A GEOMETRICAL Practice, named PANTOMETRIA, divided into three Books, Longimetra, Planimetra and Stereometria, containing Rules manifold for mensuration of all lines, Superficies and Solides: with sundry strange conclusions both by instrument and without, and also by Perspective glasses, to set forth the true description or exact plat of an whole Region: framed by Leonard Digges Gentleman, lately finished by Thomas Digges his son.  
Who hath also thereunto adjoined a Mathematical treatise of the five regulare Platonical bodies, and their Metamorphosis or transformation into five other equilater unifoorme solides Geometrical, of his own invention, hitherto not mentioned of by any Geometricians.  
Imprinted at London by Henry Bynneman ANNO. 1571.   

To the right honourable my singular good Lord Sir Nicolas Bacon Knight, Lord keeper of the great seal of England.  
CAlling to memory right honourable, and my singular good Lord, the great favour your Lordship bore my father in his life time, and the conference it pleased your honour to use with him touching the Sciences Mathematical, especially in Geometrical mensurations, perusing also of late certain volumes that he in his youth time long sithence had compiled in the English tongue, among other I found this Geometrical practice, which my father (if God had spared him life) minded to have presented your Honour withal, but untimely Death preventing his determination, I thought it my part to accomplish the same, aswell for the satisfaction of his desire, as also to show myself not unmindful of so many good turns as your honour from time to time most abundantly hath bestowed on me, having therefore supplied such parts of this treatise as were left obscure or unperfect, adjoining thereunto a Discourse Geometrical of the five regulare or Platonical bodies, containing sundry Theorical and practical propositions of the manifold proportions arising by mutual conference of these solides Inscription, Circumscription or Transformation, and now at the last fully finished the same, I am bold to exhibit and dedicated it to your honour, as an eternal memorial of your Lordships great favour towards the furtherance of learning, and a public testimony of my bounden duty: hoping your honour will rather respect the good will wherewith it is presented, than the worthiness of the present, not agreeable I confess to the excellent knowledge wherewith your Lordship is endued, even in the very fountains themselves whence these conclusions as springs or branches are derived. And yet such as I nothing doubt your honour will both accept in good part, and also at vacant leisure from affairs of more importance, delight yourself withal, the rather for that it containeth sundry such new invented Theorems, and other strange conclusions, as no Geometers have hitherto, in any language published. Whereby your Lordship shall not only encourage me hereafter to attempt greater matters, but also as it were with a sovereign medicine prevent the poisoned infection of envious backbiting tongues: for as the verity of these experiments and rules shall never be impugned, being so firmly grounded, guarded, and defended with Geometrical demonstration, against whose puissance no subtle Sophistry or crafty coloured arguments can prevail, so think I there is none so impudently malicious, as will or dare reprove them for vain or improfitable, when they shall perceive your Lordship, (whose learned judgement, gravity and wisdom is sufficiently known to the world) doth allow and accept them as fragrante flowers selecte and gathered out of the pleasant gardens Mathematical, meet to delight any noble, free, or well disposed mind, and profitable fruits serving most commodiously to sundry necessary uses in a public weal, and such as shall thereby receive pleasure or commodity, must of duty yield condign thanks unto your Lordship, under whose protection and patronage I have not feared to send abroad (as a wandering pilgrim) this Orphan and fatherless child, the which as I perceive of your Honour favourably accepted, so mean I, God sparing life, to employ no small portion of this my short and transitory time in storing our native tongue with Mathematical demonstrations, and some such other rare experiments and practical conclusions, as no foreign Realm hath hitherto been, I suppose, partaker of. In the mean time I leave longer to detain your honour with my rude and homely tale, from more serious and weighty affairs, committing your Lordship to the tuition of the Almighty, who grant you a long healthful honourable life, accompanied with perfect felicity,  
Your Honour's most bounden Thomas Digges.   

The Preface to the Reader.  
ALthough Geometry for the certainty thereof have such a privilege as few other Sciences, being so fortified with Demonstration that no precept or Rule thereof for the verity can be reproved, yet lest some, either of ignorance or malice should affirm it unprofitable, not serving to any necessary use in a public Weal, I thought good somewhat to say thereof, and first the sentence of Plato written at the entry into his School cometh to my remembrance, 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉. thereby excluding all such as were ignorant of Geometry as unable or unmeet to attain higher secrets or mysteries of Philosophy. Aristotle also entreating of Moral Philosophy, in the fifth Book of his ethics, with geometrical Figures most beautifully painteth out justice, discerning and severally comparing sundry parts thereof with Geometrical and Arithmetical proportions. But his Interpreters, whose works are yet extant, for want of skill in these matters, have so blemished, darkened and defaced his meaning, that scarcely any resemblance thereof shall to the Reader appear. In sundry other his works also of natural Philosophy, as the Physikes, Meteores, de Coelo & Mundo. etc. ye shall find sundry Demonstrations, that without Geometry may not possibly be understanded. And to leave Philosophy, how necessary it is to attain exact knowledge in Astronomy, Music, Perspective, cosmography and Navigation, with many other Sciences and faculties, who so meanly travaileth therein shall soon find. But to omit talking of Geometry in general, and to speak privately of this treatise, I think good first to open the order and effect thereof, and then somewhat to say of the applyances. Like as every body geometrical is environed with Superficies, and every Superficies enclosed with lines, so that it consisteth of these three, Longitude, Latitude, and Profundity, and without consideration of these three no solid may be measured: so is this Treatise also divided into three Books. The first entreating of lines showeth sundry means to measure all manner lengths, heights, distances, and profundities. The second termed Planimetra sets forth diverse means and rules to measure manifold superficies, plain Conuex and Concave, whether they be compound with straight or circular lines, or mixed of both. In the third named Stereometria, is set out the exact mensuration of sundry solides, replenished with a number of rules and precepts, gathered out of Euclid, Archimedes and Appolonius Pergeus his Conykes, wherein the Reader shall not a little delight himself with the fineness and subtlety of their inventions, especially if he endeavour himself to search out the reason, cause and demonstration of them. And now somewhat to speak of the commodity of these conclusions, as the skilful in Architecture can apply the Stereometria to serve his turn in preordinance and forecasting both of the charges, quantities and proportion of all parcels necessarily appertaining to any kind of buildings: so Planimetra may serve for disposing all manner ground plats of Cities, Towns, Forts, Castles, Palaces or other edifices. The Marshal of the field shall also most speedily thereby appoint place convenient for his Camp, distributing every part thereof according to the number of his men, horse, carriage. etc. Also in surveying, parting and dividing of Lands and woods, it is most requisite aswell for exact as speedy dispatch therein, and thereof we have notable record in Histories how much this science availed the Egyptians when as by the inundation of Nilus their whole country was so drowned that with slime of the water all their Bounds and marks were defaced: Yet certain wise men aided with knowledge in this science, found out and distributed to every man his own. The other part named Longimetra the ingenious practizioner will apply to topography, fortification, conducting of mines under the earth, and shooting of great ordinance. So that as there is no kind of man, of what vocation or degree so ever he be, but shall find matter both to exercise his wit and diversly to pleasure himself, so surely for a gentleman especially that professech the wars, aswell for discoveries made by sea, as fortification, placing of Camps, & conducting of Armies on the land, how necessary it is to be able exactly to describe the true plats, symmetry and porportion of Forts, camps, towns and countries, coasts & harbours, I think there are none so unskilful, but will confess these Geometrical mensurations most requisite. But if any there be so rude, ignorant and unlearned, or so much blinded with self liking, that can not be content to acknowledge any thing requisite in a perfit soldier that is wanting in themselves (for such surely they are that most arrogantly will maintain this fond opinion) let them but regard and mark the renowned Captain Alexander the great, who had this knowledge in such high estimation, that he seldom or never in his manifoulde conquests would attempt any thing, whether it were for't, town or country, but first he would have the exact topography thereof, and thereupon invent, devise and after execute his warlike policies. On the contrary side, for negligence of this knowledge we may read how Aniball the most worthy and famous Soldier that ever hath been (all circumstances duly weighed) was by that painful, grave and right valiant Captain Quintus Fabius Maximus led and trained into a field enclosed with steep hills and deep rivers, where this noble Consul had so environed his Host by fortifying of two mountains, that he with all his Army were in marvelous jeopardy to have perished for famine, had not the crafty wit of Aniball invented a present policy to escape such imminente calamity. Cyrus also that great King and mighty Monarch, was he not through ignorance of topography, even in pursuing of victory entrapped and discomfited with all his power by the Scythian queen Thomiris at the river of Oaxis? I pass over innumerable examples of all ages, manifestly declaring the great advantage or disadvantage a Captain may receive, aswell invading as defending by the strong or weak situation of places, and by the foreknowledge or ignorance of them: neither is there any liberal or free mind, whether he be of profession Warlike or Civil, that will not take great delight and pleasure to see how by Art a man may measure the distances of places remote and far a sunder, approaching nigh none of them, and that aswell, yea and far more exactly than if with Cord or pole he should painfully pass them over. Hereupon did the Poetes fain Atlas of such huge and mighty parsonage, sustaining and upholding the earth and mighty Mass of heavenly spheres, for that this man notwithstanding he were imprisoned in a mortal carcase, and thereby detained in this most inferior and vilest portion of the universal world, farthest distant from that passing pleasant and beautiful frame of celestial Orbs, yet his divine mind aided with this science of Geometrical mensurations, found out the quantities, distances, courses, and strange intricate miraculous motions of these resplendent heavenly Globes of Sun, Moon, Planets and Stars fixed, leaving the rules and precepts thereof to his posterity. Archimedes also (as some suppose) with a glass framed by revolution of a section parabolical, fired the Roman navy in the sea coming to the siege of Syracuse. But to leave these celestial causes and things done of antiquity long ago, my father by his continual painful practices, assisted with demonstrations Mathematical, was able, and sundry times hath by proportional Glasses duly situate in convenient angles, not only discovered things far off, read letters, numbered pieces of money with the very coin and superscription thereof, cast by some of his friends of purpose upon Downs in open fields, but also seven miles of declared what hath been done at that instant in private places: He hath also sundry times by the Sun beams fired Powder, and discharged Ordinance half a mile and more distant, which things I am the boulder to report, for that there are yet living diverse (of these his doings) Oculati testes, and many other matters far more strange and rare which I omit as impertinente to this place. But for invention of these conclusions, I have heard him say, nothing ever helped him so much as the exquisite knowledge he had by continual practice attained in Geometrical mensurations. And for science in great Ordinance especially to shoot exactly at Randons' (a quality not unmeet for a Gentleman) without rules Geometrical, and perfect skill in these mensurations, he shall never know any thing: thus have I partly declared the pleasure and commodity that any well disposed mind may reap by these three books of my fathers. But somewhat to say concerning the last treatise of the five regulare bodies and their Metamorphosis adjoined by myself, for that I know some, yea many shall there be that will not spare to say, as they have done by others, it is a fond toy, a mere curious trifle, serving to no use or commodity: surely I mean not greatly to labour to attain the good opinion of such▪ having learned how the dispraise of the ignorant and vicious is no less commendation than the praise of the wise and virtuous, being a precedent infallible from the beginning and always continuing that ignorance envieth knowledge, and vice virtue. Every man disliketh or extolleth things accordingly as they are agreeable or repugnant to his appetite, and rather would I enjoining the company of Euclid, Archimedes, Appollonius Pergeus, and other geometers writing of the like toys & curiosities be condemned, than in publishing matter agreeable to such men's fantasy and conceit, be of them highly commended, and of the learned laughed at or contemned, but howsoever Epicurus, Midas, or their like, given only to lucre and worldly pleasure list to think thereof, I nothing mistrust of such as covet the understanding of matters hard and difficile, desiring the knowledge of things somewhat passing the reach and capacity of the common sort, wherein only the nature of man surmounteth beastly kind, or by proof and assay in cases of like difficulty able to judge themselves, this treatise shall not be misliked but thankfully received, and for the rest persuasions are but vain, for as no words can add stomach or make the coward valiant, so surely such two footed moles and Toads whom destiny and nature hath ordained to crawl within the earth, and suck upon the muck, may not possibly by any vehement exhortation be reduced or moved to taste or savour any whit of virtue, science, or any such celestial influence, my hope is if any fault escape, as in such long and intricate tedious calculations of irrational numbers may happen to be some, the discrete and modest Reader will rather of courtesy amend it, than with envious cavillation, ungratefullye requited my painful travails, whereby I shall be provoked not only to publish the demonstrations of these and many more strange and rare Mathematical Theorems, hitherto hidden and not known to the world, but also to imprint sundry other volumes of my fathers, which he long sithence compiled in the English tongue, desiring rather with plain and profitable conclusions to store his native language and benefit his Country men, than by publishing in the Latin rare and curious demonstrations, to purchase fame among strangers.   

Elements of Geometry, or definitions.  
A Point I call which cannot be divided, whose part is nothing.  
A line is a length without breadth or thickness, whose extremities are two points.  
THe shortest drawn between two Points is a straight line, the contrary are crooked lines.  
A Superficies is that hath length and breadth only, being bounded or determined with lines.  
A Plain Superficies is that which lieth equally and evenly between his lines or bounds.  
A Plain Angle is the inclination of two lines lying in one plain Superficies, concurring or meeting in a point.  
IF those lines that contain the Angle be strait, it is called a right lined Angle, and those two lines his containing sides, but if a third strait line be drawn cross the former from one to the other, that shall be called the subtending side.  
OF strait lined angles there are three kinds, the Orthogonall, the Obtuse, and the Acute Angle.  
WHen any right line falleth Perpendicularly upon an other, that is to say, making the Angles on either side equal, each of those Angles is an Orthogonall or right Angle, and that falling line a Perpendicular.  
BAC the right angle contained of the Perpendicular, and one part of the ground line equal to BAD the right angle contained of the Perpendicular, and the other portion of the ground line, and therefore both Orthogonall.  
The Broad or Obtuse Angle is greater than the Orthogonal.  
The Acute or sharp, is lesser than the right angle.  
A Figure is comprehended within limits and bounds, whether it be one or many.  
A Circle is a plain figure, determined with one line, which is called a Circumference, in whose mids there is a point named his Centre. From the which all right lines drawn to the circumference are equal.  
A Semicircle or half Circle, doth contain both the Dimetient and Centre of his circle, with the precise half of his circumference.  
A Right line drawn through the Centre unto the Circumference of both sides, is named his Diameter or Dimetient, the half of it is called his Semidiameter.  
ALl Strait lines besides the Diameter in any Circle pulled from one part of the Circumference to the other, be called cords.  
THe portion of the Circumference from that cord comprehended, is named an Aroke.  
A Touch line is that toucheth a circle in a Point.  
IF the Triangles three sides be every of them of like length, it is called an Equilater Triangle.  
ISoscheles is such a Triangle as hath only two sides like, the third being unequal, and that is the Base.  
Schalenum hath three unequal sides.  
A Rightangled Triangle is such a one as hath one Right Angle.  
AN Obtusiangle Triangle hath one obtuse angle, and is called Ambligonium.  
Oxigonium hath all acute or sharp Angles.  
THere be also four sided Figures called Quadrangles, whose Opposite sides and angles are equal, such are named Paralelogrammes, whereof there are but four sorts.  
IF all the sides be equal, and all the angles right, than is that Paralelogramme called a square.  
IF one side containing the right Angle, be longer than the other containing side, then is that figure called a Rectangle.  
IF all the sides be equal, and no angle aright, then is it called Rhombus.  
BUt if it have only the Opposite sides equal, and the other that contain an Angle unequal, it shall be named rhomboids.  
All other quadrangles are Trapezia.  
POlygona, are such Figures as have more than four sides, whose angles if they be all like and equal, they are termed Equiangle Polygona.  
ALL other plain Superficies, whether they be environed with strait or crooked lines, shallbe named irregulare figures.  
WHen two right lines drawn in one plain Superficies, are so equedistantly placed, that though they were infinitely extended on either side, yet would never meet nor concur, they shall be called Parallels.  
A Quadrant is the fourth part of a Circle, included with two Semidiameters commonly divided in .90. portions, which parts are named grades or degrees.  

The first Chapter. How Perpendiculares upon any strait line are ereared,  
ADmit AB were the line to be crossed, 〈◊〉 that ye desired a Perpendiculare or plumb line in C, open your compass, put the one foot in C, make of either side the line one prick, D.E. Now extend the compass to the wideness of both, or shorter, putting the one foot in D and the other immovable, making an ark over and under C. This done discreetly, remove the compass from that Centre to E (remaining so opened) there fix one foot, with the other cross the ark afore made above and beneath C, where make two points, or these letters FG. Then take a Ruler and lay him upon both the points crossing the centre C. Thus draw your plumb or squire line FCG. In like manner any line may be divided in half, or circle in four equal parts. See the example on the side following▪  
  

The second Chapter. How perpendiculare or hanging lines are drawn from a point assigned to any right line lying in the same plain Superficies▪  
SUppose C the point from whence I would let fall a Perpendiculare to the straight line A.B. open therefore your compass of such wydnesse, that placing the one foot in C. ye may reach with the other beyond the line A. B. and drawing an Ark, note where it maketh intersections with the same line, which in this Figure are marked with the letters D. and E. then placing one foot of your compass in E draw an ark directly under C. and then fixing one foot again of your compass in D. cross the former ark in F, finally draw the straight line CF. for that is a perpendicular to the line AB.  
Pithagoras' invention might here take place, who did find these numbers 3.4. and 5. or like joined measures to make a right angle.   

The third Chapter. From any point assigned to extend a Parallel to any other right line lying in the same Superficies.  
FRom the point assigned let fall a perpendicular to that line, and from some other point in that line ereare a perpendiculare, as ye were in the last Chapters taught, then opening your compass to the length of the perpendiculare let fall from the point assigned, measure out the like length in the perpendiculare ereared, beginning from the ground line: then laying a ruler to the point assigned, and the end of that length draw a straight line, for that shall be a Paralelle to the other.  

Example.  
ADmit A the point assigned, BC the line AD. the perpendicular let fall from A, CF. the perpendiculare ereared from C. my compass opened to the wydnesse of AD, I set one foot in C, cutting CF, with the other in E, then applying my ruler, I draw the line A E, which is a Parallel to B.    

The 4. Chapter. To divide any limited right line into as many equal parts as ye list.  
FOrasmuch as hereafter in drawing of plats and mensuration of ground by instrument, the division of right lines into many equal parts is required, I thought good to give instruction thereof before I entreat of those matters: ye shall therefore upon either end or limit of the divisible line, ereare a Perpendiculare, the one upward the other downward, and opening your compass at adventure, measure out so many parts in either perpendiculare, as you would make divisions in your line, and drawing right lines from the points in one perpendiculare to the points in the other, beginning from the first of the one, to the last of the other, ye shall divide the line given, into so many equal parts as there be divisions in your perpendiculares.  

