• 
 DF, therefire hd = ha, and hf = hi"* |ab. 
 
 Hence ha or hd is half the difference of the sides AC, CB, and 
 HC *= half their sum or = ^ac + ^cb ; also hk «^ hi = ^if op 
 ^AB ; conseq. ck «=* ^ac + ice -f- ^ab half the sura of all the three 
 sides of the triangle abc, or ck = ^s, calling s the sum of those three 
 .sides. Again hk = hi= iiF=^AB,or ki.=» ab; theref. cl =5 ck — 
 KL =»s Js — AB and AK =^ CK — CA «• ^s — AC, and al =a dk =» 
 
 ck — CD = |s-CB. 
 
 Now, by the first riile. AG . co = the A ACE, and AG . to = th® 
 A ABE. theref AG . cf — /^ ACB* Also by the parallels, ag : 
 CO : : DF or lA : cf, theref. ag cf=( A acb=) cg .lA i= cg. df^ 
 conseq. ag cf . co . df ==» /^sacb. 
 
 But CO . CF i= CK . ci' =^s. ^s— AB, and ag . DF = AK . AL 
 = ii — AC ^s — BC ; theref, ag . cF • CG? • df =s= /^2 acb = ^s. 
 
 ^s — AB \s — AC. is — BC is the square of the area of the triangle 
 ABC q^E. D. 
 
 Other'oiise- 
 Because the rectangle ag . cf = the /^ abc, and since Co : 
 AG : : CF : DF drawing the first and second terms into CF, 
 and the third and fourth into Ao, the propor. becomes co . cf : 
 AG cf:: ag-gf: AG. Dp.orco.cF: A abc : : A abc : bg . df 
 that is, the ^ abc is a mean proportional between co • CF and 
 
 AG . DF, OP between |s. As — ab and |s — AC' ^s — BC, <i: e* d« 
 
 Ex. 1. 
 
4t6^ MENSURATION 
 
 Ex. 1. To find the area of the triangle whose three side* 
 are 20, 30, 40. 
 
 20 45 45 45 
 30 20 30 40 
 40 — —. — 
 26 1st rem 15 2d rem. 5 3d rem. 
 
 45 half sum 
 
 Then 45 X 25 X 15 X 5 = 84375, 
 The root of ^ which is 290-4737, the area. 
 
 Ex. 2. How many square yards of plastering are in a tri- 
 angle, whose sides are 30, 40, 60, feet ? Ans. 66f . 
 
 Ex. 3. How many acres, &c. contains the triangle, whose 
 sides are 2569, 4900, 5025 links ? 
 
 ^ Ans. 61 acres, 1 rood, 39 perches. 
 
 PROBLEM m. 
 
 To find the Area of a Trapezoid, 
 
 Add together the two parallel sides ; then multiply thetf 
 sum by the perpendicular breadth, or the distance betwcett 
 them ; and take half the product for the area. By Geom. 
 theor. 29.. 
 
 Ex. 1. In a trapezoid, the parallel sides are 750 and 1225, 
 and the perpendicular distance between them 1540 links : to 
 find the area. 
 1225 
 750 
 
 1975 X 770 = 152075 square hnks = 15 acr. 33 perc, 
 
 Ex. 2. How many square feet are contained in the plank, 
 whose length is 12 ieet 6 inches, the breadth at the greater 
 end 15 inches, and at the less end 11 inches ? 
 
 Ans. 13if feet. 
 
 Ex. 3. In measuring along one side ab of a quadrangular 
 field, that side, and the two perpendiculars let fall on it from 
 the two opposite corners, measured as below, required the 
 content. 
 

 OF PLANES. 
 
 kf = no links 
 
 
 A^ = 746 
 
 
 AB = 1110 
 
 
 CP = 352 
 
 
 Dft = 595 
 
 
 Ans. 4 acres, 1 rood, 6-792 perches. 
 
 AP 
 
 407 
 
 (I B 
 
 PROBLEM IV. 
 
 To find the Area of any Trapezium. 
 
 Divide the trapezium into t'\'0 triangles by a diagonal ; 
 then find the areas of these triangles, and add them together. 
 
 Or thus, let fall two perpendiculars on the diagonal from 
 the other two opposite angles ; then add these two perpen- 
 dirulars together, and multiply that sum by the diagonal, tak- 
 ing half the product for the area of the trapezium. 
 
 Ex* 1. To 6nd the area of the trapezium, whose diagonal 
 is 42, and the two perpendiculars on it 16 and 18. 
 Here 16 + 18 = 34, its half is 17. 
 Then 42 X 11 = 114 the area. 
 
 Ex. 2. How many square yards of paving are in the tra- 
 pezium, whose diagonal is 65 feet, and the two perpendiculars 
 
 let fall on it 28 and 33i feet ? 
 
 Ans. 222 -JJ yards. 
 
 Ex. 3. In the quadrangular field abcd, on account of ob- 
 structions there could only be taken the following measures, 
 viz. the two sides b€ i^66 and ad ^^20 yards, the diagonal ag 
 378, and the two distances of the perpendiculars from. the 
 ends of the diagonal, namely, ae 100, and cf 70 yards,- 
 Re':|uiredthe construction of the figure, and the area in acres, 
 when 4840 square yards make an acre ? 
 
 Ans. 17 acres, 2 roods, 21 perches. 
 
 PROBLEM V^. 
 
 To find the Area of an Irregular Polygon, 
 
 Draw diagonals dividing the proposed polygon into trape- 
 ziunis and triangles. Then find the areas of of all these sepa- 
 rately and add them together for the content of the whole 
 polygon. 
 
 EXAMP1*E. 
 
408 MENSURATION^ 
 
 Example. To find the content of the irregular figure 
 ABCDEFGA, in which are given the following diagonals and per- 
 pendiculars : namely, 
 
 A- 
 
 AC 56 
 FD 62 
 cc 44 
 cm 13 
 Bn 18 
 60 12 
 Ep 8 
 Dq 23 
 
 Ans. 1878J^ 
 
 PROBLEM VI. 
 
 Tojlnd the Area of a Regular Polygon. 
 
 Rule I. Multiply the perimeter of the polygon, or sum 
 «f its sides, by the perpendicular drawn from its centre on 
 •ne of its sides, and take half the product for the area*. 
 
 Ex. I. To find the area of the regular pentagon, each side 
 being 26 feet, and the perpendicular from the centre oa 
 each side is 17-2047737. 
 
 Here' 26 X 6 == 125 is the perimeter. 
 And 17-2047737 X 126 = 2150"6967126. 
 Its half 1076'29835625 is the area sought. 
 
 Rule II. Square the side of the polygon ; then multiply 
 that square by the' tabular area, or multiplier set against its 
 name in the following table, and the product will be the 
 areat. 
 
 No. 
 
 * This is only in effect resolving the polygon into as many equal 
 triangles as it has sides, by drawing lines irom the centre to all the 
 angles ; then finding their aicas, and adding them ail together. 
 
 f This rule is founded on the property, that like polygons, being 
 similar figures, are to one another as the equates of tlieir like sides ; 
 •which is proved in the Geom. theor. 89. Now, the multipliers in the 
 table, are the areas of the respective polygons to the side 1. W hence 
 iha rule is manifest. . 
 
OF PLANES. 
 
 405 
 
 Sides. 
 
 Names. 
 
 Areas, or 
 Multipliers. 
 
 3 
 
 Trigon or triangle 
 
 0-4330^27 
 
 4 
 
 Tetragon or square 
 
 1-0000000 
 
 5 
 
 Pentagon 
 
 1-7204774 
 
 6 
 
 Hexagon 
 
 2-5980762 
 
 7 
 
 Heptagon 
 
 3-6339124 
 
 8 
 
 Octagon 
 
 48284271 
 
 9 
 
 Nonagon 
 
 6-1818242 
 
 10 
 
 Decagon 
 
 7-6942088 
 
 11 
 
 Undecagon 
 
 9-3656399 
 
 12 
 
 Dodecagon 
 
 11-1961524 
 
 Exam. Taking here the same example as before, namely, 
 a pentagon, whose side is 25 feet. 
 Then 25^ being = 625, 
 And the tabular area 1-7204774 ; 
 Theref. 1-7204774 X 625 = 1075-298375, as before. 
 
 Ex. 2. To find the area of the trigon, or equilateral tri- 
 angle, whose side is 20, Ans. 173-20508. 
 
 Ex. 3. To find the area of the hexagon whose side is 20. 
 
 Ans. 1039-23048, 
 
 Ex. 4. To find the area of an octagon whose side is 20. 
 
 Ans. 1931-37084. 
 
 Ex. 6. To find the area of a decagon whose side is 20. 
 
 Ans. 3077-68352. 
 
 ^ote. The areas in the table to each side 1 
 thay be computed in the f )Uowing manner : 
 From the centre c of the polygon draw lines 
 to every angle, dividing the whole figure into 
 as many equal triangles as the polygon has 
 sides ; and let ABC be one of those triangles, 
 
 the perpendicular of which is cd. Divide . 
 
 360 degrees by the number of sides in the po- A D B 
 
 lygon, the quotient gives the angle at the centre acb. The half 
 of this gives the angle acd ; and Uiis taken from 90°. leaves the 
 IJ:T11 ^*'"" -^^^r^ihe, as radius is to ad, so is t.ng. angle 
 fh h^lf^'holPfP^''^'''"'^!;''^' '^i?'" perpendicular, multiplied by 
 '!« u • \^t^ ^''' ^'-'"^^ ^^« *'^* of the triangle abc : which bein? 
 Cl7 ^^?vi7 >'' "''";'"* "^ '^^ '^'""^'^^' "^ «f tne sides of the ^? 
 fifures.^ "^"^^ *'^*' *' ^" '^^ **^^^» f^^ every one of th« 
 
 ^*'" ^' 53 PROBLEM 
 
410 
 
 MENSURATION^ 
 
 PROBLEM VII. 
 
 To find the Diameter and Circumference of any Circhy the one \ 
 from the other. 
 
 This may be done nearly by either of the two follomD|;^ 
 proportions, ' 
 
 viz. As 7 is to 22, so is the diameter to the circumference#|^ 
 Or, As 1 is to 3-1416, so is the diameter to the circumfe*| 
 
 rpnr.f»*- 
 
 
 Ex. 1. To find the circumference of the circle whose di2L*7i 
 
 meter is 20. ;| 
 
 By the first rule, as 7 ; 22 : : 20 : 62#, the answer. 1 
 
 Ex. St J 
 
 * For, let ABCD be any circle, whose centre 
 is B, and let AB, bc be any two equal arcs. 
 Draw the several chords as in the figure, and 
 join BE ; also draw the diameter DA, which 
 produce to f, till bb be equal to the chord bd. 
 
 Then the two isosceles triangles deb, dbf, 
 are equiangular^ because they have the angle at 
 D Cf)mmon ; consequently oe : db : : DB : df. 
 But the two triangles afb, dcb are identical, 
 or equal in all respects, because they have the 
 angle f « the angle bdc, being each equal to 
 the angle ADB, these being subtended by the 
 equal arcs AB, bc ; also the exterior angle FAB of the quadrangle 
 ABCD, is equal to the opposite interior angle at c j and the two tri- 
 angles have also the side bf =^ the side bd ; theref'-'re the side af 
 is also equal to the side DC Hence the proportion above, viz. DE : 
 DB : : DB : DF ■=« DA + AF, becomes de : db : : db : 2de + do. 
 Then, by taking the rectangles of the extremes and means, it is db2 
 =" 2de2 -|- de . DC. 
 
 Now, if the radius DE be taken «■ 1, this expression becomes 
 
 DB» -=« 2 + DC, and hence the root db => V2 -I- nc That is. If 
 the measure of the supplemental chord of any arc be increased by the 
 number 2, the square root of the sum will be the supplemental chord 
 of half that arc. 
 
 Now, to apply this to the calculation of the circumference of 
 the circle, let the arc AC be taken equal to ^ of the circumfe- 
 rence, and be successively bisected by the above theorem : thus, 
 the* chord AC of ^ of the circumference, is the side of the in- 
 scribed regular hexagon, and is therefore equal to the radius ae 
 or 1 : hence, in the right-angled triangle acd, it will be dc -■ 
 
 V'ad»-ac* 
 
OF PLANES. 
 
 411 
 
 Ex. 2. If the circumference of the earth be 26000 miles, 
 what is its diameter ? 
 
 By the 2d rule, as 3-1416 : 1 : : 26000 : 7967^ nearly 
 the diameter. 
 
 Note by R. Adrain. Hanag applied my new theory of 
 most probable values to the determination of the magnitude 
 and Sgure of the earth, I found the true mean diameter of the 
 earth, taken as a globe, to be 7918 7 Enghsh miles, and conse- 
 quently its circumference 24877'4 E. miles, and a degree of 
 a great circle equal to 69-1039 miles. 
 
 y/3 =» 1-7320508076, the supplemental 
 
 , 
 
 
 a 
 
 12 
 
 i"s 
 
 ^ 
 
 I 
 
 
 48 
 
 I 
 
 <» y 
 
 192 
 
 0) ^ 
 
 -a 42 
 
 7k 
 
 
 T*T 
 
 fc- 
 
 <B 
 
 Lti'sj J 
 
 , ^ AD2 -AC2=s^23 — 16 - 
 
 chord of 1 of tlie |>eriphery. 
 
 Then, by the foregoing theorem, by always bisectuig the arcs, 
 and adding 2 to the last square root, there will be found the supple- 
 metjtai chords of the I2th, the 24th, the 48th, the 96th, &c. parts of 
 the periphery ; thus, 
 
 ^3-7320608076 = 1-9318516525^ 
 
 ^3'9318516525 = 1-9828897227 
 
 !^3-9828897227 = 1 '9957178465 
 
 ^3-9957178465 = 1-9989291743 
 
 ^3-9989291743 = 1-9997322757 
 
 ^^3-9997322757 = 1-9999330678 
 
 v^3'9999330678 = 1-9999832669 
 
 v/3-9999832669 = 
 Since then it is found that 3-9i/99832669 is the square of the sup- 
 plemental chord of the 1536th part of the periphery, let this nuHiber 
 be taken from 4, which is the square of the diameter, and the remain- 
 der 0'0000167331 will be the square of the chord of the said 1536th 
 part of the periphery, and consequently the root ^^0-0000167331 = 
 0-0040906112 is the length of that chord ; this number then being 
 multiplied by 1536, gives 6*2831788 for the perimeter of a regular 
 polygon of 1536 sides inscribed in the circle j which^ as the sides of 
 the polygon nearly coincide with the circumference of the circle, must 
 also express the length of the circumference itself, very nearly. 
 
 But now, to show how near this determination 
 is to the truth, let aq.p = 0-0040906112 represent 
 one side of such a regular poly/;on of 1586 sides, 
 and SRT a side of another similar polygon de- 
 scribed about the circle ; and from the centre e 
 let the perpendicular Eq,R be drawn, bisecting ap 
 and ST in q,and r. Then since aq^s = |ap = 
 0-0020453056, and EA = 1, therefore eq.» = ea^ 
 — Aq3 =s '9999958167, and consequently its root 
 gives Eq. = -9999979084 ; then because of the 
 parallels ap, st, it is eq.: er : : ap : st : : as the 
 whole inscribed perimeter : to the circumscribed one, that is, as 
 ■9999979084 : 1 : : 6 2831788 : 6-2831920 the perimeter of the circum- 
 jBcribed polygon. Now, the circumference of the circle being greater 
 
 than 
 
412 MENSURATION 
 
 PROBLEM VIU. 
 
 To find the Length of any Arc of a Circle, 
 
 Multiply the decimal -01745 by the degrees in the giveii 
 arc, and that product by the radius of the circle, for the 
 length of the arc*. 
 
 Ex. 1. To find the length of an arc of 30 degrees, the 
 radius being 9 feet. Ans. 4 7115. 
 
 Ex. 2. To find the length of an arc of 12° 10', or 12«>i, 
 the radius being 10 feet. Ans. 2-1231, 
 
 PROBLEM IX. 
 
 To find the Area of a Circle]. 
 
 Rule I. Multiply half the circumference by half the 
 diameter. Or multiply the whole circumference by the 
 whole diameter, and take | of the product. 
 
 Rule 
 
 than the perimeter of the inner polygon, but less than that of the 
 outer, it must consequently be greater than 6-2831788, 
 
 but less than 6-2831920, 
 and must therefore be nearly equal ^ their sum, or 6-2831854, 
 which in fact is true to the last figure, which should be a 3 instead 
 of the 4. 
 
 Hence, the circumference being 6-2831854 when the diameter is 
 2, it will be the half of that, or 3-1415927, when the diameter is 1, 
 to which the ratio in the rule, viz. 1 to S-14i6 is very near. Also the 
 other ratio m the rule, 7 to 22 or 1 to 3^ = 3-1428 &,c. is another near 
 approximation, " 
 
 * it having been f^und, in the demonstration of the foregoing- prob- 
 lem, that when the radius of a circle is 1, the length of the whole cir- 
 Gumftrence s 6*2831854, which consists of 360 degrees ; therefore as 
 3,60*^ : 6 2831854 ::!*>: 01745 &c. the length of the arc of 1 degree. 
 Hence the decimal -01745 multiplied by any number of degrees, will 
 give the length of ^he arc of those degrees. And because the cir- 
 cumferences and arcs are in proportion as the diameters, or as the 
 1-adii of the circles, therefore as the radius 1 is to any other radius r, 
 so is the lenffth of the arc above mentioned, to -01745 x degrees in 
 the arc X *^t which is the length of that arc, as in the rule. 
 
 t The first rule is proved in the Geom. theor. 94. 
 
 And the 2d and 3d rules are deduced f;om the first rule, in this 
 manner.— By that rule, c^c •—• 4 is the area, when d denotes the diame- 
 ter. 
 
OF PLANES. '4Vd 
 
 Rule II. Square the diameter, and moltiply that square by 
 the decimal -7864, for the area. 
 
 Rule IH. Square the circumference, and multiply that 
 square by the decimal 07958. 
 
 Kx. 1. To find the area of a circle whose diameter is 10, 
 and its circumference 31-416. 
 
 By Rule 3. 
 31-416 
 31-416 
 
 By Rule 1, 
 
 By Rule 2. 
 
 31-416 
 
 •7854 
 
 10 
 
 102= 100 
 
 4)314-16 „« . . 986-965 
 
 78-54 -07958 
 
 78-54 
 
 So that the area is 78-54 by all the three rules. 
 
 Bx. 2 To find the area of a circle, whose diameter is 7, 
 and circumference 22. Ans. 38|. 
 
 Ex. 3. How many square yards are in a circle whose dia- 
 meter is 3i feet ? Ans. l-d69. 
 
 Ex. 4. To find the area of a circle whose circumference 
 is 12 feet. Ans. 11-4595. 
 
 PROBLEM X.^ 
 
 To find the Area of- a Circular Ringy or of the Space included 
 between the Circumferences of two Circles ; the one being 
 contained within the other. 
 
 Take the difference between the areas of the two circles, 
 as found by the last problem, for the area of the ring. — Or, 
 
 ter, and c the circumference. But, by prob. 7, c is =* 3-1416t// there- 
 fore the said area dc -j- 4, becomes d X 3*1416^/ -i- 4 =» -7854^3, 
 which gives the 2d rule. — Also, by the same prob. 7, d \s «— c -r- 
 3-1416 ; therefore again the same first area dc -f- 4, becomes c 4- 
 3-1416 X c -r 4 =» c2 -7- 12*5664, which is =-. c^ x -07958, by taking 
 the reciprocal of 12-5664, or changing that divisor into the multiplier 
 •07958 ; which gives the 3d rule. 
 
 Carol. Hence, the areas of different circles are in proportion to one 
 another, as the square of their diameters, oras the square of their 
 circumferences ; as before proved in the Geom. theor. 93. 
 
 which 
 
414 MENSURATION 
 
 which is the same thing, subtract the square of the less dia- 
 meter from the square of the greater, and multiply their diflf- 
 erence by -7854. — Or lastly, multiply the sum of the dia- 
 meters by the difference of the same, and that product by 
 •7854 ; which is still the same thing, because the product of 
 M;he sum and difference of any two quantities, is equal to the 
 difference of their squares. 
 
 Ex. 1. The diameters of two concentric circles being 10 
 and 6, required the area of the ring contained between their 
 circumferences. 
 
 Here 10 -f 6 = 16 the sum, and 10 — 6 = 4 the diff. 
 
 Therefore -7864 X 16 X 4 = -7854 X 64 = 60-2656, 
 
 the area. 
 
 Ex. 2. What is the area of the ring, the diameters of 
 
 whose bounding circles are 10 and 20 ? Ans. 235* 62. 
 
 PROBLEM XI. 
 
 To find the Area of the Sector of a Circle. 
 
 Rule I. Multiply the radius, or half the diameter, by 
 half the arc of the sector, for the area. Or, multiply the 
 whole diameter by the whole arc of the sector, and take I 
 of the product. The reason of which is the same as for the 
 first rule to problem 9, for the whole circle. 
 
 Rule II. Compute the area of the whole circle : then say,^ 
 as 360 is to the degrees in the arc of the sector, so is the 
 area of the whole circle to the area of the sector. 
 
 This is evident, because the sector is proportional to the 
 length of the arc, or to the degrees contained in it, 
 
 Ex. 1. To find thie area of a circular sector, whose an^ 
 contains 18 degrees ; the diameter being 3 feet ? 
 
 1. By the 1st Rule. 
 
 First, 3- 14 16 X 3 = 9-4248, the circumference. 
 
 And 360 : 18 : ; 9-4248 : -47124, the len2;th of the arc. 
 
 Then -47124 X 3 -r- 4 = 1-41372 ~ 4 =^ -35343, the area, 
 
 2. By the 2d Rule. 
 
 First, -7854 X 3^ = 7-0686, the area of the whole circle. 
 Then, as 3$0 : 18 : : 7-0686 : -36343, the area of the 
 sector 
 
 Ex. 2. 
 
OF PLANES. 415 
 
 Ex. 2. To find the area of a sector, whose radius is 10, and 
 arc 20. Ans. 100. 
 
 Ex. 3. Required the area of a sector, whoseT radius is 35, 
 and its arc containing 147° 29*. Ans. 804 3986 
 
 PROBLEM Xn. 
 
 To find the Area of a Segment of a Circle. 
 
 Rule I. Find the area of the sector having the same arc 
 with the segment, by the last problem. 
 
 Find also the area of the triangle, formed by the chord of 
 the segment and the two radii of the sector. 
 
 Then add these two together for the answer, when the 
 segment is greater than a semicircle ; or subtract them when 
 it is less than a semicircle. — As is evident by inspection. 
 
 Ex. 1. To find the area of the segment acbda, its chord ab 
 being 12, and the radius ae or ce 10. 
 
 First, As AE : sin. ^d 90** : : ad : sin. C 
 
 36*> 62'i = 36-87 degrees, the degrees in the A /^T^ B 
 ^ AEc or arc ac. Their doubie, 73*74, / \^. '' 
 
 are the degrees in the whole arc acb. \ 
 
 Now ••7864 X 400 = 314-16, the area of 
 
 the whole circle. ^ 
 
 Therefore 360«» : 73-74 : : 314 16 : 64-3504, area of the 
 sector ACBE. 
 
 
 ^gain, ^ AE3-ADa = ^ 100-36 = ^ 64 = B = de. 
 Theref. ad X de = 6 X 8 = 48, the area of the trian- 
 gle AEB. 
 Hence sector acbe — triangle aeb = 16-3504, area of seg. 
 
 Rule II. Divide the height of the segment by the diameter, 
 and find the quotient in the column of heights ia the following 
 tablet : Take out the corresponding area in the iv^xi cohiniu 
 on the right hand ; and multiply it by the square of the cir- 
 cle's diameter, for the area of the segment*. 
 
 Note. 
 
 * The truth of this rule depends on tlie principle of r-imllar 
 plane figures, which are to one another as tlie square of (.heh* 
 like linear dimensions. The segments in the table are thost nf a 
 
 circle 
 
41S 
 
 MENSURATION 
 
 Note. When the quotient is not found exactly in the table, 
 proportion may be made between the next less and greater 
 area, in the same manner as is done for logarithms, op any 
 other table. 
 
 Table of the Areas of Circular Segments. 
 
 
 ^1 
 
 
 
 0) CCl 
 
 
 
 % 
 
 0) 
 
 o ^ 
 
 as 
 
 <% 
 
 ffi 
 
 ^ 0) 
 
 <3^ / 
 
 IX 
 
 •11990 
 
 X 
 •31 
 
 -^1 
 
 •'^0738 
 
 •41 
 
 ^ 0) 
 
 -303 r9 
 
 •01 
 
 •00133 
 
 •T7 
 
 •04701 
 
 .21 
 
 •02 
 
 •00376 
 
 •12 
 
 •06339 
 
 •22 
 
 •12811 
 
 •32 
 
 •21667 
 
 •42 
 
 •31304 
 
 •03 
 
 •00687 
 
 ■13 
 
 •06000 
 
 •23 
 
 13646 
 
 •33 
 
 •22603 
 
 43 
 
 •32293 
 
 •04 
 
 01064 
 
 •14 
 
 •06683 
 
 •24 
 
 14494 
 
 •34 
 
 •23647 
 
 •44 
 
 •33284 
 
 •05 
 
 •01468 
 
 •16 
 
 •07387 
 
 •26 
 
 •16354 
 
 •36 
 
 •24498 
 
 •45 
 
 •34278 
 
 •06 
 
 •01924 
 
 •16 
 
 •08111 
 
 26 
 
 •16226 
 
 •3fi 
 
 •2545f> 
 
 •46 
 
 •36274 
 
 •07 
 
 •02417 
 
 17 
 
 •08863 
 
 •27 
 
 •17109 
 
 •37 
 
 •26418 
 
 47 
 
 •36272 
 
 •08 
 
 •02944 
 
 -18 
 
 •09613 
 
 28 
 
 •18002 
 
 38 
 
 •2738(> 
 
 •48 
 
 •3727D 
 
 •09 
 
 •03602 
 
 •19 
 
 10:^90 
 
 •29 
 
 18906 
 
 •39 
 
 •2835^; 
 
 •49 
 
 •38270 
 
 •10 
 
 •04088 
 
 •20 
 
 •11182 
 
 •30 
 
 •19817 
 
 •40 •29337) 
 
 •50 
 
 •39270 
 
 Ex. 2. Taking the same example as before, in which- are 
 given the chord ab 12, and the radius 10, or diameter 20. 
 
 And having found, as above, de = 8 ; then ce — de = ci> 
 =: 10 — 8 = 2. Hence, by the rule, cd -r- of = 2 -f- 20 =ar 
 1 the tabular height. This being found in the first column of 
 of the table, the corresponding tabular area is •04088. Then 
 •04O88 X 202 = -04088 X 400 = 16-362, the area, nearly 
 the same as before. 
 
 Ex. 3. What is the area of the segment, whose height is 18^ 
 and diameter of the circle 60 ? Ans. 636-375. 
 
 Ex. 4. Required the area of the segment whose chord is 16, 
 the diameter being 20 ? Ans. 44*728. 
 
 circle -whose diameter Is 1 ; and the first column contains the cor- 
 responding heights or versed sines divided by the diameter. Thus 
 then, the area of the similar segment, taken from the table, and mul- 
 tiplied by the square of the diameter, gives the area of the segment t» 
 this diameter. 
 
 PROBLEM 
 
OF PLANES. 
 
 4J7 
 
 PROBLEM Xia 
 
 To measure long Irregular Figures, 
 
 Take or measure the brea<lth at both ends, and at se- 
 veral places at equal distances. Then add together all these 
 intermediate breadths and half the two extremes, which 
 sum multiply by the length, and divide by the number of 
 parts for the area*. 
 
 JSToie, If the perpendiculars or breadths be not at equal 
 distances, compute all the parts separately, as so many tra- 
 pezoids, and add them all together for the whole area. 
 
 Or else, add all the perpendicular breadths together, and 
 divide their sum by the number of them for the mean breadth, 
 to multiply by the length ; which will give the whole area, 
 not far from the truth. 
 
 Ex. 1. The breadths of an irregular figure, at five equi- 
 distant places, being 8-2, 7-4, 9-2, 10-2, 86 ; and the whole 
 length 39 ; required the area ? 
 
 8*2 35'2 sum, 
 
 8-6 39 
 
 2) 16-8 sum of the extremes. 
 
 8-4 mean of the extremes. 
 
 7-4 
 9-2 
 10-2 
 
 35-2 sum. 
 
 3168 
 105b 
 
 4) 1372-8 
 
 343-2 the area. 
 
 Ex. 
 
 ♦ This rule is made out as follows : 
 —Let ABCD be the irregular piece , 
 having- ihe several breadths ad EF gh, 
 IK, Bc, at tlie equal distances ae, eg, 
 oi, IB. Let the several breadths ui or- 
 4ep be denoted by the cori-esponding let- 
 ters a, b, c, d, e, and the whole length 
 
 ab by / ; then compute the areas of he parts into which the figure is 
 divided by the perpendiculars, as so many trapezo ds, by prob. 3, and 
 add them all together Thus, the sum of the parts is, 
 a-^ b b-\' c c + d d +e 
 
 ' X AE -1 X EG 4- X GI -f- X IB 
 
 2 2 2 2 
 
 a ■\. b A-fc c + d d ■{• e 
 
 « X i/ + X i/ + X i/ -f. X H 
 
 2 2 2 2 
 
 «=- (i« + 6 + c -h rf + le) X i/ -= (m + A + c + </) J/ 
 
 Vol. I. 
 
 wbicJi 
 
 i4 
 
U8 MENSURATION 
 
 Ex. 2. The len^h of an irregular figure being 84, and the 
 breadths at six equidistant places 17*4, 20*6, 14-2, 16*5, 20- 1, 
 24-4 J what is the area ? Ans. 1560-64. " 
 
 PROBLEM XIV. 
 
 To find the Area of an Ellisis or Oval. 
 
 Multiply the longest diameter, or axis, by the shortest j 
 then multiply the product by the decimal 7854, for the area. 
 As appears from cor. 2, theor. 3, of the Ellipse, in the Conic 
 Sections. 
 
 Ex. 1. Required the area of an ellipse whose two a\es 
 are 70 and 60 Ans. 2746-9. 
 
 Ex. 2. To find the area of the oval whose two axes are 
 24 and 18. Ans. 339-2928. 
 
 PROBLEM XV. 
 
 Tpfind the Area of any Elliptic Segment. 
 
 Find the area of a corresponding circular segment, having 
 the same height and the same vertical axis or diameter. Thea 
 say, as the said vertical axis is to the other axis, parallel to 
 the segment's base, so is the area of the circular segment 
 before found, to the area of the elliptic segment sought. 
 This rule also comes from cor. 2, t|»eor. 3 of the Ellipse. 
 
 Otherwise thus. Divide the height of the segment by the 
 vertical axis of the ellipse ; and find, in the table of circular 
 segments to prob. 12, the circular segment having the above 
 quotient for its versed sine : then multiply all together, this 
 segment and the two axes of the ellipse, for the area. 
 
 Ex. 1. To find the area of the elliptic segment, whose 
 height is 20, the vertical axis being 70, and the parallel 
 axis 50. 
 
 which is the whole area, agreeing* with the rule : m being the arith- 
 inetical mean between the txtremes, or half the sum of them both, 
 and 4 the number of the patts. And the same for any other number 
 of paits whatever. 
 
 Here 
 
OF SOLIDS. 419 
 
 Here 20 -f- 70 gires -284 the quotient or versed sine ; to 
 which io the table answers the seg. '18518 
 then 70 
 
 12-9tt260 
 60 
 
 648-13000 the area. 
 
 Ex. 2. Required the area of an elliptic segment, cut otf 
 parallel to the shorter axis ; the height being 10, and the 
 two axes 25 and 35. Ans. 162 03. 
 
 Ex. 3. To find the area of the elliptic segment, cut oflf 
 parallel to the longer axis ; the 'height being 6, and the axes 
 25 and 35. Ans. 97-8425. 
 
 PROBLEM XVI. 
 
 To find the Area of a Parabola, or its Segment, 
 
 Multiply the base by the perpendicular height ; then take 
 two-thirds of the product for the area. As is proved in theo- 
 rem 17 of the Parabola, in the Conic Sections. 
 
 Ex. 1 . To find the area of a parabola ; the height being 2, 
 and the base 12. 
 
 Here 2 X 12 = 24. Then | of 24 = 16, is the area. 
 
 Ex. 2. Required the area of the parabola, whose height 
 is 10, and its base 16. Ans. 106§, 
 
 MENSURATION OF SOLIDS. 
 
 BY the Mensuration of Solids are determined the spaces 
 included by contiguous surfaces ; and the sum of the measures 
 of these including surfaces, is the whole surface or superficies 
 of the body. 
 
 The measure of a solid, is called its. solidity, capacity, or 
 content. 
 
 Solids are measured by cubes, whose sides are inches, or 
 feet, or yards, &c. And hence the solidity of a body is said 
 to be so many cubic inches, feet, yards, &c. as will fill its ca- 
 pacity or space, or another of an equal magoitude. 
 
 The 
 
420 MENSURATrON 
 
 The least solid measure is the cubic inch, other cubes being 
 laken from it according to the proportion in the following 
 table, which is formed by cubing the linear proportions. 
 
 Table of Cubic or Solid Measures. 
 
 1728 cubic inches make 1 cubic foot 
 
 27 cubic feet - 1 cubic yard 
 
 166f cubic yards - 1 cubic pole 
 
 64000 cubic poles - 1 cubic furlong 
 
 512 cubic furlongs - 1 cubic mile. 
 
 PROBLEM I. 
 
 To Jind the Superficies of a Prism or Cylinder. 
 
 Multiply the perimeter of one end of the prism by the 
 length or height of the solid, and the product will be the sur- 
 face of all its sides. To which add also the area of the 
 two ends of the prism, when required*. 
 
 Or, compute the areas of all the sides and ends separately, 
 and add them all together. 
 
 Ex. 1. To find the surface of a cube, the length of each 
 side being 20 feet. Ans. 2400 feet. 
 
 Ex. 2. To find the whole surface of a triangular prism, 
 whose length is 20 feet, and each side of its end or base 18 
 inches. Ans. 91-948 feet. 
 
 Ex. 3. To find the convex surface of a round prism, or 
 cylinder, whose length is 20 feet, and the diameter of its base 
 is 2 feet. Ans. 126-664. 
 
 Ex. 4. What must be paid for lining a rectangular cistern 
 with lead, at 2d. a pound weight, the thickness of the lead 
 being such as to weigh 7lb. for each square foot of surface ; 
 the inside dimensions of the cistern being as follow, viz. the 
 length 3 feet 2 inches, the breadth 2 feet 8 inches, and depth 
 2 feet 6 inches ? Ans. 2/. 3s. lOirf. 
 
 * The truth of this will easily appear, by considering that the sides 
 of any prism are parallelograms, whose common length is the same 
 as the lenpjth of the solid, and their breadths taken all together make 
 up the perimeter of the ends of the same. 
 
 And the rulo is evidently the same for the surface of a cylinder. 
 
 PROBLEM 
 
OF SOLIDS. 421 
 
 PROBLEM IL 
 
 To find the Surface of a regular Pyramid or Cone, 
 
 Multiply the perimeter of the base by the slant height, or 
 perpendicular from the vertex on a side of the base, and half 
 the product will evidently be the surface of the sides, or the 
 sum of the areas of all the triangles which form it. To which 
 add the area of the end or base, if requisite. 
 
 Ex. 1. What is the inclined surface of a triangular pyra- 
 mid, the slant height being 20 feet, and each side of the base 
 3 feet? Ans. 90feet. 
 
 Ex. 2. Required the convex surface of a cone, or circular 
 pyramid, the slant height being 50 feet, and the diameter of its 
 base ^ ft^i, Ans. 667-39. 
 
 PROBLEM m. 
 
 To find the Surface of the Frustum of a regular Pyramid or 
 Cone ; being the lower part when the top is cut off hy a plane 
 parallel to the base. 
 
 Add together the perimeters of the two ends, and multiply 
 their sum by the slant height, taking half the product for the 
 answer —As is evident, because the sides of the solid are tra- 
 pezoids, having the opposite sides parallel. 
 
 Ex. 1. How many square feet are in the surface of the 
 frustum of a square pyramid, whose slant height is 10 feet ; 
 also, each side of the base or greater end being 3 feet 4 inches, 
 and each side of the less end 2 feet 2 inches ? Ans. 110 feet. 
 
 Ex. 2. To find the convex surface of the frustum of a cone, 
 the slant height of the frustum being 12^ feet, and the cir- 
 cumferences of the two ends 6 and 8*4 feet. Ans. 90 feet. 
 
 PROBLEM IV. 
 
 To find the Solid Content of any Prism or Cylinder. 
 
 Find the area of the base, or end, whatever the figure 
 of it may be ; and multiply it by the length of the prism or 
 cylinder, for the solid content*. 
 
 * This rule appears from the Geom. theor. 110, cor. 2- The same 
 ift more particularly shown as follows ; Let the annexed rectangular 
 
 parallelopipedon 
 
422 
 
 MENTSURATION 
 
 Note. For a cube, take the cube of its side by multiplying 
 this twice by itself ; and for a paraJlelopipedon, multiply the 
 length, breadth and depth all together, for the content. 
 
 Ex. 1. To find the solid content of a cube, whose side is 24 
 inches. Ans. 13824. 
 
 Ex. 2. How many cubic feet are in a block of marble, it* 
 length being 3 feet 2 inches, breadth 2 feet 8 inches, and 
 thickness 2 feet 6 inches ? Ans. 21|. 
 
 Ex. 3. How many gallons of water will the cistern con- 
 tain, whose dimensions are the same as in the last example, 
 when 282 cubic inches are contained in one gallon ? 
 
 Ans. 129if 
 
 Ex. 4. Required the solidity of a triangular prism, whose 
 length is 10 feet, and the three sides of its triangular end or 
 base are 3. 4. 6. feet. Ans. 60. 
 
 Ex. 5. Required the content of a round pillar, or cylinder, 
 whose length is 20 feet, and circumference 6 feet 6 inches. 
 
 Ans. 48-1459 feet. 
 
 % 
 
 parallelopipedon be the solid to be mea- 
 sured, and the cube p the solid measu- 
 ring unit, its side being 1 inch, or 1 foot, 
 &c. i also, let the length and breadth of 
 the base, abcd and also the height au, 
 be each divided into spaces equal to the 
 length of the base of the cube p name- 
 ly/here 3 in the length and 2 in the 
 breadth, making 3 times 3 or 6 squares 
 in the base AC, each equal to the base of 
 the cube p. Hence it is manifest that 
 the parallelopipedon will contain the -"* -O 
 
 cube p, as many times as the base ac contains the base of the cube, 
 repeated as often as the height AH contains the height of the cube. 
 That is, the content of any parallelopipedon is found, by muLiplymg 
 the area of <he base by the altitude of that solid. 
 
 And, because all prisms and cylinders are equal to parallelopipe- 
 dons of equal bases and altitudes> by Geom- theor. 108, it follows that 
 the rule is general for all such solids, whatever the figure of the base 
 may be. 
 
 F 
 
 FEOBUi^ 
 
OF SOLIDS, 4^3 
 
 PROBLEM V. 
 
 Ihjlfid the Content of any Pyramid or Cone, 
 
 Find the area of the base, and multiply that area by the 
 perpendicular height ; then take ^ of the product for the 
 content*. 
 
 Ex. 1. Required the solidity of the square pyramid, each 
 side of its base being 30, and its perpendicular height 26. 
 
 Ans. 75G0. 
 
 Ex. 2. To find the content of a triangular pyramid, whose 
 perpendicular height is 30, and each side of the base 3. 
 
 Ans. 38-97117. 
 
 Ex. 3. ' To find the content of a triang^ilar pyramid, its 
 height being 14 feet 6 inches, and the three sides of its base 
 5, 6, 7 feet. Ahs. 71-0352. 
 
 Ex. 4. What is the content of a pentagonal pyramid, its 
 height being 12 feet, and each side of its base 2 feet ? 
 
 Ans. 27-5276. 
 
 Ex. 5. What is the content of the hexagonal pyramid, 
 whose height is 6*4 feet, and each side of its base 6 inches ? 
 
 Ans. 1 38664 feet. 
 
 Ex. 6. Required the content of a cone, its height being 
 10^ feet and the eircumference of its base 9 feet. 
 
 Ans. 22-56093. 
 
 PROBLEM VI, 
 
 To find the Solidity of the Frustum of a Cone or Pyramid, 
 
 Add into one sum, the areas of the two ends, and the 
 mean proportional between them ; and take ^ of that sum 
 for a mean area ; which being multiplied by the perpendicu- 
 lar height or length of the frustum will give its contentt. 
 
 fiote. 
 
 * Thisxule follows from that of the prism, because any pyramid is 
 §■ of a prism of equal base and altitude j by Geom. theor 1 15, cor. 1 
 and 2. 
 
 f Let ABCD be any pyramid, of whxh bcdgfe is a frustum. 
 And pat a^ for the area of th^ base BCD, b^ the area of the top 
 
424 MENSURATION 
 
 Hote. This general rule may be otherwise expressed, as 
 follows, when the ends of the frustum are circles or regular 
 polygons. In this latter case, square one side of each poly- 
 gon, and also multiply the one side by the other ; add all 
 these three products together ; then multiply their sum by the 
 the tabular area proper to the polygon, and take one-third of 
 the product for the mean area to be multiplied by the length, 
 to give the solid content. And in the case of the frustum 
 of a cone, the ends being circles, square the diameter or the 
 circumference of each end, and also multiply the same two 
 dimensions together ; then take the sum of the three products 
 and multiply it by the proper tabular number, viz. by '7864 
 when the diameters are used, or by 07958 in using the cir- 
 cumferences ; then taking one- third of the product to multiply 
 by the length, for the content. 
 
 Ex. 1. To find the number of solid feet in a piece of timber, 
 whose bases are squares, each side of the greater end being 
 16 inches, and each side of the less end 6 inches ; also, the 
 length or perpendicular altitude 24 feet. Ans. 19^. 
 
 Ex. 2. Required the content of a pentagonal frustum, whose 
 height is 6 feet each side of the base 18 inches, and each side 
 of the top or less end 6 inches. Ans. 931 925 feet. 
 
 EFG, A the height ih of the frustum, and c the 
 height A I of the top part above it. Then 
 c -f A = AH is the height of the whole pyra- 
 mid. 
 
 Hence, by the last prob. i as (c -f A) is the 
 content of the whole pyramid abcd, and ^b^c 
 the content of the top part AEFg ; therefore 
 the difference laS (c 4- h)—^b'^ c is the content 
 of the frustum bcdgfe. But the quantity c 
 being no dimension of the frustum, it must be ex- 
 pelled from this formula, by substituting its value, found in the 
 following manner. By Geom. theor. IIJ, a^ : 62 : : (c -|- A)3 : c2, or 
 d : ^ : : c + A : c, hence (Geom. th. 69) a — 6 : 6 : : h : e, and 
 
 bh ah 
 
 a —6 : a\:b\ c+A ; hence therefore c <=3 and c+A «= — — — ; 
 
 a — 6 a-—b 
 
 then these values of c and c -f- A being substituted for them 
 in the expression for the content of the frustum, gives that con- 
 ah bh a3 ^bz 
 
 tent = 1 a2 X ^^2 x = ^AnX — ='AA x (^^ 
 
 a— b a — b a^^b 
 
 4- «3 + 62 ) ; which is the rule above given j ab being the mean be- 
 tween «2 and ^2 : 
 
 Ex, 6. 
 
SOLIDS. 
 
 42^ 
 
 Ex. 3. To find thp content of a conic frustuna, the altitude 
 being 18, the greatest JiameLer 8, and trie least dia:neter 4. 
 
 Ans. 627-7888, 
 
 Ex. 4. What is the solidity of the frustum of a cone, the al- 
 titude being 25, also the circumference at the greater end be- 
 ing 20, and at the less end 10 ? Aos. 464-216. 
 
 Ex. 5. If a cask, which is two equal conic frustums joined 
 together at the bases, have its j^ung diameter 28 inches, the 
 head diameter 20 inches, and length 40 incoes ; how many 
 gallons of wine will it hold. Ans. 79*0613, 
 
 PROBLEM VII. 
 
 To find the Surface of a Sphere ^ or any Segment, 
 
 Rule 1. Multiply the circumference of the sphere by 
 its diameter, and the product will be the whole surface oJt 
 
 it*. 
 
 Rule II, 
 
 * These rules come from the following theorems for the sur- 
 fece of a sphere, viz. That the said surfsce is equal to the cu)^^ 
 surface of its circumscribing cylinder; or that it is equal to 4 great 
 circ es of the same sphere, or of the same diameter : which are thuij 
 proved. 
 
 Let A BCD be a cylinder, circumscribing 
 tJie sphere ErcHi tlie former generated 
 by the rotation of the rectangle FBClf 
 abo'it the axi.<i or d'ameter fh ; and the 
 latter by the rotation of the seroic-rcle 
 ■pca aboat the same diamete:- fh. Draw 
 tw> lines kl, mx, perp-ndiciilar to the 
 axis intercepting the parts ln, op, of the 
 cylinder and sphere ; then will the ring 
 or cylindric surface g-eneruted by the ro- 
 taiion of LM, be <qual to the nng or spherical surface generated 
 by the arc op. For first, suppose the parallels kt. and m.v to be in- 
 definitely near together; drawing" lo, and also oq. paraUel to ln. 
 Then the two triane-les iko, oq.p, being eqtiiansf-lar, it is, as op ; 
 o<j. )r ln: ; lo or kl : ko :; ciscumf ren .v:> described by kl: cir- 
 cunf. lescribed by ko , theTef>re the rect^ns^le op X circumf of ko 
 is equal to the rectangle Ln X circumf. of kl ; that is, the r.nc;' des- 
 cribed by OP on tlie sphere* is equal to the ring described by l.n on 
 the cyMnder. 
 
 Yot. I. 56 Am€ 
 
426 MENSURATION 
 
 Rule II. Square the diameter and multiply that square by 
 3«1416, for the surface. 
 
 Rule III. Square the circumference ; then either multi- 
 ply that square by the decimal -3183, or divide it by 3-1416, 
 for the surface. 
 
 Note. For the surface of a segment or frustum, multiply the 
 whole circumference of the sphere by the height of the part 
 required. 
 
 Ex. 1. Required the convex superficies of a sphere, whos6 
 diameter is 7, and circumference 22. Ans. 154. 
 
 Ex! 2. Required the superficies of a globe, whose diameter 
 is 24 inches. Ans. 1809-5616. 
 
 Ex. 3. Required the area of the whole surface of the earth, 
 its diameter being 7957| miles, and its circumference -25000 
 miles. Ans. 198943750 sq. miles, 
 
 Ex. 4. The axis of a sphere being 42 inches, what is the 
 convex superficies of the segment whose height is 9 inches ? 
 
 Ans. 1187-5248 inches. 
 
 Ex. 6. Required the convex surface of a spherical zone, 
 whose breadth or height is 2 feet, and cut from a sphere of 
 12i- feet diameter. Ans. 78-54 feet. 
 