Example.  
ADmit the line AB which I would divide into 7 equal portions, I ereare upon A and B, the perpendiculars AC, BD, as you may behold in the Figure: and opening my compass at adventure, I measure out 7 parts ending at EF. then drawing lines from the divisions of the one to the divisions in the other (beginning from the last in one perpendicular, to the first in the other) you may behold in the Figure the line A B, parted into 7 equal portions, in this manner may you proceed infinitely to divide it into as many portions as you list.    

The 5. Chapter How equal angles are made.  
Fix one foot of your compass upon the concourse or meeting of those two right lines that contain the angle whose like or equal you would make, and opening your compass at pleasure, describe an ark cutting the two containing sides of the angle: then draw an other right line & placing one foot of the compass (remaining immovable) thereon, with the other describe an ark rising from that last drawn line, then resort to your angle, and open your compass to the wydnesse or distance of the two intersections made by the ark in the two containing sides, and transporting the same distance to your second ark, set one foot of the compass at the beginning thereof, I mean where it riseth from the line, with the other cut the last described ark, then laying your ruler to that intersection and to the centre of the ark, draw a right line till it concur with the other: thus have you a new angle equal to the former.  

Example.  
SVppose BAC, the angle whose like or equal I desire, DE, the ark drawn with one foot of my compass, while the other remained in A, the compass immovable, I set one foot in the line FG, drawing with the other the ark KI. This done, I open my compass to the distance of DE, and placing one foot in K, with the other I cross the ark in I, finally, laying the ruler to I and F, I draw the line FH. and thus have I made the angle HFG. equal to the first angle BAC.    

The 6. Chapter. To make a triangle equal to any other right lined triangle assigned.  
first, as ye were taught in the last proposition make an angle equal to some one angle in that triangle, it forceth not which of them it be. Then extend out those straight lines that contain this angle till they be of equal length with the containing sides of his corresponding angle in the triangle. This done, couple the ends of those two right lines together with a third, and so have you framed a Triangle equal to the former.  

Example.  
THe triangle assigned is ABC, to his angle at B by the former problem I frame an equal or like, contained with the lines DEF, extending ED and OF, till they become of equal length with AB and BC, which I may easily do, by extending my compass first to AB, BC, and after transferring those distances or lengths to DE and EF. Finally applying my ruler to the ends or limits of those lines, I draw the subtending side DF, and thus have I framed a new triangle DEF, equal to the other ABC.  
I Think it not amiss before I entreat of Geometrical mensurations, to premise certain Theorems whereby the ingenious may readily conceive the ground, reason, and demonstration of such rules as shall ensue.  

The first Theorem.  
ENy two right lines crossing one an other, make the contrary or vertical angles equal.   

The second Theorem.  
IF any right line fall upon two Parallel right lines, it maketh the outward angle on the one, equal to the inward angle on the other, and the two inward opposite angles on contrary sides of the falling line also equal.   

The third Theorem.  
IF any side of a triangle be produced, the outward angle is equal to the two inward opposite angles, and all three angles of any triangle joined together, are equal unto two right angles.   

The fourth Theorem.  
IN equiangle triangles, all their sides are proportional aswell such as contain the equal angles, as also their subtendente sides.   

The fifth Theorem.  
IF any four quantities be proportional, the first multiplied in the fourth, produceth a quantity equal to that which is made by multiplication of the second in the third.   

The sixth Theorem.  
THE visible beams falling on plain conveses or concave glasses, are reflected in equal angles.   

The seventh Theorem.  
IN right angled triangles the square of the side subtending the right angle, is equal to both the squares of his containing sides.   

The eight Theorem.  
ALL Parallelogrammes are double to the triangles that are described upon their bases their altitudes being equal.   

The ninth Theorem.  
ALL like or equiangle Figures retain double the proportion of their correspondent sides.   

The tenth Theorem.  
IF from any angle of a triangle to his subtendent side, a perpendiculare descend the square of that subtendente side, or basis added to one of the containing sides square, surmounteth the square of the third side, by a rightangled Parallelogramme contained of the whole base and double that his portion, which lieth between the perpendiculare and the last named third sides subtendente angle.     

The .7. Chapter. The description of the Quadrant Geometrical.  
FIrst ye must make a common simple large quadrante thus, with your compass draw an Ark or Circumference, that may be more, or at the lest sufficient for a quadrant, then put both the feet of your compass in that ark, making two Pricks. Now the distance of these two points divided in two equal parts, adding one portion to the aforesaid circumference or distance, showeth a precise quadrant, ye aught then to pull of each side from the centre a line to the uttermost points, which be the extremes of your quadrant. Again draw a line from your centre A to the mids of the quadrants circumference C, and if ye list ye may divide that quadrant into 90. grades thus: First in 3. parts, than every third in 3. so have ye 9 portions: Now every of them in 2. riseth .18. Then each in 5. equal parts maketh .90. degrees.  

Of the Scale.  
FOr the Scale ye shall draw from either side of your quadrant a right hanging line (as is declared) touching the middle line in one point, so have ye the sides of your scale each to be divided in 12.60.100.1000 points all marked from the centre A, the more the more commodious. Forget not to have two equal fine plates of brass pierced in the mids (for your sights) and placed on the side AD, as ye see OF, with a line and plummet falling out of the centre A. I call the scale in this quadrant the two sides within divided in certain portions or parts. And those .12. parts next to your sights I name points of right shadow: tother sided of the scale portions or points of contrary shadow: better it were, yea and more for the purpose if each side had .60.100. or, 1000 divisions.    

The .8. Chapter. The use of the Scale, showing perpendiculare or direct heights by their shadows.  
 
Convey the left side of your quadrant Geometrical toward the Sun, the thread and Plummet having their free course moving it up or down, until both your sights have received the sun beams. Then if your thread be found in the twelfth part, shadows of all things (being perpendiculare elevated) be equal with their bodies, if the plummet with the thread be perceived cutting the parts next to the sights, which I call right shadows, than every thing direct is more than his shadow. By that proportion which 12. exceedeth the parts where the thread was found: if it fall on the first part of right, take the shadow 12. times to make the height, if it chance on the second portion six times, on the third four times, on the fourth thrice, in the fifth twice and two fifth of the shadow, in the sixth point twice, in the seventh once, and five sevenths of the shadow, in the eight portion once and a half, in the ninth once and the third part, in the tenth once and the fift part: in the eleventh point ye shall take the shadow once and the eleventh part of that shadow: or in few words, multiply the length of the shadow by 12. and the producte divide by the parts in which you found the thread, your quotient showeth the height: but and if it be in the parts of contrarye-shadowe, augment the length of the shadow with the parts declared by the Plummet, and the increase divide by 12. so cometh the altitude also. Ensample, in the figure that goeth before it is plain to be perceived. When the thread falleth on 12 portions, the shadow is equal with the thing itself. In 6 of right it is but half, in six of contrary it is twice the height: So to conclude, ye may see as the side in right exceedeth the parts, so doth the altitude or body the shadow, and contrary in contrary shadow, behold your figure how the thread cutteth 6 parts of contrary shadow in the Quadrant next to the right hand, the shadow BC then being 210 foot, multiply (as I have said) the length of the shadow 210 feet, with 6 the parts cut by the thread, increaseth 1260, that divided by twelve, riseth 105: the altitude of such a body which had a shadow then 210 feet. Thus of all such like.   

The .9. Chapter. Of Vigetius concerning heigthes.  
HE affirmeth by any certain measure directly standing (whose shadow is known) the height of any other thing the shadow than measured not to be hid, performed by the rule of proportion. Example.  
Suppose the shadow of any thing 210 feet. Now say 20 the shadow of a thing known giveth 10, what shall 210, riseth 105: the height.   

The .10. Chapter. Without shadow or any supputation by your quadrant geometrical to take heigthes approchable.  
Lift up ingeniously your quadrant exactly made toward the thing to be measured, looking diligently through both the sights, going back or forward as occasion is given until ye see the top, so that your fine or subtle thread fall justly upon the twelfth point. Now if you measure the distance from you to the base (which base here I call the point directly under the top, then have ye the altitude of the highest summitie to the right point or base in height equal with your standing, adjoining unto it the height of your eye downward. Ensample, The line and plummet in the figure afore falleth precisely in the twelfth portion, the space then being from you to the base, which is from A to C 15. foot. To this ye must add the height of your eye (here imagined 5. foot) so have ye 20 feet, the true altitude from A to B. As the length of the cord from the eye to C showeth the measure to be laid back: so doth the touch of the line and plummet in C, declare where ye must begin to lay the measure back.   

The .11. Chapter. With the aid of two places to search out improcheable heights.  
Seek two stations going hither and thither, yea toward or from the thing ye intend to measure, so that in the one place the thread may fall in 12, the other station in 6. points of right shadow, then if ye double the distance of both places, the summitie shall appear from that part of the thing measured which is equal in height with your eye, or if your standing be even with the base, joining to that doubled distance the height of your eye, ye have the whole altitude from the ground, etc. If the one room 'cause the thread to fall in 12. the other in 8. of right shadow, then triple the space, so have ye the height also. Or the one in 12. the other in 9 of right shadow, then quadruplate the distance: yea the one under 12. that other 6. of contrary shadow, than the space between both stations is equal with that you measure, ever understanding from your eye upward. Or if the plummet be enforced to fall under 6 points of the contrary shadow, the other under 4 parts of the same, or in 4 and 3 of contrary, in all these the distance of the place is equal with the altitude. So then in measuring the space between the two places ye have gotten the height from your eye up, putting unto it (as I have said) the length from your sight downward, the just altitude of the whole appeareth: the Base being even with your standing. Example, This Figure declareth the falling of the thread under 6 of right, and 12. Also under 12 and 6 of contrary, by doubling the space between the two first places, the altitude appeareth. In like manner the distance from the middle standing to the last, bringeth the height from your sight.  

How lengths in height are known.  
I Would not have you ignorant here how to know lengths which be in height not easy to come unto, first by your art afore mentioned get either height, subduce the less height out of the more, of force your desired length remaineth. Or thus, let the plummet fall under 12 portions, mark your place, go in toward the thing (the thread as it was) until ye see the base of that length, so the distance between the two standings is undoubtedly the length: here needeth no Example.    

The 12. Chapter. How by your Quadrant with calculation speedily to know all heights accessible.  
YOur Quadrant as tofore is said handsomely elevated against or toward the thing to be measured, perceiving thorough the sights not more than the top. Mark well the divisions of points touched in your scale, if they be of right shadow, multiply the distance from you to the base by 12, and divide by the parts of your Scale which your thread made manifest. But and if they be of contrary shadow, work contrarily: that is augment by the parts, and make partition by 12, remembering ever to add the height of your eye downward to your Quotient, so have ye your desire: the Base being equal with your standing.  

Example.  
Admit the thread with the plummet note 6 parts of contrary, as ye may see in this Figure: the distance from the base A to your standing B 115 foot, multiply that by 6, so have ye 690, the which divided by 12, yieldeth 57 ½ foot: to this adjoin five foot (being the height of your sight to the ground) conclude the Altitude 62 ½ foot.  
Hear ye shall note that in mensuration of heights with this instrument, it shall be requisite first to find what part of the Altitude is level with your eye, which you may thus do:  
'Cause the plummet and thread to fall upon the side line of your quadrant where the degrees begin, and then searching through the sights, that part which you can espy of your Altitude, (the plummet hanging upon the foresaid lateral line) is level with your eye, the height whereof from the base compared with the altitude of your eye, discovereth the unequality or difference of the ground: that is to say, how much higher or lower the base of the thing to be measured, is than the ground at your station which difference as ye shall see cause added or detracted from your heights found as is before declared, yieldeth most exactly the true altitude, and thus shall you be assured never to err, how uneven or unlevell so ever the ground be.    

The .13. Chapter. To get inaccessible heights by supputation (with the help of two places) supposing either side of the Scale divided, 100 parts.  
IF your thread in the first station fall upon 50 points of contrary, with those divide 100, so have ye 2. In the other place (going right back or forward no way declining) admit it note 25 of contrary, now 100 divided with 25 riseth 4, withdraw 2 from 4, 2 is left your divident, meet the space between both standings, and divide that by 2, your divisor, so have ye the height from the eye up. Note, if the difference of the Quotient be 1, the space between the standings shallbe equal with the desired height, adding your stature. If 2, the space is double to the altitude, if 3, threefold, etc.  
Or thus work: Reduce the parts of contrary shadow unto portions of right, and then do as you would with points of right: that reduction is made thus, multiply 100 in himself, so have ye 10000, the which divided by every part of contrary shadow, so shall they be as points of right shadow. And if you have made two stations, pull the less Quotient from the great, the rest weigh as you have been instructed. No end hath the Geometer in finding true measures, many I might say infinite more ways heights are found, by any two equal things orthogenally joined, by staff, cord, squire, triangle glass, etc. as briefly followeth.  

By any two things of one length joined thus in right Angle, Altitudes are found.  
THe end C applied to your eye, go backward or forward as you shall see cause, till you can espy the top and base of your Altitude, by the extremes AB: so doing the distance between your foot and the base, is equal to the height, without adjoining the altitude of your eye, which in all the rest before showed is required, only here ye shall take heed so to couple AB and CD, that in beholding the altitude, your line AB may depend perpendiculare or equedistant to the height. In like sort may you meet the distance of any two things in sight, and that exactly, if you use discretion in placing AB, that it be always parallel to the line measured.    

The 14. Chapter. heights are ingeniously searched out by a staff.  
IF any staff be erected, the measurer upon his back beholding the top of the thing, the distance of the eye from the foot of the thing showeth the height. Or thus receive my mind more largely, prepare a right staff divided in 12 or more equal parts, that done, set it right up a certain distance (as ye list) from the height which you will measure. Now go right from that staff some space at pleasure: laying your eye to the ground equal with the base of the thing to be measured, moving back or drawing near to that staff, until ye may rightly and plainly see the very summitie or upmost part of the thing to be measured, by the top of your staff, which performed make a mark where your eye had his place: Now measure the distance or space from the staff to your eye with the staff itself, and note what proportion the staff hath to the distance, the same shall your height have to the length from your eye to the base of that Altitude. Ensample. The staff CD (in this figure) and the distance CE are equal, therefore affirm the height AB to be equal yea so long as the distance between your eye and the base of that required height which is A, if otherwise according to the proportion afore mentioned, ye may by the rule (called the golden precept,) bring the just height thus, meet the ground between your eye & the staff suppose it 12 foot, than the distance from your eye to the base 200 foot, your staff 5 foot, say of 12 cometh 5, what shall come of 200, so have ye 83 feet and ⅓ your exact height.   

The .15. Chapter. Ye may also hereby readily measure all lengths standing in height as the altitude of any Tower standing on a hill, or the length of a steeple above the battlements, or the distance between story and story in edefices, yourself standing on the ground.  
FOr Example I would measure the distance between B and G, suppose the line visual EG cut your staff (which I would wish divided in 12 parts) upon the fourth part from the top. Then work by the rule of proportion: saying 20 parts the distance between the eye and the staff giveth 200 foot, the distance of the tower, what yieldeth 4 parts, thus shall you find the fourth number proportional 40 foot, which is the exact length of BG. And thus may you measure only by a staff divided in 12 equal parts (without any other instrument) any altitude, how so ever it be situate.  

Behold the Figure on the other side.  

The like is brought to pass by the Squire, behold the Figure.   
Always DC will retain the same proportion to DE, that BF doth to AB, so that BF measured, you may by the golden rule attain the height of AB, or contrary if you know the Altitude AB, you may proportionally learn the longitude FB: Pleasanter to practise is this than the former and most exact for Altitudes. But where as some do use it as well for longitudes and distances, making the staff a side common of the greater and lesser triangles: though demonstration bear it, I allow not of it, the angle made with the squire and staff, groweth so acute, and unsensible, that great error ensueth the lest mistaking.    

The .16. Chapter. By a Glass heighthes may be pleasantly practised and found on this wise.  
CAst a glass on the ground, so it may lie equal, yea even in height with the base of the thing to be measured, your eye on the glass fixed, go from it until ye see no more than the very top of that thing of which ye require the height. Then let a Plummet with a line come from your eye to the ground, mark the fall of that Plummet, look what proportion the distance beareth (from the said Plummet to the Centre of the glass) compared to the length of the cord, the same shall the space between the glass and the base of the height, have to the altitude desired.  
THe precept of proportion may be as well here used as afore, and so to attain heights. Behold the figure, as the distance DC from the fall of the Plummet to the centre of the glass, is equal with the line ED, falling from the sight E so the length CE from the glass to the base of the tower, swerveth not from the desired Altitude AB.   

The .17. Chapter. To measure the deepness of any well by the Quadrant Geometrical.  
Now to return to the first instrument Geometrical and so to end, the breadth or Diameter known, set your Quadrant on the end of your Well in the very top, then lift up or put down this Quadrant until ye see the bottom on the contrary side, mark the points cut, look what proportion the parts have to the whole side, the same shall the Diameter have to the length or deepness. Therefore by the number of the points form divide 12, so the Quotient uttereth how often the 〈◊〉 of the Well maketh the depth, or multiply the breadth by 12, the 〈◊〉 divide by the points found, the depth also appeareth. Ensample. In 〈◊〉 figures following there be 3 points of the Quadrant cut etc. 3 in 1● is contained four times, so is the Diameter in the length or profundity. In like manner of proportion (as is declared) ye may gather the length of things ascending some high Turret. Farther note whatsoever I have said of the Quadrant appertaining to heights know that to be spoken of the square Geometrical, which here shallbe put forth for lengths only, one reason one ground serveth them both. As easy also is it to know how much water, I mean how many quarts, gallons, or other measure are contained in these Welles, or in any other Regular excavate body, which shall plainly appear in the last kind of Geometry in place due.  
By this instrument also ye may know from any spring or fountain, whether the water may be conveyed by Pipes or otherwise, to any other place or places how far distant so ever they be, yea though there be many hills and valleys between.  
I Would not have you ignorant that the nature of water is such, as by pipes it may be reared above the fountain he, and carried over hills or mountains how high so ever they be, so that the vent or end where the water must issue out of the Pipe be inferior to the Fountain whence it is derived: neither need you care though there be many valleys in the way, for experience teacheth that the lower your water falleth, the more freely it will run, and the more pure and wholesome it shall be. Always it behoveth you to have consideration of the Fountain whence it proceedeth, and the Sestarne, Well, or other place whereinto it falleth▪ that this be lower than the other. It is also to be weighed how this difference of highness and lowness is to be accounted, some suppose that all places lying in a strait level line from the spring head, are of one height, which opinion is erroneous, because the water (being an heavy body as the earth is) presseth and tendeth always to the centre. And in all his courses (being not violently forced contrary to his nature) moveth downward, or at the least uniformly and equodistantly from that centre. Whereby it is manifest that no spring can of his own nature run in level or right line from his head, for this equidistant course to the centre, is an ark or portion of a circle. But every level right line (considering it is a contingent or touch line) is carried above the circumference, and the farther it is extended, the farther distant it is from the Centre, so that either the water must make his course upward utterly contrary to his nature, or else it shall decline from this level right line. In deed the globe of the earth & water being so great, any small Ark or portion of their Circumference will not sensibly differ from a strait line. And yet in conveying of waters any great distance, very experience will bewray an error. The mean therefore to attain perfection herein, is to find the difference between this strait and Circular line, whereof hereafter I shall entreat more at large: only here will I open how without any error sensible, (sufficient for any Mechanical operation) by the aid of sundry stations ye shall accomplish this purpose.  
First it behoveth ye diligently to mark at the fountain head the Superficies of the water, and if the ground so serve that ye may place your eye even therewith, take your Quadrant, and turning yourself toward the place whether you mean to convey this water, (if it be within sight) espy the same through the sights of your Quadrant, meeting diligently the fall of the Plummet, which if it cut any of the Degrees, ye may conclude it is not possible for the water naturally to run thither. But if your Sestourne or place be not to be seen at the Spring head, then espy some other mark through the sight of your Quadrant on either side toward your place, always causing the thread and Plummet to fall directly upon the lateral line where your Degrees begin, then removing and situating your eye at the point or mark which you last espied, find through your sights a new mark, causing your Plummet and thread to fall upon the aforesaid lateral line, and thus proceed from station to station, till you come to the sight of your last place, then if your Plummet and thread cut the Degrees, ye may conclude as afore▪ but if the ground at the Spring head be such, as you cannot conveniently place your eye as I have said, then let fall from your eye, or some place of equal height with your eye, a string and Plummet to the brim of the water, measuring the length thereof. If at your last station your Plummet and thread hanging as I have tofore said, your visual line passing through the sights of your Quadrant fall above the Sestourne where this water should issue out, erreare a Pole or such like thing to the height or length that the string was at the Fountain head: and if your visual line reach higher than the top or summitie of that Pole also, ye may conclude that this water may be derived thither. And if from that part of the Pole your visual line cutteth, ye abate the perpendicular from your eye to the water at the fountain head, and for every Mile travailing 4 inches, the point where you leave is exactly level with the Superficies of the water, and so high it may be brought, and not above. It behoveth ye also to take order that your stations be not above 200 or 300 pace at the most a sunder, otherwise error sensible may ensue. Ye may also (if need require) at every station erect an high Pole, and so may you pass over both mountains and valleys, always noting at every station, what portion of the Pole your line visual doth cut, reserving them to be added or subtracted as you see cause, at your last station. Superfluous, yea rather tedious should it be to use more words in so plain a matter, the ingenious practitioner will find sundry ways to help himself as occasion requires by sight of the ground. etc.   

The .18. Chapter. To get the length or distance of any place or mark in sight, be it never so far, and that without instrument.  
AMong many practices I find six ways principally be had in estimation, the first ensuing without any instrument, other requiring aid of instruments, whose name compositions and use follow as seemeth meet. Although in measuring of lengths after the minds of many writers the avoiding of hills, & in few words most plain hath been desired, least great errors should ensue, here such things shall not be required: Only it shallbe needful at the time of your measuring to have ground at liberty on the one side. This commodity had the ground level or otherwise work thus, at the beginning of your length set up a staff or mark which may be seen a far of, then go from it Orthogonally ● squirewise of which side ye will 200 foot, or as ye list, the more ground● the better, put up there a staff also: Now convey yourself to the first staff or mark going back from it 300 foot more or less at your pleasure, set up there the third staff, so that the first mark or staff and it agrée● in a right line from your sight to the farthest point of your length by the judgement of your eye. Now go sidewise from thence as afore in a right angle until the second mark offer itself aright between the extreme part of your length and sight, there put up the fourth staff. All this performed, seek out the distance between the first staff and the second that name your first distance, than the length between the first & third staff, call that the second distance. Again the space between the third and the fourth staff is the third distance, subduce the first distance from the third, so remaineth your Divisor, then multiply the third distance by the second, & the product divide by your divident or divisor, the Quotient showeth the true length from the third staff to the fortress or mark desired. For more plainness behold the Figure.  
 

Example.  
HEre this letter A representeth the fortress Castle or mark which is the extreme or farthest part of length to be measured, B is your first staff, C the second staff, differing from B the first orthogonally 100 foot, D the third staff being distant back from the first in a right line with the mark 133 foot. E is the fourth staff running sydewise orthogonally or in a squire from the third, until the farthest part of your length or mark is perceived in a right line with the second staff, being distant from D the third staff 120 foot. Now by subtraction subduce 100 from 120, there remaineth your divisor 20. Then multiply 133 with 120, so riseth 15960, which divided by 20 cometh 798 foot the true distance between D and A.  
But forasmuch as this conclusion is to be done without instrument, and here orthogonall motion sidewise is required, it shall be requisite also to declare how an orthogonal or right angle is upon the sudden to be made, ye shall therefore (according to Pythagoras' invention mentioned among the definitions at the beginning of this Book) take 3 staves, cords or such like, making the one 4 such parts as the other is 3 and the third 5: This done conjoin their ends together and the angle subtended of the longest staff is a right, which first placed at B, and after at D, directing one of his comprehending sides to A, the other shall guide you to C and E, or if you desire with more expedition to dispatch and not tarry the proportioning of the cords or such like to this Pythagorical rule, take any 3 staves, sticks or threads, and conjoin them, making a triangle, it forceth not of what form or fashion it be: then placing one angle thereof at B, turning one side to A, direct yourself sydewise by the other, always remembering to place the same angle at D, and departing sydewise again in like manner, in all the rest do as before is declared. Thus using any mean diligence, ye shall most exactly measure any distance.    

The 19 Chapter. With halberds, pikes or staves having no other instruments, you may measure the distance between any two marks lying in a right line from you, not approaching any of them.  
YOu shall first (as was declared in the last Chapter) prepare an angle with joining any 3 staves or such like together, which you must (at your standing) place in such sort that one of the sides containing the Angle, may lie directly toward the mark: them setting up a staff, pike or other mark there, depart sidewise as the other side of your angle shall direct you, so far as you list, the more ground the better. And there set up your second staff or mark, then go directly back from your first staff (always keeping it exactly between your sight and the marks) as many score again or pike length as ye list, setting up a third staff. This done, you shall place the same angle you used at your first staff, now again at your third staff, in all points as it was before: The one side of the angle lying directly toward the first staff, the other side will show you whither you shall go to place your fourth staff, for passing on still in a right line with that side of your angle, you shall at the last find the second staff justly situate between you and the farthest mark, there set up the fourth staff, then remove your angle again to the second staff, and placing there as before the one side even with the first staff pass on in a right line with the other till you come directly between your nearest mark and the fourth staff, there pitch up the fift. Now shall you measure how many pace, halberd or pike length is between your first and second staff, deducting that from the distance between the third & fourth, and this remain you shall reserve for a divisor. Then multiply the distance between the second and fift staff in the distance between the third and fourth staff, the producte divided by your reserved divisor, yieldeth in the quotient the true distance between these two marks.  

Example.  
AB the two marks whose distance I would measure, C my standing place where I set up my first staff, I my triangle made of three stavos, halberdes, bills, or any such like things, KLM, the angle which I will now use in this practise N, first placed at C, secondly at D, thirdly at E, at C and D, the situation is all one, but at E, it somewhat differeth as you may behold in this figure, which I would have you note jest happily ye be deceived in your practise, the first staff C, the second E, the third staff D, F the 4, G the fift staff, CE the distance between the first and third deducted from DF, the distance between the second and fourth there remaineth HF your divisor, which measured, I admit 50 halberd lengths. The distance between GE 3● halberd lenghtes, the space between DF 100 pike length. Now, 100 multiplied in 30 produceth 3000, which divided by 50 leaveth in the quotient 60, I conclude therefore the distance between AB 60 pike lengths. This one thing is specially to be noted, that whatsoever you meet the space GE withal, whether it be halberd, bill, arrow or staff that ye use the same in measuring HF, and as for DF, it forceth not what you measure it withal, marry what soever it be, your quotient shall bear the same denomination: preciseness is to be used in placing of your triangle, and in measuring EG and HF, otherwise error may ensue, especially if DF be but a small distance, and the angle at B very sharp, there needeth in this matter no further admonition, small practise will resolve all doubts.  
   

The .20. Chapter. To measure the distance between any two marks howsoever they be situate, though there be rivers or such like impediments between you, as ye cannot approach nigh any of them, and that without instrument also.  
Your angle as before hath been said, prepared of any three staves, you shall first at pleasure set up one staff, and applying thereunto your angle in such sort, that the one containing side lie directly to one of your marks (which here for distinction I will call the first) go backward to and fro till you find your second mark precisely covered with your staff, noting what part of the side subtending the angle is cut by your line visual, and there make a fine notche, or mark upon that subtending staff, which done you shall go sidewise from the first erreared staff, as the other containing side of your triangle will direct you so far as ye list, and then set up your second staff, yet pass on from thence in a right line as many foot pace or other measure, as you will, setting up again the third staff, now at your second staff situate your triangle in all respects as it was at the first, and pass on from thence in a right line with that containing side of your angle that riseth from your staves, and cometh somewhat toward the mark going so far till you espy yourself justly between your third staff and the first mark, there set up the fourth staff, then resort to your angle again, and standing behind that second staff, note whether a right line from the angle to that notche (before made on the subtendent staff or side of the triangle) will direct you, for that way precisely shall you go on till you come in a right line with the second mark and third staff erecting there the fift staff. This done you must measure the distance between your second and third staff, reserving that for a divisor, then multiply the distance between your first and third staff in the distance between the fourth and fift staff, the producte divided by your reserved divisor, yieldeth in the quotient the true distance between the two marks.  

Example.  
AB the two marks whose distance I would meet C, my first staff, I my triangle made of 3 staves placed thereat, as you may see in the figure directing with the one containing side to A, the first mark, and with the other to D and E my second and third staves, H is the fine notche or mark upon the side subtendent to my angle, where the line visual from C to the second mark B passeth, my triangle I situate now at D as it was before at C, the one containing side lying even with the errered staves, the other directeth to my forth staff F, placed in aright line with E, the third staff, and A the first mark. Again my line visual proceeding from D to H the subtle notche in the subtendente side of the angle, extendeth to my fift staff G, situate exactly between E the third staff, and B the other mark: This done, I measure the distance between my second and third staff, finding it 20 foot, likewise between the fourth and fifth staff 72 foot, finally between the first and third staff 65. pace, so that according to the rule before given, multiplying 65 in 72, I have 4680 which divided by 20 yieldeth in the quotient 234, so many pace is there between A and B. I have not here set out the figure in just proportion, answering to these numbers, for that is not requisite, but in such form as may best open and make manifest the situation of the staves and triangle, wherein consisteth all the difficulty of this practice.  
A proof or demonstration Geometrical of the former conclusion.  
〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉. And may thus be englished, if any right line falling or passing through two other right lines, making the outward angle equal to the inward opposite on the same side, or the two inward angles joined together equal unto two right angles, those two right lines are parallel: but here the line EC passing through the lines AC, DF maketh the outward angle FDE equal to the inward opposite on the same side ACD by supposition, because they were both made with one angle of the triangle, I may therefore conclude by this Theorem, that DF is equidistante to CA, and farther infer by the second Theorem of the sixth book, that OF to FF hath the same proportion that CD hath to DE, the proposition is this. 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉.  
If a parallel line be drawn to any side of a triangle it shall proportionally cut the two other sides, and if two sides of a triangle be proportionally divided the line that coupleth those divisions shall be parallel to the other side. It is manifest by the first part of this theorem that DF being as it was before proved parallel to AC one side of the triangle ACE that it doth proportionally divide the two other sides CE, EA, in the points FD the like shall be proved of DG, for seeing the right line CE falleth on the two right lines DG, CB making the outward angle HDE equal to the inward and opposite angle HCD on the same side of the line CE which in the construction of the figure was supposed, it must needs follow by the 28 proposition of the first book of Euclid tofore recited, that DG is parallel to CB, and forasmuch as in the triangle BE, DG is drawn parallel to the one side CB, it shall by the second proposition of the 6 book of Euclid (before also recited) divide the two other sides CE, EBB proportionally, so that BG shall retain the same proportion to GE that CD doth to DE, and so consequently the same that OF doth to FE, as it is plain by the eleventh theorem of the fifth book of Euclid: his words be these.  
〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉. The sense thereof is this, that any two proportions being equal or agreeable to any one are also equal between themselves, as here first it was proved that OF to FE bore the same proportion that CD to DE, and now that BG to GE retaineth the same proportion that CD to DE: therefore by the theorem last recited OF to FE and BG to GE, shall be proportional: so have you now two sides of the triangle AEB proportional with two sides of the triangle FEG, and the angles contained of those sides equal. Therefore by the 6 theorem of the 6 book those two triangles are equiangle, the words of that theorem are these.  
〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉.  
If two triangles have an angle in the one equal to one angle in the other, and about those equal angles the sides proportional, those two triangles shall be equiangle, and those angles, equal whose subtendent sides are proportional, Euclid also in his 4 proposition of the same book saith thus.  
〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉.  
In equiangle triangles aswell the containing as the subtending sides of equal angles are proportional, I may therefore affirm (seeing the triangles ABE, EFG are equiangle) that AB hath the same proportion to FG, that A hath to FE, but before it was proved that OF and FE were proportional to CD and DE, conjoinedly therefore A to FE, shall retain the same proportion that CE doth to DE, by the 18 proposition in the fift Book of Euclid, saying thus: 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉.  
If magnitudes disjoinedly or separately be proportional, conjoinedly or compounded, they shall also be proportional, whereupon I may finally infer by the 11 proposition of Euclides fift Book tofore recited, that AB to GF retaineth the same proportion that CE doth to DE, because they both observe the same proportion that A doth to EF. And thus to conclude, it is manifest that AB, FG, CE, DE, are 4 quantities proportional: whereof three known, the fourth AB by the rule of proportion is to be found. Very like unto this is the Demonstration of the other, and by the self-same propositions to be proved, and therefore superfluous to use more words.    

The .21. Chapter How ye may most pleasantly and exactly with a plain glass from an high cliff, measure the distance of any ship or ships on the sea as followeth.  
THE best kind of glass for this purpose is of steel finely polished, so that the Superficies thereof be smooth, neither convexe nor concave, but flat and plain as may be possible. This glass it behoveth ye to hang up above the top of the cliff with the pullished side downward equedistant to the Horizon wherein you must use great diligence, for if there happen any error in the situation thereof, great inconvenience may follow in your mensurations. This done, let a plumbeline fall from the centre of your glass to the Superficies or ground plat on the top of the cliff: (which ground plat) also you must use some diligence in the choice thereof that it be as level and plain as ye can find any, but if it be not altogether even or exact level, ye shall supply that want as I shall hereafter show you: but to return to the purpose, your Glass thus situate, turning your face toward the ship or other mark on the sea, whose distance ye desire, go backward, always having your eye fixed on the glass till such time as ye can see the ship, or rather the very hull next to the water therein, that done let an other plumb line fall from your eye to the ground, then circumspectly measure both the length of these plumb lines, and also the distance between this plumb line and the other that fell from the centre of the glass, this done ye shall deduct the length of the perpendiculare from your eye, out of the length of the other perpendiculare from your glass, and the difference reserve for a divisor, then multiply the distance of the two perpendiculare lines, by the height of the cliff, I mean from the water upward to the glass, and the producte divide by your reserved divisor, the quotient will show you the exact distance to the ship or mark. But if your ground be not level, ye shall by your quadrant search the difference or inequality thereof, & if it be lower at the glass than at the viewing station, you shall deduct the difference from your divisor, but if contrariwise, the difference shall be added to the divisor, & then shall ye work as I have before declared.  
 

Example.  
SVppose a ship on the sea C, whose distance I desire to know standing on the cliff GFE, your glass A equidistantly elevate, AB the perpendiculare line 72 inches, E your viewing station, DE the altitude of your eye 69 inches, BE the distance between your two perpendiculare lines 20 foot, FG the height of the cliff from the sea 96 pace. Now by deductings 69 inches the altitude of your eye from 72 the length of AB, the plumb line from the glass, there remaineth 3 your Divisor; if the ground be level, otherwise ye must add or detract the difference or unequality according to the rules to fore given, but here supposing your ground level your divisor remaineth 3. then resolve 20 foot into inches, and multiply the same in the altitude of the cliff up to the glass 98, so have ye 23520, which divided by 3 yieldeth in the quotient 7840, which is the exact distance of that part of the ship which ye did espy from the centre of your glass.  
YOu may on this manner from an high hill or mountain, having any plain or level ground on the top, not only measure the distance of any mark that ye can see, but also set forth the true plat and proportion of an whole Country, with all the Towns, Coasts, Harboronghes. etc.  
For if you mean circularely about your glass, always when you espy any mark, setting up a staff, writing thereupon the name of the place ye see, whether it be village, port, road, or such like, ye shall in the end situate as it were the whole country in due proportion upon your plaltform, so that measuring the distance of every staff set up from the middle line perpendiculare falling from the glass, and the distance likewise of every staff from other, ye may (working by the golden rule) find out the exact distance of every town, village, port, road or such like from your plaltform, and also how far every one is distant off from other. Thus much I thought good to open concerning the effects of a plain Glass, very pleasant to practise, yea most exactly serving for the description of a plain champion country. But marvelous are the conclusions that may be performed by glasses concave and convex of circular and parabolical forms, using for multiplication of beams sometime the aid of glasses transparent, which by fraction should unite or dissipate the images or figures presented by the reflection of other. By these kind of glasses or rather frames of them, placed in due angles, ye may not only set out the proportion of an whole region, yea represent before your eye the lively image of every town, village, etc. and that in as little or great space or place as ye will prescribe, but also augment and dilate any parcel thereof, so that whereas at the first appearance an whole town shall present itself so small and compact together that ye shall not discern any difference of streets, ye may by application of glasses in due proportion 'cause any peculiar house, or room thereof dilate and show itself in as ample form as the whole town first appeared, so that ye shall discern any trifle, or read any letter lying there open, especially if the son beams may come unto it, as plainly as if you were corporally present, although it be distant from you as far as eye can descry: But of these conclusions I mind not here more to entreat, having at large in a volume by itself opened the miraculous effects of perspective glasses. And that not only in matters of discovery, but also by the sun beams to fire, powder, or any other combustible matter, which Archimedes is recorded to have done at Syracuse in Sicily, when the Roman Navy approached that Town. Some have fond surmised he did it with a portion of a section Parabolical artificially made to reflect and unite the son beams a great distance of, and for the construction of this glass take great pains with high curiosity to writ large and many intricate demonstrations, but it is a mere fancy and utterly impossible, with any one glass whatsoever it be to fire any thing, only one thousand pace off, no though it were a 100 foot over, marry true it is, the Parabola for his small distance, most perfectly doth unite beams, and most vehemently burneth of all other reflecting glasses. But how by application of more glasses to extend this unity or concourse of beams in his full force, yea to augment and multiply the same, that the farther it is carried the more violently it shall pierce and burn. Hoc opus hic labor est, wherein God sparing life, and the time with opportunity serving, I mind to impart with my country men some such secrets, as hath I suppose in this our age been revealed to very few, no less serving for the security and defence of our natural country, than surely to be marveled at of strangers.    