 And as this is every where the case, therefore the sums of any 
 corresponding number of these are also equal j that is, the whole 
 surface of the sphere, described by the whole semicircle fgh, is 
 equal to the whole curve surface 6f the cylinder, described by 
 the height BC ; as well as the surface of any segment described by 
 Fo, equal to the surface of the corresponding segment described 
 byBL. 
 
 CoroU 1. Hence the surface of the sphere is equal to 4 of its great 
 circles, or equal to the circumference e fgh, or of DC, multiplied by 
 the height bc, or by the diameter fh. 
 
 Corol. 2. Hence also the surface of any such part as a segment or 
 frustum, or zone, is equal to the same circumference of the sphere, 
 multiplied by the height of the said part. And consequently such 
 spherical curve surfaces are to one another in tl\e same proportion as 
 their altitudes. 
 
 PROBLEM 
 
OF SOLIDS. 
 
 427 
 
 PROBLEM VIIL 
 
 To find the Solidity of a Sphere or Globe, 
 
 Rule I. Multiply the surface by the diameter, and take 
 I of the product for the content*. Or, which is the same thing, 
 multiply the square of the diameter, by the circumference, and 
 take ^ of the product. 
 
 RuLB II. Take the cube of the diameter, and multiply it 
 by the decimal '5236, for the content. 
 
 Rule III. Cube the circumference, and multiply by 
 •01688. 
 
 Ex. 1. To find the content of a sphere whose axis is 12. 
 
 Ans. 904-7808. 
 
 Ex. 2. To find the solid content of the globe of the earth 
 supposing its circumference to be 23000 miles. 
 
 Ans. 263858149120 miles. 
 
 PROBLEM IX. 
 
 To find the Solid Content of a Spherical Segment, 
 t Rule I. From 3 times the diameter of the 
 
 sphere 
 take 
 
 • For, put d= the diameter, c — the circumference, and s = the 
 surface of the sphere, or of its circumscribing cylinder : also, a = 
 the number 3-1416. 
 
 Then, i « is = the base of the cylinder, or one great circle of the 
 sphere ; and d is the height of the cylinder ; therefore ^ds is the 
 content of the cylinder. But |. of the cylinder is the sphere, by th. 11 T, 
 Geom. that is, |. of |cfe, or ids is the sphere ; which is the first rule. 
 
 Again, because the surface » is = ad^ ; therefore Xds = ^ad^ = 
 *'S236d^, is the content, as in the 2d rule. Also d being = c -r- a 
 therefore Xad^ "^ ics -i. a2 = -01688^ the 3d rule for the content, 
 
 I By corol. 3, of theor. 117* Geom.it 
 appears that the spheric segment pfn, is 
 equal to the difference between the cylinder 
 ABLo, and the conic frustum abmq^. 
 
 But, putting rf= AB or TH the diameter 
 of the sphere or cyUnder,A=rK the height 
 of the segment, r = pk the radius of its 
 base, and a = 3*1416 ; then the content of 
 
 the cone ABi is =:iad* X ^ Fi = 
 
 ,ad^'. 
 
 and by the similar cones abi, <i.Mi, as 
 
4gS MENSURATION OP SOLIDS. 
 
 take double the height of the segment : then multiply the 
 remainder by the square of the height, and the product by 
 the decimal -6236, for the content. 
 
 Rule II. To 3 times the square of the radius of the 
 Segment's base, add the square of its height ; then multiply 
 the sum by the height, and the product by 'Si^Se, for the 
 content. 
 
 Ex. 1. To find the content of a spherical segment, of 2 
 feet in height, cut from a sphere of 8 feet diameter. 
 
 Ans. 41-888. 
 
 Ex. 2. What is the solidity of the segment of a sphere, 
 its height being 9, and the diameter of its base 20 1 
 
 Ans. 1795-4244. 
 
 JVofe. The general rules for measuring all sorts of figures 
 having beep now dehvered, we may next proceed to apply 
 the© to the sereral practical uses in life, as i' Hows. 
 
 1Pi3 : Ki3 : : 2\«c?3 : 2V«<^^ X C-^-ry )^== the cone q.mi ; therefore 
 
 the cone abi — the cone <5_mi =• -^^ad^ — .-Lac^s x,-— — -)3 ea 
 
 \ad^ h—^adh^ "^-^ah^ is = the conic frustum abmq.. 
 
 And \aa^k 's = the cylinder ablo. 
 
 Then the difierence of these two is ladh'^ — icAs = Xak^ X 
 (3^— 2A)i fot the spheric segment pfn ; which is the first rule. 
 
 Again, because pk« = fk X kh (cor. to theor. 87, Geom.) or rs 
 
 a=! A ((i — A), therefore d e=s j- A, and Zd — 2A = — + A = 
 
 h h 
 
 ■ I ; which being substituted in the former rule, it becomes 
 h 
 
 ^A* X ■ = ^a^^ X (3r2 -f- 7*2), which is the 2d rule. 
 
 h 
 
 Note. By subtracting a segment from a half sphere, or from another 
 «esin«nt, the content of any frustum or zone may be found. 
 
 LAND 
 
£ 429 j 
 
 LAND SURVEYING. 
 
 SECTION I. 
 
 DESCHIPnON AND USE OF THE INSTRDMENTS. 
 
 1. OF THE CHAIN. 
 
 LAVD is measured with a chain, called Gunter's Chain, 
 fl-om its inventor, the length of which is 4 poles, or 22 yarcjs, 
 or ^^ feet. It consists of 100 equal links ; and the length 
 of each link is therefore y^/^ of a yard, or -jVo of a foot, or 
 7*92 inches 
 
 Land is estimated in acres, roods, and perches. An acre 
 is equal to 10 square chains, or as much as 10 chains in h ngth 
 and 1 chain in breadth. Or, in yards, it is 220 X 22 = 4840 
 square yards. Or, in poles, it is 40 X 4 = 160 square poles. 
 Or, in links, it is 1000 X 100 = 100000 square links : these 
 being all the same quantity 
 
 Also, an acre is divided into 4 parts called roods, and a 
 rood into 40 parts called perches, which are square poles or 
 the square of a pole of b\ yards long, or the square of | of a 
 chain, or of 25 links, which is 626 square links. So that the 
 divisions of land measure, will be thus : 
 
 625 sq. links = 1 pole or perch 
 40 perches =t 1 rood 
 4 roods = 1 acre. 
 The length of lines, measured with a chain, are best set 
 down in links as integers, every chain in length being 100 
 links ; and not in chains and decimals. Therefore after the 
 content is found, it will be in square hnks ; then cut oif five 
 of the figures on the right-hand for decimals, and the rest 
 will be acres. These decimals are then multiplied by 4 for 
 roods, and the decimals, of these again by 40 for perches. 
 
 Exam. Suppose the length of a rectangular piece of ground 
 be 792 links, and its breadth 386 ; to find the area in acres, 
 roods, and perches. 
 
 792 3-04920 
 
 sas 4 
 
 3960 -19680 
 
 6336 40 
 
 2376 . 
 
 7-87200 
 
 3-04920 
 
 Ahs. 3 acres, roods, 7 perches. 3. OF 
 
430 LAND 
 
 2. OF THE PLAIN TABLE. 
 
 This instrument consists of a plain rectangular board, of 
 any convenient size : the centre /of which, when used, is fixed 
 by means of screws to a three-legged stand, having a ball 
 and socket, or other joint, at the top, by means of which, 
 when the legs are fixed on the ground, the table is inclined in 
 any direction. 
 
 To the table belong various parts, as follow. 
 
 1. A frame of wood, made to fit round its edges, and to 
 be taken oflf, for the convenience of putting a sheet of paper 
 on the table. One side of this frame is usually divided into 
 equal parts, for drawing lines across the table, parallel or 
 perpendicular to the sides ; and the other side of the frame 
 is divided in 360 degrees, to a centre in the middle of the 
 table ; by means of which the table may be used as a theo- 
 dohte, &c. 
 
 2 A magnetic needle and compass, either screwed into the 
 side of the table, or fixed beneath its centre, to point out the 
 directions, and to be a check on the sights. 
 
 3. An index, which is a brass two-foot scale, with either a 
 small telescope, or open sights set perpendicularly on the 
 ends. These sights and one edge of the index are in the same 
 plane, and that is called the fiducial edge of the index. 
 
 To use this instrument, take a sheet of paper which will 
 cover it, and wet it to make it expand ; then spread it flat on 
 the table, pressing down the frame on the edges, to stretch 
 it and keep it fixed there ; and when the paper is become 
 dry, it will, by contracting again, stretch itself smooth and 
 flat from any cramps and unevenness. On this paper is to be 
 drawn the plan or form of the thing measured. 
 
 Thus, begin at any proper part of the ground, and make a 
 point on a convenient part fif the paper or table, to repre- 
 sent that place on the ground ; then fix in that point one leg 
 of the compasses, or a fine steel pin, and apply to it the 
 fiducial edge of the index, moving it round till through the 
 sights you perceive some remarkable object, as the corner of 
 a field, &c. ; and from the station-point draw a line with the 
 point of the compasses along the fiducial edge of the index, 
 which is called setting or taking the object : then set another 
 object or corner, and draw its line ; do the same by another ; 
 and so on, till as many objects are taken as may be thought 
 tit. Then measure from the station towards as many of 
 the objects as piay be necessary, but not more, taking the re- 
 <Juisite offsets to corners or crooks in the hedges, laying 
 the measures down on their respective lines on the table. 
 
 Then 
 
SURVEYING. 431 
 
 Then at any convenient place measured to, fix the table in the 
 same position, and set the objects which appear from that 
 place ; and so on, as before. And thus continue till the work 
 is finished, measuring such hnes only as are necessary, and 
 determining as many as may be by intersecting lines of direc- 
 tion drawn from different stations. 
 
 Of shifting the Paper on the Plain Table. 
 
 When one paper is full, and there is occasion for more ; 
 draw a line in any manner through the farthest point of the 
 last station line, to which the work can be conveniently laid 
 down ; then take the sheet off the table, and fix another on, 
 drawing a line over it, in a part the most convenient for the 
 rest of the work ; then fold or cut the old sheet by the line 
 drawn on it, applying the edge to the line on the new sheet, 
 and, as they lie in that position, continue the last station line 
 on the new paper, placing on it the rest of the measure, be- 
 ginning at where the old sheet left off. And so on from sheet 
 to sheet. 
 
 When the work is done, and you would fasten all the sheets 
 together into one piece, or rough plan, the aforesaid lines are 
 to be accurately joined together, in the same manner as when 
 the lines were transferred from the old sheets to the new ones. 
 But it is to be noted, that if the said joining lines, on the old 
 and new sheets, have not the same inclination to the side of 
 the table, the needle will not point to the original degree when 
 the table is rectified ; and if the needle be required to respect 
 still the same degree of the compass, the easiest way of draVvr- 
 ing the lines in the same position, is to draw them both paral- 
 lel to the same sides of the table, by means of the equal divi- 
 sions marked on the other two sides. 
 
 3. OF THE THEODOLITK 
 
 The theodilite is a brazen circular ring, divided into 360 
 degrees, &ic. and having an index with sights, or a telescope, 
 placed on the centre, about which the index is moveable ; also 
 a compass fixed to the centre, to point out courses and check 
 the sights ; the whole being fixed by the centre on a stand of 
 a convenient height for use. 
 
 In using this instrument, an exact account, or field-book, of 
 of all measures and things necessary to be remarked in the 
 plan, must be kept, from which to make out the plan on re- 
 turning home from the ground. 
 
 Begin at such part of the ground, and measure in such di- 
 rections as are judged most convenient ; taking angles or di- 
 rections to objects, and measuring such distances as appear 
 
 necessary, 
 
43* LANB 
 
 necessary, under the same restrictions as in the use of the 
 plain table. And it is safest to fix the theodolite in the origi- 
 nal pos^ition at every station, by means of fore and back ob- 
 jects, and the compass, exactly as in using the plain table ; 
 registering the number of degrees cut off by the index when 
 directed to each object ; and, at any station, placing the index 
 at the same degree as when the direction towards that station 
 was taken from the last preceding one, to fix the theodolite 
 there in the original position. 
 
 The best method of lying down the aforesaid lines of di- 
 rection, is to describe a pretty large circle ; then quarter it, 
 and lay on it the several numbers of degrees cut off by the in- 
 dex in each direction, and drawing lines from the centre to all 
 these marked points in the circle. Then, by means of a pa- 
 rallel ruler, draw from station to station, hnes parallel to the 
 aforesaid lines drawn from the centre to the respective points 
 in the circumference. 
 
 4. OF THE CROSS. 
 
 The cross consists of two pair of sights set at right angles 
 to each other, on a staff having a sharp point at the bottom, 
 to fix in the ground. 
 
 The cross is very useful to measure small and crooked pieces 
 of ground. The method is, to measure a base or chief line, 
 usually in the longest direction of the piece, from corner to 
 corner, and while measuring it, finding the places where 
 perpendiculars would fall on this line, from the several cor- 
 ners and bends in the boundary of the piece, with the cross, 
 by fixing it, by trials, on such parts of the line, as that 
 through one pair of the sights both ends of the line may 
 appear, and through the other pair the corresponding bends 
 or corners ; and then measuring the lengths of the said per- 
 pendiculars. 
 
 REMARKS. 
 
 Besides the fore-mentioned instruments, which are most 
 commonly used, there are some others : as, 
 
 The perambulator, used for measuring roads, and other 
 great distances, level ground, and by the sides of rivers. It 
 has a wheel of V,\ feet, or half a pole in circumference, by the 
 turning of whirJj the machine goes forward : and the distance 
 measured is pointed out by an index, which is moved round by 
 clock work. 
 
 Levels, with telescopic or other sights, are used to find the 
 level between place and place, or how much one place is 
 higher or lower than another. And in measuring any slop- 
 ing or oblique line, either ascending or descending, a small 
 
 pocket 
 
SURVEYING. 433 
 
 pocket level is useful for showing how many links for each 
 chain are to be deducted, to reduce the line to the horizon- 
 tal length. 
 
 An offset-staff is a very useful instrument, for measuring 
 the offsets and other short distances. It is 10 links in length, 
 being divided and marked at each of the 10 links. 
 
 Ten small arrows, or rods of iron, or wood, are used to 
 mark the end of every chain length, in measuring lines. 
 And sometimes pickets, or staves with flags, are set up as 
 marks or objects of direction. 
 
 Various scales are also used in protracting and measuring 
 on the plan or paper ; such as plane scales, line or chords, 
 protractor, compasses, reducing scale, parallel and perpen- 
 dicular rules, &c. Of plane scales there should be several 
 sizes, as a chain in 1 inch, a chain in f of an inch, a chain 
 in ^ an inch, &c. And of these, the best for use are those 
 that are laid on the very edges of the ivory scale, to mark off 
 distances, without compasses. 
 
 SECTION 11. 
 
 THE PRACTICE OF SURVEYING. 
 
 This part contains the several works proper to be done 
 in the field, or the ways of measuring by all the instruments, 
 and in all situations. 
 
 PROBLEM I. 
 
 To Measure a Line or Distance, 
 
 To measure a line on the ground with the chain : Having 
 provided a chain, with 10 small arrows, or rods, to fix one 
 into the ground, as a mark, at the end of ev^^ry chain ; two 
 persons take hold of the chain, one at each end of it ; and 
 all the iO arrows are taken by one of them, who goes fore- 
 most, and is called the leader ; the other being called the 
 follower, for distinction's sake. 
 
 A picket, or station-staff being set up in the direction of 
 the line to be measured, if there do not appear some marks 
 naturally in that direction, they measure straight towards it, 
 the leader fixing down an arrow at the 'end of every chain, 
 which the follower always takes up, as he comes at it, till all 
 the ten arrows are used. They are then all returned to the 
 leader, to use over again. And thus the arrows, are chang- 
 ed from the one to the other at every 10 chains' length, 
 till the whole line is finished ; then the number of changes 
 
 Tct. I. h% of 
 
434 LAND 
 
 of the arrows shows the number of tens, to which the fol- 
 lower adds the arrows he holds in his hand, and the number 
 of hnks of another chain over to the mark or end of the 
 line. So, if there have been 3 changes of the arrows, and 
 the follower hold 6 arrows, and the end of the line cut off 
 45 links more, the whole length of the line is set down in links 
 thus, 3646 
 
 When the ground is not level, but either ascending or de- 
 scending ; at every chain length, lay the offset-staff, or link- 
 staff, down in the slope of the chain, on which lay the small 
 pocket level, to show how many links or parts the slope line 
 is longer than the true level one ; then draw the chain for- 
 ward so many links or parts, which reduces the line to the 
 horizontal direction. 
 
 PROB!uEM II. 
 
 To take Angles and Bearings, 
 
 Let b and c be two objects, or 
 two pickets set up perpendicular; 
 and let it be required to take their 
 bearings, or the angles formed be- 
 tween them at any station a. 
 
 !. With the Plain Table, 
 
 The table being covered with a paper, and fixed on its 
 stand ; plant it at the station a, and fix a fine pin, or a foot 
 of the compasses, in a proper point of the paper, to repre- ' 
 sent the place a : Close by the side of this pin lay the fiducial 
 edge of the index, and turn it about, still touching the pin, 
 till one object b can be seen through the sights : then by the 
 fiducial edge of the index draw a line. In the same manner 
 draw another line in the direction of the other object c. 
 And it is done. 
 
 2. With the Theodolite, ^c. 
 
 Direct the fixed sights along one of the lines, as ab, by 
 turning the instrument about till the mark b is seen through 
 these sights ; and there screw the instrument fast. Then 
 turn the moveable index round, till through its sights the 
 other mark c is seen. Then the degrees cut by the index, 
 on the graduated limb or ring of the instrument, show the 
 quantity of the angle. 
 
 3. With 
 
SURVEYING, 43$ 
 
 3. With the Magnetic Needle and Compass. 
 
 Turn the instrament or compass so, that the north end 
 ©f the needle point to the flower-de-luce. Then direct the 
 sights to one mark as b, and note the degrees cut by the 
 needle. Next direct the sights to the other mark c, and 
 note again the degrees cut by the needle. Then their Sum 
 or difference, as the case may be, will give the quantity of 
 the angle bag. 
 
 4. By Measurement with the Chaiuy «$*c. 
 
 Measure one chain length, or any other length, along 
 both directions, as to b and c. Then measure the distance 
 b^ c. and it is done. — This is easily transferred to paper, by 
 making a triangle Abe with these three lengths, and then 
 measuring the angle a. 
 
 PROBLEM ni. 
 
 To Survey a Triangular Field abc. 
 ' I. By the Chain. 
 
 AP 794 
 AB 1321 
 PC 826 
 
 c 
 
 A P B 
 
 Having set up marks at the corners, which is to be done 
 in all cases where tuere are not marks naturally ; measure 
 with the chain from a to p, where a perpendicular would 
 fall from the angle c, and set up a mark at p, noting down 
 the distance ap. Then complete the distance ab, by mea- 
 suring from p to B. Having set down this measure, return 
 to p, aud measure the perpendicular pc. And thus, having 
 the base and perpendicular, the area from them is easily 
 found. Or having the place p of the perpendicular, the 
 triangle is easily constructed. 
 
 Or, measure all the three sides with the chain, and note 
 them down. From which the content is easily found, or the 
 figure is constructed. 
 
 2. By taking som£ of the Angles. 
 
 Measure two sides ab, ac, and the angle a between them. 
 Or measure one side ab, and the two adjacent angles a and 
 B. From either of these ways the figure is easily planned ; 
 then by measuring the perpendicular cp on the plan, and 
 multiplying it by half ab, the content is found. 
 
 PBGBLBM 
 
436 
 
 LAND 
 
 PROBLEM IV. 
 To Measure a Four-sided Fields 
 
 AE214 
 
 AF 362 
 AC 692 
 
 210 DE 
 
 306 BF 
 
 1. By the Chain, 
 
 Measure along one of the diaponaU, as ac ; and either 
 the two perpendiculars de, bf, as in the last problem ; or 
 else the sides ab, bc, cd, da. From either of which the 
 figure ma}? be planned and computed as before directed. 
 
 Otherwise by the Chain. 
 
 AP 
 
 no 
 
 352 PC 
 
 Aq, 
 
 745 
 
 595 QD 
 
 AB 
 
 1110 
 
 
 .AP 
 
 Measure, on the longest side, the distances ap, 
 and the perpendiculars pc, qd. 
 
 2. 15y taking some of the Angles. 
 
 Measure the diagonal ac (see the last fig. but one), and 
 the angles cab, cad, acb, acd. — Or measure the four sides, 
 and any one of the angles, as bad. 
 
 Or thus. 
 ab 48G 
 BC 394 
 CD 410 
 DA 462 
 BAD 78<> 35' 
 PROBLEM V. 
 To Survey any Fiela by the Chain only. 
 Having set up marks at the corners, where necessary, of 
 the proposed field abcdefg, walk over the ground, and con- 
 sider how it can best be divided in triangles and trapeziums ; 
 and measure them separately, as in the last two problems. 
 Thus, the following figure is divided into the two trapeziums 
 ABCG, GDEF, and the triangle gcd. Then, in the first tra-- 
 pezium, beginning at a, measure the diagonal ac, and the 
 
 two 
 
 Thus. 
 
 
 AC 591 
 
 
 CAB 37° 
 
 20' 
 
 CAD 41 
 
 15 
 
 ACB 72 
 
 25 
 
 ACD 54 
 
 40 
 
SURVEYING. 
 
 437 
 
 two perpendiculars cm, sn. Then the base cc, and the 
 perpendicular Dq. Lastly, the diagonal df, and the two 
 perpendiculfirs pE, og. All which measures write agjainst 
 the corresponding parts of a rough figure drawn to resemble 
 the figure surveyed, or set them down in any other form you 
 choose. 
 
 
 Thus. 
 
 
 Am 
 
 135 130 
 
 mo 
 
 An 
 
 410 
 
 180 
 
 nB 
 
 AC 
 
 650 
 
 
 
 cq 
 
 162 
 
 230 
 
 qo 
 
 CG 
 
 440 
 
 
 
 FO 
 
 237 
 
 120 
 
 OG 
 
 Fp 
 
 288 
 
 80 
 
 PE 
 
 FD 
 
 520 
 
 
 
 Or thus. 
 
 Measure all the sides ab, bc, cd, de, ef, fg, ga ; and the 
 diagonals ac, cg, gd, df. ^ 
 
 Otherwise, 
 
 Many pieces of land may be very well surveyed, by mea- 
 suring any base line, either within or without them, with the 
 perpendiculars let fall on it from every comer. For they arc 
 ' by those means divided into several triangles and trapezoids, 
 all whose parallel sides are perpendicular to the base hue ; 
 and the sum of these triangles and trapeziums will be equal 
 to the figure proposed if the base line fall within it ; if not 
 the sum of the parts which are without being taken from the 
 sum of the whole which are both within and without, will leave 
 the area of the figure proposed. 
 
 In pieces that are not very large, it will be sufficiently ex- 
 act to find the points, in the base line, where the several per- 
 pendiculars will fall, by means of the cross^ or even by judg- 
 ing by the eye only, and from thence measuring to the corners 
 for the lengths of the perpendiculars. — And it will be most 
 convrenient to draw the line so as that all the perpendiculars 
 may fall within the figure. 
 
 Thus in the following figure, beginning at a, and measuring 
 along the line ag, the distances and perpendiculars on the 
 right and left are as below. 
 
 Ab 
 
438 
 
 LANB 
 
 Ab 
 
 316 
 
 350 bB 
 
 AC 
 
 440 
 
 70 cc 
 
 Ad 
 
 686 
 
 320 do 
 
 AC 
 
 610 
 
 60 eE 
 
 a£ 
 
 990 
 
 470 fF 
 
 Ag 
 
 1020 
 
 
 
 PROBLEM VI. 
 
 To Measure the Offsets. 
 Ahiklmn being a crooked hedge, or brook, &c. From A 
 measure in a straight direction along the side of it to b. .'^nd 
 in measuring along this line ab, observe when you are direct- 
 ly opposite any bends or corners of the boundary, as at c, d, 
 e, &c. ; and from these measure the perpendicular offsets ch, 
 di, &c. with the oflfset-staff, if they are not very large, other- 
 wise with the cham itself ; and the work is done. The regis- 
 ter, or field book, may be as follows : 
 
 Uffs. 
 
 left. 
 
 Base line ab 
 
 
 
 
 A 
 
 ch 
 
 62 
 
 45 AC 
 
 di 
 
 84 
 
 220 Ad 
 
 ek 
 
 70 
 
 340 AC 
 
 fl 
 
 98 
 
 610 ■ Af 
 
 :^m 
 
 67 
 
 634 Ag 
 
 Bn 
 
 91 
 
 786 AB 
 
 AC 
 
 PROBLEM Vir. 
 
 To survey any Field with the Plain Table, 
 1. From one Station. 
 
 Plant the table at any angle as 
 G, from which all the other angles, 
 or marks set up, can be seen ; turn 
 the table about till the needle point 
 to the flower-de-luce : and there 
 screw it fast. Make a point for 
 c on the paper on the table, and 
 lay the edge of the index to c, 
 turning it about c till through the A B 
 
 sights you see the mark d : and by the edge of the index 
 draw a dry or obscure line : then measure the distance cd, 
 and lay that distance down on the line cd. Then turn the 
 index about the point c, till the mark e be seen through the 
 
 ' 'its, 
 
SURVEYING. 43» 
 
 sights, by which draw a line, and measure the distance to e, 
 laying it on the hne from c to e. In like manner determine 
 the positions of ca and cb, by turning the sights successively to 
 A and B ; and lay the lengths of those lines down. Then 
 connect the points, by drawing the black lines cd, de, ea, ab^ 
 Bc, for the boundaries of the field. 
 
 2. From a Station Within the Field. 
 
 When all the other parts cannot - 
 
 be seen from one angle, choose some 
 place within, or even without, if 
 more convenient, from which the 
 other parts can be seen. Plant the 
 table at 0, then fix it with the needle 
 north, and mark the point on it. 
 Apply the index successively to 0, _ 
 
 turning it round with the sights to A 
 
 each angle, a, b, c, d, e, drawing dry lines to them by the edge 
 of the index ; then measuring the distance oa, ob, &,c. and 
 laying them down on those lines. Lastly, draw the bounda- 
 ries AB, bc, CD, DE, EA.. 
 
 . 3. By going Round the Figure. 
 
 When the figure is a wood, or water, or when from some 
 other obstruction you cannot measure lines across it : begia 
 at any point a, and measure around it either within or 
 without the figure, and draw the directions of all the sides, 
 thus ; Plant the table at a ; turn it with the needle to the 
 north or flower-de-luce ; fix it, and mark the point a. Apply 
 the index to a, turning it till you can see the point e, and 
 there draw a line : then the point b, and there draw a line ; 
 then measure these lines, and lay them down from a to e and 
 B. Next move the table to b, lay the index along the line 
 AB, and turn the table about till you can see the mark a, and 
 screw fast the table ; in which position also the needle will 
 again point to the flower-de-luce, as it will do indeed at every 
 station when the table is in the right position. Here tura 
 the index about b till through the sights you see the mark c ; 
 there draw a line, measure bc, and lay the distance on that 
 line after you have set down the table at c. Turn it then 
 again into its proper position, and in like manner find the next 
 line CD. And so on quite around by e, to a again. Then the 
 proof of the work will be the joining at a ; fdr if the work be 
 all right, the last direction ea on the ground, will pass exactly 
 through the point a on the paper ; and the measured distance 
 will also reach exactly to a. If these do not coincide, or nearly 
 so, some error has b^ea cemBOiitted, aqd the work must be ex- 
 aiQined over again. 
 
 PROBLEM 
 
440 - LAND 
 
 PROBLEM VIII. 
 
 To Survey a Field with the Theodolite, ^c, 
 1 . From One Point or Station. 
 When all the angles can be seen from one point, as the 
 angle c (first fig. to last prob.) place the instrnmenf at c. and 
 turn it about, till through the fixed sights you see the mark 
 B, and there fix it. Then turn the moveable index about 
 till the mark a be seen through the sights, and note the de- 
 grees cut on the instrument. Mextturn the index successively 
 to E and D, noting the degrees cut off at each ; \^hich gives ill 
 the angles bca. bce, bcd. Lastly measure the lines cb, ca, ce, 
 CD ; and enter the measures in a field-book, or rather against 
 the corresponding parts of a rough figure drawn by guess to 
 resemble the field. 
 
 1 . From a point Within or Without, 
 
 Plant the instrument at (last fig.) and turn it about till the 
 fixed sights point to any object, as a ; and there screw it fast. 
 Then turn the moveable index round till the sights point suc- 
 cessively to the other points e, d, c, b, noting the degrees cut 
 off at each of them ; which gives all the angles round the point 
 0. Lastly measure the distances oa, ob, oc, od, oe, noting 
 them down as before, and the work is done. 
 
 3. By going Round the Field. 
 
 By measuring round, either 
 within or without the field, pro- 
 ceed thus. Having set up marks 
 at B, c, kc. near the corners as 
 usual, plant the instrument at 
 any point a, and turn it till the 
 fixed index be in the direction 
 AB, and there screw it fast : then 
 turn the moveable index to the 
 direction af ; and the degrees cut off will be the angle a. 
 Measure the line ab, and plant the instrument at b, and 
 there in the same manner observe the angle a. Then measure 
 Bc, and observe the angle c. Then measure the distance cd, 
 and take the angle d. 'Phen measure de, and take t*^e angle e. 
 Then measure ef, and take the angle f. And lastly measure 
 the distance fa. 
 
 To prove the work ; add all the inward angles a, b, c, 
 &c. together ; for ,when the work is right, their sum will be 
 equal to twice as many right angles as the figure has sides, 
 wanting 4 right angks. But when there is an angle, as f, 
 that bends inwards, and you measure the external angle, 
 
 which 
 
SURVEYING. 
 
 441 
 
 which is less than two right angles, subtract it from 4 right 
 angles, or 360 degrees, to give the internal angle greater 
 than a semicircle or 180 degrees. 
 
 Olherwise, 
 Instead of observing the internal angles, we may take the 
 external angles, formed without the figure by producing the 
 sides farther out. And in this case, when the work is right, 
 their sum altogether will be equal to 360 degrees. But when 
 one of them, as f, runs inwards, subtract it from the sum of 
 the rest, to leave 360 degrees. 
 
 PROBLEM IX. 
 
 To Survey a Field with Crooked Hedges, S^c. 
 
 WITH any of the instruments, measure the lengths and 
 positions of imaginary lines running as near the sides of the 
 field as you can ; and, in going along them, measure the 
 offsets in the manner before taught ; then you will have the 
 plan on the paper in using the plain table, drawmg the 
 crooked hedges through the ends of the offsets ; but in sur- 
 veying with the theodolite, or other instrument, set dowa 
 the measures properly in a field-book, or memorandum- 
 book, and plan them after returning from the field, by laying 
 down all the lines and angles. 
 
 So, in surveying the piece abcde, set up marks a, b, c, d, 
 dividing it so as to have as few sides as may be. Then begin at 
 any station, a, and measure the lines ab, be, cd, da, taking 
 their positions, or the angles a, b, c, d ; and, in going along 
 the lines, measure all the offsets, as at m, n, o, p, &jc. along 
 every station-line. 
 
 And this is done either within the field, or without, as 
 may be most convenient. When there are obstructions 
 within, as wood, water, hills, &c. then measure without, as 
 in the next following figure. 
 
 ^OL. I. 57 PROBLEM 
 
442 
 
 LAND 
 
 PROBLEM X. 
 
 To Survey a Field, or any other Things by Two Stations. 
 
 This is performed by choosing two stations from which 
 all the marks and objects can be seen ; then measuring the 
 distance between the stations, and at each station taking the 
 angles formed by every object from the station line or dis- 
 tance. 
 
 The two stations may be taken either within the bounds, 
 or in one of the sides, or in the direction of two of the objects, 
 or quite at a distance and without the bounds of the objects 
 or part to be surveyed. 
 
 In this manner, not only grounds may be surveyed, with- 
 out even entering them, but a map may be taken of the 
 principal parts of a county, or the chief places of a town, 
 or any part of a river or coast surveyed, or any other inacces- 
 sible objects ; by taking two stations, on two towers, or two 
 hills, or such-like. 
 
 PROBLEM XL 
 
 To Survey a Large Estate. 
 If the estate be very large, and contain a great number of 
 fields, it cannot well be done by surveying all the fields 
 
 singly 
 
SURVEYING. 443 
 
 singly, and then putting them together ; nor can it be done 
 by taking all the angles and boundaries that enclose it. For 
 in these cases, any small errors will be so much increased, as 
 to render it very'raach distorted. But proceed as below. 
 
 1. Walk over the estate two or three times, in order to 
 get a perfect idea of it, or till you can keep the figure of it 
 pretty well in mind. And to help your memory, draw an 
 eye-draught of it on paper, or at least of the principal parts 
 of it, to guide you ; -setting the names within the fields in 
 that draught. 
 
 2. Choose two or more eminent places in the estate, for 
 stations, from which all the principal parts of it can be seen : 
 selecting these stations as far distant from one another as 
 convenient. 
 
 3. Take such angles, between the stations, as you think 
 necessary, and measure the distances from station to station, 
 always in a right line : .these things must be done, till you 
 get as m^ny angles and lines as are sufficient for determiumg 
 all the points of station. And in measuring any of these 
 station-distances, mark accurately where these lines meet 
 with any hedges, ditches, roads, lanes, paths, rivulets, &c. ; 
 and where any remarkable object is placed, by measuring its 
 distance from the station-line ; and where a perpendicular 
 from it cuts that line. And thus as you go along any main 
 station-line, takft offsets to the ends of all hed<res, and to any 
 pond, house, mill, bridge, &.c. noting every thing down that 
 is remarkable. 
 
 4. As to the inner parts of the estate, they must be deter- 
 mined, in hke manner, by new station-lines : for, aft^^r the 
 main stations are determined,' and every thing adjoining to 
 them, then the estate must be subdivided into two or three 
 parts by new station-lines ; taking inner stations at proper 
 places, where you can have the best view. Mv^^asure these 
 station-lines as you did the first, and all their intersections 
 with hedges and offsets to such objects as appear. Then 
 proceed to survey the adjoining fields, by taking the angles 
 that the sides make with the station-line, at the intersections, 
 and measuring the distances to each corner, from the inter- 
 sections. For the station- lines will be the bases to all the 
 future operations ; the situation of all parts being entirely 
 dependent on them ; and therefore they should be taken of 
 as great length as possible ; and it is best for them to run 
 along some of the hedges or boundaries of one or more fields, 
 or to pass through some of their angles. All things being 
 
 .determined for these stations, you must take more inner sta- 
 tions, and continue to divide and subdivide till at last you 
 r.ome to single fields ; repeating the same work for the inner 
 
 stations 
 
444 LAND 
 
 stations as for the outer ones, till all is done ; and close the 
 work as often as you can, and in as few lines as possible. 
 
 5. An estate may be so situated that the whole cannot be 
 surveyed together ; because one part of the estate cannot be 
 seen from another. In this case, you may divide it into 
 three or four parts, and survey the parts separately, as if 
 they were lands belonging to different persons ; and at last 
 join them together. 
 
 6. As it is necessary to protract or lay down^the work as 
 you proceed in it, you must have a scale of a due length to 
 do it by. To get such a scale, measure the whole length of 
 the estate in chains ; then consider how many inches long 
 the map is to be ; and from these will be known how many 
 chains you must have in an inch ; then make the scale accor- 
 dingly, or cho6se one already made. 
 
 PROBLEM Xn. 
 
 To Survey a County, or large Trad of Land. 
 
 1. Choose two, three, or four eminent places, for stations : 
 such as the tops of high hills or mountains, towers, or church 
 steeples which may be seen from one another ; from which 
 most of the towns and other places of note may also be seen ; 
 and so as to be as far distant from one another as possible. 
 On these places raise beacons, or long poles, with tiags of 
 different colours flying at them, so as to be visible from all 
 the other stations. 
 
 2. At all the places which you would set down in the 
 map, plant long poles, with flags at them of several colours, 
 to distinguish the places from one another; fixinav. them on 
 the tops of church steeples, or the tops of houses ; or in the 
 centres of smaller towns and villages. 
 
 These marks then being set up at a convenient number of 
 places, and such as may be seen from both stations ; go to 
 one of these stations, and, with an instrument to take angles, 
 , standing at that station, take all the angles between the other 
 station and each of these marks. Then go to the other 
 station and take all the angles between the first station and 
 each of the former marks, setting them down with the others, 
 each against its fellow with the same colour. You may, if 
 convenient, also take the angles at some third station, which 
 may serve to prove the work, if the three lines intersect in 
 that point where any mark stands. The marks must stand till 
 the observations are finished at both stations ; and then they 
 may be taken down, and set up at new places. . The same 
 operations must be performed, at both stations, for these 
 new places ; and the like for others. The instrument for 
 
 taking 
 
SURVEYING. 445 
 
 taking angles must loe an exceeding good one, made on pur- 
 pose with telescopic sights, and of a good length of radius. 
 
 3. And though it be not absolutely necessary to mea^su^e 
 any distance, because a stationary line being laid down from 
 any scale, all the other lines will be proportional to it ; yet it 
 is l3etter to measure some of the lines, to ascertain the dis- 
 tances of places in miles, and to know how many geometrical 
 miles there are in any length ; as also from thence to make a 
 scale to measure any distance in miles. In measuring any 
 distance, it will not ,be exact enough to go along the high 
 roads ; which, by reason of their turnings and windings, hard- 
 ly ever he in a right line between the stations ; which must 
 cause endless reductions, and require great trouble to make 
 it a right line ; for which reason it can never be exact. But 
 a better way is to measure in a straight line with a chain, be- 
 tween station and station, over hills and dales, or level fields, 
 and all obstacles Only in case of water, woods, towns, rocks, 
 banks, &c. where we cannot pass, such parts of the line must 
 be measured by the methods of inaccessible distances ; and 
 besides allowing for ascents and decents, when they are met 
 with. A good compass, that shows the bearing of the two 
 stations, will always direct us to go straight, when the two 
 stations cannot be seen ; and in the progress, if we can go 
 straight, offsets may be taken to any remarkable places, 
 likewise noting the intersection of the station-line with all 
 roads, rivers, &c. 
 
 4. From all the stations, and in the whole progress, we 
 must be very particular in observing sea coasts, river-mouths, 
 towns, castles, houses, churches, mills, trees, rocks, sands, 
 roads, bridges, fords, ferries, woods, hills, mountains, rills, 
 brooks, parks, beacons, sluices, floodgates, locks, &c. and in 
 general every thing that is remarkable. 
 
 5. After we have done with the first and main station-lines, 
 which command the whole county ; we must then take inner 
 stations at some places already determined ; which will di- 
 vide the whole into several partitions : and from these stations 
 we must determine the places of as many of the remaining 
 towns as we can. And if any remain in that part, we must 
 take more stations, at some places already determined, from 
 which we may determine the rest. And thus go through all 
 the parts of the county, taking station after station, till we 
 have determined the whole. And in general the station-dis- 
 tances must always pass through such remarkable points as 
 have been determined before, by the former stations. 
 
 PROBLEM 
 
446 
 
 LAND 
 
 PROBLEM XUI. 
 
 To Survey a Town or City. 
 
 This may be done with any of the instruments for taking 
 angles, but best of all with the plain table, where every mi- 
 nute part is drawn while in sight. Instead of the common 
 surveying or Gunter's chain, it will be best, for this purpose, 
 to have a chain 50 feet long, divided into 50 links of one foot 
 each, and an offset-staff of 10 feet long. 
 
 Begin at the meeting of two or more of the principal 
 streets, through which we can have the longest prospects, to 
 get the longest station-lines : there having fixed the instru- 
 ment, draw lines of direction along those streets, using two 
 men as marks, or poles set in wooden pedestals, or perhaps 
 some remarkable places in the houses at the farther ends, 
 as windows, doors, corners, &c Measure these lines with 
 the chain, taking offsets with the staff, at all corners of 
 streets, bendings, or windings, and to all remarkable things, as 
 churches, markets, halls, colleges, eminent houses, &c. Then 
 remove the instrument to another station, along one of these 
 lines ; and there repeat the same process as before. And so 
 on tillthe whole is finished. 
 
 Thus, fi3f the instrument at a, and draw lines in the direc- 
 tion of all the streets meeting there ; then measure ab, noting^ 
 the street on the left at m. At the second station b, draw the 
 directions of the streets meeting there ; and measure from b 
 to c, noting the places of the streets at n and o as you pass by 
 them. At the third station c, take the direction of all the 
 streets meeting there, and measure cd. At d do the same, 
 and measure de, noting the place of the cross streets at p. And 
 in this manner go through all the principal streets This done, 
 proceed to the smaller and intermediate streets ; and lastly to 
 the lanes, alleys, courts, yards, and every part that it may be 
 thought proper to represent in the plan. 
 
SURVEYING. 447 
 
 PRCWBLEM XIV. 
 
 To lay down the Plan of any Survey. 
 
 Ip the survey was taken with the plain table we have a rough 
 plan of it already on the paper which covered the table. But 
 if the survey was with any other instrument, a plan of it is to 
 be drawn from the rneasuies that were taken in the survey ; 
 and first of all a rough planon paper. 
 
 To do this, you must have a set of proper instruments, for 
 laying down both lines and angles, kc. ; as scales of various 
 sizes (the more of them, and the more accurate, the better), 
 scales of chords, protractors, perpendicular and parallel rulers, 
 &c. Diagonal scales are best for the lines, because they ex- 
 tend to three figures, or chains, and links, which are 100 parts 
 of chains. But in using the dia^ronal scale, a pair of compasses 
 must be employed, to take off the lengths of the principal 
 lines very accurately. But a scale with a thin edge divided, is 
 much readier for laying down the perpendicular offsets to 
 crooked hedges, and for marking the places of those offsets on 
 the station-line ; which is done at only one application of the 
 edge of the scale to that line, and then pricking off all at once 
 the distances along it. Angles are to be laid down either with 
 a good scale of chords, which is perhaps the most accurate 
 way, or with a large protractor, which is much readier when 
 many angles are to be laid down at one point, as they are prick- 
 ed off all at once round the edge of the protractor. 
 
 In general, all lines and angles must be laid down on the 
 plan in the sam^ order in which they were measured in the 
 field, and in which they are written in the field-book ; lay- 
 ing down first the angles for the position of lines, next the 
 lengths of the lines, with the places of the offsets, and then 
 the lengths of the offsets themselves, all with dry or obscure 
 lines ; then a black line drawn through the extremities of all 
 the offsets, will be the hedge or bounding line of the field, 
 &c. After the principal bounds and lines are laid down, and 
 made to fit or close properly, proceed next to the smaller 
 objects, till you have entered every thing that ought to appear 
 in the plan, as houses, brooks, trees, hills, gates, stiles, roads, 
 lanes, mills, bridges, woodlands, &c. &c. 
 
 The north side of a map or plan is commonly placed 
 uppermost, and a meridian is some where drawn, with the 
 compass or flower-de-luce pointing north. Also, in a vacant 
 part, a scale of equal parts or chains is drawn, with the title 
 .of the map in conspicuous characters, and erat)ellished with 
 a compartment. Hills are shadowed to distinguish them in 
 the map. Colour the hedges vv^th different colours ; repre- 
 sent 
 
448 LAND 
 
 sent billy grounds by broken bills and valleys ; draw single 
 dotted lines for foot-paths, and double ones for horse or car- 
 riage roads. Write the name of each field and remarkable 
 place within it, and, if you choose, its content in acres, roodSj 
 and perches. 
 
 In a very large estate, or a county, draw vertical and ho- 
 rizontal lines through the map, denoting the spaces between 
 them by letters placed at the top, and bottom, and sides, for 
 readily finding any field or other object mentioned in a table. 
 
 In mapping counties, and estates that have uneven grounds 
 of hills and valleys, reduce all oblique lines, measured up- 
 hill and down-hill, to horizontal straight lines, if that was 
 not done during the survey, before they were entered in the 
 field-book, by making a proper allowance to shorten them. For 
 which purpose there is commonly a small table engraven on 
 some of the instruments for surveying. 
 
 THE NEW METHOD OF SURVEYING. 
 
 PROBLEM XV. 
 
 To Survey and Plan by the JNTea: Method. 
 
 In the former method of measuring a large estate, the ac- 
 curacy of it depends both on the correctness of the instru- 
 ments, and on the care in taking the angles. To avoid the 
 errors incident to such a multitude of angles, other methods 
 have of late years been used by some few skiful surveyors : 
 the most practical, expeditious, and correct, seems to be the 
 following, which is performed, without taking angles, by mea- 
 suring with the chain only. 
 
 Choose two or more eminences, as grand stations, and mea- 
 sure a principal base Hne from one station to another ; noting 
 every hedge, brook, or other remarkable object, as you pass 
 by it ; measuring also such short perpendicular lines to the 
 bends of hedges as may be near at hand. From the extre- 
 mities of this base line, or from any convenient parts of the 
 same, go off with other lines to some remarkable object situa- 
 ted towards the sides of the estate, without regarding the 
 angles they make with the base Ime or with one another ; 
 still remembering to note every hedge, brook, or other object, 
 that you pass by. These lines, when laid down by inter- 
 sections, will with the base line, form a grand triangle on 
 the estate ; several of which, if need be, being thus mea- 
 sured and laid down, you may proceed to form other smaller 
 triangles and trapezoids on the sides of the former j and so on 
 lill you finish with the enclosures individually. By which 
 means a kind of skeleton of the estate may first be obtained, 
 
 and 
 
SURVEYING. 449 
 
 and the chief lines serve as the bases of such triangles and 
 trapezoids as are necessary to fill up all the interior parts. 
 
 The field-book is ruled into three columns, as usual. In 
 the middle one are set down the distances on the chain-line, 
 at which any mark, offset, or other observation, is made ; and 
 in the right and left hand columns are entered the offsets and 
 observations made on the right and left band respectively of 
 the chain-line ; sketching on the sides the shape or resem- 
 blance of the fences or boundaries. 
 
 It is of great advantage, both for brevity and perspicuity, 
 to begin at the bottom of the leaf, and write upwards ; deno- 
 ting the crossing of fences, by lines drawn across the middle 
 column, or only a part of such a line on the right and left op- 
 posite the figures, to avoid confusion ; and the corners of 
 fields, and other remarkable turns in the fences where offsets 
 are taken to, by lines joining in the manner the fences do ; as 
 will be best seen by comparing the book with the plan annex- 
 ed to the field-book following, p. 450. 
 
 The letter in the left-hand corner at the beginning of every 
 line, is the mark or place measured /rom; and that at the 
 right-hand corner at the end, is the mark measured to : But 
 when it is not convenient to go exactly from a mark, the 
 place measured from is described such a dUtanct from one. 
 mark towards another ; and where a former mark is not mea- 
 sured to, the exact place is ascertained by saying, turn to the 
 right or left hand, such a distance to such a mark^ it being al- 
 ways understood that those distances are taken in the chain-line. 
 
 The characters used are, f for turn to the right hand^^ for 
 turn to the left hand, and-^placed over an offset, to show that it 
 is not taken at right angles with the chain-line, but in the di- 
 rection of some straight fence ; being chiefly used when cross- 
 ing their directions ; which is a better way of obtaining their 
 true places than by offsets at right angles. 
 
 When a line is measured whose position is determined, 
 either by former work (as in the case of producing a giveo 
 line, or measuring from one known place or mark to another) 
 or by itself (as in the third side of the triangle), it is called 
 ^fast line, and a double line across the book is drawn at the 
 conclusion of it ; but if its position is not determined (as in 
 the second side of the triangle), it is called a loose line, and a 
 single line is drawn across the book. When a line becomes 
 determined in position and is afterwards continued fartheti 
 a double line half through the book is drawn. 
 
 When a loose line is measured, it becomes absolutely ne- 
 cessary to measure some other line that will determine its 
 position. Thus, the first line ah or hh^ being the base of a 
 triangle is always determined ; but the position of the second 
 
 Vol. F. 58 side 
 
450 • , LAND 
 
 side hj, does not become determined, till the third side J6 is 
 measured ; then the position of both is determined, and the 
 triangle may be constructed. 
 
 At the beginning of a line, to fix a loose line to the mark or 
 place measured from, the sign of turning to the right or left 
 hand must be added, as at h in the second, and j in the third 
 line ; otherwise a strs^nger, when laying down the work, may 
 as easily construct the triangle hjb on the wrong side of the line 
 ahf as on the right one ; but this error cannot be fallen into, 
 if the sign above named be carefully observed. 
 
 In choosing a line to fix a loose one, care must be taken 
 that it doe§ not make a very acute or obtuse angle ; as in the 
 t^^iangle pBr, by the angle at b being very obtuse, a small de- 
 viation from truth, even the breadth of a point at p or r, 
 would make the error at b, when constructed, very consider- 
 able ; but by constructing the triangle pBg, such a deviation is 
 of no consequence. 
 
 Where the words leave off" are written in the field-book, it 
 signifies that the taking of ofisets is from thence discontinued ; 
 and of course something is wanting between that and the next 
 offset, to be afterwards determined by measuring gome other 
 line. 
 
 The field-book for this method, and the plan drawn froai 
 it, are contained in the four following pages, engraven on cop- 
 per-plates ; answerable to which, the pupil is to draw a plan 
 from the measures in the field-book, of a larger size, viz. to a 
 scale of a double size will be convenient, such a scale being 
 also found on most instruments. In doing this, begin at the 
 commencement of the field-book, or bottom of the first page 
 and draw the first, line ah in any direction at pleasure, and then 
 the next two sides of the first triangle bhj by sweeping inter- 
 secting arcs ; and so all the triangles in the same manner, after 
 each other in their order ; and afterwards setting the perpendi- 
 cular and other offsets at their proper places, and through the 
 ends' of them drawing the bounding fences. 
 
 J^ote. That the field-book begins at the bottom of the first 
 page, and reads up to the top ; hence it goes to the bottom of 
 the next page, and to the top ; and thence it passes from the 
 bottom of the third page to the top which is the end of the 
 field-book. The several marks measured to or from, are here 
 denoted by the letters of the alphabet, first the small ones 
 «, 6, c, df &c, and after them the capitals A^ B, C, Z>, he. But, 
 instead of these letters, some surveyors use thq numbers in 
 order, 1, 2, 3, 4, &c. 
 
 OF 
 
SURVEYING. 
 
 45} 
 
 OP THB OLD KIND OF FIELD-BOOK. 
 
 In surveying with the plain table, a. field-book is not used, 
 as every thing is drawn on the table immediately when it is 
 measured. But in surveying with the theodolite, or any other 
 instrument, some kind of a field book must be used, to write 
 down in it a register or account of all that is done and occurs 
 relative to the survey in hand. 
 
 This book every one contrives and rules as he thinks fittest 
 for himself. The following is a specimen of a form which has 
 been formerly used. It is ruled into three columns, as below. 
 
 Here ® 1 is the first station, where the angle or bearing is 
 10r)*» 25'. On the left, at 73 links in the distance or principal 
 line, is an ofi*set of 92 ; and at 610 an offset of 24 to a cross 
 hedge. On the right at 0, or the beginning, an offset 26 to 
 the corner of the field ; at 248 Brown's boundary hedge com- 
 mences ; at 610 an offset 35 ; and at 954, the end of the first 
 line, the denotes its terminating in the hedge. And so on 
 for the other stations- 
 
 A line is drawn under the work, at the end of every station 
 line, to prevent confusion. 
 
 Form of this Field-Book. 
 
 
 Stations, 
 
 
 Offsets and Remarks 
 
 Bearings, 
 
 Offsets and Remarks 
 
 on the left. 
 
 and 
 
 on the right. 
 
 
 Distances. 
 
 
 
 ® 1 
 
 
 
 105° 25' 
 
 
 80 
 
 00 
 
 25 corner 
 
 92 
 
 . 73 
 
 
 
 248 
 
 Brown's hedge 
 
 a cross hedge 24 
 
 610 
 
 35 
 
 
 954 
 
 00 
 
 
 ® 2 
 
 
 
 53«» 10' 
 
 
 house corner 51 
 
 23 
 
 21 
 
 
 120 
 
 29 a tree 
 
 34 
 
 734 
 
 40 a stile 
 
 
 3 
 
 
 
 670 20' 
 
 
 
 61 
 
 35 
 
 a brook 30 
 
 248 
 
 
 
 639 
 
 16 a spring 
 
 foot path 16 
 
 810 
 
 
 <:ross hedge 18 
 
 973 
 
 20 a pond Th( 
 
45t 
 
 tANB 
 
 Then the plan, ob a small scale drawn from the above field- 
 book, will be as in the following figure. But the pupil may 
 draw a plan of 3 or 4 times the size on his paper book. The 
 dotted lines denote the 3 chain or measured lines, and the 
 black lines the boundaries on the right and left. 
 
 But some skilful surveyors now make use of a different 
 method for the field-book, namely, beginning at the bottom 
 of the page and writing upwards ; sketching also a neat 
 boundary on either hand resembling the parts near the 
 measured lines as they pass along ; an example of which will 
 be given further on, in the method of surveying a large estate. 
 
 In smaller surveys and measurements, a good way of setting 
 down the work, is, to draw by the eye on a piece of paper, a 
 a figure resembling that which is to be measured ; and so 
 writing the dimensions, as they are found, against the corres- 
 ponding parts of the figure. And this method may be practis- 
 ed to a considerable extent, even in the larger surveys. 
 
 Another specimen of a field-book, with its plan, is as fol- 
 lows ; being a single field, surveyed with the chain, and the 
 theodolite for taking angles ; which the pupil will likewise 
 draw of a larger size. 
 
 
 e A 
 
 
 
 ©CI' 
 
 
 82*» 65' 
 
 
 
 67° 10' 
 
 
 
 
 
 
 
 35 
 
 268 
 
 
 40 
 
 230 
 
 
 60 
 
 470 
 
 
 48 
 
 672 • 
 
 
 
 
 846 
 
 
 30 
 
 860 
 
 
 30 
 
 1140 
 
 
 
 © B 
 
 
 
 © D 
 
 
 
 130<' 35' 
 
 
 40 
 
 
 
 
 
 
 238 
 
 
 25 
 
 117 
 
 ^ 
 
 20 
 
 520 
 
 
 45 
 
 
 312 
 664 
 
 
 
 . .. , 
 
 
 
 
 
 
 SECTIOir 
 

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 404 
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SURVEYING. 453 
 
 SECTION III. 
 
 OF COMPUTING AND DIVIDING. 
 
 PROBLEM XVI. 
 
 To Compute the Contents of Fields, 
 
 1. Compute the contents of the figures as divided into 
 triangles, or trapeziums, by the proper rules for these figures 
 laid down in measuring ; multiplying the perpendiculars by 
 the diagonals or bases, both in links, and divide by 2 ; the 
 quotient is acres, after having cut oflf five figures on the right 
 for decimals. Then bring these decimals to roods and perches, 
 by multiplying first by 4, and then by 40. An example of 
 which is given in the description of the chain, pag. 429. 
 
 2. In small and separate pieces, it is usual to compute their 
 contents from the measures of the lines taken in surveying 
 them, without making a correct plan of them. 
 
 3. In pieces bounded by very crooked and winding hedges, 
 measured by offsets, all the parts between the offsets are most 
 accurately measured separately as small trapezoids. 
 
 4. Sometimes such pieces as that last mentioned, are com- 
 puted by finding a mean breadth, by adding all the offsets 
 together, and dividing the sum by the number of them, ac- 
 counting that for one of them where the boundary meets 
 the station-line, (which increases the number of them by 1, 
 for the divisor, though it does not increase the sum or 
 quantity to be divided) ; then multiply the length by that 
 mean breadth. 
 
 5. But in larger pieces and whole estates, coosistiDg of 
 many fields, it is the common practice to make a rough plan 
 of the whole, and from it compute the contents, quite inde- 
 pendent of the measures of the lines and angles that were 
 taken in surveying. For then new lines are drawn in the 
 
 fieldi! 
 
454 LAND 
 
 fields on the plan, so as to divide them into trapeziums and 
 triangles, the bases and perpendiculars of which are mea- 
 sured on the plan by means of the scale from which it was 
 drawn, and so multiplied together for the contents. In this 
 way, the work is very expeditiously done, and sufficiently 
 correct ; for such dimensions are taken as afford the most 
 easy method of calculation ; and among a number of parts, 
 thus taken and applied to a scale, though it be likely that 
 some of the parts will be taken a small matter too little, and 
 others too great, yet they will, on the whole, in all probability, 
 very nearly balance one another, and give a sufficiently ac- 
 curate result. After all the fields and particular parts are 
 thus computed separately, and added all together into one 
 sum ; calculate the whole estate independent of the fields, by 
 dividing it into large and arbitrary triangles and trapeziums, 
 and add these also together. Then if this sum be equal to 
 the former, or nearly so, the work is right ; but if the sums 
 have any considerable difference, it is wrong, and they 
 must be examined, and re-computed, till they nearly agree. 
 . 6. But the chief art in computing, consists in finding 
 the contents of pieces bounded by curved or very irregular 
 lines, or in reducing such crooked sides of fields or boun- 
 daries to straight lines, that shall inclose the same or equal 
 area with those crooked sides, and so obtain the area of the 
 curved figure by means of the right-lined one, which will 
 commonly be a trapezium. Now this reducing the crooked 
 sides to straight ones, is very easily and accurately performed 
 in this manner : — Apply the straight edge of a thin, clear 
 piece of lanthorn-horn to the crooked line, which is to be 
 reduced, in such a manner, that the small parts cut off from 
 the crooked figure by it, may be equal to those which are 
 taken in : which equality of the parts included and excluded 
 you will presently be able to judge of very nicely by a little 
 practice ; then with a pencil, or point of a tracer, draw a 
 line by the straight edge of the horn. Do the same by the 
 other sides of the field or figure. So shall you have a straight- 
 sided figure equal to the curved one ; the content of which, 
 being computed as before directed, will be the content of the 
 crooked figure proposed. 
 
 Or, instead of the straight edge of the horn, a horse hair, 
 or fine thread, may be applied across the crooked sides in the 
 same manner ; and the easiest way of using the thread, is to 
 string a small slender bow with it, either of wire or cane, or 
 whale-bone, or such-Hke slender elastic matter ; for the bow 
 keeping it always stretched, it can be easily and neatly ap- 
 plied with one hand, while the other is at liberty to make 
 two marks by the side of it, to draw the straight line by. 
 
 EXAMPLE. 
 
SURVEYING, 4^5 
 
 EXAMPLE. 
 
 Thus, let it be required to find the contents of the same 
 figure as ia Prob. ix, page 4U, to a scale of 4 chains to aa 
 inch. 
 
 A 
 
 B 
 
 C 
 
 Draw the 4 dotted straight lines ab, bc, cd, da, cutting ofl" 
 equal quantities on both sides of them, which they do as near 
 as the eye can judge : so is the crooked figure reduced to an 
 equivalent right-lined one of 4 sides abcd. Then draw the 
 diagonal bd, which, by applying a proper scale to it, measures 
 suppose 1256. Also the perpendicular, or nearest distance 
 from A to this diagonal, measures 456 ; and the distance of c 
 from it, is 428. 
 
 Then, half the sum of 456 and 428, multiplied by the dia- 
 gonal 1256, gives 555152 square links, or 5 acres, 2 roods, 
 8 perches, the content of the trapezium, or of the irregular 
 crooked piece. 
 
 Asa general example of this practice, let the contents be 
 computed of all the fields separately in the foregoing plan ia 
 page 452, and by adding the contents altogether, the whole 
 sum or content of the estate will be found nearly equal to 
 103| acres. Then, to prove the work, divide the whole plan 
 into two parts, b}^ a pencil line drawn across it any way near 
 the middle, as from the corner I on the right, to the corner 
 near s on the left ; then by computing these two large parts 
 separately, their sum must be nearly equal to the former sura, 
 when the work is all right. 
 
 PROBLEM XVU. 
 
 To Transfer a Plan to Another Paper ^ ^c. 
 
 After the rough plan is completed, and a fair one is want- 
 ed ; tkis may be done by any of the following methods. 
 
 First 
 
456 LAND 
 
 First Method, — Lay the rough plan on the clean papei 
 keeping them always pressed flat and close together, by 
 " weights laid on them. Then, with the point of a fine pin oi 
 pricker, prick through all the corners of the plan to be copi- 
 ed. Take them asunder, and connect the pricked points, on 
 the clean paper, with lines ; and it is done. This method 
 is only to be practised in plans of such figures as are small 
 and tolerably regular, or bounded by right lines. 
 
 Second Method. — Rub the back of the rough plan over with 
 black-lead powder ; and lay this blacked part on the cleaa 
 paper on which the plan is to be copied, and in the proper 
 position. Then, with the blunt point of some hard substance, 
 as brass or such-like, trace over the lines of the whole plan ; 
 pressing the tracer so much, as that the black lead under the 
 lines may be transferred to the clean paper : after which, 
 take oflT the rough plan, and trace over the leaden marks 
 with common ink, or with Indian ink — Or, instead of blacking 
 the rough plan, we may keep constantly a blacked paper to 
 lay between the plans. 
 
 Third Method. — Another method of copying plans, is by 
 means of squares. This is performed by dividing both ends 
 and sides of the plan which is to be copied into any conve- 
 nient number of equal parts, and connecting the correspond- 
 ing points of division with lines : which will divide the plan 
 into a number of small squares. Then divide the paper, on 
 which the plan is to be copied, into the same number of 
 squares, each equal to the former when the plan is to be co- 
 pied of the same size, but greater or less than the others, in 
 the proportion in which the plan is to be increased or dimin- 
 ished, when of a different size. Lastly, copy into the clean 
 squares the parts contained in the corresponding squares of 
 the old plan ; and you will have the copy, either of the same 
 size, or greater or less in any proportion. 
 
 Fourth Method. — A fourth method is by the instrument call- 
 ed a pentagraph, which also copies the plan in any size re- 
 quired. 
 
 Fifth Method. — But the neatest method of any, at least in 
 copying from a fair plan, is this. Procure a copying frame 
 or glass, made in this manner ; namely, a large square of the 
 best window glass, set in a broad frame of wood, which can be 
 raised up to any angle, when the lower side of it rests on a 
 table. Set this frame up to any angle before you, facing a 
 strong light ; fix the old plan and clean paper together, with 
 several pins quite around, to keep them together, the clean 
 
 paper 
 
SURVEYING. 457 
 
 paper being laid uppermost, and over the face of the plan to 
 be copied Lay them, with the back of the old plan, on the 
 glass ; namely, that part which you intend to bejj;in at to copy 
 first ; and by means of the light shining through the papers 
 you will very distinctly perceive every line of the plan 
 through the clean paper. In this state then trace all the lines 
 on the paper with a pencil Having drawn that part which 
 covers the glass, slide another part over the glass, and copy it 
 in the same manner. Then another part. And so on, till 
 "the whole is copied. Then take them asunder, and trace 
 all the pencil lines over with a ^ne pen and Indian ink, or 
 with common ink. And thus you may copy the finest plan 
 without injuring it in the least. 
 
 OF ARTIFICERS' WORKS, 
 
 AND 
 
 TIMBER MEASURING. 
 
 I. OF THE CARPENTER'S OR SLIDING RULE. 
 
 THE Carpenter's or Sliding Rule, is an instrument much 
 tised in measuring of timber and artificers' works, both for 
 taking the dimensions, and computing the contents. 
 
 The instrument consists of two equal pieces, each a foot in 
 length, which are connected together by a folding joint. * 
 
 One side or face of the rule, is divided into inches, and 
 eighths, or half-quarters. On the same face also are several 
 plane scales, divided into twelfth parts by diagonal lines ; 
 which are used in planning dimensions that are taken in feet 
 and inches. The edge of the rule is commonly divided de- 
 cimally, or into tenths ; namely, each foot into ten equal 
 parts, and each of these into ten parts again : so that by 
 me£lns of this last scale, dimensions are taken in feet, tenths, 
 and hundredths, and multiplied as common decimal numbers, 
 which is the best way. 
 
 On the one part of the other face are four lines, marked 
 A, B, c, D ; the two middle ones b and c being on a slider, 
 which runs in a groove made in the stock. The same num- 
 bers serve for both these two middle lines, the one being 
 above the numbers, and the other below. 
 
 Vol. i. 59 These. 
 
458 BRICKLAYERS' WORK. 
 
 These fonr lines are logarithmic ones, and the three a, h, 
 c, which are all equal to one another, are double lines, as 
 they proceed twice over from 1 to 10. The other or lowest 
 line, D, is a single one, proceeding from 4 to 40. It is also 
 called the girt line, from its use in computing the contents 
 of trees and timber ; and on it are marked wg at 17- 15, and 
 AG at 18'95, the wine and ale gage points, to make this in*- 
 strument serve the purpose of a gaging rule. 
 
 On the other part of this face, there is a table of the value 
 of a load, or 30 cubic feet, of timber, at all prices, from 6 
 pence to 2 shillings a foot. 
 
 When 1 at the beginning of any line is accounted 1, then 
 the 1 in the middle will be 10, and. the 10 at the end 100 ; 
 but when 1 at the beginning is counted 10, then the 1 in 
 the middle is 100, and the ^0 at the end ioOO ; and so on. 
 And all the smaller divisions are altered proportionally. 
 
 II. ARTIFICERS' WORK. 
 
 Artificers compute the contents of their works by several 
 different measures. As, 
 
 Glazing and masonry, by the foot ; Painting, plastering, 
 paving, &c. by the yard, of 9 square feet : Flooring, 
 partitioning, roofing, tiling, &c. by the square of 100 
 square feet : 
 
 And brickwork, either by the yard of 9 square feet, or by 
 the perch, or square rod or pole, containing 212\ square 
 feet, or 30| square yards, being the square of the rod 
 or pole of 161 feet or 5^ yards long. 
 
 As this number ?72i is troublesome to divide by, the ^ is 
 of\en omitted in practice, and the content in feet divided only 
 by the 272. 
 
 All works whether superficial or solid, are computed by 
 the rules proper to the figure of them, whether it be a triangle, 
 or rectangle^ a parallelopiped, or any other figure. 
 
 III. BRICKLAYERS' WORK. 
 
 Brickwork is estimated at the rate of a brick and a half 
 thick, bo that if a wall be more or less than this standard 
 thickness, it must be reduced to it, as follows : 
 
 Multiply the superficial content of the Wall by the number 
 of half bricks in the thickness, and divide the product by 3. 
 
 The 
 
MASONS' WORK. 469 
 
 The dimensions of a building may be taken by measuring 
 half round on the outside and half round it on the inside ; the 
 sum of these two gives the compass of the wall, to be multi- 
 plied by the height, for the content of the materials. 
 
 Chimneys are commonly measured as if they were solid, 
 deducting only the vacuity from the hearth to the mantle, 
 oq account of the troubl<» of them. All windows, doors, &c. 
 are to be deducted out of the contents of the walls in which 
 they are placed. 
 
 EXAMPLES. 
 
 Exam. 1 . Hour many yards and rods of standard brick-work 
 are in a wall whose length or compass is 57 feet 3 inches, and 
 height 24 feet 6 inches ; the wall being 2^ bricks or 5 half- 
 brirks thick ? Ans. 8 rods, 17f yards. 
 
 Exam. 2. Required the, content of a wall 62 feet 6 inches 
 long, and 14 feet 8 inches high, and 2^ bricks thick ? 
 
 Aiis. 169-763 yards. 
 
 Exam. 3. A triangular gable is raised 17i feet high, on 
 an end wall whose length is 24 feet 9 inches, the thickness 
 being 2 bricks : required the reduced content ? 
 
 Ans. 32-08^ yards. 
 
 Exam. 4. The end wall of a house is 28 feet 10 inches 
 long, and nb feet 8 inches high, to the eaves ; 20 feet high is 
 Q\ bricks thick, other 20 feet high is 2 bricks thick, and the 
 remaining 15 feet 8 inches is H brick thick ; above which is 
 a triangular gaSle, of 1 brick thi<k ; which rises 42 courses of 
 bricks, of which every 4 courses make a foot. What is the 
 whole content in standard measure ? Ans. 253626 yards. 
 
 IV. MASONS' WORK. 
 
 To masonry belong all sorts of stone-work ; and the mea- 
 sure made use of is a foot, either superficial or solid. 
 
 Walls, columns, blocks of j^tone or marble, &c. are mea- 
 sured by the cubic foot ; and pavements, slabs, chimney-pieces, 
 &c. by the sui)erficial or square foot. 
 
 Cubic or solid measure is used for the materials, and square 
 measure for the workmanship. 
 
 In the sohd measure, the true length, breadth, and thick- 
 ness are taken and multiplied continually together. In the 
 superticial, there must be taken the length and breaith of 
 every part of the projection which is seen without the gene- 
 ral upright face of the building. 
 
 EXAMPUES. 
 
46© CARPENTERS' AND JOINERS' WORK. 
 EXAMPLES. 
 
 Exam. 1. Required the solid content of a wall, 63 feet tf 
 inches long, 12 feet 3 inches high, and 2 feet thick ? 
 
 Ans. 1310f feet. 
 
 Exam. 2. What is the solid content of a wall, the length 
 being 24 feet 3 inches, height 10 feet 9 inches, and 2 feet 
 thick? Ans. 621-375 feet. 
 
 Exam. 3. Required the value of a marble slab, at 8s. per 
 foot ; the length being 6 feet 7 inches, and breadth 1 foot 10 
 inches ? Ans. 4/. Is. lO^d, 
 
 Exam, 4- In a chimney-piece, suppose the 
 length of the mantle and slab, each 4 feet 6 inches 
 breadth of both together ;- 3 2 
 length of each jamb - - 4 4 
 breadth of both together - 1 9 
 
 Required the superficial content ? Ans. 21 feet 10 inches. 
 
 V. CARPENTERS' AND JOINERS' WORK, 
 
 To this branch belongs all the wood-work of a house, such 
 as flooring, partitioning, rooting, &c. 
 
 Large and plain articles are usually measured by the 
 square foot or yard, &.c. ; but enriched mouldings, and ^ome 
 other articles, are often estimated by running or iineal mea- 
 sure ; and some things are rated by the piece. 
 
 la measuring of Joists, take the dimensions of one joist, 
 and multiply its content by the number of them ; considtMing ■ 
 that each end is let into the wall about | of the thickness, 
 as it ought to be. 
 
 Partitions are measured from wall to wall for one dimen- 
 sion, and from floor to floor, as far as they extend, for the 
 other. 
 
 The measure of Centering for Cellars is foiind by making a 
 string pass over the surface of the arch for the breadth, and 
 taking the length of the cellar for the length : but in groin 
 centering, it is usual to allow double measure, on account of 
 their extraordinary trouble. 
 
 In Roofing, the dimensions as to length, breadth, and depth, 
 are taken as in flooring joists, and the contents computed the 
 same way. 
 
 In Floor-boarding, take the length of the room for one di- 
 mension, and the breadth for the other, to multiply together 
 for the content. 
 
 For Stair-cases, take the breadth of all the steps, by ma>nng 
 
 a line 
 
CARPENTERS' AND JOINERS' WORk. 461 
 
 a line ply close over them, from the top to the bottom, and 
 multiply the length of this line by the length of a step, for 
 the whole area. — By the length of a step is meant the length 
 of the front and the returns at the two ends ; and by the 
 breadth is to be understood the girts of its two outer surfaces, 
 er the tread and riser. 
 
 For the Balustrade^ take the whole length of the upper part 
 of the hand-rail, and girt over its end till it meet the top of 
 the newel post, for the one dimension ; and twice the length 
 of the baluster on the landing, with the girt of the hand-rail, 
 for the other dimension. 
 
 For Wainscoting^ take the compass of« the room for the 
 one dimension ; and the height from the floor to the ceiling, 
 making the string ply close into all the mouldings, for the 
 other. 
 
 For Doors y take the height and the breadth, to multiply 
 them together for the area. — If the door be panneled on both 
 sides, take double its measure for the workmanship ; but 
 if one side only be panneled, take the area and its half for 
 the workmanship. — For the Surrounding Architrave^ girt it 
 about the uppermost part for its length ; and measure over it, 
 as far as it can be seen when the door is open, for the 
 breadth. 
 
 Window-shutters ^ Bases, &c. are measured in like manner. 
 
 In measuring of Joiners' work, the string is made to ply 
 close into all the mouldings, and to every part of the work 
 over which it passes. 
 
 EXAMPLES. 
 
 Exam. 1. Required the content of a floor, 48 feet 6 inches 
 long, and 24 feet 3 inches broad ? Ans. 1 1 sq. 16} feet. 
 
 Exam. 2. A floor being 36 feet 3 inches long, and 16 feet 
 6 inches broad, how many squares are in it ? 
 
 Ans. 5 sq. 98i feet. 
 
 ExAivr. 3. How many squares are there in 173 feet 10 
 inches in length, ^and 10 feet 7 inches height, of partition- 
 ing ? Ans. 18-3973 squares. 
 
 BxAM. 4. What cost the roofing of a house at 10s. 6d. 
 a square ; the length within the walls being 62 feet 8 inches, 
 and the breadth 30 feet 6 inches ; reckoning the roof f of 
 the flat ? Ans. 121, ns. life/. 
 
 Exam. 
 
462 SLATERS* AND TILERS' WORK. 
 
 Exam. 5. To how much, at 6s. per square jard, amounte 
 the wainscoting of a room ; the height, taking in the cor- 
 nice and mouldings, being 12 feet 6 inches, and the whole 
 compass 83 feet 8 inches ; also the three window- shutters 
 are each 7 feet 8 inches by 3 feet 6 inches, and the door 
 7 feet by 3 feet 6 inches ; the doors and shutlers, being 
 worked on both sides, are reckoned work and half work ? 
 
 Aas. 36/. 12s. 2^d. 
 
 VI. SLATERS' AND TILERS' WORK. 
 
 In these articles, the content of a roof is found by multi- 
 plying the length of the ridge by the girt over from eaves to 
 eaves ; making allowance in this girt for the double row of 
 slates at the bottom, or for how much one row of slates or 
 tiles is laid over another. 
 
 When the roof is of a true pitch, that is, forming a right 
 angle at top ; then the breadth of the building, with its half 
 added, is the girt over both sides nearly. 
 
 In angles formed in a roof, running from the ridge to the 
 eaves, when the angle bends inwards, it is called a valley j 
 but when outwards, it is called a hip- 
 Deductions are made for chimney shafts or window holes. 
 
 EXAMPLES. 
 
 Exam. 1. Required the content of a slated roof, the length 
 being 45 feet 9 inches, and the whole girt 34 feet 3 inches ? 
 
 Ans. 174j^g yards. 
 
 Exam. 2. To how much amounts the tihng of a house, 
 at 2)5. 6d. per square ; the length being 43 feet 10 inches, 
 and the breadth on the flat 27 feet 6 inches ; also the eaves 
 projecting 16 inches on each side, and the roof of a true 
 pitch ? Ans. 24^. 9s. 6-3</. 
 
 VII. PLASTERERS' WORK. 
 
 Plasterers' work is of two kinds ; namely, ceiling, which 
 is plastering on laths ; and rendering, which is plastering on 
 vifails : which are measured separately. 
 
 Th^ 
 
PAINTERS' WORK. 466 
 
 The contents are estimated either by the foot or the yard, 
 or the sfjuare» of 100 feet. Enriched mouldings, Lc. are rate* 
 by mulling or lineal measure. 
 
 Deductions are made for chimneys, doors, windows, &c. 
 
 EXAMPLES. 
 
 Exam, 1. How many yards contains the ceiling which is 43 
 feet 3 inches long, and 26 feet 6 inches broad ? 
 
 Ans. 122i. 
 
 Exam. 2. To how much amounts the ceiling of a room, at 
 10^. per yard ; the lensfth being 21 feet 8 inches, and the 
 breadth !4 feet 10 inches. Ans. 11. 9s. 8fc?, 
 
 Exam. 3. The leni^th of a room is 18 feet 6 inches, the 
 breadth 12 feet 3 inches, and heis^ht 10 feet 6 inches ; to how 
 much amounts the ceiling and rendering, the former at 8d and 
 the latter at Sd. per yard ; allowing for the door of 7 feet by 
 3 feet 8, and a fire-place of 5 feet square ? 
 
 Ans. 1/. 13s. 3^d, 
 Exam. 4. Required the quantity of plastering in a room, 
 the length being 14 feet 5 inches, breadth 13 feet 2 inches, 
 and height 9 feet 3 inches to the under side of the cornice, 
 which girts 8i inches, and projects 5 inches from the wall on 
 the upper part next the ceiling ; deducting only for a door 7 
 feet by 4 ? 
 
 Ans. 63 yards 6 feet 3^ inches of rendering 
 18 6 6 of ceihng 
 
 39 Of i of cornice. 
 
 VIII. PAINTERS' WORK. 
 
 Painters' work is computed in square yards. Every part 
 is measured where the colour lies ; and the measuring line is 
 forced into all the mouldings and corners. 
 
 Wiiviows are done at so much a piece. And it is usual t© 
 allow double measure for carved mouldings, &c. 
 
 EXAMPLES. 
 
 Exam. 1. How many yards of painting contains the room 
 which is 65 feet 6 inches in compass, and 12 feet 4 inches 
 high ? . Ans. 89fi yards. 
 
 Exam. 2, The length of a room being 20 feet, its breadth 
 
 14 feet 
 
464 GLAZIERS' WORK. 
 
 H feet 6 inches, and height 10 feet 4 inches ; how many 
 yards of painting are in it, deducting a fire-place of 4 feet by 
 4 feet 4 inches, and two windows each 6 feet by 3 feet 2 in- 
 ches ? An&. IS-^j yards. 
 Exam. 3. What cost the painting of a room, at 6d. per 
 yard ; its length being 24 feet 6 inphes, its breadth 16 feet .3 
 inches, and height 12 feet 9 inches ; also the door is 7 feet by 
 3 feet 6, and the window-shutters to two windows each 7 feet 9 
 by 3 feet 6 ; but the breaks of the windows themselves are 8 
 feet 6 inches high, and 1 foot 3 inches deep ; including also 
 the window sills or seats, and the soffits above, the dimensions 
 of which are known from the other dimensions : but deduct- 
 ing the fire-place of 6 feet by 5 feet 6 ? 
 
 Ans. 3/. 3s. lO^d- 
 
 IX. GLAZIERS' WORK. 
 
 Glaziers take their dimensions, either in feet, inches and 
 parts, or feet, tenths, and hundredths. And they compute 
 their work in square feet. 
 
 In taking the length and breadth of a window, the cross 
 l»ars between the squares are included. Also windows of 
 round or oval forms are measured as square, measuring them 
 to their greatest length and breadth, on account of the waste 
 in cutting the glass. 
 
 EXAMPLES. 
 
 Exam. 1. How many square feet contains th6 window which 
 
 is 4-25 feet long, and 2-75 feet broad ? Ans. 1 If. 
 
 Exam. 2. What will the glazing a triangular sky -light come 
 
 to, at lOd. per foot ; the base being 12 feet 6 inches, and the 
 
 perpendicular height 6 feet 9 inches ? 
 
 Ans. 11. 15s. Ifd. 
 Exam. 3. Ther^ is a house with three tiers of windows, 
 three windows in each tier, their common breadth 3 fe€t 1 1 
 inches : 
 
 now the height of the first tier is 7 feet 10 inches 
 of the second 6 8 
 
 of the third 5 4 
 
 Required the expense of glazing at I4d. per foot ? 
 
 Ans. 13/. lis. lO^d. 
 Exam. 
 
PLUMBERS' WORK. 465 
 
 RxAM. 4. Required the expense of glazing; the windows 
 ef a hoase at 13i.^a foot ; there being three stories, and three 
 windows in each storjr : 
 
 the height of the lower tier is 7 feet 9 inches 
 of the middie 6 6 
 
 of the upper 5 3^ 
 
 and of an oval window over the door 1 10^ 
 the common breadth of ail the windows being 3 feet 9 inches ? 
 
 Ans, 12L OS, 6d, 
 
 X. PAVERS' WORK. 
 
 Pavers' work is done by the square yard. And the con- 
 tent is found by multiplying the length by the breadth. ' 
 
 EXAMPLES. 
 
 Exam. 1. What cost the paving a foot-jpath, at 3s. 4c?, 
 a yard ; the length being 35 feet 4 inches, and breadth 
 8 feet 3 inches ? ' Ans. 5/. 7s. I Urf. 
 
 Exam. 2. What cost the paving a court, at 3s. 2d. per 
 yard ; the length being 2 7 feet 10 inches, and the breadth 
 14 feet 9 inches ? . Ans. 11. 4s. b^d. 
 
 Exam. 3. What will be the expense of paving a rectan- 
 gular court-yard, whose length is 63 feet, and breadth 45 
 feet ; in which there is laid a foot-path of 5 feet 3 inches 
 broad, running the whole length, with broad stones, at 3s. 
 a yard ; the rest being paved with pebbles at 2s. 6d. a yard ? 
 
 Ans. 40/. 5s. iOic/. 
 
 XI. PLUMBERS' WORK. 
 
 Plumbers' work is rated at so much a pouifd', or else by 
 
 the hundred weight of 112 pounds- 
 Sheet lead, used in roofing, guttering, 5lc. is from 6 to 
 
 10 lb. to the square foot. And a pipe of an inch bore is 
 
 aommonly 13 or 14 lb. to the yard in length. 
 
 EXAMPLES. 
 
 • Exam. 1. How much weighs the lead which is 39 feet 
 Vol.. I. 69 6 inches 
 
466 TIMBER MEASURING. 
 
 6 inches long, and 3 feet 3 inches broad, at Silfe. to the 
 square foot ? Ans. 109lyyb, 
 
 Exam. 2. What cost the covering and guttering a roof 
 with lead, at 18s. the cwt. ; the length of the roof being 43 
 feet, and breadth or girt over it 32 feet ; the guttering 57 
 feet long, and 2 feet wide ; the former 9*831 lb. and the lat- 
 ter 7-37^ ib. to the square foot ? Ans. 116/. 9s, Urf. 
 
 XII. TIMBER MEASURING. 
 
 PROBLEM I, 
 To find the Area^ or Superficial Content^ of a Board or Plank. 
 
 Multiply the length by the mean breadth. 
 
 J\'ote. When the board is tapering, add the breadths at 
 the two ends together, and take half the sum for the mean 
 breadth. Or else take the mean breadth in the middle. 
 
 By the Sliding Rule. 
 
 Set 12 on b to the breadth in inches on a : then against 
 the length in feet on b, is the content on a, in feet and 
 fractional parts. 
 
 EXAMPLES. 
 
 Ej^m. 1. What is the value of a plank, at l^d. per foot, 
 whose length is 12 feet 6 inches, and mean breadth 11 inches ? 
 
 Ans. Is. 5d. 
 Exam. 2. Required the content of a board, whose length 
 is 11 feet 2 inches, and breadth 1 foot 10 inches ? 
 
 Ans 20 feet b inches 8''. 
 Exam. S. What is the value of a plank, which is 12 feet 
 9 inches long, and 1 foot 3 inches broad, at 2irf. a foot ? 
 
 Ans. 3s 3|d. 
 Exam. 4. Required the value of 5 oaken planks at 3d. 
 per foot, each of them being 17| feet long ; and their several 
 breadths as follows, namely, two of 1.3^ inches in the middle, 
 one of 14i inches in the middle, and the two remaining 
 ones, each 18 inches at the broader end, and 1 4 at the nar- 
 rower ? Ans. 1/. 5s. 9id. 
 
 PROBLEM 
 
TIMBER MEASURING. 467 
 
 PROBLEM U. 
 
 To find the Solid Content of Squared or Four- sided Timber, 
 
 Multiply the mean breadth by the mean thickness, and the 
 product again by the length, for the content nearly. 
 
 By the Sliding Rule. 
 
 CD DC 
 
 As length : 12 or 10 : : quarter girt : solidity. 
 
 That is, as the length in feet on c, is to 12 on ©, when 
 the quarter girt is in inches, or to 10 on d, when it is in 
 tenths of feet ; so is the quarter girt on d, to the content 
 on c. 
 
 JVote 1. If the tree taper regularly from the one end to 
 the other ; either take the mean breadth and thickness in the 
 middle, or take the dimensions at the two ends, and half 
 their sum will be the mean dimensions ; which multiplied as 
 above, will give the content nearly. 
 
 2. If the piece do not taper regularly, but be unequally 
 thick in some parts and small in others ; take several diffe- 
 rent dimensions, add them all together, and divide their sum 
 by the number of them, for the mean dimensions. 
 
 EXAMPLES. 
 
 Exam. 1. The length of a piece of timber is 18 feet 6 in- 
 ches, the breadths at the greater and less end 1 foot 6 inches 
 and 1 foot 3 inches, and the thickness at the greater and less 
 end 1 foot 3 inches and 1 foot ; required the solid content ? 
 
 Ans. 28 feet 7 inches. 
 
 Exam. 2. What is the content of the piece of timber, 
 whose length is 24| feet, and the mean breadth and thickness 
 each 1-04 feet ? . Ans. 26^ feet. 
 
 Exam. 3. Required the content of a piece of timber, 
 whose length is 20 38 feet, and its ends unequal squares, the 
 sides of the greater being 191 inches, and the side of the less 
 91 inches ? Ans. 28-7562 feet. 
 
 Exam. 
 
468 TIMBER MEASURING. 
 
 Exam. 4. Required the content of the piece of timber, 
 whose length is 27 -36 feet ; at the greater end the breadth is 
 1-78, and thickness 1 23 ; and at the less end the breadth is 
 1-04, a^d thickness 0'91 feet ? Ans. 41*278 feet. 
 
 PROBLEM III. 
 
 To find the Solidity of Round or Unsquared Timber. 
 
 Multiply the square of the quarter girt, or of ^ of the 
 mean circumference, by the length, for the content. 
 
 By the Sliding Rule. 
 
 As the length upon c : 12 or 10 upon d : : 
 quarter girt, in 12ths or lOths, on d : content on c* 
 
 JVoi^e 1. When the tree is tapering, take the mean dimen- 
 sions as in the former problems, either by girting it in the 
 middle, for the mean girt, or at the two ends, and take half 
 the sum of the two ; or by girting it in several places, then 
 adding all the girts together, and dividing the sum by the num- 
 ber of them, for the mean girt. But v.'hen the tree is very 
 irregular, divide it into seyeral" lengths, and tind the content 
 of each part separately. 
 
 2. Tiiis rule, which is commonly used, gives the answer 
 about i less than the true quantity in the tree, or nearly what 
 the quantity would be, after th(i tree is hewed square in the 
 usual way : so that it seems intended to make an allowance 
 for the squaring of the tree. 
 
 EXAMPLES. 
 
 Exam. 1. A piece of round timber being 9 feet inches 
 long, and its mean quarter girt 42 inches ; what is the con- 
 tent ? Ars. 11 6J feet. 
 
 Exam. 2. The length of a tree is 24 feet, its girt at the 
 thicker end 14 feet, and at the smaller end 2 feet ; required 
 the content ? Ans. 96 feet. 
 
 Exam. 3. What is the content of a tree, whose mean girt 
 is 3-15 feet, and length 14ieet 6 inches ? 
 
 Ans. 8-9922 feet. 
 
 Exam. 4. Required the' content of a tree, whose length 
 is 17i feet, which girts in tive different places as follows, 
 namely, in the tirst place 9-43 feet, in the second 7-92, in the 
 (bird 615. in the fourth 4-74, and in the fifth 3- 16 ? 
 
 Ans. 42-619525. 
 CONIC 
 
[469] 
 
 eONIC SECTIONS. 
 
 DEFINITIONS. 
 
 1. Conic Sections are the figures made by a plane cuttiag 
 a cone. 
 
 2. According to the diflferent positions of the cuttins: plane, 
 there arise five different iigures or sections, namely, a triiangle, 
 a circle, an ellipsis, an hyperbola, and a parabola : the three 
 last of which only are peculiarly called Conic Sections. 
 
 vT 
 
 3. If the cutting plane pass through 
 the vertex of the cone, and any part of 
 the base, the section will evidently be a 
 triangle ; as vab. 
 
 4. If the plane cut the cone parallel t« 
 the base, or make no angle with it, the 
 section will be a circle ; as abd. 
 
 5. The section dab is an ellipse when 
 the cone is cut obliquely through both 
 sides, or when the plane is inclined to 
 the base in a less angle than the side of 
 the cone is. 
 
 6. The section is a parabola, when 
 the cone is cut by a plane parallel to 
 the side, or when the cutting plane and 
 the side of the cone make equal angles 
 with the base. 
 
 7. The 
 
470 
 
 GONIC SECT[0NJS. 
 
 7. The section is an hyperbola, when 
 the cutting plane makes a greater angle 
 with the base than the side of the cone 
 makes. 
 
 8. And if all the sides of the cone 
 be continued through the rertex, form- 
 ing an opposite equal cone, and the 
 plane be also continued to cut the op- 
 posite cone, this latter section will be 
 the opposite hj'perbola to the former ; 
 as dee. 
 
 And further, if there be four cones 
 GMN, COP, CMP. cNo, having all the same 
 vertex c, and all their axes in the same 
 plane, and their sides touching or coin- 
 ciding in the common intersecting lines 
 Mco, NOP ; then if these four cones be 
 all cut by one plane, parallel to the com- 
 mon plane of their axes, there will be 
 formed the four hyperbolas rgq, sft, 
 KVL, HWi, of which each two opposites 
 are equal, and the other two are con- 
 jugates to them ; as here in the annexed 
 figure, and the same as represented in the 
 two following pages. 
 
 9. The Vertices of any section, are the points where the 
 cutting plane meets the opposite sides of the cone, or the sides 
 of the vertical triangular section ; as a and b. 
 
 Hence the ellipse and the opposite hyperbolas, have each 
 two vertices ; but the parabola only one ; unless we consider 
 the other as at an infinite distance.- 
 
 10. The Axis, or Transverse Diameter, of a conic section, 
 is the line or distance ab between the vertices. 
 
 F Hence the axis of a parabola is infinite in length, Ab being 
 only a part of it. 
 
 Ellipse. 
 
DEFINITIONS. 
 
 471 
 
 Ellipse. 
 
 Hyperbolas. 
 
 Parabola. 
 
 11. The Centre c is the middle of the axis. 
 
 Hence the centre of a parabola is infinitely distant from the 
 rertex. And of an ellipse, the axis and centre lie within the 
 curve ; but of an hyperbola, without. 
 
 12 A Diameter is any right line, as ab or de, drawn 
 through the centre, and terminated on each side by the curve; 
 and the extremities of the diameter, or its intersections with 
 the curve, are its vertices. 
 
 Hence all the diameters of a parabola are parallel to the 
 axis, and infinite in length. And hence also every diameter 
 of the ellipse and hyperbola have two vertices ; but of the 
 paraboli, only one ; unless we consider the other as at an in- 
 finite distance. 
 
 13. The Conjugate to ariy diameter, is the line drawn 
 through the centre, and parallel to the tangent of the curve 
 at the vertex of the diameter. So, fg, parallel to the tangent 
 at D, is the conjugate to de ; and hi, parallel to the tangent 
 At A, is the conjugate to ab. 
 
 Hence the conjugate hi, of the axis ab, is perpendicular 
 to it. 
 
 14. An Ordinate to any diameter, is a line parallel to its 
 conjugate, or to the tangent at its vertex, and terminated by the 
 diameter and curve. So dk, el, are ordinates to the axis ab ; 
 and MN, NO, ordinates to the diameter de. 
 
 Hence the ordinates of the axis are perpendicular to it. 
 
 15. An Absciss is a part of any diameter contained be- 
 tween its vertex and an ordinate to it ; as ak or bk, or dn 
 dr EN. 
 
 Hence, in the ellipse and hyperbola, every ordinate has two 
 determinate abscisses ; but in the parabola, only one ; the other 
 vertex of the diameter being infinitely distant. 
 
 16. The Parameter of any diameter, is a third proportion- 
 al to that diameter and its conjugate. 
 
 17. The 
 
472 
 
 CONIG SECTION^: 
 
 17. The Focus is the point in the axis where the ordinate 
 is equal to half the parameter. As k and l, where dk or el 
 is equal to the semi- parameter. 'J he name focus heing given 
 to this point from the pecuHar properly of it mentioned in the 
 corol to theor. 9 in the Ellipse and Hyperbola following, and 
 to theor. 6 in the Parabola. 
 
 Hence, the ellipse and hyperbola have each two foci ; but 
 the parabola only one. «. 
 
 18. If DAK, FBG, be two opposite hyperbolas, having ae 
 for their first or tran*vei^e axis, and ab for their second or 
 conjugate axis. And if dae, fbg, be two other opposite hy- 
 perbolas having the same axes, but in the'contrary order, 
 namely, ab their first axis, and ab their second ; then these two 
 latter curves dae, fbg, are called the conjugate hyperbolas to 
 the two former dae, fbg, and each pair of opposite curves 
 mutually conjugate to the other ; being all cut by one plane, 
 from four conjugate cones, as in page 470, def. 8. 
 
 19. And if tangents be drawn to the four vertices of the 
 curves, or extremities of the axes, forming the inscribed 
 rectangle hikl ; the diagonals hck, icl, of this rectangle, 
 are called the asymptotes of the curves. And if these asymp- 
 totes intersect at right angles, or the inscribed rectangle be 
 a square, or the two axes ab and ab be equal, then the hyper- 
 bolas are said to be right-angled, or equilateral: 
 
 SCHOLIUM. 
 
 The rectangle inscribed between the four conjugate hy- 
 perbolas, is similar to a rectangle circumscribed about an 
 ellipse, by drawing tangents, in like manner, to the four ex- 
 tremities of the two axes ; and the asymptotes or diagonals 
 in the hyperbola, are analogous to those in the ellipse, cut- 
 ting this curve in similar points, and making that pair of 
 conjugate diameters which are equal to each other Also, 
 the whole figure formed by the four hyperbolas, is, as it 
 were, an ellipse turned inside out, cut open at the extre- 
 mities D, E, F, G, of the said equal conjugate diameters, and 
 those four points drawn out to an infinite distance ; the cur- 
 vature being turned the contrary way, hnt the axes, and the 
 rectangle passing through their extremities, continuing fixed. 
 