The .22. Chapter. The making of an Instrument named the Geometrical square.  
YE shall prepare a fine plain Plate, or a clean four square planed board, yea, for want of them, four equal, smooth, and well tried rulers, of what length, breadth or thickness ye list, the longer the better, yet in my fantasy to avoid painful carriage, it is most commodious that every of them be but a foot and an half in length, one inch in breadth, a quarter or more thick, ye must join them by the help of some Artificers squirewise, upon those ruler's plate or board justly joined, draw four lines perpendiculare or squire the one to the other: Now divide the two farthest sides from the centre, each having 1200 portions at the last, mark all from your centre, forget not to have an index, not with common sights, but thus, let the nearest be a thin plate half an inch broad, and 3 inches in height, and in the mids a fine slit, the second and farthest from the centre of that length, a streigth pin with a little knob in the top. These sights must be justly set upon the line fiducial of your Index. This Index I would wish also marked with 2000 such divisions as the scale side hath 1200, it hath place in the centre, and there made to tarry, so that with ease it may be turned from the first to any point. The exact handling of this instrument with most comeliness framed, I commit to the diligent maker. For more instruction, behold the Figure, ye may commodiously describe a quadrant in your square the making of which is declared.   

The .23. Chapter. You may readily hereby without Arithmetic meet the distance of any mark.  
THis instrument handsomely placed upon his staff or otherwise, lay the line fiducial of your index upon the beginning of the degrees in your Quadrant, and turn your whole instrument (the index not moved) till ye may espy through the sights your mark then remove the index to the contrary side of the quadrant, placing the line fiducial on the side line where the degrees end, and looking through the sights, ye shall espy a mark sidewise, some certain number of scores, the more the better. This done, set up a staff where the centre of your instrument was, and placing it again at the mark last espied, set your index on the beginning of the degrees, moving your whole instrument, till you find through the sights the ●●af at the first station, then remove your index (the quadrant keeping his place) till ye may again espy through the sights your mark, which done, note the degrees cut by the line fiducial: and then work thus, upon some even smooth superficies whether it be board plate or paper: draw first a straight line, & opening your compass to some small distance, call that space a score and make so many such divisions in your line as there is scores between your stations. Then ereare upon the one end of your line a perpendicular, and fixing the one foot of your compass at the other end, opening it to what wydnesse ye think good, draw an ark rising from the same line that representeth your stationary distance, and dividing it into degrees (as you were taught in making the quadrant) extend from the centre to the number of grades cut by your line fiducial a right line, till it concur with the perpendiculare before ereared. Then see how oft that space (which represented the score in dividing your stationary distance) is contained in the perpendiculare, so many score is the mark off from your first station, and by dividing the Hypothenusal line, you may in like manner find the distance from the second station.  

Example.  
A the first station, C the second, D the mark, AC four score pace: degree● of the quadrant cut at the second station, 71 ½: H the unity or measure representing one score, OF 4 parts, GF 12, GE 12 ⅔, or near there about. Thus may you conclude the distance of the mark from the first station 12 score pace. The Hypothenusal line or distance of the mark from the second station 12 score & 13 pace.  
   

The .24. Chapter. How with your square Geometrical to tell any length so far as ye may see by supputation.  
YOu shall disagree from those writers that have declared the use of the square in the Latin tongue, other than this which I shall now open unto you. So order your Geometrical square, that all sides may be of like height from the ground, to avoid grass, mole hills and such other impediments, preciseness in this ordering is not so requisite, call to remembrance that your square hath two principal lines, one squire unto the o●her running from the centre of either side to the beginning of your points: ●et your index▪ yea the line fiducial upon one of those lines, that side lying ●long with the index towards the mark. Now, the extreme part of ●our length perceived through your sights, turns the index (the square ●ot moved) to the other principal line, squire to the first, looking again through the sights, and noting some mark a good distance from you the ●ore ground the surer: This done and a staff pitched up where the cen●re of your instrument stood, convey that instrument to the second mark, ●urning it and your index to the place where you first were, the index being in the principal line as afore, even so soon as ye can espy your first station through the sights, remove that index until you may see the extreme ●art of your length, your sight receiving it diligently note the points touched: Now if the index fall on the left side of your scale, I mean the side which falleth perpendicular to that side of the square issuing from the cen●re, whereon your index was first placed, then must ye multiply the space between the first and second place by these parts cut, and divide by 1200, the quotient is your desire. But if the index fall on the right side of the scale, then shall you work contrary, multiplying the space between your stations in 1200, and dividing by the parts cut, or ye may reduce the parts of the right side, to parts proportional of the left, and work with them according to the first rule thus: Divide the square of 1200 by the parts cut in the right side of your scale, the quotient is the parts proportional, which increased by the distance of your stations, making partition by 1200, the quotient is the true distance of the mark from your first station.  

Example.  
A Is the place to be measured, B, the mark where I first disposed mine instrument, from it I go orthogonally to C the index cutting there 400 in the right side of your square, the distance between B and C, I have supposed 80 pace, wherefore multiply 1200 by 80, so there cometh 96000, which divided by 400, declareth unto me 240 pace, the true length from B to A. Or by dividing 1440000 the square of 1200, with 400 the parts cut, you shall produce in the quotient 3600, your proportional parts found by the rule of reduction, which augmented in 80, yieldeth 288000, and that divided by 1200, bringeth in the quotient 240, which is the longitude AB, agreeing with the former operation. This manner of reduction, I would wish you diligently to note, for it shall hereafter in diverse conclusions be used.  
   

The .25. Chapter. How to meet any line hypothenusal as the distance from your eye standing in a valley to the top of an hill or high turret. etc.  
FIrst, if the hill or turret be steep up, so that the foot be visible lying perpendicularly under the top, ye shall first measure the distance of the base, either by this instrument or otherwise as was before declared: and also the height of the top or summitie of the same hill or four: which done, ye shall square aswell the longitude as the altitude, joining together the productes, the root quadrat of the whole number, is the desired distance or line Hypothenusal: you may in this manner (approaching nigh any town of war) tell the just length of the scaling laders that shall reach from the brim of the ditch or edge of the counterscarp, to the top of the wall or curtain, by adding the square of the ditches latitude, to the height of the curtain above the level of the outward bank, for the root of the producte will be the true length of the scaling ladder.  
A the top of the hill, B the foot, C my station or the place of mine eye, A B 60 pace, CB 200 pace, the square of 60 is 3600, the square of 200 is 40000, these two joined together make 43600, whose Quadrate root being about 208 pace 3 foot is the hypothenusal line AC. Likewise AB the breadth of the Ditch being 30 foot, and BC the altitude of the curtain 20 foot, there two squares added together bring forth 1300, whose Quadrate root being 36 foot very nigh, is the length of the scaling ladder AC. But if the base of your mountain be not visible, then ereare up your Geometrical square, the index placed (as was before declared) toward the top of the Hill A, and removing the Index (your square standing immovable) espy your second station Orthogonally at D where ye must place the Centre of your Instrument, and so situate your square again, that you may behold both your station and the mountain top without stirring of the square, only removing the Index: in all the rest do as is before already sufficiently declared, behold the Figure, there needeth no other Example. The last Chapter well understood, openeth this most plainly.  
To measure the distance between any two marks that lie in one right line from your eye.  
YOu may resolve this by the former, measuring how far either is Distant from yourself, and then deduct the one from the other, Or thus an other way, the side of your Geometrical square directed towards them, depart Orthogonally (as is tofore declared) 100 or 200 paces as ye list, the more the better, then place your Instrument again, turning the side toward your first station, removing the index to either of the marks, noting what parts at either place the index doth cut of the scale. And if the index at both times fall on the left side, deduct the less from the greater, with the number remaining, augment the distance between your stations, and divide by the whole side of the Scale, your Quotient is the distance. If the index at either time fall on the right side, then must you by the rule afore given, reduce them into parts proportional, or if at one time the index fall on the left, at an other time on the right, then shall you only reduce the parts cut on the right side, which done, deduct as before is said the lesser from the greater, and with the remainder multiply your distance stationary, the product divided by 1200 yieldeth how far one mark is beyond the other.  

Example.  
Admit AB the marks in a right line from C your first station, D the second station Orthogonally situate from C, where your square being placed, suppose your Index first cut 800 parts on the left side, and after 900 parts on the right, these 900 of right ye must reduce, dividing the square of 1200 by 900, as was taught in the former Chapter: so will your Quotient amount to 1600, from which if ye withdraw 800 the parts cut on the left side, there will remain 800 which multiplied in 200 pace the distance stationary CD, there amounteth 160000 This divided by 1200 yieldeth in the Quotient 133 ⅓ the distances therefore of AB your marks is 133 pace, one foot 8 inches.    

The .26. Chapter. To measure the distance between any two marks lying in one plain level ground with your eye or station how so ever they be situate without supputation.  
MEasure by the rules tofore given, how far either mark is of from you, then placing the index upon the side of the square, turn your instrument till you can espy through the sights one of your marks, the square so remaining steady, move your index toward the other mark, & when you have found that mark also through your sights, note what degrees of the Quadrant, the line fiducial cutteth, this done, ye shall upon some plain board, plate, or such like, draw a strait line, then open your Compass some mean wideness, and fixing one foot at the beginning of the line, with the other make an intersection. Now if the marks be many Miles off, you may term that portion a Mile, or if the distance of the marks be small, a score, but if they be very near at hand, this little line shall represent a pace only, herein you must use discretion, respecting the distance of the marks, and so proportioning your line which is the unity of the work, that your Plate or board may receive the rest of your operations: then proceed with your compass, making so many Divisions in your drawn line, as there are Miles, scores or paces in the distance of one of your marks from your standing, this finished, open your Compass at pleasure, fixing the one foot at the end of your line, with the other draw a Circumference or Ark, and this Ark you shall divide into Degrees, as was taught before in the making of the Quadrant, beginning at your drawn line, and so passing on, till you come to such number of Degrees, as was cut by the line Fiducial, then laying a Ruler to the Centre, applying it to that end of the Ark, draw an other strait line, and your Compass again opened to the length of your little line, (which I term the unity of your work) begin at the foresaid Centre, making again so many Divisions in that latter line, as there was Miles, scores or paces in the distance of the other mark from your standing: And if it fall out that in those Distances there be any odd score, pace, or feet, ye may divide one of those little lines or unities of your work, into more parts accordingly, and so set forth proportionally the exact Distances of the two marks according to the measure first found: this done, ye shall couple together the ends of the two strait lines with an other strait line, finally opening your compass to the length of the unity, beginning at the one end of this last drawn line, measure how many of these unities is therein contained, for so many Miles, scores or pace (according to the Denomination of the unity) may ye say there is certainly between those two marks. But if at the end of this latter mesuring, there be any portion left less than the unity, you must as I have tofore said by Division of the unity search out what portion it is. For more plainness behold the Example.  

Example.  
Admit I would measure the distance between AB two churches, myself standing on a hill at C, first I suppose CA 10 mile ½, CB 13 ¼, the ark of the quadrant cut by the line fiducial .30. degrees, then resorting to some plain board such like, I draw the right line OF: My unity or mile I make D, and opening my compass to that measure, I tell on 13 ¼, making an end at F, then opening my compass at pleasure, I make the ark HIGH, one foot of my compass ●●stned in E, and beginning at H, I number toward I 30 grades, ending at K: ●●en draw I the line EKE, forth to G. which with my compass extended to the length of D, I divide into 10 parts, and ½, finishing at G, and with a right line enjoining GF, I measure again how many unities is therein contained, I find ●⅔ or thereabout, herefore I conclude between these two churches 6 ⅔ miles, this ●ude of measuring is good for the unlearned, but such as have Arithmetic, o●er rules most exact shall ensue.    

The .27. Chapter. The composition of the instrument called Theodelitus.  
IT is but a circle divided in 360 grades or degrees, or a semicir●● parted in 180 portions, and every of those divisions in 3 or rather 6 smaller parts, to it ye may add the double scale, whose single composition is mentioned. The sides of that scale divided in 〈◊〉 60, or 100 parts. The index of that instrument with the sights etc. are not unlike to that which the square hath: In his back prepare a vice 〈◊〉 to be fastened in the top of some staff if it be a circle as here: let you● instrument be so large that from the centre to the degrees may be a 〈◊〉 in length, more if ye list, so shall you not err in your practises, the ba●● side must be plain and smooth to draw circles and lines upon, as shall 〈◊〉 declared: for a farther declaration of that I have said, behold this fy●gure following.  
  

The .28. Chapter. To search the best proportion or simetrie of many places with the true distance approaching near none of them by the instrument named Theodelitus.  
THis instrument upon his staff or otherwise in the field placed the index being in his diameter, let it direct your sight to some one place ●hich ye will measure. Truly in my fantasy it were more commodious if his dimetient or diameter were first laid in a strait line, bringing the sight to the uttermost place toward the left hand, so conveying your index to every 〈◊〉 or mark on the right side, noting diligently the angle or angles of 〈◊〉 upon some state stone or table prepared, which angles here I call grades or degrees from the dimetient apparently cut by the line 〈◊〉 whilst he is brought to every mark. This performed, resort to back side of your instrument where necessity requireth a circle or a ●●●icircle to be made, divided exactly in 360, or if it be an half circle in degrees or portions, even as your Theodelitus here is, from whose 〈◊〉 must finely draw those angles of position noted before in your 〈◊〉 taken by your instrument, so that after your purpose had, they 〈◊〉 be clean put out, then pull the index the instrument unmoved to●●●d the right hand, at pleasure observing through the fights some mark 〈◊〉 yards from you or less as ye list. There shall be your second 〈◊〉, noting upon your slate the angle of position from the dimetient 〈◊〉 line fiducial, directing to the second place or mark, which sword must draw in the back side from the Centre at large even as you 〈◊〉 it in your instrument, then convey your Theodelitus from thence to second mark or standing place, causing the diameter justly to note 〈◊〉 first abiding. And here even as tofore ye must search Angles of 〈◊〉 again, and mark them in the table or slate, which done, resort to the 〈◊〉, and upon the last angle being the line directing to the second standing place, draw a circle as far or near to the other as ye list: Or a 〈◊〉 duided in 180 degrees. Whose divisions must take their begyn●ge at the line which is Diameter of the semicircle. Now draw 〈◊〉 visual lines or angles of position last taken by your instrument ●arge, and see where the lines meet, or a like toucheth his like. So 〈◊〉 you the due proportions, for the distance ye shall work thus, 〈◊〉 the line that goeth from the centre of the one circle or semicircle to the other, in as many portions as ye think meet, or rather in so many 〈◊〉 find certain measure, and by those parts divide the lines betwixt eue●ry place of which ye require the length. Then multiply the portions that are between any two sections or places in the distance of your two stati●ons, which I imagined here .300 pace, and make partition by those part● that are betwixt the two centres, so have ye the true length or distance 〈◊〉 two such places. In like manner ye must do of the rest.  
 

Example.  
Whereby all thing may better appear, ABC are the marks in the field to be measured, D the first abiding or standing place, where ye shall set the centre of your instrument, his diameter lying directly against A the first mark EFGH the four visual lines running by the angles of position of the instrument unto all the marks: the first noting no degree or portion: the second 20, the third 40, the fourth 90 degrees of the instrument, which directeth to the second station M, where ye shall now set the centre of your instrument, the diameter lying right against your first abiding. here the lines visual IKL running to the marks: out new angles of position again. The first noteth 55 degrees, the second 74, the third 85 grades. Now, if ye mark diligently where these lines cross the other, there is the true proportion of such places, from those sections or crossings draw right lines as appeareth by the Figure. Now to get the distance, ye shall seek out the space between the two stations DM: being 300 pace or yards, notwithstanding it is divided but into 18 parts: also between the marks or places A and B, are contained 11 such parts. Seeing that I am ignorant what number of yards be contained in those 11 portions, I am compelled to work by the rule of proportion thus, 18 bringeth 300 yards, what shall 11 bring me? your quotient showeth 183 and 2/6 that is ⅓, which maketh a foot: so between A and B are contained 183 paces and a foot. Thus of all other aswell of DA, DB, DC, MC, MB, MA, as of CB, CA, the Reader must not be ignorant that even as I have supposed by this figure, the instrument in the field placed and all things performed by it with all diligence marked upon a slate or such like, so aught it to be exactly drawn on the backside of your instrument, even as the Figure afore declareth.  
Certes most excellent and far passing all other is this kind of measuring, requiring great exercise, the distance of one place had I am brought into the knowledge of many with the best proportion.    