 ©F 
 
OF THE ELLIPSE. 473 
 
 OF THE ELLIPSE, 
 
 THEOREM I, 
 
 The Squares of the Ordinates of the Axis are to each other 
 as the Rectangles of their Abscisses. 
 
 Let avb be a plane passing through V 
 
 the axis of the cone ; agih another /V 
 
 flection of the cone perpendicular to / \ 
 
 the plane of the former ; ab the axis /^^""y^ 
 
 of this elliptic section ; and fg, hi, or- ^^.JI^JJj 
 
 dioates perpendicular to it. Then it wS^^l^^AjT 
 
 will be, as fg2 : hi* : : af . fb : ah . hb. ■^^^^^^^^^'^^^^ 
 
 For, through the ordinates fg, hi, 
 draw the circular sections kgl, min, 
 
 parallel to the base of the cone, having kl, mn, for their di- 
 ameters, to which fg, hi, are ordinates, as well as to the axii& 
 of the ellipse. 
 
 ]N^ow, by the similar triangles afl, ahn, and bfe, shjS^ 
 
 it is AF : ah : : pl : hn, • 
 and FB : HE : : kf : mh ; 
 
 hence, taking the rectangles of the corresponding termji, 
 it is, the rect. af . fb : ah . hb : : kf . fl : mh . hn. 
 
 3ut, by the circle, kf . fl = fg^, and mh hn = hi* ; 
 Therefore the rect. af . fb : ah . hb ; : f«* : hi*. ^. £. ». 
 
 THEOREM 11. 
 
 As the Square of the Transverse Axis ; 
 Is to the Square of the Conjugate : ; 
 ISo is the Rectangle of the Abs.'.issejS :» 
 To the Square ef their Ordinate. 
 
 ?e». }. 61 . Tkat 
 
4f4 
 
 CONIC SECTIONS. 
 
 That is, ab2 : ab^ or 
 Ac2 : ac^ : : AD . db : de^. 
 
 For, by theor. 1 , ac . cb : ad . db : : ca^ : de^ ; 
 
 But, if c be the centre, then ac . cb =» ac^ , and ca is the 
 
 semi-conjugate. 
 Therefore ac^ : ad . db : : ac^ : de^ ; 
 
 or, by permutation, ac^ : ac^ : : ad . db : de^ ; 
 or, by doubling, ab^ : ab« : : ad . db : de^. q. e. ». 
 
 ab» 
 Corol. Or, by div. ab : • — : : ad . db or ca^ — cd^ : de* , 
 ab 
 that is, AB : 2? : : AD . db or ca^ — cd^ : de^ ; 
 aba 
 where p is the parameter -^-^j by the definition of it. 
 
 AB 
 
 That is. As the transverse. 
 Is to its parameter, 
 So is the rectangle of the abscisses, 
 To the square of their ordinate. 
 
 THEOREM m. 
 
 As the Square of the Conjugate Axis : 
 
 Is to the Square dof the Transverse Axis : : 
 
 So is the Rectangle of the Abscisses of the Conjugate, or 
 the Difference of the Squares of the Semi-conjugate and 
 Distance of the Centre from any Ordinate of that Axis : 
 To the Square of their Ordinate. 
 
 ca3 : cp" 
 
 That is, 
 : : ad . db or ca^—cd* : dE^. 
 
 For, draw the ordinate ed to the transverse ab. 
 Then, by theor. 2, cas : ca^ : : de^ : ad . db ©r ca^ — cd^, 
 or - - - - ca? ; ca3 : : cd^ : cA^—dE^. 
 
 But - - - - ca2 : ca^ : : ca^ : ca^, 
 theref. by subtr. ca^ : ca^ : : ca* —cd^ or ad . db : de^. 
 
 d. E. d. 
 Corol. 
 
OF THE ELLIPSE. 475 
 
 CoroL 1. If two circles be described on the two axes as 
 diameters, the one inscribed within the ellipse, and the other 
 circumscribed about il ; then an ordinate in the circle will 
 be to the corresponding ordinate in the ellipse, as the axia of 
 this ordinate, is to the other axis. 
 
 That is, CA : ca : : dg : de, 
 and CA : CA : : dg : dE. 
 For, by the nature of the circle, ad . db = dg^ ; theref. 
 by the nature of the ellipse, ca^ : ca3 : : ad . db or dg^ : de^, 
 or CA : ca : : dg : de. 
 In like manner - , ca : ca : : dg : ds. 
 
 Also, by equality, - dg : de or cd : : dE or dc : dg. 
 Therefore cgG is a continued strai|ht line. 
 
 Corol. 2. Hence also, A the ellipse and circle are made up 
 of the same number of corresponding ordinates, which are 
 all in the same proportion of the two axes, it follows that 
 the areas of the whole circle and ellipse, as also of any like 
 parts of them, are in the same proportion of the two axes, 
 or as the square of the diameter to the rectangle of the two 
 axes ; that is, the areas of thfi two circles, and of the ellipse, 
 are as the square of each axis and the rectangle of the two"^ 
 and therefore the ellipse is a mean proportional between the 
 two circles. 
 
 THEOREM rV. 
 
 The Square of the Distance of the Focus from the Centre, 
 is equal to the Difference of the Squares of the Semi- 
 axes ; 
 
 Or, the Square of the Distance between the Foci, is equal to 
 the Difference of the Squares of the two Axes. 
 
 That is, cf2 = ca^ — ca^, 
 or Ff3 = ab2 — ab2. 
 
 b 
 
 For, to the focus f draw the ordinate fe ; which, by the 
 definition, will be the semi-parameter. Then by the nature 
 of the curre - - ca^ : ca^ : : ca^ — cf^ : fb^ ; 
 and by the def. of the para, ca^ : ca^ :: ca^ : fe^ ; 
 therefore - - ca^ =ca2 — cf^ ; 
 
 and b) addit. and subtr. cf^ =ca2 — ca* ; 
 or, by doubling, - fP=ab3 —abs ; <i. e. d. 
 
 Corol. 
 
476 CO^tC SECTIONS. 
 
 Corot. 1. The two semi-axes, and the focal distance from 
 the centre, are the sides of a right angled triangle era ; and 
 the distance Fa from the focus to the extremity of the conju- 
 gate axis, is = AC the semi-transverse. 
 
 Corol. 2. The conjugate semi-axis ca is a mean propor- 
 tional between af, fb, or between Af, fe, the distances of ei- 
 ther focus from the two vertices. 
 
 For ca3 s= ca^ -cf^ = (ca -|- cf) . (ca - cf) s= af . fb'V 
 THEOREM V. 
 
 The sum of two Lines drawn from the two Foci to meet at 
 any Point in the Curve, is equal to the Transverse Axis. 
 
 That is, 
 f E -H fe = ^ 
 
 For, draw ag parallel and equal to ca the semi-conjugate ; 
 »nd join cg meeting the ordinate de in h ; also take ci a 4th 
 proportional to ca, cf, cd. 
 
 Then, by theor. 2, ca2 : ag^ : : ca^ — cd^^ : de^ ; 
 
 and, by sim. tri. ca^ : ag2 : : ca^ — cd^ : ag^ — dh^ * 
 
 consequently de^ = ag^ — dh^ =ca2 — dh^. 
 
 Also FD = cf (/3 CD, and fd^ = cf^ — 2cf . cd 4" cd' ; 
 and, by right-angled triangles, fe^ = fd^ -f- de^ ; 
 therefore fe^ = cf^ + ca^ — 2cf . cd -j- cd^ — dh^. 
 
 But by theor. 4, cf2 + ca2 = ca^, 
 
 and by supposition, 2cf . cd= 2ca . ci ; 
 theref. fe^ = ca^ — 2ca . ci -f- cd^ — dh^ 
 
 Again, by supp. ca^ : cd^ : : cf^ or ca^ — ag2 ci^ ; 
 
 and, by sim. tri. ca^ : cd^ : : ca2 — ag2 : cd* •— dh^ : 
 
 therefore - ci2 = cd2 — dh* ; 
 
 consequently fe^ = ca2 — 2ca . ci -f- ci®. 
 
 And the root or side of this square is fe = ca »— ci = ai. 
 
 In the same manner it is found that fE = ca -f- ci = bi. 
 Conseq. by addit. fe -f- fE = ai -|- bi := ab. q. e. p*. 
 
OP THE ELLIPSE. 
 
 477 
 
 €orol. 1. Hence ci or ca 
 
 FE is a 4th proportional to ca, 
 
 OF, CD. 
 
 Corol. 2 And fc — fe = 2oi ; that is, the difference be- 
 tween two lines drawn from the foci, to any point in the curve, 
 is double the 4th proportional to ca, cf, cd. 
 
 Corol. 3. Hence is derived the common method of de- 
 icribing this curve mechanically by points, or with a thread 
 thus : 
 
 In the transverse take the foci f, f, 
 and any point i. Then with the radii 
 Ai, Bi, and centres f, f, describe arcs 
 intersectinj; in e^, which will be a 
 point in the curve. In like manner, 
 assuming other points i, as many 
 other points will be found in the 
 curve. Then with a steady hand 
 the curve line may be drawn through all the points of inter- 
 section E. 
 
 Or, take a thread of the length ab of the transverse axis, 
 and fix its two ends in the foci f, f, by two pins. Then carry 
 a pen or pencil round by the thread, keeping it always stretch- 
 ed, and its point will trace out the curve line. 
 
 THEOREM VL 
 
 Tf from any Point i in the Axis produced, a Line il be 
 drawn touching the curve in one point l ; and the Ordi- 
 nate LM be drawn ; and if c be the Centre or Middle of 
 ab : Then shall cm be to ei as the Square of abi to the 
 Square of ai. 
 
 That is, 
 CM : ci : : am^ 
 
 For, from the point i draw any other line ieh to cut the 
 curve in two points e and h ; from which let fall the perpen- 
 diculars ED and HG ; and bisect dg in k. 
 
 Then, by theo 1, ad . db : ag . gb : de^ : gh^, 
 and by sim. triangles, id^ : ig^ : : de^ : gh^ ; 
 theref. by equality, ad , db : ag . gb : : id* : ig^. 
 
 But DB = CB 4- CD = AC 4- CD = AG -f" DC — CG = 2cK -f AG, 
 
 and GB = CB — CG = ac— cg = ad 4" dc — cg = 2ck 4- ad ; 
 
 theref.. AD . 2ck 4" ad . ag : ag . 2ck 4- ad , ag : : id^ : ig^, 
 
 and,bydiv. DG . 2ck : lo^-^ips ^^y p^ . gjg ; . ad - 2ck 4* 
 
 ad . AG : i»3, or 
 
476 
 
 CONIC SECTIONS. 
 
 or - 2eK : 2iK : : AD . 2ck 4* ad . aq : id^ , 
 
 or AD . 2cK : ad . 2ik : : ad : 2ck -f ad . ag : id^ ; 
 
 theref. by div. ck : ik : : ad ag : id^— ad : ik 
 
 and, by comp ck : ic : : ad ag : id^-^ad . id -f- ia, 
 or - CK : CI : : AD . AG : ai^. 
 
 But, when the line m, by revolving about the point i, comes 
 into the position of the tangent il, then the points e and h 
 meet in the point l, and the points d, k, g, coincide with the 
 point M ; and then the last proportion becomes cm : ci : : am^ : ai^. 
 
 a. E. D* 
 
 THEOREM VII. 
 
 If a Tangent and Ordinate 1?e drawn from any Point in the 
 Curve, meeting the Transverse Axis ; the Semi-transverse 
 will be a Mean proportional between the Distances of the 
 said Two Intersections from the Centre. 
 
 Th^itis, 
 CA is a mean proportional be- 
 tween CD and CT ; 
 or CD, CA, CT, are continued 
 proportionals. 
 
 For,^ by theor. 6, cd : ct : : ad^ : at^. 
 that is, CD : cT : : (ca — cd)2 : (cT— Oa)^, 
 
 or - CD : CT : : cd^ -{-ca^ : ca^ -{- ct^ , ' 
 
 and - CD : DT : : cD^-f-CA^ : ct^ — cd^, 
 or - CD : DT : : cd^ -f ca^ : (ct -{- gd) . dt, 
 
 or - cd2 : CD . dt : : cd^ + ca^ : cd . dx + ct . dt, 
 
 hence - cd^ : ca^ : ; cd . dt : ct . dt, 
 and - cd2 : ca^ : : cd f ct. 
 
 therefore (th. 78, Geom.) c© : ca : : ca : ct. ft. e. d. 
 
 Corol. Since ct is always a third proportional to cd, ca ; 
 if the points d, a, remain constant, then will the point t be 
 constant also ; and therefore all the tangents will meet in 
 this point t, which are drawn from the point e, of every 
 ellipse described on the same axis ab, where they are cut. by 
 the common ordinate dee drawn from the point p. 
 
 THEOREM 
 
OF THE ELLIPSE. 
 
 47S 
 
 THEOREM Vin. 
 
 If there be any Tangent meeting Four Perpendiculars to the 
 Axis drawn from these four Points, namely, the Centre, the 
 two Extremites of the Axis, and the Point of Contact ; those 
 Four Perpendiculars will be Proportionals. 
 
 11 
 
 That is, 
 AG : DE : : CH : bi. 
 
 For, by theor. 7, tc : ac : : ac : dc, 
 theref by div. ta : ad : : Tq : ac or cb, 
 
 and by comp. ta ; td : : tc : tb, 
 
 and by sim. tri. ag : de : : ch : bi. , q; e. ». 
 
 Carol. Hence ta, td, tc, tb > , *• i 
 
 and TG, TE, TH, T, | "« ^^'^ ProportioDaU. 
 For these are as ag, ©e, ch, bi, by similar triangles. 
 
 THEOREM IX. 
 
 If there be any Tangent, and two Lines drawn from the Foci 
 to the Point of Contact ; these two Lines will make equal 
 Angles with the Tangent* 
 
 7e 
 
 That is, 
 the ^ FET = ZfEe. 
 
 For, draw the ordinate de, and fe parallel to fe 
 By cor. 1, theor. 5, ca : cd : : cf : ca ^ — fe, 
 CA : CD : : CT : CA ; 
 CT : cp : : CA ; ca — ^ fe ; 
 TF : Tf : : FE : 2ca — fe or Te by th. 5. 
 TF : Tf : : FE : fe ; 
 
 fE = fe, and conseq. Ze = ^fse. 
 
 FE is parallel to fe, the ^e = Z^et ; 
 Zfet = Zft^G. Q. E. D. 
 
 CQrol. 
 
 and by theor. 7, 
 
 therefore 
 
 and by add. and sub. 
 
 But by sim. tri. 
 
 therefore 
 
 But, because 
 
 therefore the 
 
480 CONIO SECTIONS. 
 
 Corol. As opticians find that the angle of incidence is equal 
 to the angle of reflection, it appears from this theorem, that 
 rays of light issuing from the one focus, and meeting th^ curve 
 in every point, will be reflected into lines drawn from those 
 points to the other focus. So the ray {e is reflected into fe. 
 And this is the reason why the points f, f, are called the/oa, 
 or burning points. 
 
 THEOREM X. 
 
 All the ' Parallelograms circumscribed about an Ellipse arc 
 equal to one another, and each equal to the Rectangle of 
 the two Axes. 
 
 That is, 
 the parallelogram pqrs = 
 the rectangle ab . ab. 
 
 S 
 Let EG, eg, be two conjugate diameters parallel to the sides 
 of the parallelogram, and dividing it into four less and equal 
 parallelograms. Also, draw the ordinates de, de, and ck per- 
 pendicular to PQ ; and let the axis ca produced meet the sides 
 of the parallelogram, produced if necessary, in t and t. 
 Then by theor 7, 
 and 
 
 theref by equality, 
 but, by sim. triangles, 
 theref by equality, 
 and the rectangle 
 Again, by theor. 7, 
 or, by division, 
 and by composition, 
 conseq. the rectangle 
 
 But, by theor. 2, ca^ : ca^ : : (ad . db or) cd^ : de^, 
 
 therefore 
 
 *Corol Because c^^ ■« AD . db — ca* -cd*, 
 therefore ca^ «» gd* 4- cd^. 
 In like maimer, ca^ <= jde^ -f de®. 
 
 Id 
 
 CT 
 
 CA : 
 
 CA : CD, 
 
 ct 
 
 CA : 
 
 CA : cd ; 
 
 CT 
 
 ct : 
 
 cd : CD ; 
 
 CT 
 
 Ct : 
 
 TD : cd, 
 
 TD 
 
 cd : 
 
 CD : CD, 
 
 TD 
 
 DC is 
 
 = the square cd^. 
 
 CD 
 
 CA : 
 
 CA : CT, 
 
 CD 
 
 CA : : 
 
 DA : AT, 
 
 CD 
 
 DB : 
 
 AD : DT ; 
 
 CD 
 
 DT = 
 
 - Cd2 = AD . DB*. 
 
 CA2 
 
 : ca2 
 
 : : (ad . DB or) cd^ 
 
 ca 
 
 ca : : 
 
 cd : DE ; 
 
OF THE ELLIPSE. 
 
 48t 
 
 Iq like manner, 
 
 «r 
 
 But, by theor. 7, 
 
 theref. by equality, 
 
 But, by sim. tri. 
 
 theref. by equality, 
 
 and the rectangle ck . ce = ca . ca. 
 
 But the rect. ck . ce = the parallelogram cepc, 
 
 theref the rect. ca . ca = the parallelogram cEpe, 
 
 eonseq. the rect. ab . ab = the parallelogram pqrs. 
 
 CA 
 
 : ca : 
 
 : CD 
 
 : de, 
 
 ca 
 
 de : 
 
 : CA 
 
 CD. 
 
 CT 
 
 : CA : 
 
 : ca 
 
 CD , 
 
 de. 
 
 CT 
 
 CA : 
 
 : ca 
 
 CT 
 
 CK : 
 
 : ce 
 
 . de 
 
 CK 
 
 CA : 
 
 : ca 
 
 ce, 
 
 THEOREM XL 
 
 The Sum of the Squares of every Pair of Conjugate Diame- 
 ters, is equal to the same constant (Quantity, namely, the 
 Sum of the Squares of the two Axes. 
 
 That is, 
 AB^+ab^ = eg2 -f- eg2 ; 
 where eg, eg, are any pair of con- 
 jugate diameters. 
 
 For, draw the ordinates ed, ed. 
 Then, by cor to theor. 10, ca^ = cd^ -f* cd*» 
 and . - - ca2 = de- -j- de^ ; 
 
 therefore the sum ca^ -\- ca* = cd* + de^ -f. cd^ 
 But, by right-angled As, ce* = cd* -f- de*, 
 and . - - ce* = cd* -f de* ; 
 
 therefore the sum ce* + ce* = cd* + de* -f- cd* 
 consequently - ca* + ca* = ge* -f ce* ; 
 or, by doubling, ab* -j- ab* = eg* -j- eg*. 
 
 + de*. 
 
 -f de* 
 
 THEOREM XU. 
 
 The difference between the Semi-transverse and a Line drawn 
 from the Focus to any point in the Curve, is equal to a 
 Fourth Proportional to the Semi-transverse, the Distance 
 from the Centre to the Focus, and Distance from the 
 Centre to the Ordinate belonging to that Point of the 
 Curve. 
 
 T©fc. I, 
 
 62 
 
 That 
 
48S 
 
 eONIC SECTIONS. 
 
 That is, 
 
 AC FE = CI, or FE = AI ; 
 
 and /e — AC = ci, of/e = m. 
 Where CA i cf : : cd : ci the 4th 
 proportional to ca, cf, cd. 
 
 For, draw ag parallel and equal to ca the semi-conjugate j 
 and join cg meeting the ordinate de in h 
 Then, by theor. 2 ca^ : ag^ : : ca* — cd* : de* : 
 and, by sim. tri. oa* : ag* ; : ca* -— cd* : ag* — dh* ; 
 consequently de* = ag* — dh* = ca* — dh*. 
 Also Fb = CF cc CD. and fd* = cf* — 2gf . cd -f- cd* ; 
 but by right-angled triangles, fd* -{- de* = fe* ; 
 therefore fe* = cf* -f- ca* — 2cf . cd -j- cd* — dh*. 
 But by theor. 4, ca* -f- cf = ca* ; 
 and, by supposition, 2cf . cd = 2ca . ex ; 
 theref. fe* = ca* — 2ca . ci -f- cd* — dh* ; 
 But b}' supposition, ca* : cd* : : cp^ or ca* — ^ ag* : ci* ; 
 and, by sim. tri. ca* : cd* : : ca* — ag* : cd* — dh* j 
 therefore - - ci* = cd* — dh* ; 
 consequently - fe* = ca* — 2ca . ci 4- ci*. 
 And the root or side of this square is fe = ca — ci = ai. 
 In the same manner is found^k: = ca -|- ci = bi. q. e. d, 
 
 ,, THEOREM XIII. 
 
 If a Line be drawn from either Focus, Perpendicular to a 
 Tangent to any I'oint of the curve ; the Distance of their 
 Intersection from the Centre will be equal to the Semi- 
 transverse Axis. 
 
 That is, if fp, fp 
 be perpendicular to 
 the tangent Tpp, 
 then shall cp and 
 cp be each equal to 
 
OP THE ELLIPSE. 483 
 
 For, through the point of contact e draw pe, and /e meet- 
 ing:, FP produced in g. Then, the ^gkp = ^Ifep, being each 
 equal to the Z/k^» and the angles at p being I'ight, and the side 
 PE being common, the two triangles gep, fep are equal in all 
 respects, and so ge = fe, and gp = fp. Therefore, since 
 Fp = ipG, and Fc = ^f/, and the angle at f common, the side 
 cp will be = ^fo or ^ab, that is cp = ca or cb. And in the 
 same manner cp = ca or cb. q. e. d. 
 
 Carol. 1. A circle described on the transverse axis, as a 
 diameter, will pass through the points p, p ; because all the 
 lines cA, cp, cjo, cb, being equal, will be radii of the circle. 
 Corol 2. cp is parallel to/fi, and cp parallel to fe. 
 
 Corol. 3. If at the intersections of any tangent, with the 
 circumscribed circle, perpendicular to the tangent be drawn, 
 they will meet the transverse axis in the two foci. That is, 
 the perpendiculars pf ,/j/'give the foci f,/. 
 
 THEOREM XIV. 
 
 The equal Ordinates, or the Ordinates at equal Distances 
 from the Centre, od the opposite Sides and Ends of an 
 Ellipse, have their Extremities connected by one Right 
 Line passing through the Centre, and that Line is bisected 
 by the Centre. 
 
 That is, if CD = cg, or the ordinate de = gh ; 
 
 then shall ce = ch, and ech will be a right line # 
 
 For when cd = cg, then also is de = gh by th. 1. 
 But the /_D = Zg, bdng both right angles ; 
 therefore the third side ce = ch, and the ^dce == ^gch, 
 and consequently ech is a right line. 
 
 Corol 
 
484 
 
 CONIC SECTIONS. 
 
 CoroL 1. And, conversely, if ech be a right line passing 
 through the centre ; then shall it be bisected by the centre, 
 or have ce = ch ; also de will be = gh, and cd = cg. 
 . Corol. 2. Hence also, if two tang;ents be drawn to the two 
 ends, E, H ©f any diameter eh ; they will be parallel to each 
 other, and will cut the axis at equal angles, and at equal dis- 
 tances from the centre For, the two cd, ca being equal to 
 the two CG, CB, the third proportionals ct, cs will be equal 
 also ; then the two sides ce, ct being equal to the two ch, 
 ■cs, and the included angle ect equal to the included angle hcs,. 
 all the other corresponding parts are equal : and so the 
 ^T= ^s, andiE parallel to hs. 
 
 Corol 3. And hence the four tangents, at the four extrem- 
 ities of any two conjugate diameters form a parallelogram 
 circumscribing the ellipse, and the pairs of opposite sides are 
 each equal to the corresponding paralleL.conjugate diameters. 
 IPor, if the diameter eh be drawn parallel to the tangent te 
 or HS, it will be the conjugate to eh by the definition ; and the 
 tangents to e, h will be parallel to each other, and to the diam- 
 eter EH for the same reason. 
 
 THEOREM XV. 
 
 If two Ordinates ed, ed be drawn from the Extremities e, e 
 of two Conjugate Diameters, and Tangents be drawn to 
 the same Extremities, and meeting the Axis produced in t 
 and R ; 
 
 Then shall cd be a mean proportional between cd, dR, 
 and cd a mean proportional between c©, dt. 
 
 For, by theor. 7, 
 and by the same, 
 theref. by equality, 
 But by sim. tri. 
 theref. by equality, 
 In like manner, 
 
 CD 
 
 CA 
 
 : c A 
 
 CT, 
 
 cd 
 
 CA 
 
 : CA 
 
 CR 
 
 CD 
 
 cd 
 
 : CR 
 
 CT, 
 
 DT 
 
 cd 
 
 : CT 
 
 CR 
 
 CD 
 
 cd 
 
 : cd 
 
 DT. 
 
 cd 
 
 ► CD 
 
 : CD 
 
 dK, 
 
 <l. t. 
 
 Corel 
 
or THE ELLIPSE. 
 
 485 
 
 ct>E = A cde. 
 
 dc 
 
 dB, 
 
 CA 
 
 CT, 
 
 AD 
 
 AT, 
 
 AD 
 
 DT, 
 
 DC 
 
 DB. 
 
 Carol. 1. Hence cd : cd : : ck : ct. 
 Carol. 2. Hence also cd : cd : ; de : db. 
 And the rectangle cd . de = cd . de, or A 
 Coral, 3. Also cd^ = cd . dt, 
 and cd^ = cd . dft. 
 Or cd a mean proportional between cd, dt ; 
 and CD a mean proportional between cd, da. 
 
 THEOREM XVL 
 
 The same Figure being constructed as in the last Theorem 
 each Ordinate will divide the Axis, and the Semi-axis added 
 to the external Fart, in the same Ratio. 
 [See the last fig.] 
 That is, DA : DT : : DC : DB, 
 and dA : dR : : 
 For, by theor. 7, cd : ca : : 
 and by div. cd : ca : : 
 
 and by comp. cd : db : : 
 or, - - - - DA : DT : : 
 In like manner, dA : da : : dc : do. ^. e. d. 
 
 Coral. 1. Hence, and from cor. 3 to the last, it is, 
 
 c82 = CD . dt = ad . DB = CA^— CD2, 
 
 CD^ = cd . dR = Ad . dfi = ca2 —cd^. 
 Carol. 2. Hence also, ca^ =r cd^ -f cd^, 
 and ca2 = de^ -f de^. 
 
 Coral. 3. Further, because ca^ : ca^ : : ad . db or cd^ ; de^, 
 therefore ca : ca : : cd : de. 
 likewise ca : ca : : cd : de. 
 
 THEOREM XVn. 
 
 If from any Point in the Curve there be drawn an Ordinate, 
 and a Perpendicular to the Curve, or to the Tangent at 
 that point : Then, the ^ 
 
 Dist. on the Trans, between the Centre and Ordinate, cd : 
 
 Will be to the Dist. pd : : ^^ 
 
 As Sq of the Trans. Axis : ^ 
 
 To Sq. of the Conjugate. 
 
 That is, X 
 
 CA^ : ca' : : DC : dp. 
 
486 
 
 GONIC SECTIONS. 
 
 For, by theor. 2, ca^ : ca^ : : ad . db : de^ , 
 
 But, by rt. angled As, the rect. td - up = de^ ; 
 
 and, b}^ cor. 1, theor. 16. cd . dt = ad db ; 
 
 therefore - - ca^ : ca^ : : td . dc : td . dp, 
 
 or - - - - ac2 : ca2 : : dc : dp. q. e. d. 
 
 THEOREM XVIII. 
 
 li there be Two Tangents (Jrawn, the One to the Extremity 
 of the Transverse, and the other to the Extremity of any 
 other Diameter, each meeting the other's diameter pro- 
 duced ; the two Tangential Triangles so formed will be 
 equal. 
 
 That is, 
 the triangle get = the 
 triangle can. 
 
 For, draw the ordinate de. Then 
 By sim. triangles, cd : ca : : ce : cn ; 
 but, by theor. 7, cd : ca : ; ca : ct ; 
 theref. by equal, ca : ct : : ce : cn. 
 
 The two triangles cet, can have then the angle e common, 
 and the sides about that angle reciprocally proportional ; those 
 triangles are therefore equal, namely, the A cet = A can. 
 
 Corol. 1. From each of the equal tri. cet, can, 
 take the common space cape, !; 
 
 « and there remains the external A pat =: A pke. 
 
 Corol. 2. Also from the equal triangles cet, can, 
 take the common triangle ced, 
 
 and there remains the A ted = trapez. aned. 
 
 THEOREM XIX. 
 
 The same being supposed as in the last Proposition ; then 
 any Lines kq. qg drawn parallel to the two Tangents, shall 
 also cut off equal Spaced. That is, 
 
 Akqg 
 
OF THE ELLIPSE 
 
 3sr 
 
 4«7 
 
 Akqg = trapez. anhg, 
 and A ity^ =trapez. Atthg. 
 
 For draw the ordinate de. Then 
 The three sim triangles can, cde, cgh, 
 are to each other as ca^, cd^, cg ; 
 
 th. hy div the trap, aned : trap, anhg : : ca^ — cds : ca* — cg*. 
 But, by theor. 1, de^ : gq^ ; . ca2-.cd2 cca^ — cg^, 
 
 theref. by equ. trap, aned : trap, anhg : : de^ : oq,^. 
 
 but, by sim. As, tri. ted : tri. k€ig : : de^ : gq^ j 
 
 theref. by equahty, aned : ted : : anhg : kqg. 
 
 But, by cor. 2, theor. 18, the trap, aned = A ted ; 
 and therefore the trap anhg = A kqg. 
 In like manner the trap. Anhg = A Ky^f. q. e. d. 
 Corol. 1. The three spaces anhg, tehg, kqg are all equal. 
 Corol. 2. From the equals anhg, kqg, 
 take the equals AN/ig, Kqg^ 
 and tliere remains ghna = gq^G» 
 
 Corol. 3. And from the equals ghna^ g^QQ, 
 ' take the common space g^LHo, 
 and there remains the A L€tH = A Lqh, 
 
 Corol, 4. Again from the equals kqg, tehg, 
 take the common space klhg, 
 and there remains telk == A lqh. 
 
 Corol. 5. Aiad when, 
 by the lines kq, gh, 
 moving with a parallel 
 motion, kq comes into 
 the position ir, where 
 CR is the conjugate to 
 CA ; then 
 
 the triangle kqg becomes the triangle irc, 
 and the space anhg becomes the triangle anc ; 
 and therefore the A iRc = A anc = A tec. 
 
 Corol. 6. Also when the lines k^ and nq, by moving 
 with a parallel motion, come into the position ce, Me, 
 
 the 
 
488 
 
 eONIC SECTIONS; 
 
 the triangle lqh becomes the triangle c«m, 
 
 and the space telk becomes the triangle tec ; 
 
 and theref. the A ccm = A tec = a anc = A iRc. 
 
 THEOREM XX. 
 
 Any Diameter bisects all its Double Ordinates, or the Line« 
 drawn Parallel to the Tangent at its Vertex, or to its Con- 
 jugate Diameter. 
 
 That *s. if q,q be parallel „ 
 to the tangent te, or to cc, -^ 
 then shall lq = hq. 
 
 For, draw qa, qh perpendicular to the transverse. 
 Then by cor. 3, theor. 19, the A lqh = A J-qh ; 
 but these triangles are also equiangular ; 
 consequently their like sides are equal, or lq =■ t.q. 
 
 Corol. Any diameter divides the ellipse into two equal 
 parts. 
 
 For, the ordinates on each si4e being equal to each other, 
 and equal in number ; all the ordinates, or the area, on one 
 side of the diameter, is equal to all the ordiiiates, or the area, 
 on the other side of it. 
 
 THEOREM XXI. 
 
 As the Square of any Diameter : 
 Is to the Square of its Conjugate : : 
 So is the Rectangle of any two Abscisses : 
 To the Square of their Ordinate. 
 
 That is, ce2 : cc : : el . lg or ce^ — cl^ : lq^ . 
 
 For, draw the tangent 
 te, and produce the or- 
 dinate QL to the trans- 
 verse at K, Also draw 
 ^H, CM perpendicular 
 to the transverse, and 
 meeting eg in h and m. 
 
 Then similar triangles 
 
 being as the squares of their like sides, it is, 
 
OF THE ELLIPSE. 
 
 48d 
 
 by sim. triangles, A get : A clk : : ce^ : cl^ ; 
 
 or, by divisioa, A c^t ' trap, telk : : ce^ : ce^ — cl^. 
 
 Again, by sim. tri. A ccm ; A lqh : : cc^ ; lq,^. 
 But, by cor. 5 theor. 19, tbe A ccm = A cet, 
 and, by cor. 4 theor, 19, the A lqh = trap, telk ; 
 theref. by equality, ce^ : cc^ : ; ce^ — cl^ : l^^^ 
 or .- - - CE* : cc* : : el . lg : l^^. q. e. d. 
 
 Corol. 1. The squares of the ordinates to any diameter, 
 are to one another as the rectangles of their respective 
 abscisses, or as the difference of the squares of the semi- 
 diameter and of the distance between the ordinate and centre. 
 For they are all in the same ratio of ce^ to ce^ . 
 
 Corol. 2. The above being the same property as that be- 
 longing to the two axes, all the other, properties before laid 
 down, for the axes, may be understood of any two conjugate 
 diameters whatever, using only the oblique ordinates of these 
 diameters, instead'of the perpendicular ordinates of the axes ; 
 namely, all the properties ii theorems 6, 7, 8, 14, 15, 16, 
 18 and 19. 
 
 THEOREM XXU. 
 
 If any Two Lines, that any where intersect each other, meet 
 the Curve each in Two Points ; then 
 The Rectangle of the Segments of the one : 
 Is to the Rectangle of the Segments of the other : : 
 As the Square of the Diam. Parallel to the former : 
 To the Square of the Diam. Parallel to the latter. 
 
 That is, if or and cr, be 
 Parallel to any two Lines 
 PHQ, pnq ; then shall 
 Cft2 : cr2 : : PH . HQ : pH . ny. 
 
 For, draw the diameter che, and the tangent te, and iti 
 parallels pk, ri, mh, meeting the conjugate of the diameter 
 GR in the points t, k, i, m. Then, because similar triangles 
 are as the squares «f their like sides, it is, 
 
 VeL. I. 
 
 63 
 
 bj 
 
490 CONIC SECTIONS. 
 
 by sim. triangles, ch^ : g?^ : : ^ cri : A gfic, 
 
 and - - - cr2 ; gh^ : : /s, chi : A i-hm ; 
 
 theref by division, cr^ : gp^ — gh^ ; : cri : kfhm. 
 
 Again, by <iim. tri. ce^ : ch^ : : A cte : A cmh ; 
 
 and by division, ce^ : ce^ ~— ch^ : : A cte : tehbi. 
 But, by cor. 5 tbeor. 1 9, the A cte ^ A cm, 
 and by cor.'l theor. 19, tehc = kph». or tehm = kphm ; 
 theref. by equ, ce^ : ce^ — ch^ : : ch^ : gp^ — gh* or ph . hq,. 
 In like manner ce^ : ce^ — ch^ : : cH : pu . Hg» 
 Theref. by equ. cr^ : cr^ : : ph . uq. : pu . h^^ q. e. b. 
 
 Corol. 1. In like manner, if any other lines p'b'^', parallel 
 to cr or to pq^ meet phq ; ^Jince the rectangles ph'q, p'u^q' 
 are also in the same ratio of cr2 to cr^ ; therefore rect. 
 PHQ : puq : : ph'q : p'u'^'. 
 
 Also, if another line v'hq.' be drawn parallel to pq or gr : 
 because the rectangles p'/jq' p'hq are still in the same ratio, 
 therefore, in general, the rect. phq, : puq : : phq,' : phq'. 
 
 That is, the rectangles of the parts of two parallel lines, 
 are to one another, as the rectangles of the parts of two other 
 parallel lines, any where intersecting the former. 
 
 Corol. 2. And when any of tbe Jines only touch the curve, 
 instead of cutting it, the rectangles of such become squares, 
 and the general property still attends them. 
 
 That is, 
 cr2 : cr^ : : te^ : tc* , 
 or CR : cr : : te : ie. 
 and CR : cr : : tE : te. 
 
 Corol. 3. And hence te : Te : : tr, : te. 
 
 OF 
 
[ 491 ] 
 
 OF THE HYPERBOLA. 
 
 THEOREM I. 
 
 The. Squares of the Ordinates of the Axis are to each other 
 as the Rectangles of their Abscisses. 
 
 Let AVB be a plape passing 
 through the vertex and axis of 
 the opposite cones ; a»jih ano- 
 ther section of them perpendicu- 
 lar to the plane of the former ; 
 AB the axis of the hyperbolic 
 lections ; and fg, hi, ordinales 
 perpendicular to it. Then it will 
 ho as Fu^ : hi^ : : af, fb : ah.hb. 
 
 F'or, through the oniinates fg, 
 HI, draw the circular sections 
 K. L, Misr, parallel to the ba«<e of 
 . the cone, having kl, m.v, for their 
 diameters, t;> which fg, hi, are ordinates, as well to the axis 
 of the hyperbola. 
 
 ^Jow, by tile similar triangles afl, ahn, and bfk, bhm, 
 it is AF : AH : : fl : hn, 
 and EB : UB : : KF : MH ; 
 
 hence, takin» the rectangles of the corresponding terms, 
 
 it is, the rect. af . fb : kh . hb : : kf . fl : mh , h.n. 
 But, by the cir: l^, kf . fl = fj^, and mh . hn = bi^ ; 
 Therefore the feet, af . fb : ah . hb : ; fg^ : hi^. 
 
 THEOREM H. 
 
 As the Square of the Transverse Axis 
 Is to the Si|uare of the Conjji;ate : 
 So is the Rectangle of the Abscisses 
 To the Square of their Ordinate. 
 
 That is, ab« : ab^ or 
 
 ag2 : ac^ : : AD . db : de^. 
 
 For, 
 
%9t 
 
 CONIC SECTIONS. 
 
 For, by theor. 1 , ac . cb : ad . db : : ca^ : de^ ; 
 
 But, if c be the centre, then ac . cb = ac^, and ca is the 
 
 semi-conj. 
 Therefore - ac^ : ad . db : : ac^ : de^ ; 
 or, by permutation, ac^ : ac^ : : ad . db : de^ ; 
 or, by doubling, ab^ : ab^ : : ad . db : de^ . q. e. d. 
 
 ab3 
 CoroL Or, by Aiv. ab : — ; : ad . db or cd^— ca^ : de^, 
 
 AB 
 
 that is, AB : j9 : : AD . db or cd^ — ca^ : de^ ; 
 ab2 
 where p is the parameter — by the definition of it^ 
 
 AB 
 
 That is, As the transverse, 
 Is to its parameter, 
 So ii? the rectangle of the abscisses. 
 To the square of their ordinate. 
 
 THEOREM III. 
 
 As the Square of the Conjugate Axis : 
 
 To the Square of the Transverse Axis : :. 
 
 The Sum of the Squares of the Semi-conjugate, and 
 
 "Distance of the Centre from any Ordinate of the Axi^ \ 
 
 The Square of their Ordinate. 
 
 That is, 
 ca^ : ca2 ; : ca2 + cd2 dE^ . 
 
 For, draw the ordinate ed te the transverse ab. 
 
 Then, by theor. 1. ca* : ca^ : : de^ : ad . be or cd^ ^ ca^, 
 
 or - - ca2 : ga^ : : cd^ : dE^ — ca^. 
 
 But r - ca2 : ca3 : : ca2 : ca2. 
 
 theref. by compos, ca^ : ca^ : : ca« + cd^ : ds^ . 
 
 In like manner, ca^ : ca^ : : ca^ -f- cd^ : be^. *. e. ». 
 
 Carol. By the last theor. ca^ : ca2 : : cd3-— ca^ : de*, 
 and by this theor. ca^ : ca^ : : cd2-|-ca : ^De^ 
 therefore - de2 : dc^ : : cd^— ca^ icD^-f-CA^. 
 In like manner, dea : dE* : : cd^— ca^ cd^ : -fca*. 
 
 theorem; 
 
OF THE HYPERBOLA. 
 TIffiOREM IV. 
 
 495 
 
 The Square of the Distance of the Focus, from the Centre, ia 
 equal to the Sum of the Squares of the Semi axes. 
 
 Or, the Square of the Distance between the Foci, is equal to 
 the Slim of the Squares of the two Akcs. 
 
 That is, 
 
 cp2 = cas -|- ca3, or 
 Ffa = ab3 -f. ab3 
 
 For, to the focus f draw the ordinate fe ; which, by the 
 definition, will be the semi-parameter. Then, by the nature 
 of the curve - ca^ : ca^ : : cf^ — ca^ : fe^, 
 
 and by the def. of the para, ca^ : ca^ : : ca^ : fe^ ; 
 therefore - - ca^ = cf^ — ca^ j 
 
 and by addition, - cf^ = ca^ -}- ca^ ; 
 
 or, by doubling, - rf^ =AB2 4-ab2. q, e. Do 
 
 Corol. 1. The two semi-axes, and the focal distance from 
 the centre, are the sides of a right-angled triangle CAa ; and 
 the distance Aa is = cf the focal distance. 
 
 Corol. 2. The conjugate semi-axes ca is a mean propor- 
 tional between af, fb, or between Af, fs, the distances of either 
 focus from the two vertices. 
 
 For ca2 = cf^ — ca^ = (cf -f- ca) . (cp — ca) = af . fb- 
 
 THEOREM V. 
 
 The Difference of two Lines drawn from the two Foci, to 
 meet at any Point in the Curve, is equal to the Transverse 
 Axis. 
 
 That H, 
 
 fe FE = AB. 
 
 For, draw ag parallel and equal to ca the semi-conjugate ; 
 and join cg, meeting the ordinate de produced in h : also take 
 ci a 4th proportional to ca, cf, cb. 
 
 Then, 
 
494, CONIC SECTIONS. 
 
 Then by th. 2, ca^ : ag^ : : cd2 — ca^ : de2 ; 
 and, by sim. As» ca^ : ag^ : : cd^ — ca* : dh» — ag j 
 consequently de^ = dh2 — ag^ = dh^ — ca*. 
 Also, FD = CF i/: CD, and fd^ = cf^— 2cf . cd -H cd^ j 
 and, by right-angled triangles, fe^ = fd^ -f- de^ 
 therefore fe^ = cf2 — ca^ — 2cf . cd -f* cd^ 4" dH^v 
 But by theor. 4, cf^ —cas = ca^^ 
 and, by supposition, 2cf . c» = 2ca ci ; 
 theref. fe^ = ca2— . 2ca . ci -f cd^ + dh^ ; 
 Again, by suppos. ca^ : cd^ : ; cf^ or ca^ -{- ag^ : ci^ ; 
 and, by sim. tri. ca^ : cd^ : : ca^ -f- ag^ : cd^ -f* dh* ; 
 
 therefore - ci^ = cd^ -f- dh^ = ch^ ; 
 
 consequently fe^ = ca^ - 2ca . ci,+ ci*. 
 
 And the root or side of tl is square is ff = ci — ca = ai. 
 In the same manner , it it; found that fE = ci + ca = bi. 
 Conseq. by subtract, fs — fe = bi — ai = ab. q,. e. d. 
 
 Corol. 1. Hence ch = ci is a 4th proportional to ca, of, 
 
 CD. 
 
 Corol. 2. And fs -f- fe = 2ch or 2ci ; orFE, ch, fE, are in 
 continued arithmetrcal progression, the common diflference 
 being ca the semi-transverse. 
 
 Corol. 3. Hence is derived the common method of describ- 
 ing this curve mechanically by points, thus ; 
 
 In the transverse ab, produced, take the foci f, f, and any 
 point I. Then with the radii ai, bi, and centres f, f, describe 
 arcs intersecting in e, which will be a point in the curve. In 
 like manner, assuming other points i, as many other points will 
 be found in the curve. 
 
 Then, with a steady hand the curve line may be drawn 
 through all the points of intersection e. 
 
 In the same maner are constructed the other two or conju- 
 gate hyperbolas, using the axis ab instead of ab. 
 
 THEOREM VI. 
 
 If from any Point i in the Axis, a line il be drawn touching 
 the Curve in one point l ; and the Ordinate lu he drawn : 
 and if c be the Centre or the Middle of ab : Then shall cm 
 be to ci as the Square of am to the Square of ai. 
 
 H 
 
 That is, y:^^\ \ 
 
 CM : ci : : am^ : ai^ . J^/Y" ' 
 
 "~ll C lA^DMKG 
 
 For, 
 
OF THE HYPERBOLA. 496 
 
 for, from the point i draw any line ieh to cut the curve in 
 two points E and h ; from which let fall the perps. ed, hg ; and 
 bisect ou in K. 
 
 Then by theor. 1, ad . db : ag . gb : : Dt^ : gh^, 
 and by sim. triangles, id^ : ig^ : : de^ : gh^ ; 
 
 therof. by equality; ad . db : ag gb : : id^ : ig^ ; 
 
 But db = CB H- CD = CB 4- CD = CG -j- CD — AG = 2cK — AG, 
 
 and GB = 4- CG = cA -f- cg = cg + cd — ad = 2ck — ad ; 
 theref ad . 2ck — ad . ag : ag . 2ck— ad . ag : : id^ : ig^, 
 and, by div» dg . 2ck : ig^ — id^ or dg . 2ik : : ad . 2gk 
 
 AD . AG : ID^ . 
 
 or - 2cK : 2ik : : ad . 2ck — ad . ag : id^ ; 
 or AD . 2cK : ad 2ik : : ad . 2ck--ad . ag : id^ ; 
 theref. by div. ck : ik : : ad , ag : ad . 2ik — io^, 
 
 and, by div. ck : ci : : ad . ag : id^ — ad . id -|- ia, 
 or - CK : CI : : ad . ag : ai^. 
 
 But, when the line m, by revolving about the point, i, 
 comes into the position of the tangent il, then the points e 
 and H meet in the point l, and the points d, k, g, coincide 
 with the point m ; and then the last proportion becomes gm : 
 ci : : am2 : ai^. q. e. d. 
 
 THEOREM VII. " 
 
 If a Tangent and Ordinate be drawn from any Point in the 
 Curve, meeting the Transverse Axis ; the Semi-transverse 
 V will he a Mean Proportional between the Distances of the 
 jaid Two Intersections from the centre. 
 
 ^B 
 That is, \\ ^E 
 
 CA is a mean proportional between 
 CD and CT ; or cd, ca, ct, are con- ■ — jg — ^ 
 tinued proportionals. 
 
 \° 
 
 ^ For, by th. 6, cd : ct : : ad^ : at^ , 
 that is, - CD : CT : : (cd — ca)^ : (ca — ct)^, 
 or - - CD : CT : : cd^ -f* ca^ : ca^ -f ct^, 
 and - - CD : DT : : cd^ -f- ca^ : cd^ — ct^, 
 or - - CD : DT : : CD^ -f" ca^ : (cd + ct) dt, 
 or ' CD^ : CD . DT : : cd^ -|- ca^ : cd . dt -f- ct . td ; 
 hence cd^ : ca^ : : cd . dt : ct . td, 
 and cd3 : ca : : cd : ct, 
 
 theref. (th. 78, Geom.) cd : ca : : ca : cd. *ft. e. d.~ 
 
 Carol 
 
496 CONIC SECTIONS. 
 
 CoroL Since ct is always a third proportional to cb, ca ; 
 if the points d, a, remain constant, then will the point t be 
 constant also ; and therefore all the tangents will meet in thii 
 point T, which are drawn from the point e, of every hyper- , 
 bola described on the same axis ab, where they are cut by the 
 common ordinate dfe drawn from the point d. 
 
 tueJorem vm. 
 
 If there be any Tangent meeting Four Perpendiculars to the 
 Axis drawn from these four Points, namely, the Centre, the 
 two Extremities of the Axis, and the Point of Contact ; 
 those Four Perpendiculars Will be Proportionals. 
 
 That is, 
 AC : DE : : CH : Bi. 
 
 ^ Tl 
 
 For, by theor, 7, to : ac : : ac : do, 
 
 theref. by div. ta : ad : : tc ; ac or cb, 
 
 and by comp. ta : td : : tc : tb, 
 
 and by sim. tri. ag : de : : ch : bi. q. 
 
 Carol. Hence ta, td, tc, tb, ? i -a- i 
 
 ^„ 1 „ ' } are also proportionals, 
 
 and tg, te, th, ti, ^ f r 
 
 For these are as ag, de, ch, bi, by similar triangles. 
 
 THEOREM IX. 
 
 If there be any Tangent, and two Lines drawn from the Foci 
 to the Point of Contact ; these two Lines will make equal 
 Angles with the Tangent. 
 
 That is, 
 the Z, fet = j^fEe, 
 
 For, draw the ordinate de, and fe parallel to ee. 
 By cor. 1 , theor. 6, ca : cd : ; cf : ca -f- fe, 
 and by th. 7. <;a : €» : ct : ca ; 
 
 therefore 
 
OF THE HYPERBOLA. 
 
 497 
 
 therefore - ct : cp : : ca : ca -|- fe ; 
 
 and by add. and sub. tp : xf : : fe : 2ca4- fe orfeby th. 6. 
 But by sim. tri. tf : xf : : fe : tE ; 
 
 therefore - fe = fe, and conseq ^e = ^fee. 
 
 But, because fe is parallel to fe, the Ze = Z*»2T ; 
 
 tiierefore the ^fet = i^fEe. q,. e. »• 
 
 Corol. As opticians find that the angle of incidence is equal 
 to the angle of reflexion, it appears, from this proposition, 
 that ray« of light issuing from the one focus, and meeting the 
 curve in every point, will be reflected into lines drawn from 
 the other focus. So the ray fE is reflected into fe. And tliis 
 is the reason why the points f, f, are called /oa, or burning 
 points. 
 
 THEOREM X. 
 
 All the Parallelograms inscribed between the four Conjugate 
 Hyperbolas are equal to one another, and each equal to the 
 Rectangle of the two Axes. 
 
 That is, 
 the parallelogram p^rs = 
 the rectangle ab . ab. 
 
 Let EG, eg be two cogjugate diameters parallel to the sides 
 of the parallelogram, and dividing it into four less and equal 
 parallelograms. Also, draw the ordi«ites de, de, and ck 
 perpendicular to pq ; and let the axis produced meet the sides 
 of the parallelograms, produced, if necessary, in t and t. 
 Then, by theor. 7, ct : ca 
 
 CA : CD, 
 
 ; CA : cd ; 
 
 cd : CD ; 
 
 TD : cd, 
 
 cd : CD, 
 
 and - - ct 
 
 theref by equality ex : ct 
 
 but, by sim, triangles, ex : ct 
 
 theref by equality, xd : cd 
 
 and the rectangle xd : dc is = the square cd^. 
 
 Again, by theor. 7, cd ; ca : : ca : ex, 
 
 or, by division, cd : ca : : da : ax, 
 
 and, by composition, cd : db : : da : dx ; 
 
 eonseq the rectangle cd : dx = cd^ = ad . db*. 
 
 * Carol. Because cJ> = ad . db =3 cd« «^ CA>. 
 
 therefore ca* = cd^ — cd^. 
 In like manner ca* =s de* — dez. 
 Vol. K 64 
 
 But, 
 
4^^ 
 
 CONIC SECTIONS. 
 
 But, by theor. 1, 
 
 eA3 
 
 : ca3 : : (ad . db or) cd* : be», 
 
 therefore 
 
 CA 
 
 : ca : : cd : DE ; 
 
 In like manner, 
 
 CA 
 
 : ca : : CD : de ; 
 
 or - - 
 
 ca 
 
 : de : : ca : cd. 
 
 But, by theor. 7, 
 
 CT 
 
 : ca t: CA : cd ; 
 
 theref. by equality, 
 
 CT 
 
 : CA : : ca : de. 
 
 But, by sim. tri. 
 
 CT 
 
 : CK : ; ce : de ; 
 
 theref by equality, 
 
 CK 
 
 : CA : : ca ; ce. 
 
 and the rectangle 
 
 CK 
 
 . ce r= CA . ca. 
 
 Byt the rect. 
 
 CK 
 
 . ce — the parallelogram cepc, 
 
 theref. the rect. 
 
 CA 
 
 . ce = the parallelogram cepc, 
 
 conseq. the rect. 
 
 AB 
 
 . ab = the paral. p^rs. q. e. 
 
 
 THEOREM XL 
 
 The Diflference of the Squares of every Pair of Conjugate 
 Diameters, is equal to the same constant Quantity, namely, 
 the Difference of the Squares of the two Axes. 
 
 That is, 
 ab2 — ab2 = eg2 — eg3 ; \ d 
 
 where eg, eg are any conjugate 
 diameters. 
 
 For, draw the ordinates ed, ed. 
 Then, by cor. to theor. 10, ca^ = cd^ — cd^, 
 and - - - - ca2 = de^ — de^ ; 
 
 theref. the difference ca« — ca^ = cd^ -f- de* — cd^ — de*. 
 But, by right-angled As, ce* = cd* -{- de* ; 
 
 and - - - - ce* = cd* -f de* ; 
 
 theref. the difference ce^ — ce* = cd* -{- de* — cd* — de 
 consequently - ca* — ca* = ce* — ce* ; 
 or, by doubling, . ab* — ab^ = eg*— eg*. q. e. ». 
 
 THEOREM XU. 
 
 All the Parallelograms are equal which are formed between 
 the Asymptotes and Curve, by Lines drawn Parallel to the 
 Asymptotes. 
 
 That is, the lines ge, ek, ap, aq, 
 being parallel to the asymptotes ch, cl ; 
 then the paral. cgek = paral. cpaq. 
 
 CXQ 
 
OP THE HYPERBOLA. 499 
 
 For, let A be the vertex of the curve, or extremity of the 
 semi-transverse axis ac, perp. to which draw al or a1, which 
 will be equal to the serai-conjugate, by definition 19. Also, 
 draw HEoeh parallel to l1, 
 
 Then, by th'eor. 2, ca^ : al* :: cd^-caS : m^, 
 and, by parallels, ca* : al* ; : cd^ : dh^ ; 
 theref. by subtract, cas : al« z : ga^ : dh? — de or 
 
 rect. HG . Eh ; 
 coQseq. the'square ai:« = the rect. he . Eh. 
 
 But, by sim. tri. pa : al : : ge : eh, 
 and, by the same, qa : a\ : : ek : Eh ; 
 theref. by comp. pa : a^ : al^ : : ge - ek : he . Eh ; 
 and, because al^ = he . Eh, theref. pa A(i=GE.EK. 
 
 But the parallelograms cgek, cpaq, being equiangular, are 
 as the rectangles ge . ek and pa . aq. 
 
 Therefore the parallelogram gk = the paral. pq. 
 
 That is, all the inscribed parallelograms are equal to one 
 another. Q- e. d. 
 
 Corol. 1. Because the rectangle gek or cge is constant, 
 therefore ge is reciprocally as cg, or cg : cp : : pa : ge. 
 And hence the asymptote continually approaches towards the 
 curve, but never meets it : for ge decreases continually as 
 CG increases ; and it is always of some magnitude, except 
 when cG is supposed to be infinitely great, for then ge is 
 infinitely small, or nothing. So that the asymptote cg may be 
 considered as a tangent to the curve at a point infinitely distant 
 fromc. 
 
 Corol. 2. If the abscisses cd, ce, 
 CG, &c. taken on the one asymp- 
 tote, be in geometrical progression 
 increalsing ; then shall the ordi- 
 nates dh, di, gk, &c parallel to / ^'^'^^^L.^I 
 
 the other asymptote, be a decreas- 
 ing geometrical progression, hav- ^ 
 ing the same ratio. For, all the 
 rectangles, cdh, cei, cgk, kc. being equal, the ordinates dh, 
 EI, gk, &c. are reciprocally as the abscisses, cd, ce, cg, kc» 
 which are geometricals. And the reciprocals of geornetricals 
 are also geometricals, and in the same ratio, but decreasing, or 
 in converse order. 
 
 THEOREM 
 
50t CONIC SECTIONS. 
 
 THEQREM XIII. 
 
 The three following Spaces, between the Asymptotes and 
 the Curve, are equal ; namely, the Sector or 'J rihnear 
 Space contained by an Arc of the Curve and two Radii, 
 or Lines drawn from its Extremities to the Centre ; and 
 each of the two (Quadrilaterals, contained by the said Arc, 
 and two Lines drawn from its Extremities parallel to one 
 Asymptote, and the intercepted Part of the other Asymp- 
 tote. 
 
 That is. 
 The sector cae = paeg == qaek, 
 all standing on the same arc ae. 
 
 For, by theor. 12, cpaq = cgek ; 
 subtract the conimon space CGiq, 
 there remains the paral. pi = the par. ik ; 
 to each add the trilineal iae, then 
 the sum is the quadr. paeg = qaek. 
 
 Again, from the quadrilateral caek 
 take the equal triangles caq, cek, 
 and there remains the sector cae = qaek. 
 Therefore cae = qaek = paeg. ^. e. ©,. 
 
 THEOREM XIV. 
 
 The Sum or Difference of the Semi-transverse and a Line 
 drawn from the Focus to any 1 oint in the Curve, is equal 
 to a Fourth Proportional to the Semi-transverse, the Dis- 
 tance from the Centre to the Focus, and the distance from 
 the Centre to the Ordinate belonging to that Point of the 
 Curve. 
 
 That is, 
 FE -}- AC = CI, or fe == ai; 
 and /e — AC = ci, or /k = bi. 
 Where ca : cf : : cd : ci the 
 4th propor. to ca, cf, cd. 
 
OF THE HYPERBOLA. 601 
 
 Pop, draw ag parallel and equal to ca the semi-conjugate ; 
 and join cc meeting the ordinate de produced in h. 
 Then, by theor. 2, ca^ : ag2 : : cd^ — ca^ : de2 ; 
 and, by sim. As, ca^ : ag^ : : cd^ — ca^ : dh^ — ag^ ; 
 consequently de^ = dh- — ag^ = dh^ — ca^. 
 
 Also FD = CF GO CD, and fd^ = cf^ — 2cf . cd -f cd^ ; 
 but, by right angled triangles, fd^ -\- de^ == fe^ ; 
 therefore fe^ =cf2 — ca^ ,— 2cf . cd -|- cd^ -j- dh^. 
 
 But by theor. 4, cf^ — ca^ = ca^, 
 and, by supposition, 2cf . cd = 2ca . ci ; 
 theref. fe^ =ca2 — 2ca . ci -{- cd^ -|- dh^. 
 
 But, by supposition, ca^ : cd^ : : cf^ or ca^ -\- ag^ : ci^ ; 
 and, by sim. As, ca^ : cd^ : : ca^ -f- ag^ : cd^ -|- dh^ ; 
 therefore - ci^ =cd2 -|- dh^ =ch2 ; 
 
 consequently - fe2=ca2 — 2ca . ci + ci^. 
 And the root or side of this square is fe =ci — ca = ai. 
 In the same manner is found /e = ci -|- ca = bi. q. e. d. 
 
 Corol. From the demonstration it appears, that de^ = 
 dh^ — Ao2 = dh^ — ca. Consequently dh is every where 
 greater than de ; and so the asymptote cgh never meets 
 the curve, though they be ever so far produced : but dh and 
 de approach nearer and nearer to a ratio of equality as they 
 recede farther from the vertex, till at an infinite distance they 
 become equal, and the asymptote is a tangent to the curve at 
 an infinite distance from the vertex. 
 
 THEOREM XV. 
 
 If a Line be drawn from either Focus Perpendicular to a 
 Tangent to any Point of the curve ; the Distance of their 
 Intersection from the Centre will be equal to the, Semi- 
 transverse Axis. 
 
 That is, if ep, fp be perpen- 
 flicular to the tangent tp/), then 
 shall cp and cp be each equal 
 
 to CA or CB. 
 
602 CONie SECTIONS. 
 
 For, through the point of contact e draw fe and /e, meet- 
 ing FP produced in g. Then, the ^gep = ^kep, being each 
 equal to the Z/Ep, and the angles at p being right, and the 
 side PE being common, the two triangles gep, fep are equal 
 in all respects, and so ge = fe, and gp = fp. Therefore, 
 since fp =iFG, and fc = 1f/, and the angle at f common, 
 the side cp will be = ^/g or \ab. that is cp = ca or cb. ,, 
 
 And in the same manner cp = ca of.cb. q. e. d. * 
 
 Corol. 1. A circle described on the transverse axis as a 
 diameter, will pass through the points p, p ; because all the 
 lines, CA, cp, cp, cb, being equal, will be radii of the circle. 
 
 Corol. 2. cp is parallel to/E, and cp parallel to fe. 
 
 Corol. 3. If at the intersections of any tangent, with the 
 circumscribed citcle, perpendiculars to the tangent be drawn, 
 they will meet the transverse axis in the two foci. That is, 
 the perpendiculars pf, pfgive the foci f,/. 
 
 THEOREM XVI. 
 
 The equal Ordinates, or the Ordinates at equal Distances 
 fr. m the Centre, on the opposite Sides and Ends of an 
 Hyperbola, have their Extremities connected by one Right 
 Line passing through the Centre, and that Line is bisected 
 by the Centre. 
 
 That is, if cd = cg, or the 
 ordinate dk = gh ; then shall 
 CE = cH, and ech will be a 
 right line. 
 
 For, when cd = cg, then also is de == gh by cor. 2 theor. 1, 
 But the Zd = Zg, being both right angles ; 
 therefore the third side ce,= ch, and the Zdce = Z^CH, 
 and consequently ech is a right line. 
 
 Corol. 1. And, conversely, if ech be a right line passing 
 through the centre ; then shall it be bisected by the centre, 
 or have ce = ch ; also de will be = gh, and cd = cg. 
 
 Corol. 2. Hence also, if two tangents be drawn to the two 
 ends e, h of any diameter eh ; they will be parallel to each 
 
 other, 
 
OF THE HYPERBOLA. 
 
 503 
 
 other, and will cut the axis at equal angleS) and at equal dis- 
 tances from the centre. For, the two cd, ca being equal to 
 the two CG, CB, the third proportionals ct,' cs will be equal 
 also ; then the two sides ce, ct being equal to the two ch, cs, 
 and the included angle ect equal to the included angle hcs, 
 all the other corresponding parts are equal: and so the ^t 
 . = ^s, and TE parallel to hs. 
 
 Corol. 3. And hence the four tangents, at the four ex- 
 tremities of any twe conjugate diameters, form a parallelogram 
 inscribed between the hyperbolas, and the pairs of opposite 
 sides are each equal to the corresponding parallel conjugate 
 diameters. — For, if the diameter eh be drawn parallel to the 
 tangent te or hs, it will be the conjugate to eh by the defi- 
 nition ; and the tangents to eh will be parallel to each other, 
 and to the diameter eh for the same reason. 
 
 THEOREM XVn. 
 
 If two Ordinates ED, ed be drawn from the Extremities e, c, 
 of two Conjugate Diameters, and Tangents be drawn to the 
 same Extremities, and meeting the Axis produced in t 
 and R ; 
 
 Then shall cd be a mean Proportional between cd^ Jr, 
 and cd a mean Proportional between cd, dt. 
 
 For, by theer. 7, cd : ca : : ca : ct, 
 ■ and by the same, cd! : ca : : ca : cr j 
 
 theref. by equahty, cd : cd : : cK : ct. 
 
 But by sim. tri. dt : cd : : ct : cr ; 
 
 the ef. by equality, cd : cd : : cd : dt. 
 
 In like manner, cd : cb : : cd : dK. 
 
 Corol. 1, Hence cd : cd : : CK : ct,. 
 Corol. 2. Hence also cd : cd : : de : de. 
 and the XQct. cd . de = cd . de, or A cde = A cde. 
 Corol, 3. Also cd^ = cd . dt, and cd^ = cd . dK. 
 
 Or cd a mean proportional between cd, dt ; 
 
 and CD a mean proportional between cdy dn. 
 
 Q. £. d. 
 
 THEOREM 
 
504 CONIC SECTION^^. 
 
 THEOREM XVIIL 
 
 The same Figure being constructed as in the last Proposition, 
 each Ordinate will divide the Axis, and the Semi-axis add- 
 ed to the external Part j^n the same Ratio. 
 
 [See the last fig.] 
 
 That is, DA 
 
 : DT : : dc 
 
 : DB, 
 
 
 and ^A 
 
 : dR : : dc 
 
 : dB. 
 
 
 For, by theor. 7, 
 
 CD : CA : : 
 
 CA : 
 
 CT, 
 
 and by div. 
 
 CD : CA : : 
 
 AD : 
 
 AT, 
 
 and by comp. 
 
 CD : DB : : 
 
 AD 
 
 DT, 
 
 or 
 
 DA :' DT : : 
 
 DC 
 
 DB. 
 
 In like manner, 
 
 (Ia : dR : : 
 
 dc 
 
 dB. 
 
 Corol. 1. Hence, and from cor. 3 to the last prop, it is, 
 
 cd^ = CD . DT = AD . DB = CD^ CA^ , 
 
 and cd . dR = Ad . ds = ca^ — cd^ . 
 Corol. 2. Hence also ca^ = cd^ — cri^^andca^ =c?e2— de^. 
 Corol. 3. Farther, because ca^ : ca^ : : ad . db or cd^ : be^. 
 therefore ca : ca : : cd : de. 
 likewise ca : ca : : cd : de. 
 
 THEOREM XIX. 
 
 If from any Point in the Curve there be drawn an Ordinate, 
 and a Perpendicular to the Curve, or to the Tangent at that 
 Point: Then the 
 
 Dist. on the Trans, between the Centre and Ordinate, cd : 
 
 Will be to the Dist. pa : : 
 
 As Square of Trans. Axis : 
 
 To Square of the Conjugate. 
 
 That is, 
 CA^ : ca^ : : DC : dp. 
 
 For, by theor. 2, ca^ : ca^ : : ad . db : de^ , 
 
 But, by rt. angled As? the rect. t» . bp = de^ ; 
 
 and, by cor. 1 theor. 16, cd . dt = ad . db ; 
 
 therefore - - ca^ : ca^ : : td . dc : td . dp, 
 
 or - - . _ cA^ : ca^ : : do : dp q. e. d. 
 
 THEOREM 
 
OF THE HYPERBOLA. 
 
 605 
 
 TUCO^EM XX. 
 
 ff there be Two Tangents drawn, the One to the Extremity 
 of the Transverse, and the other to the Extremity of any 
 other Diameter, each meeting the other's Diameter pro- 
 duced ; the two Tangential Tri?ingles so formed, will he 
 equal. 
 
 That is, 
 the triangle get = 
 the triangle can 
 
 For, draw the ordinate de. Then 
 By sim. triangles, cd : ca : : ce : cn ; 
 bat, by theor. 7, cd : ca : : ca : ct ; 
 theref. by equal, ca : ct : : ce : cn. 
 The two triansjles cet, cav have then the angle c common, 
 and the fiides about that angle reciprocally proportional : those 
 triangles are therefore equal, viz. the A cet = can. q. k. b. 
 Corol. 1. Take each of the equal tri. qkt, can, . 
 from ttie common space cape, 
 and -there remains the external A pat = ^ ^M^' 
 Corol. 2. Also take the equal triangles cet, can, 
 from the common triangle ced, 
 
 and there remains the A ted = trapez. aned. 
 
 THEOREM XXL 
 
 The same being supposed as in the last Proposition ; then 
 any Lines s^, gq, drawn parallel to the two Tangents » 
 shall also cot o^ equal Spaces. 
 
 That is, 
 the A K^G = trapez. anhg. 
 the A Kqg =; trapez. Ashg, 
 
 
 
 For, draw the ordinate de. Then 
 
 The three sim. triangles can, cde, cgh, " 
 
 Vol. I. 6* ar« 
 
b06 CONIC SECTIONS. 
 
 are to each other as ca^ , cd^ , cg^ ; 
 th. by dlv. the trap, aned : trap, anhg : : cd^ — ca^ : cg^ — ca- , 
 But, by theor. 1, de^ : cq,^ : : cd^ — ca^ : cg^ — ca» ; 
 
 theref. by equ. trap, aned : trap, anhg : : de* : gq,2. 
 But, by sim. As, tri. ted : tri. kqg : : de^ : gq* ; 
 theref. by equal. aned : ted : : anhg : k^g. 
 
 But, by cor. 2 theor. 20, the trap, aned = A ted ; 
 
 and therefore the trap. anhg = A k^g. 
 
 In like manner the trap. an^|: = A Kgg. q. e. d. 
 
 Corol. 1. The three spaces anhg, tehg, kqg are all 
 equal. 
 
 Corol. 2. From the equals anhg, k^g, 
 take the equals ANhg^ Kqg, 
 and there i-emains ghuG = gq<iG, 
 
 Corol. 3. And from the equals ghuG, gq^a, 
 take the common space gqLHGy 
 and there remains the A i-ftH = Lqh, 
 
 Corol. 4. Again, from the equals kqg, tehg, 
 take the common space klhg, 
 and there remains telk = A l^h. 
 
 Corol: 5. And when, by 
 the lines kq, gh, moving 
 with a parallel motion, kci 
 comes into the position ir, 
 where cr is the conjugate to 
 CA : then 
 
 the triangle kqg becomes the triaagle irc, 
 and the space anhg becomes the triangle anc ; 
 and therefore the A iRc = A anc = A tec. 
 
 Corol. 6. Also when the lines kq, and hq, by moving with 
 a parallel motion, come into the position cc, Me, 
 the triangle lqh becomes the triangle ccm, 
 and the space telk becomes the triangle tec ; 
 and theref. the A ceM = A tec = A anc = A iRc. 
 
 THEOREM XXIL 
 
 Any Diameter bisects all its Double Ordinates, or the Lines 
 drawn Parallel to the Tangent at its Vertex, or to its Con*, 
 iugate Diameter. 
 ^ ^ Tbat 
 
OF THE HYPERBOLA 
 
 That is, if q,q be paral- 
 lel to the taagent te, or 
 to ce, then shall lq = Lq. 
 
 For, draw qh, qh perpendicular to the transverse. 
 
 Then by cor 3 theor. 21, the A l^h = A i^qh ; 
 
 but these triangles are also equiangular ; 
 
 conseq. their like sides are equal, or lq = Lq. 
 
 Corol. 1. Any diameter divides the hyperbola into two 
 equal parts. 
 
 For, the ordinates on each side being equal to each other, 
 and equal in number ; all the ordinates,. or the area, on one 
 side of the diameter, is equal to all the ordinates, or the area, 
 on the other side of it. 
 
 Corol. 2. In like manner, if the ordinate be produced to 
 the conjugate hyperbolas at q', q , it may be proved that 
 L€i' = L^'. Or if the tangent te be produced, then ev = ew. 
 Also the diameter gceh bisects all lines drawn parallel to te 
 or Qy, and limited eithef by one hyperbola, or by its two con- 
 jugate hyperbolas. 
 
 THEOREM XXni. 
 
 As the Square of any Diameter : 
 Is to the Square of its Conjugate : : 
 So is the Rectangle of any two Abscisses : 
 To the Square of their Ordinate. 
 That is, ce2 : ce^ ; : el . lg or cl^ — ce^ 
 For, draw the tangent 
 te, and produce the ordi- 
 nate €tL to the transverse 
 at K Also draw qh, cm 
 perpendicular to the trans- 
 verse, and meeting eg in 
 H and M. Then similar 
 triangles being as the 
 squares of their like sides, 
 it is, 
 
 by 
 
60S CONfC SEGTIOKS. 
 
 by sim. triangles, A cet : A clk : : ce^ i cl^ ; 
 
 or, by divJsioa, A cet : trap, telk : ; ce^ : cl^—ce*. 
 
 Again, by sim. tri. . A cea : A lqh : : ce^ : l^^. 
 
 But, by cor. 5tbeor 21, the A ccm = A cet, 
 
 and, by cor. 4 tbeor. 21, the A lqH= trap, telk ; 
 
 theref. by equality, ce^ ; ce* :: cl* —ce^ : l^^, 
 
 or - - - CE^ : cc* : : el . lg : lq,* Q. e. t). 
 
 Corol 1. The squares of the ordinates to any diameter, 
 are to one another as the rectangles of their respective ab- 
 scisses, or as the difference of the squares of the semi-diame- 
 ter and of the distance between the ordinate and centre. For 
 they are all in the same ratio of cfi* to ce^ . 
 
 CoroL 2. The above being the same property as that be- 
 longing to the two axes, all the other properties before laid 
 down, for the axes, may be understood of any two conjugate 
 diameters whatever, using only the oblique ordinates of these 
 diameters instead of the perpendicular ordinates of the axes ; 
 namely, all the properties in theorems 6, 7, 8, 16, 17, 20, 21. 
 
 Corol. 3. Likewise, when the ordinates are continued to 
 the conjugate hyperbolas at q,', q\ the same properties still 
 ohtain, substituting only the sum for the difference of the 
 squares of ce and cl, 
 
 That is, ce2 : cea : : cl^ + ce^ : lq'3 . 
 
 And 80 L^3 : jt^'s : ; Qj^a __ q^s . cl^ -f- ces . 
 
 Corol. 4. When by the motion of lq' parallel to itself, that 
 line coincides with ev, the last corollary becomes 
 ces : ce2 : : gcE^ : ev^, 
 or cc3 : Ev2 : : 1 : 2, 
 orce : EV : : 1 :^2, 
 or as the side of a square to its diagonal. 
 That is, in all conjugate hyperbolas, and all their diameters, 
 any diameter is to its parallel tangent, in the constant ratio of 
 the side of a square to its diagonal. 
 
 THEOREM XXIV. 
 
 If any Two Lines, that any where intersect each other, meet 
 the Curve each in Two Points ; theft 
 The Rectangle of the Segments of the one : 
 Is to the Rectangle of the Segments of the other : : 
 As the Square of the Diam. Parallel to the former : 
 To the Square of the Diam. Parallel to the latter. 
 
 TK«t 
 
OF THE HYPERBOLA, 
 
 That is, if cr and 
 Cr be parallel to any 
 two lines vhq, p^q ; 
 then shall cr^ : cr* : : 
 FH . Hq : pu, Hq. 
 
 For, dravv the diameter che, and the tangent te, and its 
 parallels pc, ri. mh, meeting the conjugate of the diameter 
 CR in the points t, k, i. m. Then because similar triangles 
 are as the squares of their like sides, it is, 
 
 by sim. triangles, cr* : gp* : : A cri : Agpk, 
 
 and - - CR* : gh* : : A cri : Aghm ; 
 
 theref. by division, cil* : gp* — gh* : : cri : kphm. 
 
 Again, by sim. tri. ce* : ch* : : Acte : Acmh ; 
 
 and by division, ce* : ch* — ce* : : Acte : tehm. 
 
 But, by cor. 6 theor. 21, the A cte = A cir, 
 
 and by cor. 1 theor. 21, tehg = kphg, or tehm = kphm ; 
 
 theref. by equ. ce* : ch* — ce* : : cr* gp* — gh* or ph . hq. 
 
 In like manner ce* : ch* — ce* ; : cr* : joh . h^. 
 
 Theref. by equ. cr* : cr* : : ph . na •* p^ > Hy. q. e. d. 
 
 Corol. 1. In like manner, if any other line p h' q, parallel 
 to cr or to pq, meet phq ; since the rectangles ph'q, p h q' 
 are also in the same ratio of cr* to cr* ; therefore the rect. 
 PHQ ; ptLq : : PH'Ci : pviq. 
 
 Also, if another line p'^a' be drawn parallel to pq or cr ; 
 because the rectangles p'/iq' plicf are still in the same ratio, 
 therefore, in general the rectangle ph^ : pviq : : p'/ia' : ph^ . 
 That is, the rectangles of the parts of two parallel lines, are 
 to one another, as the rectangles of the parts of two other 
 parallel lines, any where intersecting the former. 
 
 CoTol. 2. And when any of the lines only touch the curve, 
 instead of cutting it, the rectangles of such become squares, 
 and the general property still attends them. 
 
 That 
 
510 
 
 CONIC SECTIONS. 
 
 That is, 
 
 cr2 : cr2 : : te^ : tc^, 
 
 or CR : cr : : TE : tCj 
 
 and CR : cr : : tE : te. 
 
 Carol. 3. And hence te : tc : : iE, : te. 
 
 THEOREM XXV. 
 
 If a Line be drawn through any Point of the Curves, Parallel 
 to either of the Axes, and terminated at the Asymptotes ; 
 the Rectangle of its Segments, measured from that Point, 
 will be equal to the Square of the Semi-axis to which it is 
 parallel. 
 
 That is, 
 the rect. hek or hck = ca^ , 
 aad rect. hEk or hek = ca^ . 
 
 For, draw al parallel to ca, and ol to ca. Then 
 ' by the parallels, ca^ : ca^ or al^ : : cd^ : dh^ ; 
 
 and by the«>r. 2, ca^ : ca^ : : cd^ — ca^ : de^ ; 
 
 theref. by subtr. ca^ : ca^ : : ca^ : dh^ — de^ or hek. 
 
 But the antecedents ea^ , ca^ are equal, 
 
 theref. the consequents ca^ ^ hek must also be equal. 
 In like manner it is again, 
 
 by the parallels, ca^ : ca^ or al^ : : cd^ : dh^ ; 
 
 and by theor. 3, ca^ : ca^ : : cd^ -|- ca^ : dc ^ ; 
 
 theref. by subtr. ca^ : ca^ : : ca^ : dc^ — dh^ or hck. 
 But the antecedents ca^ , cas are the same, 
 theref. the conseq. ca^ , hck must be equal. 
 In like manner, by changing the axes, is fiEk or hek = ca^. 
 
 Corol. 1. Because the rect. hek = the rect. hck. 
 therefore eh : ch : : ck : ek. 
 And consequently he : is always greater than ue. 
 Corol. 2. The rectangle ^ ek = the rect. uEk. 
 For, by sim. tri. eH : eh i : zk : ek. 
 
 SCHOLIUM» 
 
OF THE HYPERBOLA. 511 
 
 SCHOLIUM. 
 
 It is evident that this proposition is general for any Kne ob- 
 lique to the axis also, namely, that the rectangle of the seg- 
 ments of any line, cut by the curve, and terminated by the 
 asymptotes, is equal to the square of the s^rui-diameter to 
 which the line is parallel. Since the demonstration is drawn 
 from properties that are common to all diameters. 
 
 THEOREM XXVI. 
 
 All the rectangles are equal which are made of the Segments 
 of any Parallel Lines cut by the Curve, and hmited by the 
 Asymptotes. 
 
 That is, 
 therect. hek = hck. 
 and rect. hsk = hek. 
 
 For, each of the rectangles hek or hck is equal to the 
 square of the parallel semi-diameter cs ; and each of the rect- 
 angles hKk or hek is equal to the square of the parallel semi- 
 diameter ci. And therefore the rectangles of the segments 
 of all parallel lines are equal to one another. Q. e. d. 
 
 Corol 1. The rectangle mek being constantly the same, 
 whether the point e is taken on the one side or the other of 
 the point of contact i of the tangent parallel to hk, it follows 
 that the parts he, ke, of any line hk, are equal. 
 ' And because the rectangle hck is constant, whether the 
 point e is taken in the one or the other of the opposite hyper- 
 Solas, it follows, that the parts He, kc, are also equal. 
 
 GoroL 2. And when hk comes into the position of the tan- ■ 
 gent DiL, the last corollary becomes il = id, and im = in, and 
 
 LM = DN. 
 
 Hence also the diameter cir bisects all the parallels to dl, 
 which are terminated by the asymptote, namely rh = rk. 
 
 Corol. 
 
bit 
 
 CONIC SECTIONS. 
 
 CoroL 3. From the proposition, and the last corollary, it 
 follows that the constant rectangle hek or ehe is = il^ . And 
 the equal constant rect. hck or ene = mln or im^ — it^. 
 CoroL 4. And hence il = the parallel semi-diameter C3> 
 For, the rect. ehe = il^ , 
 and the equal rect. ene = im^ — jl^, 
 theref. il^ = im^ — il^ , or im^ = 2il^ ; 
 but, by cor. 4 theor. 23, im^ — 2gs2, 
 and therefore - - il = cs. 
 And so the asymptotes pass through the opposite angles of all 
 >the ioscribed parallelograms. 
 
 THEOREM XXVII. 
 
 The rectangle of any two Lines drawn from any Point in the 
 Curve, Parallel to two given Lines, and Limited by the 
 Asymptotes, is a Constant Qjaantity. 
 
 That is, if ap, eg, di be parallels, 
 as also AQ., EK, DM parallels, 
 then shall the rect. paq = rect. gek = rect. JP^f . 
 
 For, produce ke, md to the other asymptote at h, l. 
 Then, by the parallels, he : ge : : ld : id ; 
 but - - - BK : EK : : DM : DM ; 
 
 theref. the rectangle hek : gek : : ldm : idm. 
 But, by the last theor. the rect. hek = ldm ; 
 and therefore the rect. gek = idm = paq. 
 
 THEOREM XXVm. 
 
 Every Inscribed Triangle, formed by any Tangent and the 
 two Intercepted Parts of the Asymptotes, is equal to a 
 Constant (Quantity ; namely Double the Inscribed Paral- 
 lelogram. 
 
 That 
 
OP THE HYPERBOLA. 
 
 2 paral. ox. 
 
 518 
 
 That is, the triangle cts 
 For, since the tan«fent ts is 
 bisected by the point of contact 
 E, and EK.is parallel to to, and 
 GE to cK ; therefore ck, ks, ge 
 are all equal, as are also cg, gt, 
 
 KE. Consequently the triangle .__ 
 
 GTE = the triangle kes, and ' ^^ S 
 
 each equal to half the constant inscribed parallelogram ck. 
 And therefore the whole triangle cts, which is composed of 
 the two smaller triangles and the parallelogram, is equal to 
 double the constant inscribed parallelogram gk. q. e. d. 
 
 THEOREM XXIX. 
 
 If from the Point of ' 
 
 Intersections of the Curve witii a Line parallel to the 
 Tangent, three parallel Lines be drawn in any Direction, 
 and terminated by either Asymptote ; those three Lines 
 shall be in continued Proportion. 
 
 That is, if hkm and the 
 tangent il be parallel, then 
 are the parallels dh, ei, gk 
 in continued proportion 
 
 For, by the parallels, ei : il : : dh : H»f ; 
 
 and, by the same ei : il : : gk : km ; 
 
 theref by compos, ei^ : il^ : : dh . gk : hmk ; 
 but, by theor. 26, the rect. hmk = il^ ; 
 and theref the rect. dh gk = ei^, 
 ©r - - - dh : ei : : ei : gk. q. 
 
 THEOREM XXX. 
 
 Draw the semi-diameters ch, cin, ck *, 
 Then shall the sector ©hi = the sector cik. 
 
 Vol. I. 
 
 For, 
 
1^14 CONIC SECTIONS. 
 
 For, becaujse hk and all its parallels are bisected by cm, 
 
 therefore the triangle cnh = tri. cnk, 
 
 and the seju^ment inh = seg ink ; 
 
 consequently the sector cih = sec. cik. 
 CoroL If the geometricals dh, ei, gk be parallel to the 
 ©ther asymptote, the spnces i>hie, eikg will be equal ; for 
 they are equal to the equal sectors chi, cik. 
 
 So that by taking any geonietricals cd, ce, cg, &c. and 
 drawing dh, ei, ok, kc. parallel to the other asymptote, a^ 
 also the radii ch, ci, ck ; 
 
 then the sectors chi, cik, &c. 
 
 or the spaces DHiE, eikg, &,c. 
 
 will be all equal among themselves. 
 
 Or the sectors chi, chk, &c. 
 
 or the sv>aces dhie, dhkg, &c. 
 
 will be in arithmetical progression. 
 And therefore these sectors, or Rpaces,»will be analogous to 
 the logarithn^s of the lines or bases cd, ce, cg, kc. ; namely 
 CHI or DHIE the log. of the ratio of 
 GD to GE, or of CE to CG, &c. ; or of EI to DH, or of gk to Ei„&c. ; 
 
 and ciiK or dhkg the log. of the ratio of 
 
 CD to CG, &C. or of GK tO DH, &C. 
 
 OF THE PARABOLA. 
 
 THEOREM I. 
 
 The Abscisses are Proportional to the Squares of their 
 Ordinates. 
 
 Let avm be a section through 
 the axis of the cone, and agih a 
 parabolic section by a plane per- 
 pendicular to the former, and 
 parallel to the side vm of the 
 cone ; also let afh be the common 
 intersection of the two planes, or 
 the axis of the parabola, and fg, 
 hi ordinates perpendicular to it. 
 
 Then it will be, as af : ah : : fg^ : hi^. 
 
 t'or, through the ordinates fg, hi draw the circular sec- 
 tions, KGL, Miw, parallel to the base of the cone, having kl, 
 
OF THE PARABOLA. 
 
 £16 
 
 MN for their diameters, to which fg, hi are ordinates, as 
 well as to the axis of the parabola. *c 
 
 Then, hy similar triangles, af : ah : : fl : hw ; 
 but, because of the parallels, kf = mh ; 
 
 therefore - - - af : ah : : kf . fl : mh . hw. 
 But, by the circle, kf . fl = fg", and mh . hn = hi^ ; 
 Therefore - - - af : ah : : fg : hi^. q. e. o. 
 
 fg3 hi* 
 
 Corol. Hence the third proportional or is a constant 
 
 af ah 
 
 quantity, and is equal to the parameter of the axis by defin, 
 16. 
 
 Or AF : fg : : FG : p the parameter 
 Or the rectangle p . af = fq^. 
 
 THEOREM U. 
 
 As the Parameter of the Axis : 
 Is to the Sum of any Two Ordinates : : 
 So is the Difffirence of those Ordmates : 
 To the Difference of their Abscisses : 
 
 J 
 
 ^--n 
 
 L 
 
 '^ 
 
 / ^ 
 
 
 N 
 
 d 
 
 ] 
 
 t V 
 
 That is, 
 F : GH -|- DE : : gh — de ; 
 Or, p : Ki : : ih : is. 
 
 For, by cor. theor, 1, p . ag = gh^, 
 and - - - p . AD = de2 
 theref by subtraction, p . dg = gh" — de^ 
 Or, - - - p . dg = KI . IH, 
 therefore - - p : ki : : ih : dg or ef. <i. e. b. 
 
 Corol Hence because p . ei = ki . ih, 
 and, by cor. theor. 1, p . ag = gh^ , 
 therefore - - ag : ei: : gh^ : ki . ih. 
 
 So that any diameter ei .is as the rectangle of the. seg- 
 ments ki, ih of the double ordinate kh. 
 
 THEOREM nr. 
 
 The Distance from the Vertex to the Focus is equal to \ of 
 Hie Parameter, or to Half th« Ordinate at the Focus. 
 
 That 
 
316 
 
 CONIC SECTIONS. 
 
 That is, 
 
 where f is the focus. 
 
 / E 
 
 For, the general property is af : fe : : fe : p. 
 
 But by dt^fiuitioD 17, - ef = ^p ; 
 therefore also . - - af = ^fe = ip, 
 
 THEOREM IV. 
 
 A Line drawn from the Focus to any Point in the Curve, is 
 e(}ual to the Sum of the Focal Distance and the Absciss of 
 the Ordinate to that Point. 
 
 G 
 
 That is, ^--^■" 
 
 fe = FA -\- AD = GD, y^ ^~ 
 
 taking ag = af. / "E < 
 
 For. since fd — ad co af, 
 theref by squaring, fd^ = af^ — 2af . ad + A»^, 
 But, by cor. theor. 1, de^ = p . ad = 4af . ad ; j \ 
 
 theref by addition, fd^ -|- de= = af^ -f- 2af . ad 4" At)3, 
 But, by right- ang. tri. fd^ -{- de^ = fe^ ; 
 therefore - - fe^ = af^ -{- 2af . ad + ad^, 
 
 and the root or side is fe — af + ad, 
 or - - - i,E = GD, by taking ag = af. 
 
 ^. E. ©. 
 
 Corol 1. If, through the point G, the IIHH G- HHH 
 line GH be drawn perpendicular to the 
 axis, it is called the directrix of the pa- 
 rabola. The property of which, from 
 this theorem, it appears, is this : That 
 driiwing any line he parallel to the axis, 
 HK is always equal to fe the distance of J^ jq- ^ 
 
 the focus from the point e, 
 
 'Corol. 2. Hence also the curve is easily described by points. 
 Namely in the axis produced take ag = af the focal dis- 
 tance, and draw a number of lines ee perpendicular to the 
 axis AD ; then with the distances gd, gd, gd, &c. as radii 
 and the centre f, draw arcs crossing the parallel ordinates 
 in F, E, E, &G. Then draw the curve through all the points, 
 
 £; £; £. 
 
 THEOREM 
 
 
 y 
 
 \ 
 
 ■ 
 
 
 /E 
 
 F E\ 
 
 
 / 
 
 /B 
 
 D E 
 
 \ 
 
OF THE PARABOLA. 
 
 517 
 
 THEORKM.V- 
 
 If a Tangent be drawn to any Point of the Parabola, meet- 
 ing the Axis produced ; and if an Ordinate to the Axis 
 be drawn from the Point of Contact ; then the Ahscigs of 
 that Ordinate will be equal to the External Fart of the 
 Axis, 
 
 That is, 
 if Tc touch the curve 
 at the point c ; 
 then is at = am. 
 
 For, from the point t, draw any line cutting the curve in the 
 the two points e, h : to which draw the ordinates de, gh ; also 
 draw the ordinate mc to the point of contact c. 
 AD : AG : : de2 : gh^ ; 
 : tg3 ; : de^ : gh^ ; 
 ; AG : : TD^ : tg2 ; 
 
 pg : : TD^ : tg^ — td^ or dg.(td-|-tg), 
 : TD : : TD : td-|-tg ; 
 : AT : : td' : tg, 
 AT : : AT : AG ; 
 
 Then, by th 
 
 and by sim tri. td- 
 theref. by equality, ad 
 
 and, by division, ad 
 
 or - - ad 
 
 and, by division, ad 
 
 and again by div. ad 
 
 AT is a mean propor. between ad, ag. 
 
 Now if the line th be supposed to revolve about the 
 point T ; then, as it recedes farther from the axis, the points e 
 and H approach towards each other, the point e descending and 
 the point H ascending, till at last they meet in the point c, 
 when the line becomes a tangent to the curve at c And 
 then the points d and g meet in the point m, and the ordinates 
 DE, gh in the ordinates cm. Consequently ad, ag, becoming 
 earh equal to am, their mean proportional at will be equ^J to 
 the absciss am. That is the external part of the axis, cut off 
 by a tangent, is equal to the absciss of the ordinate to the point 
 ©f contact. Q. E. D. 
 
 THEOREM VI. 
 
 If a Tangent to the Curve meet the Axis produced ; then 
 the Line drawn frbm the Focus to the i oint of Contact, will 
 be equal to the Distance of the Focus from the Intersection 
 of the Tangent and Axis. 
 
 , That 
 
&u 
 
 aONIC SECTIONS. 
 
 That is, 
 
 FG = FT. 
 
 For, draw the ordinate dc to the point of contact c. 
 
 Then, by theor. 5, at = ad ; 
 therefore - ft = af + ad. 
 
 But, by theor. 4, fc = af -|- ad ; 
 ther^f. by equaUty, fc = ft. 
 
 Corol. 1. If CG be drawn perpendicular to the curve, or to 
 the tangent, at c ; then shall fg = fc = ft. 
 
 For, draw fh perpendicular to tc, which will alsjo bisect 
 tc, because ft = fc ; and therefore, by the nature of the 
 parallels, fh also bisects tg in f. And consequeaily tc = 
 
 FT = FC. 
 
 So that F is the centre of a circle passing through r, c, g. 
 
 Corol. 2. The tangent at the vertex ah is a mean propor- 
 tional between af and ad. 
 
 For, because fht is a right angle, 
 therefore - ah is a mean between af, at, 
 or between - af, ad, because ad = at. 
 Likewise, - fij is a mean between fa, ft, 
 or between fa, fc 
 
 Corol. 3. 
 
 The tangent tc makes equal angles with fc and 
 the axis ft. 
 
 For, because ft = fc, ^ 
 
 Therefore the ^ fit = ^ftc. 
 
 Also, the angle gcf = the angle gck, 
 
 drawing ick parallel to the axis ag. 
 Corol. 4. And because the angle of incidence gck is = 
 the angle of reflection gc f ; therefore a ray of light falling 
 on the curve in the direction kc, will be reflected to the fo- 
 cus F. That is, all rays parallel to the axis, are reflected to 
 the focus, or burning point. 
 
 THEOREM 
 
OP THE PARABOLA. 
 
 51« 
 
 THEOREM VU. 
 
 if there be any Tangent, and a DouMe Ordinate drawn from 
 the I'oint of Contact, and also any Line parallel to the Axis, 
 limited hy the Tangent and Double Ordinate : then shall 
 the Curve divide that Line in the same Ratio, as the Line 
 divides the double Ordinate. 
 
 That is, 
 ; EK : : CK 
 
 For. by sim. triangles, 
 but by the def the param 
 therefore, by equality, 
 But, by theor. 2, 
 therefore, by equality, 
 and by division. 
 
 3S, CK : KI : : cd 
 
 DT or 2da ; 
 
 •am. p : CL : : CD 
 
 : 2da; 
 
 p : CK : : cl 
 
 : KI. 
 
 - p : CK : : KL 
 
 . ke; 
 
 CL : KL : : ki 
 
 ke; 
 
 CK : KL : : ie 
 
 : EK. d. E. 
 