The .29. Chapter The construction of an instrument topographical serving most commodiously for all manner mensurations.  
Having already plainly declared the making of the Quadrant Geometrical with his scale therein contained, whose use is chiefly for altitudes and profundities: the composition also of the square and planisphere or circle named Theodelitus, for measuring lenghtes, breadthes and distances. It may seem superfluous more to writ of these matters, yet to finish this treatise, I think it not amiss to show how you may join these three in one, whereby you shall frame an instrument of such perfection, that no manner altitude, latitude, longitude, or profundity can offer itself, howsoever it be situate, which you may not both readily and most exactly measure. You shall therefore first prepare some large four square polished plate of Latin, wherein you may describe your Geometrical square, his sides divided in 1200 parts at the jest, with index and sights as was before showed: describing also within the same square the Planisphere or circle called Theodelitus, then must you upon an other fine polished plate, draw your Quadrant, or rather a semicircle divided justly into 180 grades, and within the same a double scale: every side containing at the lest an 120 parts, finally, fixing on the dimetient thereof two sights perpendicularly reared, and equedistantly pierced, so as the line visual may pass parallel to that diameter. You have a double Quadrant Geometrical with a double scale, which you must by the aid of some skilful Artificer, so place over the other plate wherein your square Geometrical and Theodelitus was described, that his centre may exactly rest in a Perpendicular line from the centre of the planisphere or circle named Theodelitus his circumference depending downward. And this double Quadrant or semicircle, must in such sort be connexed to the Perpendiculare erected from the centre of the planisphere, and alhidada at the foot thereof, that what way so ever the Diameter with sights be turned, the Alhidada may always remain exactly underneath it, directing both to one vertical circle or point of the Horizon: this perpendiculare whereunto the semicircle s centre is fastened, aught also to be marked with 200 parts equal to the divisions of the scale beginning at the centre, so proceeding downward till you come to the end of those 200 portions: more I need not say of this instrument, considering the construction, if every part hath been severally declared sufficiently before, for the placing and conjoining of them, behold the Figures.  
IKLH the square Geometrical, MN his index with sights, GEFO Theodelitus, GF his Alhidada oer index with sights AB the line perpendiculare from B downward noted with 200 parts, equal to the divisions of the scale, DRC the semicircle having on his Diameter two sights fixed as was tofore declared. This is also to be noted, that the double scale is compound of two Geometrical squares, the one serving for altitudes, the other for profundities. The square which the line perpendicular cutteth when the Diameter is directed to any marks lying lower than your station, I call the scale of profundities, the other shall for distinction be named the scale of altitudes.  
THis semicircle aught so to be placed that the centre B hung directly over the centre A and that the diameter DC with his sights may be moved up and down, and also sidewise whither you list, always carrying GF about directly under it. You must also prepare a staff piked at the end, to pitch on the ground with a flat plate on the top to set this instrument upon. It is also requisite that within Theodelitus you have a needle or fly so rectified, that being brought to his due place the cross diameters of the Planisphere may demonstrate the four principal quarters of the Horizon, East, West, North and south: And this may you do by drawing a right line making an angle (with that one diameter of your instrument representing the meridian) equal to the variation of the compass in your region: which in England is 11 ¼ grades or near thereabout, and may be readily observed in all places sundry ways. But thereof I mind not here to entreat, forasmuch as it appertaineth to cosmography, & navigation, whereof I have compiled a treatise by itself, touching the fabrication this may suffice. Now for the use great heed must you take in pitching of the staff whereupon this instrument is placed, that it stand perpendicularely, which by a line and plummet ye may try, also when the instrument is placed thereon, ye shall by a line and plummet fixed on ●he centre of the semicircle discern whither it be rightly situate: for if the thread and plummet hanging at liberty fall close by the perpendicu●are, then is it well: otherwise ye must move the staff to and fro till ye find it so. This done, it behoveth you also to set this instrument on your staff, that the needle have his due place, so as the semidimetientes of Theodelitus may direct unto the four cardines or quarters of the Horizon, than what soever mark you espy, whose distance, altitude or profundity you desire, turn the dimetient of the semicircle to and fro, ●p or down, till through the sights thereon fixed you have espied it, ●lway the circle or Theodelitus remaining immovable: finally you ●hall note both what degrees the Alhidada cutteth of the circle, and the perpendiculare of the semicircle, and also what parts of the perpendicu●are is intercepted with the scale, these numbers thus found, you shall diversely use as shall hereafter be declared. This is farther to be noted that ●he double scale is compound of two geometrical squares, the one serving for altitudes, the other for profundities. The square which the line perpendicular cutteth when the diameter is directed to any mark lying ●ower than your station, I call the scale of profundities, the other shall for distinction be named the scale of altitudes.   

The .30. Chapter. By this instrument to know how many miles or pace any Ship is distant from you, yourself standing upon an high cliff or plat form by the sea coast.  
YOur topographical instrument equedistantly situate to the Horizon, (as was before declared) turn the diameter of the semicircle towards the ship, and when you have espied through the sights the wall or lowest part of the hull next the water, note exactly what part of the scale is touched with the line perpendiculare, them measure the height of the cliff, or rather the centre of your semicircles altitude above the sea, & multiply the same in the hole side of your scale, dividing by the parts touches of the perpendiculare line in your scale, the quotient is the distance of the ship from the basis or foot of the cliff lying perpendicu●arely under the centre of your instrument, but if you multiply the parts of the perpendiculare intercepted with the Scale in the cliffs altitude before measured, and divide by the parts of the scale cut, he quotient will show the line Hypothenu●al, or distance of that part of the ship which your line visual touched from your eye, or adjoining the square of the longitude first found to the square of the altitude, the root quadrat of the product is also the true length of the line visual.  

Example.  
ADmit I stand on the cliff A, and see the ship B lying at road in the sea, I desire to know how far of she is from me, mine instrument conveniently placed at C (as is tofore declared) I turn my dimetient of my semicircle toward the ship, moving it up and down till I espy through the sights ●he wall of the hull next the sea, and therewithal I find my perpendiculare cut●●ng the third part of my Scale, then measure I the altitude of the cliff above ●he sea OF, finding it 59 pace, whereunto I add one for the altitude of my instrument DC, so have I 60 pace, the height of the semicircle above the sea, which multiplied in 120 the side of the scale, there amounteth 7200, which divided by 3 yieldeth in the quotient 2400 pace, that is two miles, 3 furlongs, 25 paces, the longitude HB. Now square 2400 pace, se have you 5760000, whereunto if you adjoin the square of HD the product will amount to 5763600, whose quadrat ●●ote is the longitude of your line visual, represented by DB, being not fully 〈◊〉 foot more than HB.  
THis conclusion serveth most commodiously for all such as shall have committed to their charge any plaltform with ordinance, for here●y you may exactly at the first view, tell the distance of any ship or ●arke, so that having a table of Randons' made, mounting your pieces accordingly, no vessel can pass by your plaltform (though it be without ●oynte blank) but you may with your ordinance at the first budge her ●nd never bestow vain shot. This instrument serveth also no less aptly to be situate on the rampire or mount within any town of war, whence you desire with shot to beaten the enemy aloof of, before he shall approach nigh. But to give full instruction for shooting great ordinance, ●nd for the variation of Randons', having respect to the length of the ●eece, weight of the Bullet, force of powder, proportion of the concave Cylinders, and distance of the mark, it would require a long discourse: I have thereof in a book by itself very largely, and I dare be bold ●o say, sufficiently entreated: for conferring and conjoining Geometrical demonstration, with my long continued painful practices, I have ●t the last reduced that most irregulare course and circuit of the Bullet (framed and compounded of violent and natural motions) within ●he bounds of numbers and arithmetical rules, which whether it be dif●iculte or not, they only know that have or shall attempt to do the like. This surely I will say, that as the ignorant in Geometrical and Arithmetical proportions, shall never attain perfection, though he turmoil ●n powder and shot all the days of his life: so the Geometer, how excellent so ever he be, lening only to discourse of reason, without practice (yea and that sundry ways made) shall fall into manifold errors, or inextricable La●erinthes. Among many that I have read concerning that matter, I note one Nicholas Tartalea an Italian, who surely for his singular invention and perfect knowledge in Geometrical demonstrations, few or none in our time or many ages before may be compared with him: and yet handling this Argument, he hath erred even in the principal, and as I might term them the veriest trifles: I mean touching the uttermost Random and circuit of the Bullet, which he affirmeth to be made of a circular and right line: others have supposed it to be compact of many right lines, making several angles proportionally to the several mounts of the piece, or hollow Cylinders, and many like opinions and manifest errors have diverse well seen in Geometry for want of experience admitted and maintained. But of these in due place. Now to this purpose.    

The 31. Chapter. To know how much higher or lower any mark is than the level from your eye, although there be such impediments between, that you can neither approach nigh unto it, nor see the base.  
HEre shall you use great diligence in the placing of this instrument, that it be situate precisely equedistant to the Horizon, which done you shall turn the Diameter of your movable semicircle to and fro, till you can through the sights espy the mark, noting therewithal the points or parts cut in the scale and perpendiculare, then measure the distance of the same mark from your eye, as you were taught before, which multiplied in the parts of the scale, if you divide by the parts of the perpendiculare, the quotient is the difference how much the mark is higher or lower than your eye. Also if you add the square of the parts cut in your Scale, to 14400. reserving the producte for a Divisor, and multiply the Square of the distance hypothenusal in the Square of the parts cut in the Scale, dividing the ofcome by the reserved Divisor, the root quadrates of the quotient is the foresaid difference or unequality of levelles. And thus may ye also find how much any one mark is higher or lower than an other, although they be far a sunder, & either of them remote from you by comparing their altitudes found in this sort together.  
Admit B, my station where I place mine instrument A the mark whose altitude I desire above the level of mine eye, though I may not by reason of sundry impediments approach nigh unto it, nor see the base. first I measure the hypothenusal line AB, by the precepts tofore given, which I suppose 500 pace. then perceiving through the sights of my semicircle the mark A, I find 10 parts in my scale of altitudes, intercepted with the perpendicular line, the square thereof joined to 14400 produceth 14500 my divisoure. Then do I multiply the square of 500 in 100, the square of the parts cut in the Scale, so have I 2●●00000, which divided by 14500, the divisor before reserved, your 〈◊〉 quadrate root will he very nigh 41 pace 31 inches, and this is the true difference or unequality of level between the mark A, and the centre of your semicircle B, so that if a well be sunk of such depth that the bottom thereof were lower than A 41 pace 31 inches, as I admit the line AC, then may you certainly affirm, that C the bottom of that well is level with B, and yet may you not thereby, infer that from a fountain head, lying of equal height with B, you may naturally d●rya● water to C, for the level of waters is circular, as I have before in the book declared: And here I think it not amiss to give you a precept how to find the diversity of these levelles, whereby ye may exactly resolve sundry questions pertaining to water works, wherein divers have greatly erred, observing n●t this difference.   

The .32. Chapter. To find the difference between the straight and circular or true water level from a fountain at any place appointed.  
THis difference is nothing else but the length of a pendicular line falling from the level right line of a fountain to the water level of the same, as in the figure following A is your station, D the fountain whence I would convey water to the fort B standing on a hill, C a point by imagination conceived directly under B within the earth in a level right line from the fountain D, or rather the Superficies of the water therein contained. E the water level, that is to say, the highest point that any water will naturally run at, being conveyed by pipe from D. The line CE is this difference of levelles which you shall thus attain. first it behoveth you to get the distance of the fountain from the place whither you would convey your water, which distance you shall multiply by itself, adding the offcome to the square of the earths semidimetiente, and out of the producte extract the root Quadrate, from which root if you withdraw the foresaid Semidiameter, the remainder is your desired difference or live CE.  

Example.  
 
Admit the distance DC 10 miles, the semidiameter of the earth, 5011 italian miles, every mile containing 1000 pace geometrical, the pace being 5 foot: the square of this semidiameter is this number of pace 25110121000000. Likewise 10 miles the distance squared yieldeth 100000000 paces, this added to the square of the earths semidiameter produceth 25110221000000 paces: Now if from the root quadrate thereof ye subtract the somidiameter, there will remain 9 pace, 4 foot, and 11 inches: so much you may assuredly say, that the water level E is under the other level at C. Now if you would know standing at A by the fountain not approaching nigh the Castle how deep it were requisite to sink a well there to receive this water you may thus do, first measure the line BC, that is to say, how high the ground plat of the Castle is above the level right line of the fountain D, for this you are taught how to do before, then search out the difference between the straight and water level of the same fountain by the rule given in the last chapter, these two joined together, do produce the profundity BE, that is how many pace foot and inches you shall sink a well at the Castle to receive water from that fountain. Herein there need no example, the premises well understand, this conclusion is manifest.    

The .33. Chapter. You may also by this instrument meet the distance between ships on the sea, or other marks on the land, how so ever they be situate, with the aid of Arithmetical supputation.  
first measure the distance of either ship from your station, how soever it be, which you may do by sundry means before declared, then taking off the perpendicular and semicircle, and fixing the Index of the square Geometrical upon his due angle first place it on the side of the square, whence the ●●●isions of the Scale begin to be numbered, and turn the whole plate (●he Index keeping his place) till you can espy one of the ships through ●he sights. That done move it toward the other ship, which when you can also espy, note the parts of the Scale touched with the line fiducial, and also the parts of the index, cut with the side of the scale, then shall you thus work. Multiply the parts of the Scale by the distance of the ship which you last espied through your sights, and divide by the parts of the index, the quotient note, for it must serve you to double use, first square it and square also the distance last viewed through the sights deduct one of these squares from the other, the root square of the remainder ye must compare with the distance of the first ship, I mean that which at the first your index lying on the side of your square Geometrical ye espied, detracting the lesser from the greater, the remainder ye shall again square, and add it to the square of your reserved quotient DE, the root quadrate of the producte is the exact distance between the two ships.  

Example.  
 
I stand in the Castle A, the two ships, whose distance I require B and C, AC measured as is tofore declared 2000 pace, AB 2500, the parts of the scale touched with the fiducial line of the Index 50, the parts of the Index cut with the scale 130, B the ship first viewed, while the Index rested on the side of the square Geometrical, C the ship last espied, when the scale was cut with the line fidutial of the index, I multiply therefore according to the rule above given, 50 in 2000, so have ye 100000, which divided by 130 yieldeth 769 3/13 whose square detracted from the square of 2000, leaveth 3408284 4/169, whose Quadrate root is 1846 2/13 which deducted from 2500 the distance of the latter ship from the Castle, there remaineth 653 31/13, whose square added to the square of 769 3/13, the Quotient tofore reserved, produceth a number, whose root is near unto 1009 ½, these two ships therefore ye may conclude 1009 ½ place a sunder.    

The .34. Chapter. To draw a plat of any coast or country, containing the true proportion and Symmetry thereof, in such sort that you may readily tell how far any place is distant from other, and that without Arithmetic.  
YOu shall ascend on some high Tower, Hill, cliff, or other place, from whence you may commodiously behold on every part the whole Country round about adjacent in your Horrizon, there set up your Instrument topographical on his staff, and in such sort place it by aid of the needle, that the four Semidiameters may lie East, West, north and south, every one answering his like quarter of the heaven, then turn the Diameter of your Semicircle, to every Town, village, Haven, Rode, or such like, espying through the sights, the middle or most notable mark in every of them, noting therewithal in some Table by itself the Degrees cut by the Alhidada in the Circle, which I call the Angles of Position, and so make you a table of your first station. Then search out your eye, viewing round about, some other lofty place, from whence you may behold again all these places, for that shall be your second station: and turning thereunto the Diameter of your Semicircle, note also what parts of the Circle is touched with the line fiducial of the Alhidada: This done, situate your Topographical instrument, in all respects as was before said, and turning the Diameter of your Semicircle, espying through the sights, all such marks as you saw before, note again the Degrees cut, or Angles of position, writing the name of every place, and his Degrees by it, so have you an other table of your second station: with these tables you shall resort to some plain smooth Superficies of board, parchment, paper, or such like, and thereon describe a large Circle, dividing it as you were before taught, into 360 parts, like to the Circle in your instrument. Then from the Centre thereof to every Degree noted in your first table extend strait lines, writing upon every of them the name of his place, and upon that line that representeth your second station fix the one foot of your Compass, opening the other at pleasure, draw an other large Circle, dividing it into 360 Grades, and from the Centre thereof, extend right lines to every Degree noted in your second Table, writing as before upon every of them the names of their places or marks, whereof they are the sight Angles. Finally you shall note diligently the concourse or crossing of every two like lines, making thereon a Star or such like mark, with the name of the place correspondent. Now if you desire to know how far every of these towns, villages. etc. are distant from other, you shall thus do, measure the Distance between your stations by instrument or otherwise, as you have been before taught, and divide the right line between the Centres of your Circles into so many equal portions, as there are miles, Furlongs, or Scores between your stations, then opening your compass ●o one of those parts, you may measure from place to place, always affirming so many Miles, Furlongs or Scores (according to the Denomination of that one part whereunto you open your Compass) to be between place and place, as you find by measuring there are of those parts. Some consideration you must have in placing the Centre of your second Circle, so conveniently distant from the other, that the concourse or meeting of semblable positionall lines, may be within the Compass of your Paper. etc.  

Example.  
There is a sea coast having sundry harbours, towns, villages, castles and such like situate thereon, whose plat in due proportion I require, with the exact distances of every place from other, having therefore elected a lofty seat, from whence I may behold all these places, (mine instrument situate as is declared) removing the index to the farthest being a Castle standing in the mouth of an Haven, having received it through my sights, the line fiducial of mine Index, cutteth 30 Grades, thence removing it to the next, being a village or fisher town, the Index cutteth 50 degrees, again at the next being a City, it cutteth 75 grades at the fourth being a great Bay I note both the entrances, at the Eastern side of the Bay it cutteth 75 Degrees, at the Western 100 Degrees, finally at the farthest place Westward, being a fort within the land, the Index doth cut 130 grades. Thus have I the Table of my first station, as followeth.  

The Table of my first Station.  The Castle 
30 Degrees  The village, 
50 Degrees  The City, 
75 Degrees  The Eastern head of the Bay, 
95 Grades  The Western head of the Bay, 
97 ½ Grades  The Fort within the land. 
130 Grades   
This done, I behold an other high Hill, from whence I may in like manner view all these places, & turning my Index thereunto, I found the line Fiducial lying upon 180 degrees. Then carrying my instrument thither, & placing it in all points there, as it was at the first station, I turn my index again to the first mark or castle, finding it to cut 15 grades, at the second 25 degrees, at the third 40 grades, and so to the rest as appeareth in this Table following. And as I have done of these few, so might I how many notable marks so ever there were, writing their names severally with their Degrees noted, as ensueth.  

The Table of my second Station.  The Castle 
15 Degrees  The village, 
25 Degrees  The City, 
40 Degrees  The Eastern head of the Bay, 
50 Grades  The Western head of the Bay, 
55 Grades  The Fort within the land. 
80 Grades   
With these Tables repair to a plain smooth Superficies, drawing therein a fair Circle as you see divided in 360 parts, and from his Centre A, I pull out right lines to every Grade noted in my first table▪ AC the line ●e castle, AD of the village, A of the city, and so forth of the rest, ending 〈◊〉 the line AB, cutting 180 degrees in my circle, then making B a centre I 〈◊〉 a circle divided as the other at A, and from his centre pull out straight 〈◊〉 to all those grades that were noted in my second table: now marking the ●●●course of semblable lines, that is to say, where the line of the castle issuing from A 〈◊〉 with the line of the Castle proceeding from B, I note it with a star as 〈◊〉 may behold at the letter C, and thus prosecuting the like in the rest, always 〈◊〉 a star or like mark upon the concourse of correspondent right lines (all 〈◊〉 intersections not regarded) I have finally situated all these places in due pro●●●tion, noting them with these letters CDEFGH: as in the figure you may see. Now to learn the distance between every of them, I divide the space between B into 5 parts, for so many miles by mensuration I find between my two 〈◊〉, then drawing right lines from C to D, from D to E, and so from every one 〈◊〉, opening my compass to one of these parts, I measure how many times it contained in every of these right lines: whereby I conclude the number of miles between every of them, as appeareth in the table following.  

The Table.  

The distance of every mark from the stations.  
C 5 miles, 1 furlong.  
D 5 miles, ½ furlongs.  
F 5 miles, 3 furlongs.  
G 6 miles, 1 furlong.  
H 6 miles, 3 furlongs.  
C 9 miles, 6 furlongs.  
D 9 miles, 1 furlong.  
E 8 miles, 3 furlongs.  
F 7 miles.  
G 7 miles, 2 furlongs.  
H 5 miles.   