 THEOREM VIIL 
 
 
 Th6 same being supposed as in theor. 7 ; then shall the Ex- 
 ternal Fart of the Line between the Curve and Tangent 
 be proportional to the Square of the intercepted Part of 
 the Tangent, or to the bquare of the intercepted Fart of 
 the Double Ordinate. 
 
 That is, IE is as ci^ or as ck^ 
 and ic, TA, ON, pl, &c. 
 are as ci^, ct^, co^, cp, ^:c. 
 •r as gk2, cd2, cm^, cl^, &e. 
 
 For, by theor. 7 
 or, by equality. 
 But, by cor. th. 2, 
 therefore 
 
 IE : EK : 
 
 IE : EK 
 
 EK is as the rect. ck . kl, 
 IE is as ck3, or as ci2. q. k. d- 
 
 Corol. As this property is common to every position of the 
 tangent, if the lines ie, ta, on. &c. be appended on the points 
 I, T, o, ^c. and moveable about them, and of surh lengths as 
 that th«ir extromities e, a, n, kc. be in the curve of a para- 
 bola 
 
529 CONIC SECTIONS. 
 
 . bola in some one position of the tangent ; then making the taa* 
 gent revolve about the point c, it appears thnt the extremities 
 E, A, N, ^c. will always fortn the curve of some parabola, in 
 every position of the tangent. 
 
 THEOREM IX. 
 
 The Abscisses of any Diamr tf^r, are as the Squares of their 
 Ordinates. 
 
 That is, CQ, or, cs, &c. 
 are as q,E^ , ra^ , sn^ , lk,c. 
 Op CQ : CR : : ^e^ : ra, 
 &c. 
 
 For, draw the tangent ct, and the externals ei, at, no, &c, 
 parallel to the axis, or to the diameter cs. 
 
 Then, because the ordinates qE, ra, sn, &c. are parallel to 
 the tangent ct, by the definition of them, therefore all the fi- 
 gures iQ, TR, OS, &c. are parallelograms, whose opposite sides 
 are equal ; 
 
 namely, - - ie, ta, on, &c. 
 are equal to - cq, gr, cs, &c. 
 
 Therefore, by theor. 8, cq, cr, cs, &c. 
 areas - - ci^, ct^, co^, &c. 
 
 or as their equals : qe^, ra^, sn^, &c. q. e. ». 
 
 CoroL Here like as in theor. 2 the difference of the abscis- 
 ses is as the difference of the squares of their ordinates. or as 
 the rectangles under the sum and difference of the ordinates, 
 the rectangle of the sum and difference of the ordinates being 
 equal to the rectangle under the difference of the abs( i^ses 
 and the parameter of that diameter, or a third proportional t© 
 any absciss and its ordinate. 
 
 THEOBExM 
 
OF THE PARABOLA. 
 
 THEOREM X. 
 
 621 
 
 tf a Line be drawn parallel to any Tangent, and cut the Curve 
 in two Points ; then it two Ordinates be drawn to the In- 
 tersections, and a third to the Point of Contact, these three 
 Ordinates will be in Arithmetical Progression, or the Sum 
 of the Extremes will be equal to Double the Mean. 
 
 That is, 
 
 BG + HI = 2CD. 
 
 .^ 
 
 A 
 
 % — 1 
 
 
 / 
 
 
 H 1 
 
 L ] 
 
 L lu 
 
 For, draw ek parallel to the axis, and produce hi to l. 
 Then, by sim. triangles, ek : hk : : td or 2ad : cd ; 
 but, by theor. 2, - ek : hk : : kl : p the param. 
 theref. by equality, 2ad : kl : : cd : p. 
 But, by the defin. 2ad : 2cd : : cd : p ; 
 
 theref. the 2d terms are equal, kl = 2cd, 
 that is, - - EG -|- HI = 2cD. q. 
 
 E. o. 
 
 CoroU When the p©int e is on the other side of ai ; then 
 
 HI — GE = 2CD. 
 
 THEOREM XI. 
 
 Any Diameter bisects all its Double Ordinates, or Lines parallel 
 io the tangent at its Vertex. 
 
 That is, 
 
 ME = MH. 
 
 For, to the axis ai draw the ordinates eg, cd, hi, and mn 
 parallel to them, which is equal to cd. 
 V9&. I. 67 Then 
 
6n 
 
 CONIC SECTIONS. 
 
 Then, by theor. 10, 2mn or 2cd = eg + hi> 
 therefore m is the middle of eh. 
 
 And, for the same reason, all its parallels are bisected. 
 
 Q. E. D. 
 
 ScHOL. Hence, as the abscisses of any diameter and their 
 ordinates have the same relations as those of the axis, namely, 
 that the ordinates are bisected by the diameter, and their 
 squares proportional to the abscisses ; so all the other pro- 
 perties of the axis and its ordinates and abscisses, before de- 
 monstrated, will likewise hold good for any diameter and its 
 ordinates and abscisses. And also those of the parameters, 
 understanding the parameter of any diameter, as a third 
 proportional to any absciss and its ordinate. Some of the 
 most material of which are demonstrated in the following 
 theorems. 
 
 THEOREM Xn. 
 
 The Parameter of any Diameter is equal to four Times the 
 Line drawn from the Focus to the Vertex of that Dia- 
 meter. 
 
 That is, 4fc = p, 
 the param. of the diam. cm. 
 
 For, draw the ordinate ma parallel to the tangent ct ; also 
 CD, MN perpendicular to the axis an, and fh perpendicular to 
 the tangent ct. 
 
 Then the abscisses ad, cm or at, being equal, by theor. 6, 
 the parameters will be as the squares of the ordinates cd, ma 
 or CT, by the definition ; 
 
 that is, - - p : p : 
 
 But, by sim. tri. - fh : ft : 
 
 therefore - - p : p : 
 
 But, by cor. 2, th. 6, fh^ =fa . 
 
 therefore - - p : p : 
 
 or, by equality - p : p : 
 
 But, 
 
 CD^ 
 
 : CT^. 
 
 : CD 
 
 CT ; 
 
 FH3 
 
 : ft2. 
 
 ft ; 
 
 
 FA . 
 
 FT : ft 
 
 FA : 
 
 ft orFc 
 
But, by theor. 3, 
 and therefore - 
 
 op THE PARABOLA. 
 
 p = 4fa, 
 
 p = 4ft or 4fc. 
 
 523 
 
 q,. E. D. 
 
 Corel. Hence the parameter p of the diameter cm is equal 
 to 4fa -f 4ad, or to p -i- 4ad, that is, the parameter of the 
 axis added to 4 Az>. 
 
 THEOREM Xni. 
 
 If an Ordinate to any Diameter, pass through the Focus, it 
 will be equal to Half its Parameter ; and its Absciss equal 
 to One Fourth of the same Parameter. 
 
 That is, CM = ^p, 
 
 and ME = |p« 
 
 For, join fc, and draw the tangent ct. 
 
 By the parallels, cm =. ft ; 
 
 and, by theor. 6, fc = ft ; 
 
 also, by theor. 12, fc = Jp ; 
 
 therefore - - cm = ip, . 
 
 Again, by the defin. cm or |p : me : : me : p, 
 
 and consequently me = ^p = 2cm. ft* e. d. 
 
 CoroL 1. Hence, of any diameter, the double ordinate 
 which passes through the focus, is equal to the parameter, or 
 to quadruple its absciss. 
 
 CoroL 2. Hence, and from cor. 1 
 to theor. 4, and theor. 6 and 12, it 
 appears, that if the directrix gh be 
 drawn, and any lines he, he, pa- 
 rallel to the axis ; then e?ery parallel 
 he will be equal to ef, or ^ of the 
 parameter of the diameter to the 
 point e. 
 
 THEOREM 
 
524 
 
 CONIC SECTIONS. 
 
 THEOREM XIV. 
 
 If there be a Tangent, and any Line drawn from the Point 
 of Contact and meetinfic the Curve in some other Point, as 
 also another Line parallel to the Axis, and limited by the 
 First Line and the Tangent : then shall the Curve divide 
 this Second Line in the same Ratio, as the Second Line 
 divides the first Line. 
 
 That is, 
 IE : EK : : ck : kl. 
 
 For, draw lp parallel to ij£, or to the axis. 
 Then by theor. 8, ie : pl : : ci2 : cp^, 
 or, by sim. tri. - ie : pl : : ck^ : cl^ . 
 Also, by sim. tri. ik : pl : : ck : cl, 
 
 or - - - IK : PL : : ck^ : ck . cl ; 
 
 therefore by equality, ie : ik : : ck : cl : cl^ ; 
 or, - - - IE : IK : : CK : CL ; 
 and, by division, ie : ek : : ck : kl. 
 
 CoroL When ck = kl, then ie = ek = ^ik. 
 
 Q. E. D. 
 
 THEOREM XV. 
 
 If from any Point of the Curve there be drawn a Tangent, 
 and also Two Right Lines to cut the Curve ; and Dia- 
 meters be drawn through the Points of Intersection e and 
 L, meeting those Two Right Lines in two other Points g,. 
 and K ; Then will the Line kg joining these last Two 
 Points be parallel to the Tangent. 
 
 For, 
 
OF THE PARABOLA. 625 
 
 Fbm, by theor. 14, ck : kl : : ei : ek ; 
 
 and by composition, ck : cl : : ei : ki ; 
 
 and by the parallels ck : cl : : gh : lh ; 
 
 But, by sim. tri. - ck : cl : : Jti : lh ; 
 
 theref. by equal. - ki : lh : : gh : lh ; 
 
 cont^equently - ki «= gh, 
 
 and therefore - kg is parallel and equal to ih. 
 
 THEOREM XVI. 
 
 If an Ordinate be drawn to the Point of Contact of any 
 Tangent, and another Ordinate produced to cut the Tan- 
 gent ; It will be, as the Difference of the Ordinates : 
 
 Is to the Difference added to the external Part : : 
 
 So is Double the first Ordinate : 
 
 To the Sum of the Ordinates. 
 
 That is, KH : KI : : KL : kg. 
 T 
 
 IH K. Ja & 
 
 For, by cor. 1 theor. 1, p : dc : : dc : da. 
 
 and - - - p : 2dc : : do : dt or 2da. 
 
 But, by sim. triangles, ki : kc : : dc : dt ; 
 
 therefore by equality, p : 2dc : ; ki : kc, 
 
 or, - - - p : KI : : KL : kc. 
 
 Again, by theor. 2, p : kh : : kg : kc ; 
 
 therefore by equality, kh : ki : : kl : kg. q. e. d. 
 
 Coroh 1. Hence, by composition and division, 
 it is, kh : KI : : GK : gi, 
 and hi : hk : : hK : kl, 
 also IH : IK : : IK : ig ; 
 that is, IK is a mean proportional between ig and ih. 
 
 €orol. 2. And from this last property a tangent can easily 
 be drawn to the curve from any given point i. Namely, draw 
 iHG perpendicular to the axis, and take ik a mean proportion- 
 al between ih, ig ; then draw kc parallel to the axis, and c 
 will be the point of contact, through which and the given 
 point I the tangent le is to be drawn. 
 
 THEOREM 
 
526 
 
 CONIC SECTIONS. 
 
 THEOREM XVn. 
 
 If a Tangent cut any Diameter produced, and if an Ordinate 
 
 • to that Diameter be drawn from the Point of Contact ; 
 
 then the Distance in the Diameter produced, between the 
 
 Vertex and the Intersection of the Tangent, will be equal 
 
 to the Absciss of that Ordinate. 
 
 For, by the last th. ie ; ek : : ck : kl. 
 But, by theor. 1 1 , CK = KL, ' 
 
 and therefore ie = ek. 
 
 Corol. 1. The two tangents ci, li, at the extremities of 
 any double ordinate cl, meet in the same point of the diame- 
 ter of that double ordinate produced. And the diameter 
 drawn through the intersection of two tangents, bisects the 
 line connecting the points of contact. 
 
 Corol, 2. Hence we have another method of drawing a 
 tangent from any given point i without the curve, Namely, 
 from I draw the diameter ik, in which take ek = ei, and 
 through K draw cl parallel to the tangent at e ; then c and l 
 are the points to which the tangents must be drawn from i. 
 
 THEOREM XVIII. 
 
 If a Line be drawn from the Vertex of any Diameter, to cut 
 the Curve in some other Point, and an Ordinate of that 
 Diameter be drawn to that Point, as also another Ordinate 
 any where cutting the Line, both produced if necessary : 
 The Three will be continual Proportionals, namely, the two 
 Ordinates and the Part of the Latter hmited by the said Line 
 drawn from the Vertex. 
 
 Ml 
 
 That is, DE, GH, Gi are 
 
 / 
 
 '^p.-.TX 
 
 i=^^r 
 
 continual proportionals^ or 
 
 /^ 
 
 
 '"^^^];^m 
 
 DE : GH : : GH : gi. / 
 
 
 Ti"-^ 
 
 O \^ 
 
 For, by theor. 9, - - - de^ 
 
 : gh2 
 
 : : AD 
 
 : AG ; 
 
 and, by sim. tri. - - - de 
 
 : GI 
 
 : : AD 
 
 : AG; 
 
 theref. by equality, - - de 
 
 ; GI 
 
 : : de3 
 
 : GH3 ; 
 
 that is, of the three de, gh, gi, 1st : 
 
 : 3d 
 
 : : lst2 
 
 : U^; 
 
 therefore » - - ist 
 
 :2d 
 
 :: 2d 
 
 : 3d, 
 
 that is, - - - - DE : 
 
 : GH 
 
 : : GH ; 
 
 : GI. Q. E^ d. 
 
 Corol, 
 
OF THE PARABOLA. 
 
 527 
 
 Corol, 1. Or tbeir equals, gk, gh, gi, are proportionals ; 
 where ek is parallel to the diameter ad. . 
 
 CoroL 2. Hence it is db : ag : : p : gi, where p is 
 the paranaeter, or ag : gi : : de : p. 
 For, by the defin. ag : gh : : gh : p. 
 Corol. 3. Hence also the three mn , aii, mo, are proportion- 
 als, where mo is parallel to the diameter, and am parallel to 
 the ordinates. 
 
 For, by theor. 9, - mn, mi, mo, 
 or their equals - ap, ag, ad, 
 are as the squares of pn, gh, de, 
 or of their equals gi, gh, gk, 
 which are proportionals by cor. 1 . 
 
 THEOREM XIX. 
 
 If a Diameter cut any Parallel Lines terminated by the Curve : 
 the Segments of the diameter will be as the Rectangle of the 
 Segments of those Lines. 
 
 That is, EK : EM : : GK . KL : NM . MO. 
 
 Or, EK is as the rectangle ck . kl. 
 
 For, draw the diameter 
 PS to which the parallels 
 CL, NO are ordinates, and ' 
 the ordinate e^ parallel to 
 them, 
 
 Then ck is the differ- 
 ence, and KL the sum of 
 the ordinates eq, cr ; also 
 NM the difference, and mo the sum of the ordinates eq, ns. 
 And the differences of the abscisses, are ^r, %s, or ek, em. 
 
 Then by cor. theor. 9, ^r : ^s : : ck . kl : nm . mo, 
 that is - - ek : em : : CK . KL ; NM . mo. 
 
 Corol. 1. The rect. ck . kl = rect. EKand the param. of ps. 
 For the rect. ck . kl = rect q.r and the param. of ps. 
 
 Corol. 2. If any line cl be cut by two diameters, ke, gh ; 
 the rectangles of the parts of the line, are as the segments of 
 the diameters. 
 
 For EK is as the rectangle ck . kl. 
 and GH is as the rectangle ch . hl ; 
 therefore ek : gh : : ck . kl : ch . hl. 
 
 CoroL 
 
528 
 
 CONIC SECTIONS. 
 
 Corol. 3. If two parallels, cl, no, be cot by two diame- 
 ters, EM, Gi ; the rectangles of the parts of the parallels, will 
 be as the segments of the respective diameters. 
 
 For - - ' - EK : EM : : CK . KL : NM . mo, 
 and - - - EK : GH : : CK . KL : CH . hl, 
 theref. by equal. em : gh : : nm . mo : ch . hl. 
 Corol. 4. When the parallels come into the position of the 
 tangent at p, their two extremities, or points in the curve unite 
 in the point of contact p ; and the rectangle of the parts be- 
 comes the square of the tangent, and the same properties still 
 follow them. 
 
 So that, Ev : pv : : fv : p the param. 
 gw : pw : : pw : p, 
 EV : gw : : pv^ : pw^, 
 EV : GH : : pv^ : ch . HLi 
 
 THEOREM XX. 
 
 If two Parallels intersect any other two Parallels ; the Rect- 
 angles of the Segments will be respectively Proportional. 
 That i&, CK . KL : DK . ke : : gi . ih : ni . lo. 
 
 For, by cor. 3 theor. 23, pk : qi : : ck . kl : gi . ih ; 
 
 and by the same, pk : qi : : dk : ke . ni . lo ; 
 
 theref. by equal, ck . kl : dk : ke : : gi . ih : ni . lo. 
 
 Corol. When one of the pairs of intersecting lines comes 
 into the position of their parallel ttingents, n»eeting and limit- 
 ing each other, the rectangles of their segments become the 
 squares of their respective tangents. So that the constant 
 ratio of the rectangles, is that of the square of their parallel 
 tangents, namely, 
 CK . KL : pK . KE : : tan^ . parallel to cl : tang^ . parallel to de, 
 
 THEOREM XXI. 
 
 If there be Three Tangents intersecting 6ach other ; 
 Segments will be in the same Proportion. 
 
 their 
 That 
 
OF THE PARABOLA. 
 
 629 
 
 That is, Gi r ih : 
 For, through the points 
 9, I, D, H, draw the diame- 
 ters GK, iL, DM, HN ; as 
 also the lines ci, ei, which 
 are doahle ordinates to the 
 diameters gk, hn, by cor. 1 
 theor. 16 ; therefore 
 the diameters gk, bm, hn, 
 bisect the lines cl, ce, le ; 
 kence km = cm — ck = ^ce — ^cl =,4le = ln or ne, 
 
 and MN = ME — NE -— iCE — iLE = ^CL = CK Or KL. 
 
 But, by parallels, gi : ih : : kl : ln, 
 and - - cg : gd : : CK : km, 
 also - - DH : HE : : MN : ne. 
 But the 3d terms ki, ck, mn are all equal ; 
 
 as also the 4th terms ln, km, ne. 
 Therefore the first and second terms, in all the lines, are 
 proportional, namely gi : ih : : cg : gd : : dh : he. q. e. d. 
 
 THEOREM XXII. 
 
 If a Rectangle be described about a Parabola, having the 
 same Base and Altitude ; and a diagonal Line be drawn 
 from the Vertex to the Extremity of the Base of the Para- 
 bola, 'forming a right-angled Triangle, of the same Base 
 and Altitude also ; then any Line or Ordinate drawn across 
 the three Figures, perpendicular to the Axis, will be cut m 
 Continual Proportion by the Sides of those Figures. 
 
 A, I I > 
 
 That is, 
 
 ep : eg : : eg : eh. 
 Or, EF, eg, eh, are in con- 
 tinued proportion. 
 
 For, by theor. 1 , ab : ae : 
 and, by sim. tri. - ab : ae : 
 
 theref. of equality, - ep : bc : 
 that is - - - EF : EH : 
 theref. by Geom. th. 78, ef, eg, eh are proportionals, 
 •r • - - - EP : EG : ; eg : eh. q,. 
 
 V©B. r. 
 
 68 
 
 THBORJEJMt 
 
53t CONIC SECTIONS, 
 
 THEOREM XXm. 
 
 The Area or Space of a Parabola, is equal to Two-Thirds of 
 its Circumscribing Parallelogram. 
 
 That is, the space abcga = | abcd ; 
 or the space adcga = i abcd. 
 ^OR, conceive the space adcga to be composed of, or 
 divided into, indetiuitely small parts, by lines parallel to 
 DC or AB, such as ig, which divide ad into like small and 
 equal parts, the number or sum of which is expressed by the 
 line AD. Then, 
 
 by the parabc 
 
 •la, Bcs 
 
 : eg2 : 
 
 : AB ; 
 
 : AE, 
 
 that is, 
 
 AD2 
 
 : Ai2 : 
 
 : DC ; 
 
 : IG. 
 
 Hence it follows, that any one of these narrow parts, as 
 ic 
 IG, is = X Ai2 ; hencej ad and dc being given or 
 
 AD3 
 
 constant quantities, it appears that the said parts ig, &c are 
 proportional to ai^, &c. or proportional to a series of square 
 numbers, whofee roots are in arithmetical progression, and the 
 
 DC 
 
 area adcga equal to drawn into the sum of such a seriqs 
 
 AD2 
 
 of arithmeticals, the number of which is expressed by ad. 
 
 Now, by the remark at pag. 227 this vol. the sum of the 
 squar es of such a series of arithmeticals, is expressed by 
 ^n .. w-fl. 2n-j-l» where n denotes the number of them. 
 In the present case, n represents an infinite number, and 
 then the two factors n -f 1, 2n -|- 1, become only n and 2n, 
 omitting the 1 as inconsiderable in respect of the infinite 
 number n : hence the expression above becomes barely 
 |n . » . 2w = in^. 
 
 To apply this to the case d^ove : n will denote ad or bc ; 
 and the sum of all the ai^'s becomes i ad^ or i bc^ ; conse- 
 
 DC DC 
 
 quently the sum of all the X Ai2's,is X | ad^ = 
 
 AD2 Ad3 
 
 l AD DC = i bd, which is the area of the exterior part adcga.^ 
 That is, the said exterior part adcga, is i of the parallelo- 
 gram ABCD : and consequently the interior part abcga is f 
 if the same parallelogram. 9.- e. d. 
 
 Corel* 
 
OF THE PARABOLA. 531 
 
 Coral. The part afcga, inclosed between the cnrve and 
 the right liae afc, is ^ of the same parall^lograoi, bcina; the 
 difference between abcga ami the triangle abcfa, that is be« 
 tween | aad i of the parallelugram. 
 
 THEOREM XXIV. 
 
 The Solid Content of a Paraboloid (or Solid generated by 
 the Rotation of a Parabola about its Axis), is equal to Half 
 its Circumscribing Cylinder. 
 
 Let ABC be a paraboloid, generated by the rotation of the 
 parabola ac about its axis ad. , Suppose the axis ad be divided 
 into an infinite number of equal parts, throu2;h which let cir- 
 cular planes pass, as efg, all those circles making up the whole 
 solid paraboloid. 
 
 Now if c = the number 
 3-1416, then 2c X fg is the 
 circumference of the circle 
 BFG whose radius is fg ; there- 
 fore c X fg2 is the area of 
 that circle. 
 
 But, by cor. theor. 1, Parabola p X af = fg*, where p 
 denotes the parameter of tke parabola ; consequently pc X a? 
 will also express the same circular section eg, and therefore 
 pc X the sum of all the af's will be the sum of all those 
 circular sections, or the whole content of the solid, para- 
 boloid. 
 
 But all the af's foroi an arithmetical progression, begin- 
 ning at or nothing, and having the greatest t<^rm and the 
 sun^ of all the terras each expressed by the whole axis ad. 
 And since the sum of all the terras of such a prog^ression, is 
 equal to 5 AD X ad or i ad^ half the product of the greatest 
 term and the number of terms ; therefore i ad^ is equal to 
 the sum of all the af's, and consequently pc X i ad^, or ^ c 
 X /? X ad2, is the sum of all the circular sections, or the 
 eontent of the paraboliod. 
 
 DC? 
 
 But, by the parabola, p : dc : : dc : ad, orp = ; con.- 
 
 AD 
 
 sequently ^ c X p ^^ ad^ becomes i c X ^d X dc^ for the 
 solid content of the paraboloid. B it c X ad X dc^ is equal 
 to the cylinder bcih ; consequently the paraboloid is the half 
 of its circumscribing cylinder. ^. k. s. 
 
 THBORKBJE 
 
5S2 
 
 CONIC SECTIONS. 
 
 THEOREM XXV. 
 
 The Solidity of the Frustum begc of the Paraboloid, is equal 
 to a Cylinder whose Height is df, and its Base Half the 
 Sum of the two Circular Bases eg, bc. 
 For, by the last theor. ^pc X ad^ = the solid abc, 
 and, by the same, z P^ ^ ^^^ ~ ^^® ^^^^^ '*^^^» 
 
 theref the diflf, , i/'c X (ad2 —AF2)=the frust. begc. 
 
 But AD2 AF2 = DF X(aD -{- Af), 
 
 theref -k P^ ^ o* ^ (ad 4- AF)=the frust begc. 
 
 But, by the parab. p X ad = rJc^ , and p X af = fgs j 
 theref. i c X df X (dc2 -j- fg2) = the frust begc. 
 
 Q. E. ©, 
 
 ON THE CONIC SECTIONS AS EXPRESSED BY ALGEBRAIC 
 EQUATIONS, CALLED THE EQUATIONS OF THE CURVE. 
 
 1 For the Ellipse, 
 
 Let d denote ab, the transverse, or any diameter ; 
 
 c = iH its conjugate ; 
 
 X = AK, any absciss, from the extremity of the diam. 
 
 y =3 DK the correspondent ordirtate. 
 Then, theor. 2, ab^ : hi^ : : ak . kb : dk^, 
 that is, d^ : c^ : : x [d^ x) : 2/^, hence d^ y^ = c^ {dar—x^)^ 
 or dy =^ c ^ {dx — x^), the equation of the curve. 
 
 And from these equations, any one of the four letters or 
 quantities, </, c, x, y, may easily be found, by the reduction of 
 equations, when the other three are given. 
 
 Or, if/? denote the parameter, = c- -^ </ by its definition ; 
 then, by cor. th. 2, d :p : : x (</— jr) : y^,or dy^=p(dx^x^), 
 which is another form of the equation of the curve. 
 
 Otherwise* 
 
EQUATIONS OF THE CURVE. 583 
 
 Otherwise, 
 
 . Or, if d = AC the semiaxis ; c = ch the semiconjugate j, 
 p =• c3 -j- d the gemiparameter ; x = ck the absciss counted 
 irom the centre ; and y = vrn. the ordinate as before 
 Then is ak = d-'X, and kb=c/ + ^y and ak . kb = (d^x) X 
 
 (rf + X) = d2 _ p2.. 
 
 Then,byth. 2,d2 :c^ ::d^ ^x^iy^ ,2indd^y^ =c^{d2 ^x^), 
 OT dy = c ^ {d^ — ^^), the equation of the curve. 
 
 Or, d : p : : d^ — x^ : y^, and dy^ = p{d^ — x^), another 
 form of the equation to the curve ; from which any one of the 
 «[uantities may be found, when the rest are given. 
 
 2. For the Hyperbola. 
 
 Because the general property of the opposite hyperbolas, 
 with respect to their abscisses and ordinates, is the same as 
 that of the eUipse, therefore the process here is the very same 
 as in the former case for the ellipse ; and the equation to the 
 curve must come out the same also, with sometimes only the 
 change of the sign of a letter or term from -{- to — , or from 
 — to -|-> because here the abscisses lie beyond or without the 
 transverse diameter, whereas they lie between or upon them 
 in the ellipse. Thus, making the same notation for the whole 
 diameter, conjugate, absciss, and ordinate; as at first in the el- 
 lipse ; then, the one absciss ak being x, the other bk will be 
 d -h JT, which in the ellipse was d—x; so the sign of x most 
 be changed in the general property and equation, by which it 
 becomes d^ : c^ : : x (^d-}- x) : y^ : hence d^y- = c^ (dx -\- x^) 
 and dy = c ^ (^dx -{- x^), the equation of the curve. 
 
 Or, usingp the parameter as before, it is, d : p : : x (d-\- x): 
 y^f or dy^ ^= p {dx '\- x^), another form of the equation to 
 the curve. 
 
 Otherwise^ by using the same letters c?, c, jo, for the halves 
 of the diameters and parameter, and x for the absciss ck 
 counted from the centre ; then is ak = x — rf, and bk = x-\-d, 
 and the property d^ : c^ : : . (x — d) y. {x -{■ d) : y^, j^ives 
 d2y2 = c3 (a;2 — d^), or dy^c^{x^ — d'^), where the signs 
 of d^ and x^ are changed from what they were in the ellipse. 
 
 Or again, using the semiparameter, d : p : : x^ — d^ : y^^ 
 and dy^ = p (x^ — rfa^ the equation of the curve. 
 
 But for the conjugate hyperbola, as in the figure to theo- 
 rem 3, the signs of both x^ and d^ will be positive ; for the 
 property in that theorem being ca^ ; cos : -. cd^ -f ca^ .- T>e^^ 
 ' '' it 
 
«34 come SECTIONS. 
 
 it is ds : cs : .- x^ -\- d^ : y^ = j>e^ , or d^y^ =c3 {x^ + d^), and 
 dy =^ c y/ {x^ -\- rf^), the equation to the conjugate hyperbola. 
 Or, as c^ : p : : a;3 -\. d^ : y^ , and dy" = p {x^ -f d^), also 
 the equation to the same curve. 
 
 On the Equation to the Hyperbola between the Asymptotes. 
 
 \ 
 
 Let CE and cb be the two as}?mptotes to 
 the hyperbola dro, its vertex being f, and 
 EF, bd, AF, BD, ordinates parallel to the 
 asymptotes. Put af or ef = a. cb = x, 
 and BD = y. Then, by theor 28, af ef 
 = ee . BD, or a^ = xy, the equation to the 
 hyperbola, when the abscisses and ordinates 
 are taken parallel to the asymptotes. 
 
 3. For the Parabola* 
 
 
 C b AB 
 
 If x denote any absciss beginning at the verteir, and y its 
 ordinate, also p the parameter. Then, by cor. theorem 1, 
 AK : KD : : KD : p, or x : y : : y : p i hence px = y^ is th^ 
 i^uation to the parabola. 
 
 4. For the Circle. 
 
 Because the circle is only a species of the ellipse, in which 
 the two axes are equal to each other ; therefore, making the 
 two diameters d and c equal in the foregoing equations to the 
 ellipse, they become i/^ =. dx — x^ , when the absciss a: beg^ins 
 at the vertex of the diameter : and y^ = d^-^x^, when the 
 absciss begins at the centre. 
 
 Scholium. 
 
 In every one of these equations, we perceive that they rise 
 to the 2d or quadratic degree, or to two dimensions ; which is 
 also the number of points in which every one of these curves 
 may be cut by a right line. Hence it is also that these four 
 curves are said to be lines of the 2d order. And these four 
 are all the lines that are of that order, ev^^ry other curve be- 
 ing of some higher, or having some higLer equation, or may 
 he cut in more points by a right line. 
 
 TELEMENTS 
 
t 63S1 
 
 fiLEMENTS OF ISOPERIMETRY. 
 
 I>cf. 1. When a rariable quantity has its mutations regu- 
 lated by a certain law, or confined within certain limits, it ii 
 called a maximum when it has reached the greatest magni- 
 tude it can possibly attain ; and, on the contrary, when it has 
 arrived at the least possible magnitude, it is called a minimum, 
 
 Def. 2. Isoperimeters, or Isoperimetrical figures j are th«se 
 which have equal perimeters. 
 
 Def. 3. The Locus of any point, or intersection, &c. is 
 the right line or curve in which these are always situated. 
 
 The problem in which it is required to find, among figures 
 of the same or of different kinds, those which within equal 
 perimeters, shall comprehend the greatest surfaces, has long 
 engaged the attention of mathematicians. Since the admira- 
 ble invention of the method of Fluxions, this problem has 
 been elegantly treated by some of the writers on that branch 
 of analysis ; especially by Maclaurin and Simpson. A much 
 more extensive problem was investigated at the time of " the 
 war of problems," between the two brothers John and James 
 Bernoulli : namely, " To find, among all the isoperimetri- 
 cal curves between given limits, such a curve, that construct- 
 ing a second curve, the ordinates of which shall be functions 
 of the ordinates or arcs of the former, the area oi the se- 
 cond curve shall be a maximum or a minimum." While, how- 
 ever, the attention of mathematicians was drawn to the most 
 abstruse inquiries connected with isoperimetry the elements of 
 the subject were lost sight of Simpson was the first who call- 
 ed them back to this interesting branch of research, by giving 
 in his neat little book of Geometry a chapter on the maxima 
 and minima of geometrical quantities, and some of the sim- 
 plest problems concerning isoperimeters. The next whcr 
 treated this subject in an elementary manner was Simon Lhuil- 
 lier, of Geneva, who in 1782, published his treatise De Rela- 
 tione mutua Capacitatis et Terminorum Figu^arum^ &c. His 
 principal object in the composition of that work was to supply 
 the deficiency in this respect which he found in most of the 
 elementary Courses, and to determine, with regard to both, 
 the most usual surfaces and solids, those which possessed the 
 minimum of contour with the same capacity ; and, recipro- 
 cally, the maximum of capacity with the same boundary. M. 
 Legendre has also considered the same subject in a manner 
 somewhat different from either Simpson or Lhuiilier, in his 
 JBUsnents dt Qeomitrie. Aa elegant geonaetriGal tract, on the. 
 
536 ELEMENTS OF ISOPERIMETRT. 
 
 same subject, was also given, by Dr. Horsley, in the Philo«. 
 Trans, vol. 76, for 1775 ; contained also in the New Abridg- 
 ment, vol 13, page 653. The chief propositions deduced by 
 these four geometers, together with a few additional proposi- 
 tions, are reduced into one system in the following theorems. 
 
 SECTION I. SURFACES. 
 
 THEOREM I. 
 
 Of all Triangles of the same Base, and whose Vertices fall 
 in a right Line given in Position, the one whose Perimeter 
 is a Minimum is that whose sides are equally inclined t* 
 that Line. 
 
 Let AB be the common base of a series of triangles abc*. 
 ABC, &c. whose vertices c', c, fall in the right line lm, given 
 in position, then is the triangle of least 
 perimeter that whose sides ac, bc, are q, 
 
 inclined to the line lm in equal angles. w ' 
 
 For, let BM be drawn from b, per- 
 pendicularly to LxM, and produced till 
 DM = BM : join AD, and from the point 
 c where ad cuts lm draw bc : also, from any other point cV 
 assumed in lm, draw c a, c'b, c'd. Then the triangles dmc, 
 BMC, having the angle dcm = angle acl (th. 7 Geom.) = 
 mob (by hyp.) dmc = bmc, and dm = bm, and mc commoa 
 to both, have also dc = bc ("th. 1 Geom.). 
 
 So also, we have c'b = c b. Hence ac -f- cb = ac -|- cb 
 paB AD, is less than ac' + c'd (theor. 10 Geom.), or than its 
 equal ac' + c'b. And consequently, ab -f- bc + ac is lesi 
 than AB 4* Bc' + Ac'. q e. d. 
 
 Cor. 1. Of all triangles of the same base and the same al- 
 titude, or of all equal triangles of the same base, the isoscele§ 
 triangle has the smallest perimeter. 
 
 For, the locus of the vertices of all triangles of the same 
 altitude will be a right line lm parallel to the base ; and when 
 lm in the above figure becomes parallel to ab, smce mcb == 
 ACL, MCB = cba (th. 12 Geom.), acl = cab ; it follow* 
 that CAB = cba, and consequently ac = cb (th. 4 Geom.). 
 
 Cor. 2. Of all triangles of the same surface, that which 
 Kas the minimum perimeter is equilateraL 
 
 For 
 
SURFACES. SSV 
 
 Pop the triangle of the smallest perimeter, with the same 
 surface, must be isosceles, whichever of the sides be consider- 
 ed as base : therefore, the triangle of smallest perimeter hag 
 each two or each pair of its sid^s equal, and consequently it is 
 ^uilateral. 
 
 Cor. 3 Of all rectilinear figures, with a given magnitude 
 and a given number of sides, that which has the smallest pe« 
 rimeter is equilateral. 
 
 For so long as any two adjacent sides are not equal, we may 
 draw a diagonal to become a base to those two sides, and then 
 draw an isosceles triangle equal to the triangle so cut off, but 
 «f less perimeter : whence the corollary is manifest. 
 
 Scholium. 
 
 To illustrate the second corollary above, the student may 
 proceed thus : assuming an isosceles triangle whose base is 
 not equal to either of the two sides, and then, taking for a nev9 
 base one of those sides of that triangle, he may construct an- 
 other isosceles triangle equal to it, but a smaller perimeter. 
 Afterwards, if the base and sides of this second isosceles tri- 
 angle are not respectively equal, he may construct a third 
 isosceles triangle equal to it, but of a still smaller perimeter : 
 and so ob, by performing these successive operations, he will 
 find that all the triangles will approach nearer and nearec* 
 to an equilateral triangle. 
 
 THEOREM n. 
 
 f)f all Triangles of the Same Base, and of Equal Perimetert^ 
 the Isosceles Triangle has the Greatest Surface. 
 
 Let ABC, ABD, be two triangles of the same 
 l^ase AB and with equal perimeters, of which 
 the one abc is isosceles, the other is not : 
 then the triangle abc has a surface (or an 
 altitude) greater than the surface (or than 
 the altitude) of the triangle abd. 
 
 Draw CD through d, parallel to ab, to A E B 
 cut CE (drawn perpendicular to ab) in c : then it is to be de- 
 monstrated that CE is greater than c'e. 
 
 The triangles ac b, adb, are equal both in base and altitude ; 
 but the triangle ac'b is isosceles, while adb is scalene : there- 
 fore the triangle ac'b has a smaller perimeter than the triangle 
 ADB (th. 1 cor. 1), or than acb (by hyp.) Consequently xff 
 
 Vol. f. ea <Ag j- 
 
538 ELEMENTS OF ISOPERIMETRY. 
 
 <AC ; and in the right-angled triangles aec', aec, having ae 
 eommon, we have c'e < ce*. q. e. i>. 
 
 Cor. Of all isoperirpetrical figures, of which the number 
 of sides is given, that which is the greatest has all its sides 
 equal. And in particular, of all isoperimetrical triangles, that 
 whose surface is a maximum, is equilateral. 
 
 For, so long as any two adjacent sides are not equal, the sur- 
 face may be augmented without increasing the perimeter. 
 
 Remark. Nearly as in this theorem may it be proved 
 that, of all triangles of equal heights, and of which the sum 
 of the two sides is equal, that which is isosceles has the great- 
 est base. And, of all triangles standing on the same base 
 and having equal vertical angles, the isosceles one is the 
 greatest. 
 
 THEOREM m. 
 
 Of all Right Lines that can be drawn through a Given Point, 
 between Two Right Lines Given in Position, that which is 
 Bisected by the Given Point forms with the other two Lines 
 the Least Triangle. 
 
 Of all right lines gd, ab, GD^that 
 can be drawn through a given point 
 p to cut the right lines ca, cd, given 
 in position, that ab, which is bisect- 
 ed by the given point p, forms with 
 CA, CD, the least triangle, abc. 
 
 For, let EE be drawn through a G.%/g -"' 
 parallel to cd, meeting dg (produ- E 
 
 ced if necessary) in e : then the triangles pbd, pae, are man- 
 ifestly equiangular ; and, since the corresponding sides pb, pa 
 are equal, the triangles are equal also. Hence pbd will be 
 less or greater than pag, according as cg is greater or less 
 than CA. In the former case, let pacd, which is common, be 
 added to both ; then will bac be less than dgc (ax. 4 Geom.), 
 In the latter case, if pgcb be added, dcg will be greater than 
 bag;' and consequently in this case also bag is less than 
 
 DCG« Q. E. D. 
 
 Cor. If PM and pn be drawn parallel to cb and ca re- 
 spectively, the two triangles pam, pen, will be equal, and 
 
 * When two ipathematical quantities are separated by the charac- 
 ter < . it denotes that the preceding quantity is less than the succeed- 
 ing one : when, on the contrary, the sepaj-ating character is >, it de- 
 notes that the preceding quantity is greater than the succeeding one. 
 
 these 
 
SURFACES. 639 
 
 these two taken together (since am = pn = mc) will be equal 
 to the parallelogram pmcn : and consequently the parallelo- 
 gram PMCN is equal to half abc, but less than half dgc. 
 From which it follows (consistently with both the algebraical 
 and geometrical solution of prob 8, Application of Algebra 
 to Geometry), that a parallelogram is always less than half a 
 triangle in which it is inscribed, except when the base of the 
 one is half the base of the other, or the height of the former 
 half the height of the latter ; in which case the parallelogram 
 is just half the triangle : this being the maximum parallelo- 
 gram inscribed in the triangle. 
 
 Scholium. 
 
 From the preceding corollary it might easily be shown; 
 that the least triangle which can possibly be described about, 
 and the greatest parallelogram which can be inscribed in, any 
 curve concave to its axis, will be when the subtangejnt is equal 
 to half the base of the triangle, or to the whole base of the 
 parallelogram : and that the two figures will be in the ratio of 
 2 to I. But this is foreign to the present enquiry. 
 
 THEOREM IV. 
 
 Of all Triangles in which two Sides are Given in Magnitude, 
 the Greatest is that in which the two Given Sides are Per- 
 pendicular to each other. 
 
 For, assuming for base one of the given sides, the surface 
 is proportional to the perpendicular let fall upon that side, 
 from the opposite extremity of the other given side : there- 
 fore, the surface is the greatest when* that perpendicular is 
 the greatest; that is to say, when the other side is not in- 
 clined to that perpendicular, but coincides with it : hence the 
 surface is a maximum when the two given sides are perpendi- 
 cular to each other. 
 
 Otherwise, Since the surface of a triangle, in which two 
 sides are given, is proportional to the sine of the angle in- 
 cluded between those two sides ; it follows, that the triangle 
 is the greatest when that sine is the greatest ; but the greatest 
 sine is the sine total, or the sine of a quadrant ; therefore the 
 two sides given make a quadrantal angle, or are perpendicular 
 to each ether. q. e. d. 
 
 THBOREAf 
 
64a ELEMENTS OF ISOPERIMETRY. 
 
 THEOREM V. 
 
 Of 1^11 Rectilinear Figures in which all the Sides except oue 
 are known, the Greatest is that which may be Inscribed in 
 a Semicircle whose diameter is that Unknown Side. 
 
 For, if you suppose the contrary to be the case, then when-, 
 ever the figure made with the sides given, and the side un- 
 known is not inscribable in 2l semicircle of which this latter 
 is the diameter, viz. whenever any one of the angles, formed 
 by lines drawn from the extremities of the unknown side to 
 one of the summits of the figure, is not a right angle ; we 
 may make a figure greater than it, in which that angle shall 
 be right, and which shall only differ from it in that respect : 
 therefore, whenever all the angles formed by right lines 
 drawn from the several vertices of the figure to the extremi- 
 ties of the unknown line, are not right angles, or do not fall 
 in the circumfereace of a semicircle, the figure is not in its 
 maximum state. ft. e. dv 
 
 THEOREM VI. 
 
 Of all Figures made with sides Given in Number and Mag- 
 nitude, that which may be Inscribed in a Circle is the 
 Greatest. 
 
 Let ABCDEFG be the y 
 
 polygon inscribed, and jS^^zr^ a ■/^"^"'^^^'^(^ 
 
 abcdefg a polygon with ^yf^^^^^^^^^^^ / 
 
 equal sides, but not in- ( ■ r\ %\ ti- ^^n 
 
 ecribable in a circle ; j^tl.......""-- -v -^i? \ /^ 
 
 ^othatAB=a6,Bc — 6c, u^ ij V 
 
 &c.; it is affirmed thafe ^''^i::^-.^,^'^ ^^^„„^b 
 
 the polygon abcdepg <-^ ^ 
 
 is greater than the polygon ahcdefg. 
 
 Draw the diameter ep ; join ap, pb ; upon al? ^^ ab make 
 the triangle abp, equal in all respects to abp ; and join ep. 
 Then, of the two figures edcbp^ p(^gf^^ one at least is not (by 
 hyp ) inscribable in the semicircle of which ep is the diame- 
 ter. Consequently, one at least of these two figures is smaller 
 than the corresponding part of the figure apbcdefg (th. 5). 
 Therefore the figure apdcdefg is greater than the figure 
 apbcdefg : and if from these there be taken away the respec- 
 tive triangles apb, apbj which are equal by construction, there 
 will remain (ax 5 Geom.) the polygoa abcdefg greater than 
 fho volygon abcdefg, q. e. d. 
 
 ■* "^ ^ THEOREM 
 
SURFACES, 041 
 
 THEOREM Vll. 
 
 *l?he Magnitude of the Greatest Polygon which can be con- 
 taiaed under any number of Unequal Sides, does not at all 
 depend on the Order in which thoae Lines are connected 
 with each other. 
 
 For, since the polygon is a maximum under given sides, it 
 is inscribable in a circle (th. 6) And this inscribed polygon 
 is constituted of as many isosceles triangles as it has sides, 
 those aides forming the bases of the respective triangles, the 
 other sides of all the triangles being radii of the circle, and 
 their common summit the centre of the circle. Consequently, 
 the magnitude of the polygon, that is, of the assemblage of 
 these triangles, does not at all depend on their disposition, or 
 arrangement round the common centre. q.. e. 0. 
 
 THEOREM VHL 
 
 If a Polygon Inscribed in a Circle have all its Sides Equal, all 
 its Angles are likewise Equal, or it is a Regular Polygon. 
 
 For, if lines be drawn from the several angles of the poly- 
 gon, to the centre of the circumscribing circle, they will 
 divide the polygon into as many isosceles triangles as it has 
 sides ; and each of these isosceles triangles will be equal to 
 either of the others in all respects, and of course they will 
 have the angles at their bases all equal : consequently, the 
 angles of the polygon, which are each made up of two angles 
 at the bases of two contiguous isosceles triangles, will be equal 
 to one another. q. e. d. 
 
 THEOREM IX. 
 
 Of all Figures having the Same Number of Sides and Equ al 
 Perimeters, the Greatest is Regular. 
 
 For, the greatest 6gure under the given conditions has all 
 >9ides equal (th. 2. cor.). But since the sum of the sides and 
 the number of them are given, each of them is given : there- 
 fore (th. 6), the figure is inscribable in a circle : and conse- 
 quently (th. 8) all its angles are equal ; that is, it is regular. 
 
 Q. E. D. 
 
 Cor. Hence we see that regular polygons possess the pro- 
 perty of a maximum of surface, when compared with any 
 «Cher figures of the same aame and with equal perimeters. 
 
 THEOREM 
 
642 ELEMENTS OF ISOPERIMETRY. 
 
 THEOREM X. 
 
 A Regular Polygon has a Smaller Perimeter than an Irregular 
 one Equal to it in Surface, and having the Same Number of 
 Sides. 
 
 This is the converse of the preceding theorem, and may be 
 demonstrated thus : Let r and i be two figures equal in surface 
 and having the same number of sides, of which r is regular, i 
 irregular : let also r' be a regular figure similar to r, and hav- 
 ing a perimeter equal to that of i. Then (th. 9) r' > i ; but 
 I «« r ; therefore r' > r. But r' and a are similar ; conse- 
 quently, perimeter of r > perimeter of r ; while per. r' = 
 per. I {hy hyp). Hence, per. i > per. r. q. e. d. 
 
 THEOREM XI. 
 
 The Surfaces of Polygons, Circumscribed about the Same or 
 Equal Circles, are respectively as their Perimeters*. 
 
 I> 
 
 Let the polygon abcd be circumscribed 
 
 about the circle efgh ; and let this polygon 
 be divided into triangles, by lines drawn 
 from its several angles to the centre o of 
 the circle. Then, since each of the tan- 
 gents, AB, Bc, &c. is perpendicular to its 
 corresponding radius oe, of, &c. drawn to the point of con- 
 tact (th. 46 Geom.) ; and since the area of a triangle is equal 
 to the rectangle of the i)erpendicular and half the base (Mens, 
 of Surfaces, pr 2) ; it follows, that the area of each of the 
 triangles abo, bco,.&c. is equal to the rectangle of the radius 
 of the circle and half the corresponding side ab, bc, &c. : and 
 consequently, the area of the polygon abcd, circumscribing the 
 circle, will be equal to the rectangle of the radius of the cir- 
 cle and half the perimeter of the polygon But, the sur- 
 face of the circle is equal to the rectangle of the radius and 
 half the circumference (th. 94 Geom.). Therefore, the sur- 
 face of the circle, is to that of the polygon, as half the cir- 
 
 ♦ This theorem, togetlier with the anViajrous ones respectirg bodies 
 circumscribing" cyiindevs and spheres, were given by Emerson in his 
 Geometry, and their use in the themy of Isoperimeters was justsug- 
 geste4 I but the full appUcation of them to that theory is due to Simon 
 Lhuillicr, 
 
 cumference 
 
SURFACES. • b4$ 
 
 cumference of the former, to half the perimeter of the latter ; 
 or, as the circumference of the former, to the perimeter of 
 the latter. Now, let p and p' be any (wo polygons circum- 
 scribing a circle c : then, by the foregoing, we have 
 
 surf c : surf p : : circum. c : perim. p. 
 
 surf c : surf p' : : circum. c : perim. p'. 
 But, since the antecedents of the ratios in both these propor- 
 tions, are equal, the consequents are proportional : that is, 
 surf, p : surf p' : : perim. p : perim. p. q. e. d. 
 
 CoroL 1. And one of the triangular portions abo, of a po- 
 lygon circumscribing a circle, is to the corresponding circular 
 sector, as the side ab of the polygon, to the arc of the circle 
 included between ao and bo. 
 
 Cor. 2. Every circular arc is greater than its chord, and 
 less than the sum of the two tangents drawn from its extremi- 
 ties and produced till they meet. 
 
 The first part of this corollary is evident, because a right 
 line is the shortest distance between two given points. The 
 second part follows at once from this proposition : for ea -\- 
 ah being to the arch eih, as the quadrangle aeoh to the cir- 
 cular sector HiEO ; and the quadr?».ngle being greater than the 
 sector, because it contains it ; it follows that ea -\- ah is great- 
 er than the arch eih*. 
 
 Cor, 3. Hence also, any single tangeiit EA,'i3 greater than 
 its corresponding arc EI. 
 
 THEOREM XII. 
 
 li a Circle and a Polygon, Circumscribable about another 
 Circle, are Isoperimeters, the Surface of the Circle is a 
 Geometrical Mean Proportional between that Polygon and 
 a Similar Polygon (regular or irregular) Circumscribed 
 about that Circle. 
 
 Let c be a circle, p a polygon isoperimetrical to that circle, 
 and circumscribable about some other circle, and p' a polygon 
 similar to p and circumscribable about the circle c ; it is af- 
 firmed that p : c : : c : p'. 
 
 * This second corollary is introduced, not because of its immed-ate 
 connection with the subject under discussion, but because, notwith- 
 standingj its simplicity, some authors have employed whole pages in 
 attempting its demonstration, and failed at last. 
 
 For, 
 
544 ELEMENTS OP ISOPERIMETRY: 
 
 For, p : !>' : : perim^ . p : : perim^ . p' : : circum^ . c : p€rW. p^ 
 by th. 89, Geom. and the hypothesis 
 
 But (th. 1 1) p' : c : : per. p' : cir. c : : per^ p' : per. p'Xcir. c. 
 
 Therefore p : c : : - - - - cira . c : per. pXcir c. 
 : : cir. c : per. p' : : c : p'. . e. d. 
 
 THEOREM Xin. 
 
 U a Circle anci a Polygon, Circumscribable about another Cir- 
 cle, are Equal in Surface, the Perimeter of that figure is a 
 Geometrical Mean Proportional between the Circumference 
 of the first Circle and the Perimeter of a Similar Polygon 
 Circumscribed about it. 
 
 Let c = p, and let p' be circumscribed about c and similar 
 to c : then it is affirmed that cir c : per. p : : per p : per. p*- 
 For, cir. c : per. p : : c : p" : : p : p' ; : per^ p : per^ . p'. 
 Also, per. p' : per p - - - - : : per^ p':perpXper p'« 
 Therefore, cir c : per. p - - - : : per^. p : per. X per t\ 
 : : per. p : per. p'. ^. b. Dv. 
 
 THEOREM XIV. 
 
 The Circle is Greater than any Rectilinear Figure of the Same 
 Perimeter ; and it* has a Perimeter Smaller than any Recti- 
 linear Figure of the Same Surface. 
 
 For, in the proportion, p : c : : c : p', (th. 12), since c < p', 
 
 therefore p < c. 
 And, in the propor cir. c : per. p : : per p r per. p' (th. 13)» 
 or, cir. c : per. p' : : cir^ . c : per^ . p, 
 cir. c < per. p' ; 
 therefore, cir . c < per^ . p, or cir c < per. p. q. e. ». 
 
 Cor. 1. It follows at once, from this ancl the two preced- 
 ing theorems that rectilinear figures which are isoperimeters» 
 and each circumscribable about a circle, are respectively in 
 the inverse ratio of the perimeters, or of the surfaces, of 
 figures similar to them, and both circumscribed about one 
 and the same circle. And that the perimeters of equal rec- 
 tilineal figures, each circumscribable about a circle, are re- 
 spectively in the subduplicate ratio of the perimeters or of 
 the surfaces, of figures, similar to them, and both circumscribed 
 about one and the same circle. 
 
 Cor. 2. Therefore, the comparisson of the perimeters of 
 ei][ual regular figures, having different numbers g£ sides, and 
 
 ttiat 
 
SURFACES. 546 
 
 that of the surfaces of regular isoperimetrical figures, is re- 
 duced to the comparison of the perimeters, or of the surfaces 
 of regular figures respectively similar to them, and circum- 
 scribable about one and the same circle. 
 
 Lemma 1. 
 
 If an acute angle of a right-angled triangle be divided into 
 any number of equal parts, the side of the triangle opposite 
 to that acute angle is divided into unequal parts, which are 
 greater as they are more remote from the right angle. 
 
 Let the acute angle c, of the right- 
 angled triangle acf, be divided into equal 
 parts, by the lines gb, cd, ce, drawn from 
 that angle to the opposite side ; then shall 
 the parts ab, bd, &c. intercepted by the A tjt) 
 lines drawn from c, be successively longer 
 as they are more remote from the right angle a. 
 
 For the angles acd, bce, &c. beisg bisected by cb, cd, 
 &c. therefore by theor. 83 Geom. ac : cd : : ab : bd, and 
 Bc : CE : : BD : de, and dc : cf : : de : ef. And by th. 21 
 6reom. CD> ca, ce> cb, cf > cd, and so on : whence it 
 follows, that db> ab, de> db, and so on. Q. e- d. 
 
 Cor. Hence it is obvious that, if the part the most remote 
 from the right angle a, be repeated a number of times equal 
 to that into which the acute angle is divided, there will re- 
 sult a quantity greater than the side opposite to the divided 
 angle. 
 
 THEOREM XV. 
 
 If two Regular Figures, Circumscribed about the Sam6 Circle, 
 differ in their Number ot Sides by Unity, that which has 
 the Greatest number of Sides shall have the Smallest Pe- 
 rimeter. 
 
 Let CA be the radius of a circle, and ab, ad, the half sides 
 •f two regular polygons circumscribed about that circle, of 
 which the number of sides differ by unity, being C^ 
 respectively n -j- 1 and n. The angles acb, acd, 
 
 therefore are respectively the ^7 and the n th 
 
 part of two right angles : consequentl}? these /^ JB 1> 
 
 angles are as n and n -|- 1 : and hence, the angle may be 
 
 conceived divided into n '\- \ equal parts, ef which bod is one. 
 
 Vo7. I. 70 «on- 
 
546 ELEMENTS OF ISOPERIMETRY. 
 
 Consequently, (cor. to the lemma) (n + I) bd> ad. Taking, 
 then, unequal quantities from equal quantities we shall have 
 
 (n4"l) AD — (w+ 1) BD<(n -f- 1) AD — AD, 
 or, (w -f- I) AB<n . AD. 
 That is, the semiperimeter of the polygon whose half side is 
 AB is smaller than the semiperimeter of the polygon whose 
 half side is ad : whence tho proposition is manifest. 
 
 Cor. Hence, augmenting successively by unity the num- 
 ber of sides, it follows generally, that the perimeters of 
 polygons circumscribed about any proposed circle, become 
 smaller as the number of their sides become greater. 
 
 THEOREM XVI. 
 
 The Surfaces of Regular Isoperimetrical Figures are Greater 
 as the Number of their Sides is Greater : and the Perime- 
 ters of Equal Regular Figures are Smaller as the Number 
 of their Sides is Greater. 
 
 For, 1st. Regular isoperimetrical figures are (cor. 1. th. 14) 
 in the inverse ratio of fignres similar to them circumscribed 
 about the same circle. And (th. 15) these latter are smaller 
 when the number of sides is greater : therefore, on the 
 contrary, the former become greater as they have more sides. 
 2dly. The perimeters of equal regular figures are (cor. 1 
 th. 14) in the subduplicate ratio of the perimeters of similar 
 • figures circumscribed about the same circle : and (th. 15) 
 these latter are smaller as they have more sides : therefore 
 the perimeters of the former also are smaller when the num- 
 ber of their sides is greater. q. e. d. 
 
 SECTION 11. SOLIDS. 
 
 THEOREM XYH. 
 
 Of all Prisms of the Same Altitude, whose Base is Given in 
 Magnitude and Species, or Figure, or Shape, the Right 
 Prism has the Smallest Surface. 
 
 For, the area of each face of the prism is proportional to 
 its height ; therefore the area of each face is the smallest 
 when its height is the smallest, that is to say, when it is equal 
 to the altitude of the prism itself : and in that case the prism 
 is evidently a right prism. q. e, d. 
 
 THEOREM 
 
SOLIDS. 547 
 
 THEOREM XVIII. 
 
 Of all Prisms whose Base is Given in Magnitude and Species, 
 and whose Lateral Surface is the same, the Right Prism has 
 the Greatest Ahitude, or the Greatest Capacity 
 This is the converse of the preceding theorem, and may 
 
 readily be proved after the manner of theorem 2: 
 
 THEOREM XIX. 
 
 Of all Right Prisms of the Same Altitude, whose Bases are 
 Given in Magnitude and of a Given numher of Sides, that 
 whose Base is a Regular Figure has the Smallest Surface. 
 
 For, the surface of a right prism of given altitude, and base 
 given in magnitude, is evidently proportional to the perime- 
 ter of its base But th 10) the base being given in magni- 
 tude, and having a given number of sides, its perimeter is 
 smallest when it is regular : whence, the truth of the propo- 
 sition is manifest. 
 
 THEOREM XX. 
 
 Of two Right Prisms of the Same Altitude, and with Irregu- 
 lar Bases Equal in Surface, that whose Base has the Grtiatest 
 Number of sides has the smallest Surface : and, in particu- 
 lar, the Right Cylinder has a Smaller Surface than any Prism 
 ,of the Same Altitude and the Same Capacity. 
 
 The demonstration is analogous to that of the preceding 
 theorem, being at once deducible from theorems 16 and 14. 
 
 THEOREM XXI. 
 
 Of all Right Prisms whosi Altitudes and whose Whole Sur- 
 faces are Equal, and whose Bases have a Given Number of 
 Sides, that whose Base is a Regular Figure is the Greatest. 
 
 Let p, p', be two right prisms of the same name, equal in 
 altitude, and equal whole surface, the first of these having a 
 regular, the second an irregular base ; then is the base of the 
 prism p, less than the base of the prism p'. 
 
 For, let p'' be a prism of equal altitude, and whose base is 
 equal to that of the prism p' and similar to that of the prism p. 
 Then the lateral surface of the prism p" is smaller than the 
 lateral surface of the prism p' (th. 19) : hence, the total sur- 
 face 
 
648 ELEMENTS OF ISOFERIMETRV. 
 
 face of p ' is smaller than the total surface of p', and therefore 
 (by hyp.) smaller than the whole surface of p. But the prisms 
 y and p have equal altitudes and similar bases ; therefore the 
 dimensions of the base of p" are smaller than the dimensions 
 of the base of p. Consequently the base of p", or that of p', 
 is less than the base of p ; or the base of p greater than that 
 of p . q,. E. D. 
 
 THEOREM XXII. 
 
 Of Two Right Prisms, having Equal Altitudes, Equal Total 
 Surfaces, and Regular Bases, that whose Base has the 
 Greatest number of Sides, has the Greatest Capacity. And, 
 in particular, a right Cylinder is Greater than any Right 
 Prism of Equal Altitude and Equal Total Surface. 
 
 The demonstration of this is similar to that of the preceding 
 theorem, and flows, from th. 20. 
 
 THEOREM XXm. 
 
 The Greatest Parallelopiped which can be contained under 
 the Three Parts of a Given Line, any way taken, will be 
 that constituted of Equal length, breadth, and depth. 
 
 For, let AB be the given line, and, 
 
 if possible, let two parts ae, ed, be [ ■[ 1 '■ — 
 
 unequal. Bisect ad in c, then will A ' C E I) B 
 the rectangle under ae (= ac -j- ce) 
 
 and ED (= AC — ce), be less than ac^ , or than ac . cd, by the 
 square of ce (th. 33 Geom.). Consequently, the solid ae . 
 ED . DB, will be less than the solid ac . cd . db ; which is re-; 
 pugnant to the hypothesis. 
 
 Cor. Hence, of all the rectangular parallelopipeds, having 
 the sum of their three dimensions the same, the cube is the 
 greatest. •:: ' ' 
 
 THEOREM XXIV. 
 
 The Greatest Parallelopiped that can possibly be contained 
 under the Square of one Part of a Given Line, and the 
 other Part, any way taken, will be when the former Part Is 
 the Double of the latter. 
 
 Let AB be a given line, and 
 
 I !■ 
 
 AC = 2cB, then is ac^ . cb the ~ » • / 1 ijiT 
 gif-eatest possible. A D D C C B 
 
 For, 
 
SOLIDS. 549 
 
 For, let Ac' and c'b be any other parts into which the given 
 line AB may be divided ; and let ac, ac', be bisected in d, d', 
 respectively. Then shall ac^ . cb = 4ad . dc . cb (cor. to 
 theor. 31 Geom.) > 4ad' . d'c . cb, or greater than its equal 
 c'a^ . c'b, by the preceding theorem. 
 
 THEOREM XXV. 
 
 Of all Right Parallelopipeds Given in Magnitude, that which 
 has the Smallest Surface has all its Faces Squares, or is a 
 Cube. And reciprocally, of all parallelopipeds of Equal 
 Surface, the Greatest is a Cube. 
 
 For, by theorems 19 and 21, the right parallelepiped hav- 
 ing the smallest surface with the same capacity, or the great- 
 est capacity with the same surface, has a square for its base. 
 But, any face whatever may be taken for base : therefore, in 
 the parallelepiped whose surface is the smallest with the same 
 capacity, or whose capacity is the greatest with the same sur- 
 face, any two opposite faces whatever are squares : conse- 
 quently, this parallelepiped is a cube. 
 
 THEOREM XXVI. 
 
 The Capacities of Prisms Circumscribing the Same Right Cy- 
 linder, are Respectively as their Surfaces, whether Total 
 or Lateral. 
 
 For, the capacities are respectively as the bases of the 
 prisms ; that is to say (th. 11), as the perimeters of their 
 bases ; and these are manifestly as the lateral surfaces : whence 
 the proposition is evident. 
 
 Cor. The surface of a right prism circumscribing a cylin- 
 der, is to the surface of that cylinder, as the capacity of the 
 former, to the capacity of the latter. 
 
 Def. The Archimedean cylinder is that which circum- 
 scribes a sphere, or whose altitude is equal to the diameter of 
 its base. 
 
 THEOREM XXVn. 
 
 The Archimedean Cylinder has a Smaller Surface than any 
 other Right Cylinder of Equal Capacity ; and it is Greater 
 than any other Right Cylinder of Equal Surface. 
 
 Let c andc' denote two right cylinder?, of which the first is 
 Archimedean, the other not : then, 
 
 Ut, 
 
«65 . ELEMENfTS OF ISOPERIMETRY. 
 
 1st, If . . c == c', surf, c < surf, c : 
 2dly, if surf, c = surf, c', c > c. 
 
 For having circumscribed about the cylinders, c, c^ tht 
 right prisms p, p', with square bases, the former will be a 
 cube, the second not : and the following series of equal ra- 
 tios will obtain, viz, c : p : : surf, c : surf, p ; : base c : base p : ; 
 base c' : base p' : : c' : p' : : surf, c' : surf. p'. 
 
 Then, 1st : when c = c. Smce c : p : : c' : p', it follows 
 that p = p ; and therefore (th. 25) surf, p < surf. p. But, 
 surf, c : surf, p : : surf, c' : surf, p ; consequently surf. c< 
 surf. c'. Q E. Id. 
 
 2dly : when surf, c = surf. c'. Then, since surf, c : surf, 
 f : : surf, c : surf, p', it follows that surf, p = surf, p' ; and 
 therefore (th. 25) p > p'. But c : p : : c' : p' ; consequently 
 G>c. 0.. E. 2i>, 
 
 THEOREM XXVIII. 
 
 Of all Right Prisms whose Bases are Circumscribable a\)out 
 Circles, and Giv€n in Species, that whose Altitude is 
 Double the Radius of the Circle Inscribed in the Base, 
 has the Smallest Surface with the Same Capacity, and the 
 Greatest Capacity with the Same Surface. 
 
 This may be demonstrated exactly as the preceding theo- 
 rem, by supposing cylinders inscribed in the prisms. 
 
 Scholium, 
 
 If the base cannot be circumscribed about a circle, the right 
 prism which has the minimum surface or the maximum capa- 
 city, is that whose lateral surface is quadruple of the surface 
 of one end, or that whose lateral surface is two-thirds of the 
 total surface This is manifestly the case with the Archime- 
 dean cylinder ; and the extension of the property depends 
 solely on the mutual connexion subsisting between the proper* 
 ties of the cylinder, and those of circumscribing prisms. 
 
 THEOREM XXIX. 
 
 The Surfaces of Right Cones Circumscribed about a Sphere, 
 are as their Solidities. 
 
 For, it may be demonstrated, in a manner ianalogous to 
 the demonstrations of theorems 11 and 26, that these cones 
 
 are 
 
SOLIDS. 55t 
 
 are equal to right cones whose altitude is equal to the radius 
 of the inscribed sphere, and whose bases are equal to the 
 total surfaces of the cones : therefore the surfaces and solidi- 
 ties are proportional. 
 
 THEOREM XXX. 
 
 The Surface or the Solidity of a Right Cone Circumscribed 
 about a Sphere, is Directly as the Square of the Cone'i 
 Altitude, and Inversely as the Excess of that Altitude over 
 the Diameter of the Sphere. 
 Let VAT be a right-angled triangle which, 
 by its rotation upon va as an axis, generates 
 a right cone ; and bda the semicircle which 
 by a like rotation upon va forms the inscrib- 
 ed sphere: then, the surface or the solidity of 
 
 VA2 * 
 
 the cone varies as . 
 
 VB 
 
 For, draw the radius cd to the point of contact of the 
 semicircle at vt. Then, because the triangles vat, vdc, are 
 similar, it is at : vt : : cd : vc. 
 
 And, by compos, at : at + vt : cd : cd -|- cv = va ; 
 Therefore at^ : (at + vt) at : : cd : va, by multiply- 
 
 ing the terms of the first ratio by at. 
 But, because vb. vd, va are continued proportionals, 
 it is VB : VA : : vd^ : va^ : : cd^ : at^ by sim. triangles. 
 But CD : VA : : at^ : (at -f* vt) at "by l»he last ; and these 
 mult, give cd . vb : va^ : : cd^ : (at -|- ¥t) at, 
 
 or VB : CD : : va^ : (at -|- vt) at = cd . . 
 
 ^ ^ VB 
 
 But the surface of the cone, which is denoted by «• . at^-J- 
 «•. AT . vt*, is manifestly proportional to the first member 
 ©f this equation, is also proportional to the second member, 
 
 cr, since cd is constant, it is proportional to , or to a third 
 
 proportional to bv and av. And, since the capacities of these 
 circumscribing cones are as their surfaces (th. 29), the truth 
 #f the whole proposition is evident. 
 Lemma 2. 
 The difiference of two right lines being given, the third 
 proportional to the less and the greater of them is a minimum 
 when the greater of those lines is double the other. 
 
 * jr being ^ 3'Ul593' 
 
 Let 
 
552 ELEMENTS OP ISOPERIMETRY. 
 
 3 V 
 
 Let Av and bv be two right 
 lines, whose diflference ab is- 
 given, and let ap be a third 
 proportional to bv and av ; 
 then is ap a minimum when av = 2bv. 
 
 For, since ap : av : : aV : bv ; 
 
 By division ap : ap — av : : av : av — bv ; 
 
 That is, AP : vp : : av ; ab. 
 
 Hence vp . av = ap . ab. 
 But vp . AV is either = or <^Ap2 (cor. to th. 31 Geom> 
 and th. 23 of this chapter.) 
 
 Therefore ap . AB<iAp3 : whence 4ab<ap, or ap > 4ab. 
 Consequently, the minimum value of ap is the quadruple of 
 ab ; and in that case pv = va = 2ab. ft. e. d.* 
 
 THEOREM XXXI. 
 
 Of all Right Cones Circumscribed about the Same Sphere, 
 the Smallest is that whose Altitude is Double the Diame- 
 ter of the Sphere. 
 
 va2 
 For, by th. 30, the solidity varies as — (see the fig. t® 
 
 vb 
 that theorem) : an^, by lemma 2, since va — vb is given, the 
 
 va2 
 third proportional — is a minimum when va = 2ab. q. e. d. 
 
 VB 
 
 Cor. 1. Hence, the distance from the centre of the sphere 
 to the vertex of the least circumscribing cone, is triple the 
 radius of the sphere. 
 
 Cor. 2. Hence also, the side of such cone is triple the radius 
 of its base. 
 
 ♦ Though the evidence of a single demonstration, conducted oa 
 sound mathematical principles, is really irresistible, and therefore needs 
 no corroboration ; yet it is frequently conducive as well to mental im- 
 provement, as to mental delight, to obtain like results from different 
 processes. In this view it will be advantageous to the student, to con- 
 firm the truth of several of the propositions in this chapter by means of 
 the fluxional analysis. Let the truth enunciated in the ab ve lemma be 
 taken for an example: and let AB be denoted by a, av by a:, b v by x — a^ 
 
 Then we shall have x— a :x ::x : : the third proportional ; which 
 
 X — a 
 is to be a minimum. Hence, the fluxion of this fraction will be equal 
 
 x^x — 2axx 
 
 to zero (Flux. art. 51). That is (Flux. arts. 19 and 30), «= 
 
 (x-.a)2 
 a. Consequently, x^ — 2ax « o, and x »2a> or av =2ab ■^ h.ve. 
 ^ ' THEOREM 
 
SOLIDS. 663 
 
 THEOREM XXXIL 
 
 The Whole Surface of a Right Cone being Given,, the In- 
 scribed Sphere is the Greatest when the Slant Side of the 
 Cone is Triple the Kddius pf its Base. 
 For, let c and c' be two right cones of equal whole sur- 
 face, the radii of their respective inscribed spheres being 
 denoted by r and r' ; let the side of the cone c be triple 
 the radius of its base, the same ratio not obtaining in c' ; 
 and let c'' be a cone similar to c, and circumscribed about 
 the same sphere with c'. Then, (by th. 31) surf, c" <surf. c' ; 
 therefore surf c" <surf. c. But c" and c are similar, there- 
 fore all the dimensions of c" are less than the corresponding 
 dimensions of c : and consequently the radius r' of the sphere 
 inscribed in c" or in c', is less than the radius r of the sphere 
 inscribed in c, or r > r'. q. e. d. 
 
 Cor. T\ie capacity of a right cone being given, the in- 
 scribed sphere is the greatest when the side of the cone is 
 triple the radius of its base. 
 
 For the capacities of such cones vary as their surfaces 
 {th. 29). 
 
 THEOREM XXXUI. 
 
 Of all Right Cones of Equal Whole Surface, the Greatest 
 is that whose side is Triple the Radius of its Base ; and 
 reciprocally, of all Right Cones of Equal Capacity, that 
 whose Side is Triple the Radius of its Base has the Least 
 Surface. 
 
 For, by th. 29, the capacity of a right cone is in the com- 
 pound ratio of its whole surface and the radius of its inscribed 
 sphere. Therefore, the whole surface being given, the ca- 
 pacity is proportional to the radius of the inscribed sphere : 
 and consequently is a maximum when the radius of the in- 
 scribed sphere is such : that is, (th. 32) when the side of the 
 cone is triple the radius of the base*. 
 
 . Again, 
 
 ♦ Here again a similar result may easily be dedi'.ced from the method 
 of fluxions. Let the radius of the base be denoted by x, the slant side 
 of the cone by Zs its whole surface by <j2 , and 3'14159j by ?r. Then 
 the circumference of the cone*s basft will bfe 2wa:, its area ttx^ and the 
 convex surface ttxz. The whole surface io, tlierefore, = Trx^-^Trxz : 
 aa 
 
 and this being s= as , we have ^ = ar. But the altitude of the 
 
 •ttx 
 Vol. I. 71 cone 
 
664 ELEMENTS OF ISOPERIMETRY. 
 
 Again, reciprocally, the capacity being given, the surface 
 is in the inverse ratio of the sphere inscribed : \thereforc, it 
 is the smallest when that radius is the greatest ; that is (th. 32) 
 when the side of, the cone is triple the radius of its base. q. e. d. 
 
 THEOREM XlXIV. 
 
 The Surfaces, whether Total or Lateral, of Pyramids Cir- 
 cumscribed about the Same Right Cone, are respectively 
 as their Solidities. And, in particular, the Surface of a 
 Pyramid Circumscribed about a Cone, is to the Surface of 
 that Cone, as the SoHdity of the Pyramid is to the Solidity 
 of the Cone ; and these Ratios are Equal to those of the 
 Surfaces or the Perimeters of the Bases. 
 For, the capacities of the several solids are respectively as 
 their bases ; and their surfaces are as the perimeters of those 
 bases : so that the proposition may manifestly be demon- 
 strated by a chain of reasoning exactly like that adopted in 
 theorem 11. 
 
 cone is equal to the square root of the difference of the squares of the 
 
 a* 2a2 
 
 side and of the radius of the base ; that is, it is = v^ (■ ), 
 
 7r2x^ sr 
 
 And this multiplied into ^ of the area of the base, viz. by i'»'*3,give» 
 
 fl4 2a» 
 iiTjra ^ (— — — )> for the capacity of the cone. Now, this being 
 
 a maximum its square must be so likewise (Flux. art. 53), that is, 
 
 " • ' f " ■ " , or, rejecting the denominator, as constant, 04x2 — 
 
 9 
 27ra2x* must be a maximum. This, in fluxions, is 2a x'x — 85ra2a:3i 
 
 a2 
 = o ; whence we have a* — 4^Jtr» = o, and consequently x= ^ — i 
 
 At 
 and fls = AiTTX^. Substituting this value of a^ for it, in the value of z 
 a2 47ra?2 
 
 above given, there results z = — - ^c = - — a: =» 4«— a? ■= Sa^v 
 
 TTX TtX 
 
 Therefore, the side of the cone is triple the radius of its base. Or, 
 the square of the altitude is to the square of the radius of the base^ 
 as 8 to 1, or, to the sqiiare of the diameter of the base, as 2 to 1. 
 
 THEOREM 
 
SOLIDS. 555 
 
 THEOREM XXXV. 
 
 'f^he Base of a Right Pyramid being Given in Species, the 
 
 Capacity of that Pyramid is a Maximum with the Same 
 
 Surface, and on the contrary, the Surface is a Minimum 
 
 with the Same Capacity, when the Height of One Face is 
 
 Triple the Radius of the Circle Inscribed in the Base. 
 
 Let p and p' be two right pyramids with similar bases, the 
 
 height of one lateral face of p being triple the radius of the 
 
 circle inscribed in the base, but this proportion not obtaining 
 
 jvith regard to p : then 
 
 1st. If surf. T = surf, p', p > p' . 
 2dly, if . . p = . . p', surf, p < surf p'. 
 For, let c and c be right cones inscribed within the pyra- 
 mids p and p' : then in the cone c, the slant side is triple 
 the radius of its base, while this is not the case with respect 
 to the cone c'. Therefore, if c = c , surf, c < surf, c' and 
 if surf c = surf c', c > c' (th. 33). 
 
 But, 1st. surf p : surf c : : surf, 1p' : surf c' ; 
 whence, if surf, p = surf, p surf, c = surf c' ; 
 therefore c > c'. But p : c : : p' : c'. Therefore p > p'. 
 
 2dly, p : c : : p' : c'. Theref. if p=p', c =c : consequently 
 surf c < surf c'. But, surf, p : surf c : : surf, p' : surf, c'. 
 Whence, surf, p < surf, p' 
 
 Q)r. The regular tetraedron possesses the property of the 
 minimum surface with the same capacity, and of the maxi-. 
 mum capacity with the same surface, relatiFely to all right 
 pyramids with equilateral triangular bases, and, a fortiorij 
 relatively to every other triangular pyramid. 
 
 THEOREM XXXVI. 
 
 A Sphere is to any Circumscribing Solid, Bounded by Plane 
 Surfaces, as the Surface of the Sphere to that of the Cir- 
 cumscribing Solid. 
 
 For, since all the planes touch the sphere, the radius drawn 
 to each point of contact will be perpendicular ta each re- 
 spective plane. So that, if planes be drawn through the cen- 
 tre of the sphere and through ail the edges of the body, the 
 body will be divided into pyramids whose bases are the re- 
 spective planes, and their common altitude the radius of the 
 sphere,. Hence, the sum of all these pyramids, or the whole 
 circumscribing solid, is equal to a pyramid or a cone whose 
 
 base 
 
656 ELEMENTS OF ISOPERIMETRY. 
 
 base is equal to the whole surface of that solid, and altitude 
 equal to the radius of the sphere. But the capacity of the 
 sphere is equal to that of a cone whose base is equal to the 
 surface of the sphere, and altitude equal to its radius. Con- 
 sequently, the capacity of the sphere, is to that of the circum- 
 scribing solid, as the surface of the former to the surface of 
 the latter : both having in this mode of considering them, a 
 common altitude. Q,. e. d. 
 
 Cor. 1. All circumscribing cylinders, cones, &c. are to" 
 the sphere they circumscribe, as their respective surfaces. 
 
 For the same proportion will subsist between their indefi- 
 nitely small corresponding segments, and therefore between 
 their wholes. 
 
 Cor, 2. All bodies circumscribing the same sphere, are 
 respectively as their surfaces. 
 
 THEOREM XXXVU. 
 
 The Sphere is Greater than any Polycdron of Equal Surface. 
 
 For, first it maybe demonstrated, by a process similar to 
 that adopted in theorem 9, that a regular polyedron has a 
 greater capacity than any other polyedron of equal surface. 
 Let p, therefore, be a regular polyedron of equal surface to . 
 a sphere s. Then p must either circumscribe s, or fall partly 
 within it and partly out of it, or fall entirely within it. The 
 first of these suppositions is contrary to the hypothesis of the 
 proposition, because in that case the surface of p could not 
 be equal to that of s. Either the 2d or 3d supposition there- 
 fore must obtain ; and then each plane of the surface of p 
 must fall either partly or wholly within the sphere s : which- 
 ever of these be the case, the perpendiculars demitted from 
 the centre of s upon the planes, will be each less than the 
 radius of that sphere : and consequently the polyedron p 
 must be less than the sphere s, because it has an equal base, 
 but a less altitude. q. e. d. 
 
 Cor. If a prism, a cylinder, a pyramid, or a cone, be 
 equal to a sphere either in capacity, or in surface ; in the first 
 case, the surface of the sphere is less than the surface of any 
 of those solids ; in the second, the capacity of the sphere is 
 greater than that of either of those solids. ; 
 
 The theorems in this chapter will suggest a variety of 
 practical examples to exercise the student in computation. 
 A few such are given in the following page. 
 
 EXERCISES. 
 
SOLIDS. 567 
 
 EXERCISES. 
 
 Ex. 1. Find the areas of an equilateral triangle, a square, 
 a hexagon, a dodecagon, and a circle, the perimeter of each 
 being 36. 
 
 Ex. 2. Find the difference between the area of a triangle 
 whose sides are 3, 4, and 5, and of an equilateral triangle of 
 equal perimeter. 
 
 Ex. 3. What is the area of the greatest triangle which 
 can be constituted with two given sides 8 and 1 1 : and what 
 will be the length of its third side ? 
 
 Ex. 4. The circumference of a circle is 12, and the pe- 
 rimeter of an irregular polygon which circumscribes it is 15: 
 what are their respective areas ? 
 
 Ex. 5. Required the surface and the solidity of the great- 
 est parallelopiped, whose length, breadth, and depth, together 
 make 18 ? 
 
 Ex. 6. The surface of a square prism is 646 : what is its 
 solidity when a maximum ? 
 
 Ex. 7. The content of a cylinder is 169-645968 : what is 
 its surface when a minimum ? 
 
 Ex. 8. The whole surface of a right cone is 201-061952 : 
 what is its solidity when a maximum ? 
 
 Ex. 9. The surface of a triangular pyramid is 43-30127 : 
 what is its capacity when a maximum ? * 
 
 Ex. 10. The radius of a sphere is 10. Required the so- 
 lidities of this sphere, of its circumscribed equilateral cone, 
 and of its circumscribed cylinder. 
 
 Ex. 11. The surface of a sphere is 28-274337, and of an 
 irregular polyedron circumscribed about it 35 ; what are their 
 respective solidities ? 
 
 Ex. 12. The solidity of a sphere, equilateral cone, and 
 Archimedean cylinder, are each 500 : what are the surfaces 
 and respective dimensions of each ? 
 
 Ex. 13. If the surface of a sphere be represented by the 
 number 4, the circumscribed cylinder's convex surface and 
 whole surface will be 4 and 6, and the circumscribed equila- 
 teral cone's convex and whole surface, 6 and 9 respectively. 
 Show how these numbers are deduced. 
 
 Ex. 14. The solidity of a sphere, circumscribed cylinder, 
 and circumscribed equilateral cone, are as the numbers 4, 6, 
 and 9. Required the proof. 
 
 PROBLEMS 
 
[ 658 ] 
 
 J>JiOBLBMS RELATIVE TO THE DlVISrOlff OF FIELDS ©E 
 ' [^. OTHER SURFACES. 
 
 PROBLEM I. 
 
 To Divide a Triangle into two parts having a Given Ratio, 
 m : n " 
 
 1st. By a line drawn from one angle 
 of the triangle. 
 
 Make ad : ab : : m : m -{- n ; draw cd. 
 So shall ADC, BDC, be the parts required. 
 
 m n 
 Here, evidently, ad = ab, db = ab. 
 
 2dly. By a line parallel to one of the sides of the triangle^ 
 
 Let ABC be the given triangle, to be 
 divided into two parts, in the ratio of m 
 to n, by a line parallel to the base ab. 
 Make ce to eb as w to n: erect ed per- 
 pendicularly to cB, till it meet the semi- 
 circle described on cb, as a diameter, in 
 ®D. Make cp = cd : and draw through f, gf jj ab. So shall or 
 divide the triangle abc in the given ratio. 
 
 CD* 
 
 For, CE : cb = : : cd* (= cf?) : cb^. ButcEtEP ::m :», 
 
 CE 
 
 or CE : cb : : m : m -f- n, by the construction : therefor© 
 cf2 : cb2 : :in :tn -fn. And since A cgf : A cab ; : CF* : cb* ;, 
 it follows that cgf : cab : : m : m + n, as required. 
 
 Computation. Since cb2 : cf* : : m -{- n : w, therefore, 
 (m + n) cf2 =r. m . cb2 ; whence cf ^ (in -f ») = cb y'm,. or 
 
 771 m 
 cf = cb ^ ^.- In like manner, cg = ca ^- . 
 
 mrl;-n 
 
 3dly. By a line parallel to a given line. 
 
 Let HI be the line parallel to which 
 a line is to be drawn, so as to divide 
 the triangle aec in the ratio of m 
 to n. 
 
 By case 2d draw gf parallel to ab, 
 so as to divide abc in the given ratio. 
 Through f draw fe parallel to hi. 
 On CE as a diameter describe a semi- 
 circle ; draw gd perp. to ac, to cut 
 the semicircle in d. Make cp = cd : 
 through p, parallel to ef, draw pq,, the line required. 
 
 The 
 
DIVISION OF SURFACES. 659 
 
 The demonstration of this follows at once from case 2 ; be- 
 cause it is only to divide fce, by a line parallel to fe, into two 
 triangles having the ratio of fce to fcg, that is, of ce to cg. 
 
 Computation, cg and cf being computed, as in case I, the 
 distances ch, ci being given, and ct being to cq. as ch to ci : 
 the triangles cgf, gpq, also having a common vertical angle, 
 are to each other, as cg . cf to c^ . cp. These products there- 
 fore are eqiial ^ and since the factors of the former are known, 
 the latter product li known. We have hence given the ratio 
 of the two lines cp (= x) to cq, (— y) as ch to ci ; say, asp to 
 q ; and their product = cf . cg, say = ab : to find x and y. 
 
 abp abq 
 
 Here we find x = ^ — , y =^ y/ — • That is, 
 q P 
 
 CF .CG . CH CF . CO .Cl 
 CP := ^ ; CQ == v/ ' 
 
 CI CH 
 
 N. B. If the line of division were to be p»^rpendicular to 
 one of the sides, as tocA, the construction would be similar : 
 
 CP would be a geometrical mean between ca and — c&, 6 
 
 being the foot of the perpendicular from b upon ac. 
 4thly. By a line drawn through a given point p. 
 
 By any of the former cases draw Im (fig* 1) to divide the 
 triangle aibc, in the given ratio of m to n : bisect cl in r, and 
 through r and m let pass the sides of the rhomboid crsm. 
 Make ca = pe, which is given, because the point p is given 
 in position : make cd a fourth proportional to ca, cr, cm ; 
 that is, make ca : cr : : cm : cd ; and let a and c?, be two 
 angles of the rhomboid cahd^ figs. 1 and 2. pe, in figure 2, 
 Imping drawn parallel to ac, describe on ed as a diameter the 
 semicircle efd, on which set off c/* = ce = ap : then set off 
 d^ or dn' on ca equal to df^ and through p and m, p and m' 
 
 draw 
 
560 
 
 DIVISION OF SURFACES. 
 
 draw the lines lm, l'm', either of which will divide the triangle 
 in the given ratio.^ — The construction is given in 2 figs, merely 
 to avoid complexness in the diagrams. 
 
 The limitations are obvious from the construction : for, the 
 point L must fall between b and c, and the point m between a 
 and c ; ap must also be less than p6, otherwise ef cannot be ap- 
 plied to the semicircle on ed. 
 
 Demon. Because cr = ic/, the rhomboid crsm = triangle! 
 elm, and because ca : cr : : cm : cc/, we have ca . cd=cwi . cr, 
 therefore rhomboid cahd = rhomboid crsm = triangle elm. 
 By reason of the parallels cb, fed, and ca, ah, the triangles 
 aLP, c?GM, feop, are similar, and are to each other as the 
 squares of their homologous sides ap, dM, fep : now ed^ = ef^ 
 4-cf/'^, by construction ; and ed = rb, ef = ap, df = dm ; 
 therefore pfe^ = ap^ -}- du^ , or, the triangle pfeo taken away 
 from the rhomboid, is equal to the sum of the triangles aPL, 
 djAGj added to the part capod : consequently clm = cafed, as 
 required, By a like process, it may be shown that aLP, ^g'm', 
 pfeo', are similar, and «l'p + dou = pfeo' ; whence Fbda' = 
 aL'p, and cl'm' = caferf, as required. 
 
 Computation, cl, cm, being known, as well as ca, ap, or 
 cc, cp, cr = ^cZ, is known : and hence erf may be found by 
 the proportion ca : cr : : cm : cd. Then cd — cc = ed, and 
 ^ed^^ej^ = ^ed2-_ap2 = d/* = ^m = ck'. Thus cm is 
 
 C/ . cm 
 
 determined. Then we have = cl. 
 
 N. B. When the point is in one of the sides, as at m 
 make cl . cm . (m -f- n) = ca . cb . m, or, cl : ca : : m 
 {m-\-n) CM, and the thing is done. 
 
 5thly. By the shortest line possible. 
 
 Draw any line p^ dividing the triangle in 
 the given ratio, and so that the summit of the 
 triangle cpq shall be c the most acute of the 
 three angles of the triangle. Make cm = cn, 
 a geometrical mean proportional between cp 
 and CQ ; so shall mn be the shortest line pos- 
 sible dividing the triangle in the given ratio. — 
 The computation is evident. 
 
 Demons. Suppose mn to be the shortest 
 line cutting off the given triangle cmn, and 
 
 CG _£_ MN . MN = MG 4" «N = CG . COt M -}- 
 CG . COt N = CG (cot M-J-COt n). But, COt M-f- 
 
 cos M COS N sin (m-^k) 
 cot n = 
 
 ; then 
 . CB : 
 
 -+- 
 
 sm M sin N sm M . sm n 
 
 And (equa. 
 
 XVIII, 
 
DIVISION OF SURFACES. 561 
 
 XVIII, Analj^t. pi. trigonom.) sin m . sin n = Jcos (m— n)— ^cos 
 
 sin.(M+N) 
 (M-|-N)=icos(M— N)-flcosc.Theref.MN=cG. ; 
 
 ^CG8(m — N)«f-iCv«C 
 
 which expression is a minimum when its denominator is a 
 maximum : that is, When cos (m— n) is the greatest possible, 
 which is manifestly when m — n = o, ofm =n, or when the 
 triangle cmn is isosceles. That the isosceles triangle must 
 have the most acute angle for its summit, is evident from the 
 consideration, that since 2Acmn = cg . mn, mn varies in- 
 versely as CG ; and consequently mn is shortest when cg is 
 longest, that is, when the angle c is the most acute. 
 
 N. B. A very simple and elegant demonstration to this 
 case is given in Simpson's Geometry : vide the book on Max. 
 and Min. See also another demonstration at case 2d prob, 
 6th, below. 
 
 PROBLEM II. 
 
 To Divide a Triangle into three Parts, having the Ratio of 
 the quantities m, n, p. 
 
 1st. By lines drawn from one angle of the triangle to the 
 opposite side. 
 
 Divide the side ab, opposite the angle c G 
 
 from whence the lines are to proceed, in the 
 given ratio at d, e ; join cd, ce, and acd, 
 
 DOE, ECB, are the three triangles required. 
 
 The demonstration is manifest ; as is also the ,^.D E B 
 computation. . -^' 
 
 If it be wished that the lines of division be the shortest the 
 nature of the case will admit of. Jet them be drawn from the 
 most obtuse angle, to the opposite or longest side. 
 
 2dly. By lines parallel to one of the sides of the triangle. , 
 
 Make cd : dh : hb : : m : « : p. Erect 
 DE, HI, perpendicularly to cb, till they meet 
 the semicircle described on the diameter 
 CB in E and i. Make cf = ce, and ck = 
 ci. Draw gf through f, and lk through k, 
 
 parallel to ab ; so shall the lines gf and lk, 
 
 divide the triangle abc as required. <"> B 
 
 The demonstration and computation will be similar to those 
 in the second case of prob. 1. 
 
 3dly. By lines drawn from a given poiut on one of the 
 sides. 
 Vol. I. 72 Fig. 
 
562 DIVISION OF SURFACES. 
 
 Fig. 2. 
 
 A fljypw) B A. a b T B 
 
 Let p (fig. 1) be the given point, a and b the points which 
 divide the side ab in the given ratio of m, n, p : the point p 
 falling between a and 6. Join pc, parallel to which draw ac, 
 bd, to meet the sides ac, bc in the points c and d : join pc, 
 p</, so shall the lines cp, prf, divide the triangle in the given 
 ratio. 
 
 In fig. 2, where p falls nearer one of the extremities of ab 
 than both a and 6, the construction is essentially the same ; 
 the sole difference in the result is, that the points c, and d, 
 both fall on one side ac of the triangle. 
 
 Demon. The lines ca, c6, divide the triangle into the given 
 ratio, by case 1st. But by reason of the parallel lines aCy Pc, 
 bd, A O.CC A ocp, and A bdc = bdF, Therefore, in fig 1, 
 Aac -{- ac? = a«c + «fc that is, acp = Aac : and Bbd -\- bdp . 
 = tibd + bdc, that is, bc?p = b6c. Consequently, the re- 
 mainder ccpd = cab. — In fig. 2, acp = Aac, and xdv = ac6 ; 
 therefore cpd = acp ; and acb— A«ifp = acb— ac6, that is, 
 CB?d =■ chb. 
 
 Computation. The perpendiculars eg, cd being demitted, 
 A ACP : A ACB : ; w : m-^-n-^rp : : ap . c^ : ab . cd. Therefore 
 
 W . AB . CD 
 
 (m-]rn-\-p) AP.cg=ni.AB.cD, and eg = ■ . The line 
 
 (»m4"«+/')ap 
 eg being thus known, we soon find ac ; for cd : ac : : c^ : 
 
 AC ' eg ?w . AB . AC 
 
 AC = = . Indeed this expression may be 
 
 CD (mi-H+/')AP 
 
 deduced more simply ; for, since acb : acp : : ac . ab ; 
 AC . AP : : m-\-n-\-p : m, we have (m-f-w-f-p) ac . AP=m . ab . ac, 
 
 m . ab AC 
 and AC = . By a like process is obtained, in 
 
 (m+w4-/))AP 
 
 /J-AB BC »n+«) AB . AC 
 
 fig. 1, Bfi = ; and, in fig, 2, Ad = . 
 
 (ra-|-n+/>) PB C^^'+w+A) AP 
 
 4ihly. By lines drawn from a given point p within the 
 triangle. 
 
 Const, 
 
 J 
 
DIVISION OF SURFACES 
 c 
 
 663 
 
 Const. Through p and c draw the line cpp, and let the 
 triangle be divided into the given ratio by lines pc, pd^ drawn 
 from p to intersect ac, bc, or either of them ; according to 
 the method described in case 3 of this problem. Through p 
 draw PC, pc?, and respectively parallel to them, from p draw 
 the lines /?m, /jn : join pm, pn ; so shall these lines with p/), 
 divide the triangle in the given ratio. 
 