The distance of every mark from other.  
CD 1 mile 6 furlongs.  
CE 4 miles ½ furlongs.  
CF 5 miles 5 furlongs.  
CG 6 miles 1 ½ furlongs.  
CH 9 miles.  
DE 2 miles 3 furlongs.  
DF 4 miles 1 furlongs.  
DG 4 miles 4 furlongs.  
DH 7 miles 4 furlongs.  
OF 1 mile 7 ½ furlongs.  
EG 2 miles 2 ½ furlongs.  
EH 5 miles 5 furlongs.  
FG 5 furlongs.  
FH 3 miles 6 furlongs.  
GH 3 miles 4 furlongs.   
Thus passing or changing your stations, you may make several plats, containing the true proportion and distances of towns, villages, ports, roads, hills, rivers, and all other notable places through ●● wholeRealm, but then how to reduce them all into one fair card or map you shall be taught hereafter.     

The .35. Chapter. How to reduce many plats into one, and to make a fair card or map of an whole province or region, and that in as large or small room as you will assign, without any arithmetical calculation.  
I Mind not here to set forth the manner how to situate places in their due longitude and latitude, neither how to furnish your map with Meridian's, Parallels, Zones Climates, and other circles correspondent to the heavenly sphere, for that appertaineth to Cosmographic, only in this place shall you learn Chorographically to make a card, whereby you may readily know the true distance and situation of places one from an other, having therefore as is tofore declared (by sundry plats made in every several Horizon or Kenning) found out the true distance of every notable place from other, you shall make one entire table of all, containing the number of miles, furlongs etc. between place and place, beginning at the East, and so proceeding on till you come to the farthest Westward. This table thus finished you shall upon your Parchment paper or other plain whatsoever it be, fit for this purpose, draw one straight line directly through the middle thereof, crossing the same with an other perpendicularly, again as you were taught at the beginning of this book, then writ at the ends of these lines the four principal quarters, East, West, North and South. Now it behoveth you to consider (conferring your plats together) how far distant the most Eastern place is from the Western, and likewise the Northern from the farthest Southward, that you may accordingly so proportion your mile, as all these places fall within the compass of your card: now by the one side of your superficies, draw a straight line of competent length, then opening your compass to the wideness of that measure which ye will call your mile, mark out 20 of them in your last drawn strait line, which you may garnish with other Parallels, dividing eue●● mile into his furlongs, this shall be called your scale, Now must you ●●eyne guess (as near as you can by comparing your plaits) which is the middlemost place of this country that you describe, and the same shall ●●u situate upon the intersection of the two former drawn lines, making ●●ere a star, and writing the name thereof, whether it be city, village, ●●stle or such like, then search out in the same plat what other notable ●●ace or mark lieth east, west, north, or south from it. And if you espy ●●y resort to your table prepared as I have tofore said, searching out the ●●stance between these places, and to so many miles or furlongs extend ●●ur compass in the scale last made, then keeping your compass immoveable set one foot thereof upon the intersection or middle place, extending the other to that quarter wherein you found the place situate, make ●n intersection with the straight line, and there likewise make a star or ●●her fine mark, writing the name of the place: But if you find no no●●ble mark lying precisely east, west, north, or south from that foresaid ●iddle place, then take some other what you list in your plat, & note both ●●e distance & also the angle that it maketh, with the middlemost already marked in your card, that is to say (if it lie not in one of these principal ●uarters from it) how many degrees it declineth, and describing upon ●he intersection a circle divided into grades, pull out from the centre a ●ight line to the like number of degrees that you found in your plat: finally opening your compass to the like number of miles in your scale that ●ou perceive in the table to be between those places, fixing one foot in ●he intersection with the other, cut this last drawn line, and at the section make a mark, writing thereby his name, thus have you two places. Now for all the rest one rule shall suffice, whatsoever it be that you will next mark, search out in your table his distance from both these already ●oted, and opening your compass to the like number of miles in your scale, placing the one foot in either of the places already described, make an ark with the other, and where those arks cross there is the situation of the third place. And thus may you proceed to all the rest, only taking heed that having opened your compass to the number of miles you fix the foot in his correspondent place, and so shall you (passing on from one to an other) exactly situate them all, as by the example following shall more plainly appear.  

Example.  
 
BDCE the Parchment, paper or other plain whereon I would describe the country whereof I have already taken the plat or plats, as was before taught, BC and DE the two cross diameters, making intersection at A, I the line which I term a mile, HG the scale containing 20 of them, this mile if I list to make a large card should be divided into furlongs, yea and every furlong into parts, and therein you must use consideration, appointing the length of your mile according as you desire to have your chart great or small. Now it behoveth you to resort to your plats, searching out as nigh as you can by estimation the middlemost place. And that in my last plat (which here I mind to use for example sake) was my first station, here placed at A, then find I directly ●este from thence 5 miles my second station. Extending therefore my Compass to so many in the Scale, and setting the one foot in A, with the o●●er, I make a section in AC at K, and that is my second station. Now will I place the Castle which I find in my plat or table made by it, ●●stant from my first station 5 ⅛ miles, from the second 9 ¾ miles. first therefore I extend my Compass in the scale to 5 ⅛ miles, and fixing the one foot in 〈◊〉, with the other I draw an ark, then opening my compass to 9 ¾ parts in ●●e scale, and setting one foot in K, I cross with the other the ark tofore slawen in L, in like manner opening my compass again to 5 ½ in the Scale, ●hich is the number of miles between the Village, and my first station. The one ●●ote in A, I draw an ark, then extending the compass again to 2 miles lacking ¼ part, the distance between the Castle and Village, I cross the foresaid ●●cke in M, there is the Village. In this manner you shall proceed to the rest, ●aking fine pricks or other little marks for the true places of them, then blemishing all the arks, circles, and other lines, which I suppose drawn with black ●●ad or such like, that you may easily put out or raze away, you shall finally ●autifie it with images and Figures, as you think most agreeable and fit to ●●presse and represent the pattern, I mean the country itself that you describe, as by drawing a Castle at L, a Village at M, a City at N, a Har●●roughe at OP, and so of the rest. Even in like manner may you describe any Town, Forte, Camp, or Palace, and set out the true ground plat of all man●er Edifices, altering your Scale, and in stead of miles, dividing it into scores, ●ace, foot, and such like small measures.  

A note for sea Cards.  
YOU may also if you will divide the circle at the mids of your map into 32 parts, pulling out strait lines fro the centre to the uttermost bounds of the chart representing the 32 winds, and upon the same centre describe an other circle so large as your map can contain, the circumference thereof will be parted into 32 equal portions with the foresaid lines representing the winds. Now, if upon every of these intersections as a centre, you describe a circle dividing every of their circumferences into 32 winds, extending from their centres straight lines throughout the whole map as before you shall make a sea card wherein you may by the former rules Coasts, Harbours, Rocks, Sands, Shelves, Channels, Rhodes, with their soundings and depth of ancorage, etc. But of these matters here I mean not particularly to discourse, referring the more ample declaration of them to an other treatise of cosmography and navigation, wherein I shall not only discover the manifold errors that Mariners fall into by using their common charts and rules, but also set forth true exact and easy precepts for them, with sundry rare conclusions hitherto not practised of any.   

A note for Mines.  
MOst commodiously also serveth this instrument to conduct mines under the earth, for noting the Angles of position in the Planisphere or Theodelitus, and also Angles of altitude or profundity in the semicircle or scales appropriate thereunto, measuring the distances from Angle to Angle, you may make by the former precepts most certain plats of your journeys, and thereby always know under what place you are, and which way to direct your Mine to approach any other place you list. Whereof an infinite number of strange and necessary questions might be moved. It were an easy matter in such sort to make great volumes. But who so well understandeth the premises, (I mean not by reading them only, but by practising also) shall not need farther instruction in these or like cases. I might enlarge this book with sundry instruments, and many more ways and rules to measure, but as the premises are of themselves sufficient, so the diligent practizioner (searching out the reason and demonstration of them) shall be able of himself to invent manifold means to resolve the like or other stranger questions, whereof infinite may be proponed.     
The end of the first Book.   

The second kind of Geometry called Planimetra.  
HAVING accomplished the first part called Longimetra, containing sundry rules to measure lengths, breadthes, heigths, depths, and distances: I think it meet now to proceed to the second kind named Planimetra, wherein ye shall have rules for the mensuration of all manner plain Figures, to know their content superficial. And forasmuch as there is no Superficies, but is environed with lines either straight or curve. And all Figures comprehended with straight lines may be resolved into Triangles: It seemeth most meet, first to teach the measuring of Triangles as followeth.  

The first Chapter. Of Triangles.  
AS there are three kind of Angles, so is there also three kind of Triangles: the first is called Orthogonium, having one of his angles a right: the other Ambligonium, containing one obtuse angle: the third kind is called Oxigonium, whose three angles are all acute. Of right angled Triangles also there are two kinds, for either it hath two equal sides, and then is it called Isoscheles, or three unequal, and that is Scalenum.  

The right angled Isoscheles is thus measured.  
Multiply one of the equal sides in itself, the half of the producte is the Area or superficial content.  

Example.  
ADmit ABC an Isoscheles right angled triangle, having the sides AB and BC that contain the right angle B equal, either of them being 10, I multiply 10 in itself, thereof ariseth 100, the half 50, is the Area of that Isoscheles right angled triangle.    

A right angled Scalenum you shall thus mete.  
MEasure the longitude of two sides containing the right angle, augmenting the one by the other, the half thereof is that triangular superficial content.  

Example.  
Suppose the right angled Scalenum ABC, his two sides that contain the right angle AB 10, BC 16, the one augmented by the other, yieldeth 160, whose half being 80, is that triangular Area.  
In every of these right angled Triangles you shall not need to measure more than two of their sides, for any two known by supputation the third may be found, whereof I shall give you rules with examples.     

The second Chapter. Any two sides of right angled triangles known by calculation to find the third.  
EUCLID in the 47 proposition of his first book of Elements, hath proved by demonstration, that the squares of the two containing sides joined together, are equal to the square of the Hypothenusa, or third side subtending the right angle: whereby we may readily (any two sides being given) found the third thus: 〈◊〉 the two sides comprehending the right Angle be known, add their tauares together, and the Radix Quadrate of the Product, is the Hypothe●●sa, but if you know the length of that Hypothenusa, and one other side, 〈◊〉 shall subtract from the square of that Hypothenusa, the square of that ●●her given side, and the root Quadrate of the remainder is the third side desired.  

Example.  
In the Triangle ABC, AB is 6, and C ●, their squares joined make 100, ●hose root is 10, and that is the Hypothe●usa. In the Triangle DEF, the Hypothe●usa is ●3 DE, 5. I desire the length of OF, 〈◊〉 the square of DE, subtracted from 169 〈◊〉 square of the Hypothenusa, leaveth 144, ●hose Quadrate root 12, is the side EF.  
Of Obtuse angled Triangles, there are also two sorts, Isoscheles ●nd Scalenum.    

The .3. Chapter. Ambligonium Isoscheles is thus measured.  
YOu must first find out the Perpendicular line, from the Obtuse Angle to the contrary side, by deducting the square of half the side subtending the Obtuse angle, from the Square of one whole side containing the same angle, the Root Quadrate of ●he Product is the Perpendicular, which multiplied in the half of the foresaid subtending side, produceth the Area.  

Example.  
ABC the Isoscheles Ambligonium, AB ●0, BC 16, 10 squared is 100, 8 squared yieldeth 64, which deducted from 100, leaveth ●6, whose Quadrate root is 6, the Perpendicular AD: which multiplied by 8, produceth 48, and that is the Area of this Obtusiangle Isoscheles.   

Ambligonium Scalenum, you shall thus measure.  
FIrst you must search out his Perpendicular line in this manner, square every side, then add the square of the side subtending the Obtuse angle, to one of the other squares, from the offcome abating the third square, the half of the remainder divide by the foresaid subtending side, the square of the Quotient you shall deduct from that square which before you did add to the square of the subtending side, and from the remain, extract the root Quadrate, for that is the line Perpendicular, which multiplied by half the forenamed subtending side, will produce the content Superficial of the triangle▪  

Example.  
ABDELLA the Triangle, AB 20, AD 34, DB 42, the square of AB 400, of AD 1156, of BD 1764, 1764 added to 400 maketh 2164 from which subtracting 1156, there remains 1008, the half being 504, divided by 42, the side subtending the Obtuse angle, yieldeth in the Quotient 12, so much is the line BC, which squared maketh 144, that subtracted from 400, leaveth 256, whose root Quadrate being 16, is the Perpendicular AC. Now multiply 16 in 21, the half of the foresaid subtending side BD, so have you 336 the superficies of that Ambligonium Scalenum.     

The 4. Chapter. Of Acutiangle Triangles called Oxigonia, there are three kinds.  
FOr either the three sides are equal, and then is it an Equilater Triangle, or two sides only equal, which is Isoscheles, or all three unequal, and that is a Scalenum.  

Of Triangles Equilater.  
You shall multiply the square of the side in itself, and the offcome in 3, the Product divide by 16 the root Quadrate of the Quotient is the Area, or multiply the Cube of half the Equiangle Triangles side in the Semiperimetrie of the Triangle, the Quadrate root of the Product is the foresaid Area also.  

Example.  
Admit the Triangles side 6, the squared square is 1296, which augmented by ● yieldeth 3888, and that divided by 16, bringeth in the Quotient▪ 243, whose root Quadrate being 15, and between 1/● and 1●/31, is the superficial content of that Equilater Triangle. The same number is found by the second rule, for the Cube of 3 is 27, which multiplied in 9 the Semiperimetry yieldeth 243, whose root Quadrate is the Area of that Triangle agreeing with the former supputation.    

Of Isoschele Acutiangle Triangles.  
from the square of one of the equal sides subtract the square of half the base or unequal side, the Quadrate root of the remainder multiplied in ●●lfe the base, produceth the content superficial.  

Example.  
AB and AC the equal sides of the Isoschele Trian●●●, either of them 6, the base BC 4, the half thereof 2, ●●ose square deducted from the square of 6, leaveth 32, ●●ose root multiplied in 2, bringeth the root of 128, which ●●ery nigh 11 7/2● the Area of that Triangle.    

Of Oxigonium Scalenum.  
To find the Area of Acutiangle triangles that have side equal to other, it behoveth you to search out 〈◊〉 line Perpendicular falling from one of the angles the contrary side, and multiply the same in half the base or side whereon falleth, the Producte is your desire. But to get the line Perpendiculare, 〈◊〉 shall thus work: square every side, then add the square of the base, to 〈◊〉 square of one side, deducting fro the Producte the third square, half of 〈◊〉 Remainder divided by the base, the Quotient you shall again square 〈◊〉 deduct this number from that square which you adjoined to the square your base, the root Quadrate of the remainder is the line perpendicular.  

Example.  
〈◊〉 Acutiangle triangle Scalenun ABC, AB 13, BC 14, CA 15, the square of AB 169, of BC 196, of CA 225, 169 added to 196, pr●duceth 365, from which 225 detracted, leaveth 140, whose medietie being 70, divided by 14, yieldeth 5 the line BD, the square thereof abated from the square of AB, leaveth 144, his root● Quadrate is 12, the Perpendiculare AD▪ which augmented by 7 the medietie of the base, bringeth 84, the Area or Superficial content of that Triangle.  
Finally I think it not amiss to give you one Rule universal to measure all manner Triangles of what sort so ever they be, and that without any regard of the Perpendicular.     

The .5. Chapter. A rule general to measure all manner Triangles according to their plain.  
Add all the sides of that Triangle together, taking half of the number which surmounteth. Now out of this half, you must pull by Subtraction every side by itself, noting diligently the differences, that is, how much every side differeth from that half, multiply the differences the one in the other, and the Product in the third, with that which riseth augment the half above mentioned, then seek of that sum the Quadrate root, so have you the true content Superficial of that Triangle, the example shall ensue, whereby all that I have said, shall the better appear.  
Admit the Triangle whose sides you have measured ABC, of whom the left side AB i● 12, BC 16, AC 20, these sides joined together, make 48, whose half is 24, from whom BC 16, differeth 8, AB 12, AC 4, so the differences are 8, 12, 4. Now multiply 8 with 12, rises 96, the which augmented with 4, cometh 384, that multiplied in 24, the half if 48, surmounteth 9216, of this sum the Radix is 96, which is the very content of this Triangle.  
Of Quadrangles (that is to say plain figures, having four Angles and four sides,) there are five sorts, as appeareth in the definitions, the Square, the Rectangle, Rhombus, rhomboids, and Trapezia.   

The .6. Chapter. Of Squares.  
FOr the square you shall only measure one side, multiplying the same in itself, so have you the Area or content superficial.  

Example.  
I find AB the one side of the square ABCD 10, which augmented by itself riseth 100 the Area of that square.   

Of rectangles or right angled Paralelogrammes.  
In right angled paralelogrammes ye must measure the two unequal sides, multiplying the one in the other, the product is the content superficial of the figure.  

Example.  
I suppose the two sides AB, AC of that right angled Paralelogramme. ABCD the one 45, the other 110, these two multiplied together yield 4950 the Area.    

Of Rhombus and rhomboids.  
THese two figures have one rule, it behoveth you to measure one side, and the perpendicular falling from one of the opposite angles to the same side, these multiplied the one in the other produceth the Area.  

Example.  
Admit ABCD the Rhombus, whose Area I desire: I measure the ●●de BD finding it 20, then must I measure also the length of a line perpendicular falling from A upon BD, which I suppose here 16, these two multiplied the one in the other bring 320, the superficial content of the Rhombus. Likewise in the rhomboids suppose I find by mensuration the side BD 42, the line perpendicularly falling from A upon BD 16, multiply these numbers, I produce 672, the Area of that Romboides ABCD.  
But because it may seem somewhat difficult to get the length of those perpendiculars, because it is uncertain on what point of BD the perpendicular line shall fall, I think good to prescribe you a rule how you may exactly and readily get all such lines perpendicularly falling, and that not only in these figures, but also in triangles, which shall be no small ease and discharge of laborsom travailing, when you shall measure great fields, or champion plains, etc.     