 Demon. The triangles cpm, crp^ are manifestly equal, as 
 are also c?pn, drp ; therefore cpm = cpc, and cpn = cpd ; 
 whence also, in fig. 1, cnpm =: cdpc, and, in fig. 2, cb/jpn = 
 CBpd. 
 
 cp cd 
 
 Compvi. Since cp . cn == cp . cd, we have cn = • 
 
 In like manner cm = 
 
 CP 
 
 cp cc 
 
 Remark. It will generally be best to contrive that the 
 smallest share of the triangle shall be laid off nearest the ver- 
 tex c of the triangle, in order to ensure the possibility of the 
 construction. Even this precaution however may sometimes 
 fail, ♦of ensuring the construction by the method above given : 
 when this happens, proceed thus : 
 
 By case 1 , draw the hues cc?, cc, from 
 the vertex c to the opposite side ab, to di- 
 vide the triangle in the given ratio Upon 
 AB set off any where mx, so that mn : ab : : 
 pp (the perp. from p on ab) : cp, the alti- 
 tude of the triangle. U mp and pn are to- 
 gether to be the least possible, then set off i mn on each side 
 the point p : so will the triangle mpn be isosceles, and its 
 perimeter (with the given base and area) a minimum. 
 
 5thly. By lines, one of which is drawn yrom a given angle 
 to a given point, which is also the point of concourse of the 
 other two lines. 
 
 Const. 
 
664 DIVISION OF SURFACES. 
 
 e 
 
 A 
 
 4 
 
 ^ 
 
 \^a & B 
 
 
 (^omt. By case 1st draw the lines ca, c6, dividing the 
 triangle in the given ratio, and so that the smaller portions 
 shall lie nearest the angles a and b (unless the conditions of 
 the divison require it to be otherwise). From p and a demit 
 upon AC the perpendiculars pp, ac ; and from ? and 6, on 
 Bc, the perpendiculars pg, hd. Make cm : ca : : .ac : vp^ and 
 CN : cB ; : 6c? : vq. Draw pm, pn, which, with cp, will divide 
 •the triangle as required. 
 
 When the perpendicular from h or from a, upon bc or ac, 
 is longer than the corresponding perpendicular from p, the 
 point N or m will fall further from c than b or a does. Sup- 
 pose it to be N : then make n' e : cb : : nc : cp, and draw pn' 
 for the line of division. 
 
 The demonstration of all this is too obvious to need trac- 
 ing here. 
 
 CA . ac 
 
 Corn-put. The perp. ca = Aa . sin a ; and cm = . 
 
 CB . bd 
 hd = Bb . sin b ; and cn = . 
 
 Fq 
 
 Gthly. By lines, one of which falls from the given po^nt of 
 concourse of ^11 three, upon a given side, in a given angle. 
 
 Suppose the given angle to be a right 
 angle, and p/" the given perpendicular : 
 which will simplify the operation, though 
 the principles of construction will be the 
 same. ^_ 
 
 Const, Let ca, c&, divide the triangle M A 
 in the given ratio. Make/N : cb : : bd : f/, 
 and /m : ca :.: ac : t/'^ and draw pn, pm, thus forming two 
 triangles p/n, p/m, equal to c6b, coa, respectively. If n fall 
 between / and b, and m between a and /, this construction 
 manifestly effects the division. But if one of the points, sup- 
 pose M, falls beyond the corresponding point a, the line pm 
 intersecting ac in e : then make m'e : ca : : cm. : cp, and draw 
 pm' : so shall p/, pm', pn, divide the triangle as required. 
 
 Compvt, 
 
DIVISION OF SURFACES. 66S 
 
 Comput. Here ca and bd, are found as in case 6th ; and 
 CB ' bd CA ' ac 
 
 l^ence> = ; and/M = . ThenPM=:v^(M/3+/p3), 
 
 p/ P/ 
 
 ^/ 
 and — = sin. m. Also I H0° — (m+a) » mca. Then sin m€a : 
 
 PM 
 
 sin M : sin a oc ma (=«/'— a/) : ac : mc. Again pe = pm-m« ; 
 A' . eM 
 
 and lastly M'e = ^- 
 
 ev 
 Here also the demonstration is manifest. 
 7thly. By lines drawn from the angles to meet in a deter- 
 minate point. 
 
 Construe. On one of the sides, as ac, set 
 off AD, so that AD : AC : : m : m -\- n -{- p. 
 And on the other, as ab, set off be, so that 
 
 BE : Bc : : n : m -{- n -\- p. Through n draw . 
 
 DG parallel to ab ; and through e, eh parallel -A. IE\ JS 
 to Bc ; to their poiut of intersection i draw lipes ai, bi, ci> 
 which will divide the triangle abc into the portions required* 
 Demon. Any triangle whose base is ab, and whose vertex 
 ialls in dg parallel to it, will manifestly be to abc, as ad to 
 AC, or as m to m + »* "f* p : so also, any triangle whose base 
 is EC, and whose vertex falls in eh parallel to it, will be to 
 ABC, as BE to ba, that is, as n to m -f- ^ 4* p« 
 
 Thus we have aib : acb : : m : m-|-n+p, 
 and . . . bic : acb : : n : in-\-n~{-py 
 therefore . aib : bic : : m : n. 
 And the first two proportions give, by composition. 
 aib 4- BIC : ACB f:m-|-n : m -\- n -^ p ; and by division, 
 acb — (aib -{-bic) : acb .•*: w -j- n + p' — {m-{-n) : m-\-n-\'p, 
 or Aic : agb -. \ p : m + n -^ p, consequently aib : bic : aic 
 <JC m : 71 : p. 
 
 n . AB tn . BC 
 Comput. be = Gi = ; bg = ; angle bgi 
 
 = 2 right angles — b. Hence, in the triangle bgi, there are 
 known two sides and in the included angle, to find the third 
 side BI. 
 
 C 
 Remark. When m=- n =p,the constraction ^//K 
 
 becomes simpler. Thus : from the vertex draw -^^^ni^ 
 GD to bisect AB ; and from b draw be in like a^-j^ 3 
 manner to the middle of ac : the point of inter- 
 section I of the lines cd, be, will be the point sought. 
 
 For, on be and be produced, demit, from the angles c and 
 
566 DIVISION OF SURFACES. 
 
 A, the perpendiculars ci, ak : then the triangles cei, aek, 
 are equal in all respects, because ae = ce, kae = ice, and 
 the angles at e are equal. Hence ak = ci. But these are 
 the perpendicular altitudes of the triangles bp( ,, bpa, which 
 have the common base bp. Consequently those two triangles 
 are equal in area. In a similar manner it may be proved, that 
 AFC = ape or CFB : therefore these three triangles are equal 
 to each other, and the lines pa, pb, pc, trisect the A abc. 
 
 PROBLEM III. 
 
 To Divide a Triangle into Four Parts, having the Proportion 
 of the Quantities m, n, p, q. 
 
 This like the former problems, might be divided into sev- 
 eral cases, the consideration of all which would draw us to a 
 very great length, and which is in a great measure unnecessary, 
 because the method will in general hf suggested immediately 
 on contemplating the method of proceeding in the analogous 
 case of the preceding problem. We shall therefore only take 
 one case, namely, that in which the lines of division must all 
 be drawn from a given point of one of the sides. 
 
 Let p be the given point in the side ab. 
 
 Let the points, /, m, w, divide the base ab 
 in the given proportion ; so will the lines cZ, 
 cm*, en, divide the surface of the triangle in 
 
 the same proportion. Join cp, and parallel ^^ — ^ 
 
 to it draw, from /, m, n, the lines II, mM, /iN, A./ 2» r ^ J5 
 to cut the other two sides of the triangle in l, m, n. Draw 
 PL, PM, PN, which will divide the triangle as required. 
 
 The demonstration is too obvious to need tracing through- 
 out : for the triangles l/p, lZc, having the same base l/, and 
 lying between the same two parallels l/, cp, are equal ; to 
 each of these adding the triangle al/, there results alp = ac/. 
 And in like manner the truth of the whole construction may 
 be shown. 
 
 The computation may be conducted after the manner of 
 that in case 3d prob. 2. 
 
 PROBLEM IV. 
 
 To Divide a Quadrilateral into Two Parts having a Given 
 Ratio, m ; n. 
 
 1st. By a line drawn from any point in the perimeter of 
 the figure. 
 
 Construe. From p draw lines pa, pb, 
 to the opposite angles a, b. Through d 
 draw DF parallel to pa, to meet ba pro- 
 duced in F : and through c draw ce pa- 
 
 rallel to pb to meet ab produced in e. c A M B E 
 
 Divide 
 
DIVISION OF SURFACES. 567* 
 
 Divide fe Iq m, in the given ratio of m to n : join p, m ; so 
 shall the line pm divide the quadrilateral as required. 
 
 Demon. That the triangle fpe is equal to the quadrangle 
 ABCD, may be shown by the same process as is used to demon- 
 strate the construction of prob. 36, Geometry ; of which, in 
 fact, this is only a modification. And the line pm evidently 
 divides fpe in the given ratio. But fpm :;= adpm, and epm = 
 BCPM : therefore pm divides the quadrangle also in the given 
 ratio. 
 
 Remark 1. If the line pm cut either of the sides ad, bc, 
 then its position must be changed by a process similar to that 
 described in the 6th and 6th cases of the last problem. 
 
 Remark 2. The quadrilateral may be divided into three, 
 four, or more parts, by a similar method, being subject how- 
 ever to the restriction mentioned in the preceding remark. 
 
 Remark 3. The same method may obviously be used when 
 the given point p is in one ol the angles of the figure. 
 
 Comput. Suppose i to be the point of intersection of the 
 sides DC and ab, produced ; and let the part of the quadrila- 
 teral laid off towards i, be to the other, as n to m. Then we 
 n(lD . lA-lB . ic) 
 
 have iM = . As to the distances di, ai, (since 
 
 {m-^n) IP 
 the angles at a and d, and consequently that at i, are known), 
 they areieasily found from the proportionality of the sides of 
 triangles to the sines of their opposite angles. 
 
 2dly. By a line drawn parallel to a given line. 
 
 Construe. Produce dc, ab, till 
 they meet, as at i. Join db pa- 
 rallel to which draw cf. Divide af 
 in the given ratio, in h. Through 
 D draw DG parallel to the given 
 line. Make ip a mean proportional \jcr^P~H~B" 
 between ih, ig ; through p draw 
 
 PM parallel to gd : so shall pm divide the quadrilateral abcd 
 as required. 
 
 Demon, It is evident, from the transformation of figures, 
 so often resorted to in these problems, that the triangle adf 
 = quadri^teral abcd (th. 36 Geom.) : and that dh divides 
 the triangle adf in the given ratio, is evident from prob. 1 
 case 1. We have only then to demonstrate that the triangle 
 iHD is equal to the trianstle ipm, for in that case hdf will 
 manifestly be equal to bcmp. Now, by construction, ih : 
 IP : : IP : IG : : (by the parallels) im : id ; whence, by making 
 the products of the means and extremes equal, we have 
 
568 DIVISION OF SURFACES. 
 
 ID . iH = IP . iM ; but when the products of the sides aboui 
 the equal angles of two triangles having a common angle are 
 equal, those triangles are equal ; therefore A ihd = A ipm. 
 
 Q. E. D. 
 
 Comput. In the triangle* adi, adg are given all the angles, 
 and the side ad ; whence ai, ag, di, and ic, = bi — dc, be- 
 come known. In the triangle ifc, all the angles and the side 
 IC are known ; whence if becomes known, as well as fh, 
 since ah : hf : : m : n. Lastly, ip = -v/(ih . ig), and ig : 
 ID : : IP : im. 
 
 Cor. 1 . When the line of division pm is to be perpendicu- 
 lar to a side, or parallel to a given side ; we have only to draw 
 DG accordingly : so that those two cases are included in this. 
 
 Cor. 2. When the line pm is to be the shortest possible, it 
 ciust cut off an isosceles triangle towards the acutest angle ; 
 and in that case ig must evidently be equal to id. 
 
 3dly. By a line drawn through a given point. 
 
 The method will be the same as that to case 4th prob. 1, 
 and therefore need not be repeated here. 
 
 Scholium. If a quadrilateral were to be divided into four 
 parts in a given proportion, m, n, p, g : we must first divide it 
 into two parts having the ratio of ?« ■+• w, to p -j- g ; and thea 
 each of the quadrangles so forined into their respective ra- 
 tios, of m to w, and p to gf. 
 
 PROBLEM V. 
 
 To Divide a Pentagon into Two Parts having a Given Ratio, 
 from a Criven Point in one of the Sides. 
 
 Reduce the pentagon to a triangle by prob. 37, Geometry, 
 and divide this triangle in the given ratio by case 1 prob. 1. 
 
 PROBLEM VL 
 
 To Divide any Polygon into Two Parts having a Giveii< 
 Ratio. 
 
 1st. From a given point in the perimeter of the polygon. 
 
 Construe. Join any two opposite ^~^C 
 
 angles a, d, of the polygon by the line A^^*^^^^ 
 
 ad. Reduce the part abcd into an ../" i\\ \\ 
 
 equivalent triangle nps, whose vertex •••'/ pV V'^; 
 
 shall be the given point p, and base ad [ j Ju /^^ 
 
 produced : an operation which may be G^^ j/jj 
 
 performed at once, if the portion abcd ^ KME 
 
 be quadrangular ; or by several opera- 
 tions. 
 
DIVISION OF SURFACES. 569 
 
 tions (as from 8 sides to 6, from 6 to 4, &c.) if the sides be 
 more than four. Divide the triangle nps into two parts bar- 
 ing the given ratio, by the line ph. In like manner, reduce 
 ADEFGA into an equivalent triangle having h for its vertex, 
 and FE produced for its base ; and divide this triangle into the 
 given ratio by a line from h, as hk. The compound line phk 
 will manifestly divide the whole polygon into two parts having 
 the given ratio. To reduce this to a right line, join pk, and 
 through H draw hm parallel to it ; join pm ; so will the right 
 line PM divide the polygon as required, provided m fall between 
 F and E. If it do not, the reduction may be completed by the 
 process described in cases 5th and 6th prob. 2d. 
 
 All this is too evident to need demonstration. 
 
 R&mark There is a direct method of solving this problem, 
 without subdividing the figure : but as it requires the compu- 
 tation of the area, it is not given here. 
 
 2dly. By the shortest line possible. 
 
 Construe. From any point f', 
 in one of those two sides of the 
 polygon which, when produced, 
 meet in the most acute angle i, 
 draw a line p'm', to the other 
 of those sides (ef), dividing the 
 polygon in the giren ratio. Find ^ ^ 
 
 the points p and m, so that ip or im shall be a mean propor- 
 tional between ip', im' ; then will *m be the line of division 
 required. 
 
 The demonstration of this is the same as has been already 
 given at case 5 prob. 1. Those, however, who wish for a 
 proof, independent of the arithmetic of sines, will not be dis- 
 pleased to have the additional demonstration below. 
 • 
 
 The shortest line which, with two other lines given in po- 
 sition, includes a given area, will make equal angles with those 
 two lines, or with the segments of them it cuts off from an 
 isosceles triangle. 
 
 Let the two triangles abc, aef, having the common angle 
 A, be equal in surface, and let the former triangle be isosce- 
 les, or have ab = ac ; then is eg shorter than ef. 
 
 Vol. I. 73 First, 
 
670 
 
 DIVISION OF SURFACES. 
 
 First, the oblique base ef cannot pass 
 through D, the rpiddle point of bc, as in 
 the annexed figure. For, drawing cg 
 parallel to ab, to meet ef produced in 
 G. Then the two triangles dbe, dcg 
 are identical, or mutually equal in all 
 respects. Consequently the triangle 
 DCF is less than dbe, and therefore abc 
 less than aef. f 
 
 ' EF must therefore cut bc in some point h between b and d, 
 and cutting the perp. ad in some point i above d, as in the 
 
 2d fig. Upon EF (produced if necessary) 
 demit the perp. ak. Then in the right- 
 angled A AiK, the perp. ak is less than 
 the hypothenuse ai, much more then is it 
 less than the other perp. ad. But^, of 
 equal triangles, that which has the greatest 
 perpendicular, has the least base. There- 
 fore the base bc is less than the base ef. q, e. d. 
 
 This series of problems might have been extended much 
 further ; bat the preceding will furnish a sufficient variety, 
 to suggest to the student the best method to be adopted in 
 almost any other case that may occur. The following practi- 
 cal examples are subjoined by way of exercise. 
 
 Ex. 1. A triangular field, whose sides are 20, 18, and 16 
 chains, is to have a piece of 4 acres in content fenced off from 
 it, by a right line drawn from the most obtuse angle to the 
 opposite side. Required the length of the dividing line, and 
 its distance from either extremity of the line on which it 
 falls? 
 
 . Ex. 2. The three sides of a triangle are 5, 12, and 13. If 
 two -thirds of this triangle be cut off by a line drawn parallel 
 to the longest side^ it is required to find the length of the 
 dividing line, and the distance of its two extremities from the 
 extremities of the longest side. 
 
 Ex. 3. It is required fo find the length and position of the 
 shortest possible line, which shall divide, into two equal parts, 
 a tjriangle whose sides are 25, 24, and 7 respectively. 
 
 Ejt:.'4. The sides of a triangle arc 6, 8, and 10 : it is re- 
 quired to cut off nine-sixteenths of it, by a line that shall pass 
 through the centre of its inscribed circle. 
 
 Ex. 
 
DIVISION OF SURFACES. 571 
 
 Ex. S. Two sides of a triangle, which include an angle of 
 70'', and 14 and J 7 respectively. It is required to divide it 
 into three equal parts, by lines drawn parallel to its longest 
 side. 
 
 Ex, 6. The base of a triangle is 1 1 2*66, the vertical angle 
 S7<» 57', and the difference of the sides about that angle is 8. 
 It is to be divided into three equal parts, by lines drawn from 
 the angles to meet in a point within the triangle. The lengths 
 of those lines are required. 
 
 Ex. 7. The legs of a right-angled triangle are 28 and 43. 
 Required the lengths of lines drawn from the middle of the 
 hypothenuse, to divide it into four equal parts. 
 
 Ex. 8. The length and breadth of a rectangle are 16 and 
 9. it is proposed to cut off one-iifth of it, by a line which 
 shall be drawn from a point on the longest side at the dis- 
 tance of 4 from a corner. 
 
 Ex. 9. A regular hexagon, each of whose sides is 12, is 
 to be divided into four equal parts, by two equal lines ; both 
 passing through the centre of the figure. What is the length 
 of those lines when a minimum ? 
 
 Ex. 10. The three sides of a triangle are 5, 6, and 7. How 
 may it be divided into four equal parts, by two lines which 
 shall cut each other perpendicularly ? 
 
 ^*^ The student will find that some of these examples will 
 admit of two answers. 
 
 On the Construction of Geometrical Problems, 
 
 Problems in Plane Geometry are solved either by means of 
 the modern or algebraical analysis, or of the ancient or geo- 
 metrical analysis. Of the former, some specimens are given 
 in the Application of Algebra to Geometry, page 369, &;c. of 
 this volume. Of the latter, we here present a few examples, 
 premising a brief account of this kind of analysis. 
 
 Geometrical analysis is the way by which we proceed from 
 the thing demanded, granted for the moment, till we have 
 connected it by a series of consequences with something an- 
 teriorly known, or placed it among the number of principles 
 known to be true, \ 
 
 Analysis 
 
572 GONSTRITCTION OF 
 
 AaslyUB may be distinguished into two kinds. In th« one, 
 which 18 named by Pappus, contemplative, it is proposed to 
 ascertain the truth or the falsehood of a proposition advanced; 
 the other is referred to the solution of problems, or to the 
 investigation of unknown truths. In the first we assume as 
 true, or as previously existing, the subject of the proposition 
 advanced, and proceed by the consequences of the hypothesis 
 to something known ; and if the result thus found be true, 
 the proposition advanced is likewise true. The direct de- 
 monstration is afterwards formed, by taking up again, in ^u 
 inverted order, the several parts of the analysis. If the con- 
 sequence at which we arrive in the last place is found fake, 
 we thence conclude that the proposition analysed is also false. 
 When a problem is under consideration, we first suppose it 
 resolved, and then pursue the consequences thence derived 
 till we come to somethi&g known. If the ultimate result 
 thus obtained be comprised in what the geometers call data, 
 the question proposed may be resolved : the demonstration 
 (or rather the construction), is also constituted by taking the 
 parts of the analysis in an inverted order. The impossibiHty 
 of the last result of the analysis, will prove evidently, in this 
 case as well as in the former, that of the thing required. 
 
 In illustration of these remarks take the following ex- 
 amples. 
 
 Ex. 1. It is required to draw, in a given segment of a 
 circle, from the extremes of the base a and b, two lines ac, 
 Bc, meeting at a point c in the circumference, such that they 
 shall have to each other a given ratio, viz. that of m to w. 
 
 Analysis. Suppose that the thing is af- 
 fected, that is to say, that ac : cb : : m : n, 
 and let the base ab of the segment be cut 
 in the same ratio in the point e. Then eg, 
 being drawn, will bisect the angle ace (by 
 th. 83Geom.); consequently, if the circle 'Tf' 
 
 be completed, and ce be produced to meet it in f, the re- 
 maining circumference will also be bisected in f, or have 
 FX = FB, because those arcs are the double measures of equal 
 angles : therefore the point f, as well as e, being given, the 
 point c is also given. 
 
 Construction. Let the given base of the segment Afe be 
 cut in the point e in the assigned ratio of m to n, and com- 
 plete the circle ; bisect the remaining circumference in f ; 
 join FE, and produce it till it meet the circumference in c ; 
 then drawing ca, cb, the thing is done. 
 
 Demonstration. Since the are fa =« the arc fb, the angle 
 ACF *» angle bcf, by theor. 49 Geom. ; therefore ac : cb ; : 
 
 AE : 
 
GEOMETRICAL PROBLEMS. 
 
 573 
 
 Az : EB, by th. 83. Bot ae : eb : : m : n, by constructioa ; 
 therefore ac : cb : : m : n. q. e. d. 
 
 Ex. 2. From a give^ circle to cut off an arc such, that 
 the suQEi of m times the sine, and n times the versed sine, may 
 be equal to a given lioe. 
 
 Anal. Suppose it done, and that aee'b is 
 the given circle, be'k the required arc, ed its 
 sine, BD its versed sine ; in da (produced if ne- 
 cessary) take HP anwth part of the given sum ; 
 join PE, and produce it to meet bf 4- to ab or || 
 Uf ED, in the point p. Then, sin^e m . eu -{■ n 
 . BD = '^ . Bf = n . Tt) -\- n . BD ; consequently 
 m . ED = n . PD ; hence pd : ed ; : m : Ji "^ 
 (by sim. tri.) pb : bf ; therefore pb : bf : 
 
 B 
 But pd 
 
 : ED : : 
 
 Now PB 
 
 is given, therefore bf is given in magnitude, and, being at right 
 angles to pb, is also given in position ; therefore the point f is 
 given and consequently pf given in position ; and therefore 
 the point e, its intersection with the circumference of the cir- 
 cle aee'b, or the arc be is given. Hence the followiog. 
 
 Const. From b, the extremity of any diameter ab of the 
 given circle, draw em at right angles to ab ; in ab (produced 
 i£ necessary) take bp an nth part of the given sum ; and on 
 BM take bf so that bf : bp : : n : m. Join pf, meeting the 
 circumference of the circle in e and e', and be or be' is the arc 
 required. 
 
 Demon. From the points e and e' draw ed and e'd' at right 
 angles to ab. Then, since bp : bp : : ra : m, and (by sim. tri.) 
 BP : BP : : de : dp ; therefore de : dp : : /t : 7/1. Hence m . de 
 = n . DP ; add to each n . bd, then will m - oe -|- n . bd = w . 
 BD -j- « • DP == w . PB, or the given sum. 
 
 Ex. 3. In a given triangle abh, to inscribe another tri- 
 angle abCf similar to a given one, having one of its sides pa- 
 rallel to a line mBu given by position, and the angular points 
 a, 6, c, situate in the sides ab, bh, ah, of the triangle abh 
 respectively. 
 
 Analysis. Suppose the thing done, 
 and that ahc is inscribed as required. 
 Through any point c in bh draw cd 
 parallel to mzn or to ab, and cutting 
 AB in D ; draw ce parallel to he, and 
 bE to ac, intersecting each other in e. 
 The triangles dec, ac6, are similar, and dc : a6 : : ce : be ; 
 also bdc, Bab, are similar, and pc : a6 : : bo : b6. Therefore 
 
674 CONSTRUCTION OF, &c. 
 
 Bc : CE : : Bh ; be; and they are about equal angles, conse- 
 quently B, E, c, are in a right line. 
 
 Construe. From any point c in bh, draw cd parallel to nm; 
 on CD constitute a triangle cde similar to the given one ; and 
 through its angles e draw be, which produce till it cuts ah in 
 € : through c draw ca parallel to ed and cb parallel to eg ; join 
 fl6, then abc is the triangle required, having its side ab parallel 
 to mn, and being similar to the given triangle. 
 
 Demon. For, because of the parallel hnes ac, de, and cb, 
 EC, the quadrilaterals bdec and Bac6, are similar ; and there- 
 fore the proportional lines dc, a6, cutting off equal angles 
 BDc, Bab ; BCD, Bba ; must make the angles edc, ecd, respec- 
 tively equal to- the angles cab. cha ; while ab is parallel to dc, 
 which is parallel to mBw, by construction. 
 
 Ex. 4. Given, in a plane triangle, the vertical angle, the 
 perpendicular, and the rectangle of the segments of the base, 
 made by that perpendicular ; to construct the triangle. ' 
 
 Anal. Suppose abc the triangle re- 
 quired, BD the given perpendicular to the 
 base AC, produce it to meet the periphery 
 ©f the circumscribing circle abch, whose 
 centre is o, in h ; then, by th. 61 Geom. 
 the rectangle bd . dh = ad . dc, the given 
 rectangle : hence, since bd is given, dh 
 and BH are given ; therefore bi = hi is given : as also 
 id = OE : and the angle eoc is = abc the given one, be- 
 cause Eoc is measured by the arc kc, and abc by half the 
 arc akc or by kc Consequently ec and ac = 2ec are given. 
 Whence this 
 
 Construction. Find dh such, that db . dh = the given 
 
 AD . DC 
 
 rectangle, or find dh = ; then on any right line 
 
 BD 
 
 GF take FE = the given perpendicular, and eg ~ dh ; bisect 
 FG in o, and. make eoc = the given vertical angle ; then 
 will oc cut EC, drawn perpendicular to oe, in c. With cen- 
 tre o and radius oc, describe a circle, cutting ce produced in 
 A : through f parallel to ac draw fb, to cut the circle in b ; 
 join AB, cb, and abc is the triangle required. 
 
 Remark. In a similar manner we may proceed, when it 
 is required to divide a given angle into two parts, the rect- 
 angle 
 
QUESTIONS IN MENSURATION. 57fi 
 
 angle of whose tangents may be of a given magnitude. Sec 
 prob, 40, Simpson's Select Exercises. 
 
 Note. For other exercises, the student may construct all 
 the problems, except the 24th, in the Application of Algebra 
 to Geometry, at page 369, &c. of this volume And that he 
 may be the better able to trace the relative advantages of the 
 ancient and the modern analysis, it will be adviseabit that he 
 solve those problems both geometrically and algebraically. 
 
 • 
 
 PRACTICAL EXERCISES IN MENSURATION. 
 
 Quest. 1. WHAT difference is there between a floor 
 28 feet long by 20 broad, and two others, each of half the 
 dimensions ; and what do all three come to at 45s. per square, 
 or 100 square feet ? 
 
 Ans. diff. 280 sq. feet Amount 18 guineas. 
 
 Quest. 2. An elm plank is 14 feet 3 inches long, and I would 
 have just a square yard slit off it ; at what distance from the 
 edge must the line be struck ? Ans. 7|i inohes. 
 
 Quest. 3. A ceiling contains 114 yards 6 feet of plastering, 
 and the room 28 feet broad ; what is the length of it ? 
 
 Ans. 36| feet. 
 
 Quest. 4. A common joist is 7 inch©* deep» and 2i 
 thick ; but wanting, a scantling just as big again that shafl 
 be 3 inches thick ; what will the other dimensions be ? 
 
 Ans. 11| inches. 
 
 Quest. 5. A wooden cistern cost me 3s. ^d. painting 
 within, at Cd. per yard ; the length of it was 102 inches, 
 and the depth 21 inches ; what was the width ? 
 
 Ans. 27i inches. 
 
 Quest. 6. If my court-yard be 47 feet 9 inches square, 
 and I have laid a foot-path with Purbeck stone, of 4 feet 
 wide, along one side of it, what will paving the rest with 
 flints come to, at 6d. per square yard ? Ans. bl. 16s. OJrf. 
 
 Quest. 7. A ladder, 26f feet long, may be so .planted, 
 that it shall reach a window 22 feet from the ground on 
 one side of the street*; and by only turning it over, without 
 moving the foot out of its place, it will do the same by a 
 
 window 
 
 i- 
 
a76 QUESTIONS IN MENSURATION. 
 
 windoiT 14 feet higk an the other side ; what is the breadth of 
 the street ? Ans. 37 feet 9| inches. 
 
 QuHST. 8. The paving of a triangular court, at 18d. 
 per foot, came to 100/. ; the longest of the three sides waa 
 88 feet ; required the sum of the oth^r two equal sides ? 
 
 Ans. 106-86 f^et. 
 
 Quest. 9. There are two columns in the ruins af Perse- 
 polis left standing upright : the one is 64 fe|t above the 
 plain, and the other 60 : in a straight line Between these 
 stands an ancient small statue, the head of which is 97 feet 
 from the summit of the higher, and 86 feet from the top of 
 the lower column, the base of which is just 76 feet from the 
 statue's base. Required the distance between the tops of the 
 two columns ? Ans. 167 feet nearly. 
 
 Quest. 10. The perambulator, or surveying wheel, is so 
 contrived, as to turn just twice in the length of 1 pole, or 
 16^ feet ; required the diameter ? Ans. 2-626 feet. 
 
 Quest. 11. In turning a one-horse chaise within a ring 
 of a certain diameter, it was observed that the outer wheel 
 made two turns, while the inner made but one : the wheels 
 were both 4 feet high ; and supposing them fixed at the dis- 
 tance of 5 feet asunder on the axletree, what was the circum- 
 ference of the track described by the outer wheel ? 
 
 Ans. 62-83 feet. 
 
 Quest. 12. What is the side of that equilateral triangle, 
 whose area cost as much paving at 8d. ai foot, as the palli- 
 sading the three sides did at a guinea a yard ? 
 
 Ans. 72-746 feet. 
 
 Quest. 13. In the trapezium abcd, are given, ab = 13, 
 Bc = 31i, CD = 24, and da = 18, also b a right angle ; re- 
 quired the area ? Ans. 410-122. 
 
 Quest. 14. A roof which is 24 feet 8 inches by 14 feet 
 6 inches, is to be covered with lead at 81b. per square foot : 
 what will it come to at 18s. per cwt. ? Ans. 22/. 19s. lOirf. 
 
 Quest. 1 5. Having a rectangular marble slab, 58 inches by 
 27, I would have a square foot cut off parallel to the shorter 
 edge ; I' would then have the like quantity divided from the 
 remainder parallel to the longer side ; and this alternately 
 repeated, till there shall not be the quantity of a foot 
 
 left.' 
 
(QUESTIONS IN MENSURATiaN. • 577 
 
 left ; what will be the dimensions^of the remaining piece ? 
 
 ' Ads. 20-7 inches by 6-086. 
 
 Q,UEST. 16. Given two sides of an ohtfise-an^led triangle, 
 which are 20 and 40 poles ; required the third side, that the 
 Irianglg may contain just an acre of land ? 
 
 Ans. 68-876 or 23-090. 
 
 QvEST. 17. The end wall of a house is 24 feet 6 inches 
 in breadth, and 40 feet to the eaves ; J of whicli is '^ bricks 
 thick, i more is 1^ brick thick, and the rest 1 brick thi( k. 
 Now the triangular gable rises 38 courses of bricks, 4 of 
 which usually make a foot in depth, and this is but 4^ inches, 
 or half a brick thick : what will this piece of work come to 
 at bl, 10s. per statute rod ? Ans. 20/. 1 Is. l^d. 
 
 Quest. 10. How many bricks will it take to build a wall, 
 10 feet high, and 500 feet long, of a brick and half thick : 
 reckoning the brick iO inches long, and 4 courses to the foot 
 in height ? Ans. 72000. 
 
 Quest. 19. How many bricks will build a square pyra- 
 mid of 100 feet on each side at the base, and als6 100 feet 
 perpendicular height : the dimensions of a brick being sup- 
 posed 10 inches long, 5 inches broad, and 3 inches thick ? 
 
 An». 3840000. 
 
 Quest. 20. If, from a right-angled triangle, whose base 
 is 12, and perpendicular 16 feet, a line be drawn parallel to 
 the perpendicular, cutting off a triangle wU^e area is 24 
 square feet ; required the sides of this trian,^le ? 
 
 Ans. 6, 8, and 10. 
 
 Quest. 21. The elhpse in Grosvenor square measures 
 840 links across the longest way, and 612 the shortest, within 
 the rails : now the walls being 14 inches thick, what ground 
 do they enclose, and what do they stand upon ? 
 
 . ^ enclose 4 ac r 6 p. 
 I stand on 1760| sq. feet. 
 
 Quest. 22. If a round pillar, 7 inches over, have 4 feet 
 of stone in it : of what diameter is the cofumn, of equal 
 length, that contains 10 times as much ? 
 
 Ans. 22-136 inches. 
 
 Quest. 23. A circular fish-pond is to be made in a gar- 
 den, that shall take up just half an acre ; what must be the 
 length of the cherd that strikes the circle ? Ans. 27f yards. 
 
 V»i.. I. 74 Quest. 
 
&n QUESTIONS IN MElNSURATION. 
 
 Quest. 24. When a roof is of a true pitch, or makiog a 
 right angle at the ridge, the rafters are nearly f of the 
 breadth of the building : now supposing the eves-boards to 
 project 10 inches 9n a side, what will the new ripping a 
 house cost, that measures 32 feet 9 inches long, by 22 feet 
 9 inches broad on the flat, at 15s. per square ? 
 
 Ans. SI. 15s. 9id. 
 
 Quest. 25. A cable, which is 3 feet long, and 9 inches in 
 compass, weighs 221b ; what will a fathom of that cable 
 weigh, which measures a foot about ? Ans. 78| lb. 
 
 Quest 26. My plumber has put 281b. per square foot 
 into a cistern, 74 inches and twice the thickness of the lead 
 long, 26 inches broad, and 40 deep : he has also put three 
 stays across it within, of the same strength, and 16 inches 
 deep, and reckons 22s. per cwt. for work and materials. I, 
 being a mason, have paved him a workshop, 22 feet 10 inches 
 broad, with Purbeck stone, at 7d. per foot ; and on the 
 balance I find there is 3s. 6d. due to him ; what was the 
 length of the workshop, supposing sheet lead of j\ of aa 
 inch thick to weigh 5'8991b. the square foot ? 
 
 Ans. 32 feet, Of inch. 
 
 Quest. 27. The distance of the centres of two circles^ 
 whose diameters are each 50, being given, equal to 30 ; what 
 is the area of the space enclosed by their circumferences ? 
 
 Ans. 659- 11 9. 
 
 Quest. 28. If 20 feet of iron railing weigh half a ton, 
 when the bars are an inch and quarter square ; what will 
 50 feet come to at 2^d. per lb. the bars being | of an inch 
 square ? Ans. 201. Os. 2d. 
 
 Quest. 29. The area of an equilateral triangle, whose 
 base falls on the diameter, and its vertex in the middle of 
 the arc of a semicircle, is equal to 100 : what is the diameter 
 of the semicircle ? Ans. 26-32148. 
 
 Quest. 30. It is required to find the thickness of the 
 lead in a pipe, of an inch and quarter bore, which weighs 
 141b. per yard in length j the eubic foot of lead weighing- 
 il325 ounces ? Ans. -20737 inches. 
 
 Quest. 31. Supposing the expence of paving a semicic- 
 cular plot, at 2s. 4d. per foot, come to 10/. ; what is the 
 iiameter «f it ? Ans. 14-7737 feef. 
 
 QvissT. 
 
viUESTIONS IN MENSURATION. 679 
 
 HuEST. 32. What is the length of a chord which cuts off 
 ^ of the area from a circle whose diameter is 289 ? ' 
 
 Ans. 278-6716. 
 
 Quest. 33. My plumher has set me up a cistern, and his 
 shop-book being burnt, he has no nieans of bringing in the 
 charge, and I do not choose to take it down to have it weigh- 
 ed ; but by measure he finds it contains 64y3_ square feet and 
 that it is precisely | of an inch in thickness. Lead was then 
 wrought at 21/. per fother of 19^ cwt. It is required from 
 these items to make out the bill, allowing 6f oz for the weight 
 of a cubic inch of lead ? Ans. 4L 1 Is. 2d. 
 
 Quest. 34. What will the diameter of a globe be, when the 
 solidity and superficial content are expressed by the same 
 number ? Ans. 6. 
 
 Quest. 35. A sack, that would hold 3 bushels of corn, is 
 22i inches broad when empty ; what will another sack con- 
 tain, which, being of the same length, has twice its breadth, 
 or circumference ? Ans. 12 bushels. 
 
 Quest. 36. A carpenter is to put an oaken curb to a 
 round well, at Sd per foot square : the breadth of the curb 
 is to be 71 inches, and the diameter within 3^ feet;* what 
 will be the expense ? Ans. bs. 2 jc?« 
 
 Quest. 37. A gentleman has a garden 100 feet long, and 
 80 feet broad ; and a gravel walk is to be made of an equal 
 width half round it ; what must the breadth of the walk be 
 to take up just half the ground ? Ans. 25-968 feet. 
 
 Quest. 38. The top of a may-pole, being broken aff by a 
 blast of wind, struck the ground at 10 feet distance from the 
 foot of the pole ; what was the height of the whole may-pole, 
 supposing the length of the broken piece to be 26 feet ? 
 
 Ans. 50 feet. 
 
 Quest. 39. Seven men bought a grinding stone, of 60 
 inches diameter, each paying } part of the expense ; what 
 part of the diameter must each grind down for his share ? 
 
 Ans. the 1st 4-4508, 2d 4-8400, 3d 5-3535, 4th 60765, 
 5th 7-2079, 6th 9-3935, 7th 22-6778 inches. 
 
 Quest. 40. A maltster has a kiln, that is 16 feet 6 inches 
 fl^aare : bat ^e wante to pull it dtwn, aad build a nevir one, 
 
 tlia^ 
 
68© QUESTIONS IN MENSURATION. 
 
 that may dry three times as much at once as the ol(i one ; what 
 must be the length of its side ? Ans. 28 feet, 7 inches. 
 
 Quest 41. How many 3-inch cubes may be cut out of a 
 12-inch cube? Ans. 64. 
 
 Quest. 42. How long must the tether of a horse be, that 
 will allow him to graze, quite round, just an acre of ground ? 
 
 Ans. 39i yards. 
 
 Quest 43. What will the painting of a conical' spire come 
 to, at 8</. per yard ; supposing the height to be 118 feet, and 
 the circumference of the base 64 feet ? Ans. 141. Os. 8|d. 
 
 Quest. 44. The diameter of a standard corn bushel is 
 18i inches, and its depth 8 ingles ; then what must the 
 diameter of that bushel be whose depth is 7^ inches ? 
 
 Ans. 19-1067 inches. 
 
 Quest. 45. Suppose the ball on the top of St. Paul's 
 church is 6 feet in diameter ; what did the gilding of it cost 
 at 3|d per square inch ? Ans. 237^ 10s. \d. 
 
 Quest. 46. What will a frustum of a marble cone come 
 to, at 125. per solid foot ; the diameter of the g|-eater end 
 being 4 feet, that of the less end H ; and the length of the 
 slant side 8 feet ? " Ans. 30/. Is. lO^d. 
 
 Quest. 47. To divide a cone into three equal parts by 
 sections parallel to the base, and to tind the altitudes of the 
 three parts, the height of the whole cone being 20 inches ? 
 
 Ans. the upper part 13*867. 
 the middle part 3*605. 
 the lower part 2*528. 
 
 Quest. 48. A gentleman has a bowling green, 300 feet 
 long, and 200 feet broad, which he would raise 1 foot higher, 
 by means of the earth to be dug out of a ditch that goes round 
 it : to what depth must the ditch be dug, supposing its breadth 
 to be every where 8 feet ? Ans. 7f | feet. 
 
 Quest. 49. How high above the earth must a person be 
 y^ised, that he may see i of its surface ? 
 
 :. Ans. to the height of the earth's diameter. 
 
 Quest, 
 
QUESTIONS IN MENSURATION. 
 
 ^1 
 
 "Quest. 60. A cubic foot of brass is to be drawn into wire, 
 ef -1^ of an inch in diameter ; what will the length of the wire 
 he, aJlowing no loss in the metal ? 
 
 Ans. 97784-7fi7 yards, or 65 miles 984'797 yards. 
 
 QjUEST. 51. Of what diameter must the bore of a cannon 
 be, which is cast for a ball of 24lb. weight, so that the diame- 
 ter of the bore may be -^^ of an inch more than that of the ball ? 
 
 Ans. 5*647 inches. 
 
 Quest. 62. Supposing the diameter of an iron 91b. ball 
 to be 4 inches, as it is very nearly ; it is required to find the 
 diameters of the several balls weighing 1, 2, 3, 4, 6, 12, 18, 
 24,32, 36, and 421b, and the caliber of their guns allowing j\ * 
 of the caliber, or -^g of the ball's diameter, for windage. 
 
 Answer, 
 
 Wt. of 
 
 Diameter 
 
 Caliber of 
 
 ball. 
 
 ball. 
 
 gun. 
 
 1 
 
 1-9230 
 
 1-9622 
 
 2 
 
 2-4228 
 
 2-4723 
 
 3 
 
 2-7734 
 
 2-8301 
 
 4 
 
 3-0526 
 
 3-1149 
 
 6 
 
 3-4943 
 
 3-5656 
 
 9 
 
 4-0000 
 
 4-oai6 
 
 12 
 
 4-4026 
 
 4-4924 
 
 18 
 
 6-0397 
 
 5-1425 
 
 24 
 
 6-5469 
 
 5-6601 
 
 32 
 
 6-1051 
 
 6-2297 
 
 36 
 
 6 3496 
 
 6-4792 
 
 42 
 
 6-6844 
 
 6 8208 
 
 Quest. 53. Supposing the windage of all mortars to be 
 ^ of the caliber, and the diameter of the hollow part of the 
 shell to be j\ of the caliber of the mortar : it is required to 
 determine the diameter and weitcht of the sh^ll, and the quan- 
 tity or weight of powder reqiiisite to fill it, foreach ol the sev- 
 eral sorts ef mortars, namely, the 13, 1^, 8^ 6*8, and 4-6 inch 
 mortar. 
 
 Answct; 
 
582 
 
 QUESTIONS IN MENSURATION. 
 
 Answer, 
 
 Ckl'ib. of 
 
 Diameter 
 
 Wt. of shell 
 
 Wt. of 
 
 Wt of shell 
 
 mort 
 
 of shell. 
 
 empty. 
 
 powder 
 
 filled. 
 
 4-6 
 
 4-623 
 
 8 320 
 
 0-583 
 
 8-903 
 
 68 
 
 5-703 
 
 16 677 
 
 1 168 
 
 17-845 
 
 
 
 43 764 
 
 3-065 
 
 46 829 
 
 10 
 
 . 9-833 
 
 85-476 
 
 6-986 
 
 91 462 
 
 13 
 
 12783 
 
 187 791 
 
 13-151 
 
 200 942 
 
 Quest. 54. If a heavy sphere, whose diameter is 4 inches, 
 be let fall into a conical glass, full of water, whose diameter 
 is '5, and altitude 6 inches ; it is required to determine how 
 much water will run over ? 
 
 Ans. 26 272 cubic inches, or neiarly ^ of a pint^ 
 
 Quest. 55. The dimensions of the sphere and cone being 
 the same as in the last question, and the cone only i full of 
 water ; required what part of the axis of the sphere is im- 
 mersed in the water ? Ans. -546 parts of an inch. 
 
 Quest. 56 The cone being still the same, and i full of 
 water ; required the diameter of a sphere which shall be 
 just all covered by the water ? Ans. 2-445996 inches. 
 
 Quest. 57. If a person, with an air balloon, ascend verti- 
 cally from London, to such a height that he can just see Oxford 
 appear in the horizon ; it is required to determine his height 
 above the earth, supposing its circumference to be 25000 
 miles, and the distance between London and Oxford 49*5^38 
 miles ? Ans. yVoV of a mile, or 547 yards I foot. 
 
 Quest. 68 In a garrison there are three remarkable objects 
 A, B, c, the distances of which from one to another are known 
 to be, AB 213, AC 424, and bc 262 yards ; 1 am desirous of 
 knowing my position and distance at a place or station s, from 
 which I observed the angle asb 13^ 30', and the angle csb 29^ 
 50' both by geometry and trigonometry. 
 Answer, 
 
 AS 605-7122 ; A C 
 
 BS 429-6814; , • V'-^Bx 
 
 ■cs 524-2365. 
 
 Quest. 69. Required the same as in the last question, 
 when the point b is on the other side of ac, supposing ab 9, 
 
qjJE9TION9 IN MENSURATION. 503 
 
 AC IS, and bc 6 furlongs ; also the angle asb 33* 45', and the 
 angle bsc 22® 30'. ^JR 
 
 Answer, 
 AS 10*64, BS 15-64, cs 1401. 
 
 QjUEST. 60. It is required to determine the magnitude of 
 a cube of gold, or the standard fineness, which shall be equal 
 to a sum of 480 millions of pounds sterling, supposing a guinea 
 to weigh 5 dwts 9| grains. Ans. 18-691 feet. 
 
 ^ Quest. 61. The ditch of a fortification is 1000 feet long^ 
 9 feet deep, 20 feet broad at bottom, and 22 at top ; how much 
 water will fill the ditch ? 
 
 Ans. 1158127 gallons nearly. 
 
 Quest. 62. If the diameter of the earth be 7930 miles, 
 and that of the moon 2160 miles : required the ratio of their 
 surfaces, and also of their solidities : supposing them both to 
 l^e globular, as they are ?ery nearly ? 
 
 Ans. the surfaces are as 13i to 1 nearly : 
 anil the soUdities as 49^ to 1 nearly. 
 
 is^m Qi^ Tm I'lRsr volvmb. 
 
:$'■ 
 
 
UNIVERSITY OF CALIFORNIA LIBRARY 
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 Return to desk from which borrowed. 
 This book is DUE on the last date stamped below. 
 
 OCTl 6 1953 Lu 
 
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 LD 21-100m-9,'48(B399sl6)476 
 
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A coursj of mathematics IBIQ 
 
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 THE UNIVERSITY OF CALIFORNIA UBRARY 
 
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