The .7. Chapter. For measuring of lines perpendicular.  
YOu shall prepare a right angle by conjoining three staves proportioned according to these numbers, 3 4 and 5, as you were taught in the former book, the one of the staves containing the right angle you must place directly upon that side of your plain figure that subtendeth the angle from whence your perpendiculare should fall, marking therewithal whither the other containing side directeth, if it lie even with the foresaid opposite angle, then is the right line between your station, & that opposite angle, the perpendicular. But if it direct not justly to the angle, you shall move to and fro in that subtendent side, till you find it so, then by the precepts given in the fornier book, you may sundry ways measure the distance of the angle opposite from your station, whereby you are brought in knowledge of this perpendiculars length.  
Admit ABCD the paralelogramme, or ABC the triangle, whose per●●ndiculares falling from A to the contrary side I desire to measure, but because 〈◊〉 know not to what point in BC, or BD, these perpendicular lines from A will 〈◊〉, I take my right angle made of the three staves GEF, placing OF in the 〈◊〉 BC, and BD, passing to and 〈◊〉 in those lines till I found EG 〈◊〉 directly with the angle A, than 〈◊〉 I that a strait line from A, to 〈◊〉 station or angle E, is the perpendicular, whose length I may mea●●re either by Quadratum geome●●icū, or otherwise, without instrument, as in the first book is declared. Or thus deducting the square 〈◊〉 EBB, the distance of my station ●o the end or angle of either field 〈◊〉 B, from the square of the line A●, rising from that subtendent side 〈◊〉 the opposite angle, the root quadrate of the remainder shall be the 〈◊〉 perpendicular, which multiplied in BD, yieldeth the Area of that paralelogramme: or in half BC, so ●aue ye the superficial content of the triangle ABC.   

The .8. Chapter. To measure Trapezia.  
YOu shall measure the length of a diagonal or cross line extended to opposite angles through the Superficies of the Trapezium and likewise the length of the two perpendicular lines, falling from the other angles upon the said cross line as you were ●aught in the last chapter, then add those two perpendiculars together, ●ultipling the half of the product in that diagonal line, so have ye the Area of that Trapezium.  

Example.  
ABCD the Trapezium, CB the diagonal or cross line extnded fro the angle C, to BE and F, the two points where the perpendiculars shall 〈◊〉 from the other angles A, D, upon the cross line CB, these points I found with my right angled triangle, as was taught in the last Chapter. Now to attain the Area, I measure the length of BC 20, A 8, FD 6: 6 and 8 joined make 14, the half is 7, multiplied in 20, the line diagonal produceth 140, and that is the content superficial of that Trapezium.    

The .9. Chapter. Rules to measure all equiangle superficies how many sides soever they have.  
first you must get the centre of your figure, then from it pull a perpendicular line to the mids of some side, see how many perches or other measures it containeth, add all the sides together, multiplying half the sum in the perpendicular or han●ing line, so have ye your purpose.  

Example.  
Imagine this figure BCDEFG and every side of length 12, the centre A found, draw a line perpendicular from it to the middle of the side BC, this line being 10 ⅖ multiplied in 36 the half number of the sides, bringeth 374 ⅖ the superficial content of that figure.   

another Example.  
FGH is here a Superficies being 5 square, every side 10, Now from your centre to the mids of the side pull the perpendicular line which is 6 22/25 this multiplied in 25, the semiperimetrie of that figure, yieldeth 172, and that is the Area of this pentagonal superficies.  

A note to find the centres of those equiangle Figures.  
THe centre is found drawing lines from one angle to the contrary, or from the middle of some side being odd to the opposite angle, the cutting or concourse of those straight lines showeth the place of the Centre: and thus as is declared, you may readily measure all equiangle figures, what capacity or number of sides soever they be of.     

The .10. Chapter. To measure the Superficial content of any rightlined Figure of what form so ever it be,  
THe best rule I can perscibe, is to resolve it into Triangles, by drawing or ymagening lines from Angle to Angle, and so measure every Triangle severally: Finally, adding all the productes together, ye shall have the Area or content Superficial of that whole Figure, which although it be of itself (the premises well understand) plain enough, yet to avoid all doubts, I shall adjoin one example.  
Admit ABCDE an irregular pentagonum, whose side AB is 20, BC 30, CD 16, DE 24, EA 18 this figure you may divide in three triangles, by drawing the two lines BE BD, thus by the rules tofore given, I measure first the Area of the triangle ABE, which I find 144, likewise BED, 312, the third triangle BDC 240: these three numbers joined together, produce 696 which is the true Area or content Superficial of that irregular Superficies.   

The .11. Chapter. A ready mean to found the content superficial of any great field, or champion plain, how irregular of form or fashion soever it be, without painful travailing about it, only by measuring one side.  
YE shall, as I have taught in Longimetra (either with Theodelitus or your topographical instrument) draw upon some level, smooth plain superficies, the exact plat and symmetry of that field, then either by your instrument, or otherwise: measure how many rod or perch, is contained in some one side of that plain, so many equal divisions shall you make, opening your compass accordingly in his correspondent side of your plat. This done, note well the fashion of the field, and drawing lines from angle to angle, or otherwise, divide it into triangles, or other regulare figures, as you shall see cause: then opening your compass to one of those divisions, you may therewith measure every side, and line, in that figure as exactly as with cord, or pole, ye should painfully pace it over, whereby with the aid of these former precepts, you shall severally measure every triangle or other regular figure, and joining together their contents the resulting sum is your desire.  

Example.  
Suppose GHIKL that irregugular pentagonum, the true plat of some great field or plain made by one of the instruments Geometrical, as is before in the first book declared and that by mensuration I find the side GL 120 perches in length, opening therefore my compass accordingly, I divide GL into 120 equal portions, and proceeding on to every side I find GH 90, HI 100, IK 150, KL 80, then beholding the form thereof, I see it may aptly be parted in three Triangles by lines drawn from I, 60 G and L. In like manner I search how many such Divisions there are in them, finding either of them 170 perches in length. Finally, by the rule general of Triangles, I search the Area of every Triangle, finding the first that is IHG 360, IGL 780, IKL 600, all three ●●yned together, maketh 1740, so many perches you may conclude the Area of ●hat Figure, which reduced 10 Acres dividing by 160 bringeth 10 Acres 3 ½ R●odes.  

A note for Woods.  
THIS way shall you most speedily and exactly measure all manner Woddelande which otherwise is very hard and tedious precisely ●o do, forasmuch as ye can not at one view behold every part thereof, nor measure such diagonal lines passing thorough it in sundry places, as were requisite in irregular Figures, but with the aid of your instrument topographical encompassing it round about, alway no●ing the Angles of Position at every station, you may first make an exact plat thereof, and after measure as you were instructed in the former Chapter, the Superficial capacity.     

The .12. Chapter. How you may from an high Hil, or Cliff, measure how many Acres, roods, or Perches, is contained in any Field, Park, Wood, or other plain Superficies, in the country round about you, not approaching nigh them.  
CALl to remembrance how you were taught in the first Book by the Instrument topographical to set forth the true plat of an whole Country, and every part thereof, which, for asmuch as it is there at large set out, it were here superfluous to recite again, admitting therefore, (by the Art there taught) an exact plat form of the Field, Wood, or other plain made, ye shall with your● Compass divide some one side thereof into 40, 60, or 100 equal parts, as you list, and keeping your Compass immovable, measure all such other lines Perpendiculares, etc. as shall seem requisite to attain the Area thereof, and by the former precepts, dividing it into Triangles, Rightangled Parallelogrammes, or other regular figures, ye shall measure the contents superficial thereof, that is to say, how many of those small squares, whereof every little division was a side, is contained in that superficies or platform. This done, ye must also with your square geometrical or other instrument from the hill or cliff, measure the length of that side in the field, that the first divided side in your plat did represent, I mean how many rod or perch it is long then square aswell the number of perches in the side of the field, as also the number of divisions in his corresponding side of your plat, the number proceeding of the perches squared, ye shall multiply in the Superficial content of your plat tofore found, and the Producte divide by the square of the Divisions in the side of your plat, the Quotient will be the number of perches, which divided by 160, and the remain by 40, the Quotientes will show the Acres and roods contained in that field, park, or wood, you measure.  

Example.  
Admit ABCDEF the plat or proportional pattern of a park which from some Hill or cliff a far off I have drawn by the aid of mine Instrument topographical, as was declared in the first Book, and for that it is a Figure of many sides, I search how it may best be resolved into regular Figures, which (as you may perceive) is readily done withdrawing the two lines, AC and FD, which parts the whole Figure in two Triangles and one right angled Parallelogramme, then opening my Compass to some small distance, I divide some one side (for Example the side OF) into 40 parts, and keeping my Compass immovable, I measure how many of those Divisions are contained in AC and FD, finding either of them 60. Now, to measure the Triangles ABC, DEF, I draw Perpendiculares from B too AC, and from E to FD, finding BG 20, and HE 16, Now 40 multiplied in 60, bringeth 2400, the right angled Parallelogramme AD, 20 the Perpendicular in 30 half the Base yieldeth 600, the Triangle ABC, 8 half the other perpendiculare, in 60, the base ariseth 480, the triangle DEF, these three joined together, produce 3480, the whole content of ABCDEF, now as was taught in Longimetra, with my square geometrical, I must measure from the hill or cliff the length of that side of the park represented here by OF, which I admit found as is before said 356 rod or perch, whose square being 126736 augmented in 2400, the content of my plat bringeth 304166400, this divided by 1600, the square of the side OF yieldeth 190104, and this divided by 160 produceth in the quotient 1188, and the remain is 24. I conclude therefore that there is in that park 1188 acres, and 24 perches.  
In this sort with small exercise using industry, in making your plats you may most exactly and speedily surueye an whole country, with all his pastors, meadows, marshes, woods, and every particular enclosure, whereof although I might propound an infinite number of examples, and filled many leaves with variety of rules, yet considering the premises to the ingenious will suffice, I think good to pass them over, referring the rest to the practitioner, who shall by his experience (well understanding the premises) invent manifold more, even as occasion shall be offered in viewing the ground with situation of places.    

The 13. Chapter. A note how to survey an whole Region or plain champion Country by the aid of a plain polished glass.  
IT behoveth you to resort unto the first book where you were taught by the aid of a plain polished glass of steel) upon an high hill or leveled plaltform) to set forth the symmetry or proportion of an whole country with all his parts. Ye shall therefore by the art there showed, get the proportion or plat of that field, wood, marsh, or other enclosure, which you desire to survey, and therewith work even in like manner as you were taught in the last● chapter, to do with the plat drawn with your instrument topographical, one ground, one reason, and like operation serveth them both, the former well understand maketh this so manifest as it needeth no other example, only this I must give you warning of, that you attempt not this kind of measuring, but only in champion and plain level countries, for in unlevell and hilly places, without most exquisite observation and learned handling great error may ensue.  
Thus having declared several rules for every kind of rightlyned Superficies, it seemeth meet somewhat to say of such plain Superficies as are environed with curve lines or mixed of both, and first of the circle and his parts.   

The .14. Chapter. Of Circles.  
BEfore I entreat of the mensuration of circles, it shall be requisite to declare Archimedes rules, concerning the proportion of the circumference to his diameter, and of the Superficies to his dimetientes square. The rules ensue.  

The first Theorem.  
〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉.  
EVery circle is equal to that rightangled triangle, of whose containing sides the one is equal to the semidiameters, the other to the perimetrie or circumference.   

The second Theorem.  
〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉.  
THe proportion of every circle to the square of his dimetiente is as 11 to 14.   

The third Theorem.  
〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉.  
THe circumference of a circle is more than triple his diameter by such a part as is less than 1/7, and more than 12/7● thereof.  

How the Area of a circle is found.  
BY the former Theorems ye may collect these rules. Multiply the circumference of any circle by 7, and divide by 22, your quotient is the circles diameter, whose medietie multiplied in half the circumference yieldeth the Area, or multiplying the square of the circles dimetient by 11, and dividing the ofcome by 14, your quotient will declare the same.  

Example.  
Admit the circumference of the circle 44, which increased by 7 yieldeth 308, this divided by 22 bringeth in the quotient 14, the diameter, whose medietie being 7 augmented with 22 the half of the circumference, produceth 154. Likewise the square of 14 multiplied in 11, maketh 2156, which divided by 14, bringeth 154, the superficial capacity of that circle.    

Of the half circle.  
Even as I have declared that in multipleing the half diameter of any circle in his half circumference, the producte to utter his content, so by augmenting the half diameter in the fourth part of the circumference, that is in half the ark of the semicircle, ye have the whole sum of that half circle.  

Example.  
ADmit the semicircle DEF were to be measured, whose diameter DF drawn by the centre C is 14, as afore, the ark DEF 22. Now multiply the semidiameter 7 in 11 riseth 77, the plain of that semicircle, even so of all portions or parts of a circle. Although to the witty this may suffice, yet to satisfy also the mean witted, I shall not think it tedious to propound an other example.    

Of the fragments or parts of a circle.  
SVppose GHIA were a portion of a circle to be moten, the whole circumference of the circle whereof this is a portion is as afore 44. Now you shall multiply 7 the semidiameter HA' in the half ark 15, so have ye 105, which is the superficial content of that portion GHIA, If you desire to know the Area of GIB, contained of the cord GI', and ark GIB, ye must add to the number before found, the Area of the triangle GIA which I suppose (found by the rules tofore given) 22, that maketh 127 the Area of the segment GHI to the cord GI', which subtracted from 154, the whole circle leaveth 27 the area of the segment GIB.  
I think none will doubt how these figures following are measured, because they are made of parts or portions of a circle, whose plain is gathered as I have already declared. In the left figure there is at each end a semicircle, every of them containing (found by the art afore mentioned) 77, which added make 154, the quadrangle hath 84, as the rules of quadrangles declareth, that joined with 154, bringeth 238, the whole sum of that figure.  
The other are but two such segments as GIB, which measured as is before showed and joined together bringeth 54, the Area of that figure. Likewise the last figure called a Lunula, ABCD is measured by deducting the segment ADC (found by the former rules) 27 from ABC known in like manner 56 4/● there remaineth 29 4/7 the Area or Superficies of the Lunula ABCD. Forasmuch as it is necessary in measuring of these portions or segments of circles to know the diameter of the whole circle, whereof they are the fragments. I think it not amiss hereunto to adjoin a rule for the same purpose.  
Admit ABCD a portion or fragment of a circle, I desire to know the longitude of his circles diameter, first I measure AC which I find 12, likewise the length of the line DB from D the middle of the straight line to B the middle or highest point of the ark which I suppose 4, Now divide the square of AD that is 36 by 4▪ so have ye 9, which added to 4 bringeth 13, and that is the length of the circles dimetient, whose part or fragment this figure ABCD is.  
These precepts well noted there can no plain Superficies offer itself whether it be regularor irregular, framed of right lines environed with circular or mixed of both, but ye shall readily by dividing it into triangles & cular fragments, find out the Area and content superficial thereof. I will now therefore only, adjoin certain questions for the partition and division of ground. And so pass on to the third kind of Geometry, where you shall have rules to measure, not only the solid, but also the superficial contents of all manner bodies, but to return to these partitions my first question shall be of triangles, and so on to figures of more difficulty.     

The .15. Chapter. There is a trianguler field having on the one side a well or fountain, this field must be equally divided between two partie●, and that in such sort that either of them may have commodity of that fountain, not coming on the others land. I demand how that partition shall be made.  
Let ABC represent the triangular field, having on the side BC a well or fountain at D, from whence I would direct a hedge or ditthe in such sort that it parteth the Triangle in two even parts, first therefore I search the middle of the side BC, which I suppose E, than my instrument topographical at the fountain situate as he ought to be, I turn the Ashidada till I can espy through the sights the opposite angle here represented by A, then removing mine instrument to the foresaid middle at E, the Ashidada remaining immovable on the degrees tofore cut, the instrument again duly situate, I view thorough the sights what part of the side AC I can espy, Admit it F. Now say I that a hedge dyke or other partition running from D the fountain to F the mark espied in the side AC, will divide that triangular figure exactly into two equal portions, and thus may you without supputation only by the aid of your instrument Topographical divide any triangle in two equal parts, and that from any part or point in any of her sides assigned.   

The .16. Chapter. To cut off from any triangular field as many acres, rod or other measures, as shall be required, and that by a line drawn from any angle assigned.  
YOu shall first measure the side subtending the Angle assigded which here I will call the base. Secondly, you shall measure the Area of the whole triangle: Then multiply the number of Acres or Rods which you would cut off from the triangle, by the length of the base, and the producte divide by the Area of that Triangle, the Quotiente showeth how many perches you shall measure in the base from one of the Angles to cut of your desired number of rods▪  

Example.  
Suppose ABC a triangular piece of ground, having his three sides AB 30▪ AC 40, CB 50, my desire is with a right line from the angle A to cut of one ●cre of ground, to perform this by the ●rte before taught, I search the Area of that figure which is 600 rods. Now because I would cut off from that figure one acre, and an acre containeth 160 rods: I multiply 160 in 50, the whole side subtending the assigned angle, the product being 8000, I divide by 600 the triangles Area, my quotient is 13 1/● so many perches I reckon in the base BC from B to D, finally extending a line from A to D, I conclude the Superficies ABDELLA one exact acre, and this is there from that triangular field one acre cut of with the line AD.    

The .17. Chapter. To cut off from any triangular piece of ground what quantity of perches ye list with a line equidistant to one of the sides.  
TO perfoorme this conclusion it shall be requisite to search out what parts of the other two sides this equidistant line shall cut, whereof I will give you two rules, forasmuch as this partition may two several ways be made: for the portion to be cut of either is adjoining to the parallel side, and then is it a quadrangular figure, or else to his subtending angle, and then is it a triangle. For the triangle ye shall thus do, square the sides that the parallel line shall cut, and the products severally multiply in the number of perches to be taken away, the surmounting sums divide by the Area of the whole triangle, the roots quadrate of the quotientes is the number of perches to be accounted in the correspondent sides, from the angle that subtendeth the Parallel side. But if ye would have the portion cut off, lie adjoining to the parallel side, then shall you first deduct that portion from the whole triangular Area, and with the remainder work in like manner as ye were taught in the former rule, so will your quadrate roots (accounted from the parallel side his subtendente angle) show to what part of either side the line shall be drawn, to cut of the forenamed portion.  

Example.  
Suppose ABC the triangle from which I would cut off one acre, that is to say 160 rod by a parallel line to the side AB, first therefore I measure the Area of that whole triangle as was before taught in this book, finding AB 50 perch, AC 120, BC 130, and so consequently the Area of that triangle 3000 rod, the square of AC is 14400, the square of BC is 16900, these augmented by 160, bring 2304000, and 2704000, and these divided by 3000, produce in the quotientes 768 and 901 ⅓, the root of 768 perches) being 27 perches 12 foot) I measure out in the side AC, beginning from C, admit it end at D, likewise the root of 901 ⅓ perches is 30 perches, and between 4 and 9 inches, measuring therefore 30 perches from C in the other side CB I set up a mark at E. Now if you draw a straight line from D to E it shall be a parallel to AB, and the piece of ground, represented by DEC, an exact acre. But if you desire to say out this acre at one of the sides as ye may see in the figure signified by the quadrilater Superficies ABGF, Then must ye deduct the aforesaid 160 rods fro the Area of that triangle, the remain is 2840, which I multiply as before in the square of AC, so have I 40896000, Likewise the same 2840 augmented by the square of BC produceth 48016000, these products severally divided by 3000, the Area of the whole triangle will yield in the quotientes 13632 and 16005 ⅓ their quadrate roots are 116 perches 12 ½ foot, the length of the line CF, and 126 perches 8 ½ foot, the line GC, Or if ye deduct those roots from the whole sides AC, and BC there will remain 3 rod 4 foot from A to F, and 3 rod 8 foot from B to G And thus may you in all Triangular pieces of ground, exactly lay forth an acre, ●r any other quantity of ground ye will require, and that either against the side 〈◊〉 Angle, even as you will desire.    

The .18. Chapter. For partition of Paralelogrammes what kind so ever they be of, note these Rules ensuing.  
Because partition may sundry ways be made according to the several situation of the line wherewith it is divided, I will first entreat of that division that is made by a Parallel unto two of the Paralellogrammes sides.  
Admit therefore ABCD the piece of ground which by mensuration I find to contain 65. Acres 16 Perches, and that I would cut of with a Parallel to AC 5 Acres, as ye may see done by the line OF, which were easily brought to pass if I knew the quantity of the lines A or CF, for knowledge of them therefore I work thus, first with my square Geometrical or otherwise as hath been declared in Longimetra, I find the length of AB 308 rod, I multiply therefore 800 for so many rod are there in 5 Acres, by 308, there amounteth 246400 which divided by 10416 the number of Perches contained in the whole quadrangle, yieldeth in the Quotient 23 Perches 11 foot, the length of the two lines A & CF. Thus by making the partition OF, the portion AECF, shall contain exactly 5 acres, this rule is general for all quadrangular pieces of ground, whose sides be parallel, whether it be square rectangle, Rhombus or rhomboids.  
But if you desire to make like partition by a right line issuing out of one angle you shall thus work: first consider whether the portion ye would cut of be greater or less than half the parallelogramme: if less multiply as before the numbered of perches, that ye would separate from the quadrangle in one of the opposite sides to that angle whence your dividing line issueth, and the product divide by the medietie of the Area, the quotient will show on what part of that opposite side your dividing line will fall. But if the aforesaid portion to be separate be greater than the medietie of the Parallelogramme ye shall deduct it from the whole Area thereof, and with the residue proceed in like manner as was before taught.  

Example.  
Admit ABCD a Quadrangular piece of ground, which being measured, is found to contain 50 Acres, and that my desire is to cut of 10 acres from that figure, with a right line passing forth from the Angle C, first therefore I measure the length of the side AB, finding it 200 rod, 200 therefore I augment by 1600 the number of Perches in 10 acres, there ariseth 320000, which divided by 4000 the number of rod contained in half that figure, your Quotient will be 80, the number of Perches from A to E, so that by the Partition CE, ye shall separate the Triangle ACE containing exactly 10 acres, but if the sum of acres had been greater than half, ye should have deducted it from the whole Area, and with the residue work in all respects as ye have done with this, the only difference is, that at the end of your operation, whereas here the Triangle ACE is the portion separate, there it should be the residue or Quadrangle ECDB, for that there is no difference in the working, it were superfluous to use more Examples.    

The .19. Chapter. To cut of from any Trapezium or Quadrangular piece of ground, what part thereof ye list.  

Example.  
Admit ABCD a Trapezium, or Irregular Quadrangle, and that I would cut of a quarter of his Area, with a line issuing from the Angle A, first by the Rules tofore given, I measure his content superficial, which I suppose 72 acres, secondly I meet the Area of the Triangle ADC, finding it 50 acres. Now because I would cut of a quarter of that Trapezium, I divide his content that is 72 by 4, my Quotient is 18, and that multiplied in DC, (which here I suppose found by mensuration 146 Perches,) produceth 2520, and this divided by 50 the number of acres in the Triangle ADC, yieldeth in the Quotient 50 Perches 6 ½ foot, which measured out from D in the line DC, showeth where the line of partition issuing forth of the angle A shall fall: and thus may you conclude that ADE, is the foresaid quarter of the whole Trapezium, and containeth 18 Acres.    

The .20. Chapter. To divide the superficies of any irregular Pollygonium, with a strait line proceeding from any one of the Angles assigned in such sort, that the parts shall retain any proportion appointed.  
WHen any proportion is given, there are two Numbers wherewithal it is expressed, and they are called Termini, those two you shall add together, reserving the Producte for a Divisor, then measure the Area of that whole Irregulare Polligonium, which you shall multiply in the lesser of those Termini, the Producte shall you divide by your reserved Divisor, the Quotient is the content of the lesser Portion, which Deducted from the whole, leaveth the Superficies of the greater part. But to draw the line or Partition that shall divide this Pollygonium, it were necessary first to learn on what side and part thereof this line should fall, which you shall thus do: Draw or imagine right lines extended from the assigned Angle to every Angle on either side, so shall ye make several Triangles, whereof by the Rules tofore given, you must search the content Superficial. This done, ye shall compare the Area of the first triangle, with the Superficial content of the lesser portion found as is before declared. If the Triangle be greater, then may you by the precepts before given, cut of from that Triangle so many Acres or Perches as that lesser Portion should contain, but if the Triangle be not greater, ye shall deduct it from the foresaid lesser portion, and the Remain compare with the next Triangle, which if it be greater than that Triangle also, subtract from it the Superficies of the triangle, & so proceed on till ye find a triangle greater than your remain, then may you say, that your partition or dividing right line proceeding from the assigned angle shall fall on the base or side subtendent of the assigned angle in that last triangle, but to know on what part thereof, you must work with your last remain and the Area of your last triangle, as you were before taught in the division of triangles. And in like manner draw your dividing line which shall exactly, separate that Irregular Polligonum, in two Superficies retaining the prescribed proportion: for more plainness peruse the example ensuing.  
Admit ABCDEF, an irregular Hexagonum which I would divide with a right line issuing forth of the angle A in such sort that the greater part might be triple to the lesser; this proportion triple may be expressed with these two numbers 3 and 1, and they are called Termini or bounds of that proportion, these two added make 4, wherewith divide the whole Superficies of that Pollygonum, which I suppose by mensuration found 64 Acres, my Quotient will be 16. Admit also I find the Area of my first triangle ABC 12 acres, & the Area of the second ACD, 18 acres. Now comparing 16 with the Triangle ABC, I find it greater, deducting therefore the one from the other, there resteth 4 Acres, which forasmuch as it is less than the Area of the Triangle ACD, I conclude the partition or dividing line shall fall on the side CD, but to learn on what part thereof, I work as I was taught in the partition of Triangles thus, first with my square Geometticall or otherwise, I measure the side CD. Admit it 36 Perches, which multiplied in 4 the last Remain bringeth 144, and that divided by 18 the acres contained in the triangle ACD, produceth in the quotient 8, so many rod length shall you meet from C to G, in the side CD, finally extending a right line from the assigned angle A, to the point G, you may conclude AFEDG Triple to AGCB.   

The 21 Chapter. To divide any irregular Pollygonium into as many equal parts as ye will desire, with right lines drawn from any point within the superficies thereof assigned.  
first get the Area of the whole irregulare Pollygonium, which you shall divide by the number of parts, whereunto ye would dissever it, and the quotient reserve, then from your point assigned deduct or imagine lines to every angle of that figure, so shall ye part it into sundry triangles, then compare your reserved quotient with some one triangle beginning where ye little, and if ye find the triangle greater, then cut of from it a portion equal to your quotient, with a straight line proceeding from the angle adjacent to the assigned point, but if your quotient be greater than the first triangle, deduct the one from the other, and compare the remain with the second triangle, which triangle if ye find greater, cut of from it a portion equal to your remain, with a line issuing out of the angle joining to your assigned point, and then compare again your reserved quotient with the remaining triangle, if your quotient be greater, separate from the triangle next ensuing a portion equal to the excess or overplus, and that always with a lyn● issuing out of the angle adjacent to the point first given. Thus proceeding till ye have circulate the figure, always drawing straight lines from the point assigned, ye shall in the end depart the whole figure into as many equal portions as ye determined. This one thing is to be noted, that so oft as your remain or reserved quotient falleth out equal to any of the triangles, then shall you not draw any line from the asigned point, for the side of that triangle serveth for the partition or dividing line.  

Example.  
LEt ABCDEFG represent an irregular Pollygonium, or seven sided piece of ground, which I would divide with straight lines issuing out of the point H into three equal parts, first therefore I imagine straight lines from H to every angle, so is it divided into 7 several triangles: now by the rules tofore given I meet the area of every triangle, finding HAB 5 acres, HBC 5 acres two roods, HCD 4 acres, HDE 6 acres 1 road, HEF 5 acres, HFG 5 acres 12 perches, HAG 4 acres 2 roods. Thus by adding all these triangles together you shall found the area of that whole figure 35 acres, 1 rood, 12 perches, which resolved into perches yieldeth this number 5652, and that divided by 3 bringeth 1884 perches, and that is the content of every portion that this figure should be parted into. First therefore comparing it with the triangle ABH, I find the triangle the lesser, the one deducted from the other, there remaineth 1084 perches, and this remain I find likewise greater than the triangle CHB, detracting therefore the one from the other, there remaineth 204, which compared with the third triangle CHD, forasmuch as it is less than the area thereof, I search by the rules tofore given in division of triangles, on what part of the base, CD the dividing line shall fall that proceedeth from the angle H: admit it cut CD in the point I, from I to H extend a right line, and that shallbe the first partition. Now seeing HID joined to DEH the fourth triangle, is less than 1884 perches my reserved quotient, I deduct the one from the other, so there remaineth 448 perches, I search therefore by the rules tofore given on what part of OF the line shall fall, proceeding from H to cut of 448 perches from HEF: but here you must consider that this portion of 448 rods or perches must be adjacent to the side HE, because the quotient doth exceed the triangles by somuch. This considered admit the partition HK. now because my desire was to divide it only into three parts, I need proceed no further, for those two lines do exactly depart that whole irregular Pollygonium into three equal parts, whereof the first is that rightlined figure IHABC, the second is IHKED, the third is KHAGF, and this partition is made with right lines issuing from the point H assigned, which was required. In this manner you may divide any Superficies into as many equal or unequal parts as you list, or into what proportion ye will desire, as by these few examples the ingenious will readily conceive of any other: manifold rules might I give you in these and like partitions, but knowing the premises to the witty are sufficient; I will only adjoin certain questions of some more difficulty, with the resolution of them, and so finish the second kind of mensuration.    

The .22. Chapter. To divide any circle whose semidiameter is known, with an other circumference concentrical, in two such parts that the one portion to the other shall retain any proportion assigned.  
WHen any proportion is given, it consisteth of two numbers, as I have before said, that are called Termini rationis, those numbers ye shall add together, reserving the producte for a divisor, then multiply the square of these midiameter known by the lesser of those Termini, and the ofcome divide by the reserved divisor, from the quotient thereof, resulting, extract the quadrate root, for that is the semidiameter of the concentrical circle, whose circumference shall divide the former given Circle in two parts, retaining the proportion assigned.  

Example.  
ADmit 120 the semidiameter of the circle, BCDE, whose area I would divide with a concentrirall circumference in such sort that one part might be triple to the other, this proportion consisteth of 3 and 1, which added together make 4, now the square of the semidiameter being 14400, augmented by 1 the lesser of those Termini, produceth the same sum again which divided by 4 tofore reserved for that purpose yieldeth 3600, whose quadrate root is 60, so much is OF the semidiameter of that inward Circle, whose circumference hath divided the Circle BCDE in two parts, the lesser is the circle FGH, and the greater is the anular Superficies contained between those circumferences, the one being triple to the other. In like manner may you divide that anular Superficies into three other, every one of them equal to the same inward circle, whereof it were superfluous to adjoin any farther example, forasmuch as the proportion of the parts once known, the operation is in all points agreeable with the former.    

The .23. Chapter. The three sides of a triangle known, by supputation to get the greatest circles semidimetiente that may be described within that circle.  
YE shall first add the two shorter sides together, reserving the producte, then subtract those sides the one from the other, the remain multiply in the former product, and the amounting sum divide by the third or longest side, the quotient detract from that longest side that was your last divisor, the medietie of this remain you shall multiply in itself, and deduct the ofcome from the square of the shortest side, the root quadrate of the remainder is the perpendiculare falling from the greatest angle to the greatest side, which retaineth the same proportion to the semidiameter of the inscribed circle, that the perimetrie of the triangle doth to the base or greatest side, working therefore by the rule of proportion by three known, ye may readily find the fourth.  

Example.  
LEt ABC be the triangle, AB 36, BC 48, AC, 60, now to get the greatest circles semidiameter that may be described within that triangle, I work thus, first 36 added to 48, bringeth 84, than 36 detracted from 48 leaveth 12, which multiplied in 84 bringeth 1008, which divided by 60, the longest side yieldeth in the Quotient 16 ⅘. This deducted from 60 the Divisor, the remain will be 43 ⅕, the medietie thereof is 21 ⅖: whose square deducted from the square of 36 leaveth 829 11/25, Therefore do I conclude √² 829 11/25, ●he perpendicular BG, for the semidiameter OF, I use the rule of proportion ●aying 144 the permetrie, giveth 60 the base, what shall √² 829 11/25 by multiplication of the latter and division with the first, ye shall find the fourth proportional number 12, and that is the line OF or semidimetient of the greatest circle that may ●e described within that triangle.  
This is also to be noted, that the medietie (whose square ye subtract from the square of the left side to get the perpendiculare) is the line AG: which deducted from the whole base, leaveth the other line GC. and those two parts of the bases are called Casus.    

The .24. Chapter. To find the greatest squares sided that may be described vithin any triangle whose sides are known.  
Sundry rules might be given to resolve this question, but to avoid confusion, I will only declare the easiest: ye shall therefore (as is before in this book taught) search the perpendicular line falling from the greatest angle to the longest side: this perpendicular it behoveth you to divide in such sort, that the parts retain such proportion as doth the base or longest side to the said perpendiculare, so doing the greater portion is the squares side. But Arithmetically to attain the quantity of this longer portion, ye shall thus work: Multiply the perpendicular in itself▪ and divide the producte by the Base or longest side of the triangle and the perpendicular joined together, the quotient detracted from the whole perpendiculare leaveth your desire. Or multiply the base in the perpendicular, and divide by them both joined together as before, so shall your Quotient produce the squares side also.  

Example.  
ABC is the triangle whose sides are known, AB 24, BC 40, AC 32, the perpendicular found, a● is tofore declared, is 19 ⅕, which divided at I, in such sort that DI to IA retain the proportion of BC to DA, DI shall be the side of the greatest square that may be made within that Triangle: but too find the length of DIEGO, you must multiply 40 the base BC in 19 ⅕ the perpendiculare, so have you 3840/5 which divided by 206/5 the perpendiculare and Base joined together, yieldeth 12 36/37, the line ID. Likewise, if ye square the perpendicular, the mounting sum will be 9216/25 which divided by 196/5 the former divisor bringeth in the quotient 6 ●2/185, which deducted fro 19 ⅕ the perpendicular, there remaineth 12 180/185, the side of the greatest square agreeing with the former operation. In like manner if ye divide BD, whose longitude you were taught by the last Chapter to find, and also AB, the left side of the Triangle in such sort, that the greater sections to the lesser retain the same proportion, that the base doth to the perpendicular, the squares of the two greater sections one deducted from the other leaveth the Area of the greatest square that may be described within that Triangle. Also, if ye divide any of the two lesser sides in two parts retaining the foresaid proportion of the perpendiculare to the Base, the lesser of those Portions augmented by the base and the product divided by the side bringeth in the quotient that greatest squares sided. Thus also an other way, you may attain the same: Divide both the Casus, that is ●o say, BD, and DC the distance of either Angle from the perpendicular, in like manner as hath been said of the Perpendicular, then add both the smaller sections together, the resulting sum is the squares side: Every of these ways working, ye may resolve this question, great pleasure shall the Arithmetrician (especially if he be seen in surde and radical numbers) receive, when he shall perceive so diverse intricate and different operations alway in fine to produce the self same certainty.    
The end of the second Book.   

The third kind of Geometry named Stereometria.  
IN THIS third book ye shall receive sundry rules to measure the Superficies and Crassitude of solid bodies, whereof, although an infinite sort of different kinds might be imagined, yet shall I only entreat of such as are both usually requisite to be moten, and also may sufficiently induce the ingenious to the mensuration of all other solides what form or figure soever they bear. And forasmuch as in setting forth their several kinds, I have chosen to use the accustomable and ancient names well known to any traveled in Geometry, rather than to forge new English words which can neither so briefly nor so aptly express the like effect, lest to the common sort any obscurity might grow, I think good to adjoin every of their definitions, 

definitions.  
1 A solid body is that hath length, breadth and thickness bounded or limited with Superficies.  
2 Like solides are such as are encompassed with superficies that are like and of equal number.  
3 A Prisma is a solid Figure comprehended of plain Superficies, whereof two are equal, like, and Parallel, the rest Parallelogrammes.  
4 A Pyramid is a solid Figure enclosed with many plain Superficies rising from one, and concurring or meeting in a point.  
5 A Sphere is a gross or solid body comprehended of one convex Superficies. In the mids whereof there is a point from whence all right lines to the same superficies extended, are equal.  
6 That point is called his Centre, and a straight line by that Centre passing thorough this solid bounded on either side with the convex superficies is called the Diameter of that Sphere.  
7 Also intellectually ye may thus conceive a Sphere to be made. Suppose a semicircle (his diameter remaining immovably fixed) to be turned round about till it return to the place whence it first began to move, the figure so described, is a Sphere.  
8 Likewise, if a right angled triangle (one of the containing sides remaining fixed) be turned circularly about the Figure so described, it is called a Cone.  
9 The right line that remaineth fixed is the Axis.  
10 The circle described by the other containing side is the Base.  
11 The third line or Hypothenusa, is the side of the Cone.  
12 If a right angled parallelogramme (the one of the sides containing a right angle remaining immovable) be circularly turned, the Figure so described, is a Cylinder, and the immovable side is his Axis.  
13 The Bases are the Circles by revolution of the two opposite Parallel sides described.  
14 The altitude or height of any solid body, is a line perpendicularly falling from the top or highest part of the solid upon the plain whereon the body lieth.  
15 This perpendicular or line of altitude in direct solides falleth within the body, and upon the base, but in declining solides, it falleth without the bodies and bases.  
16 As the concourse of lines maketh a plain angle, so the concurring or meeting of many superficies in a point, maketh a solid angle.  
17 In every solid body a right line drawn from one solid angle to one other is called a line diagonal. But if it pass between opposite angles, it is named the Diameter.    

The first Chapter. To measure the contents Superficial and solid of any Prisma.  
FOr the Superficies, ye shall add all the Parallelogrammes & Bases their Area together, the resulting sum is your desire, but for the crassitude, ye shall augment the altitude of the solid in the Area of the Base, the producte is the gross capacity.  

Example.  
For the Crassitude I search the altitude of either solid, which in the upright Prisma is, the ereared side of any of his Paralelogrammes, as AD, BC, or FE. ●or they are all equal, every one of them being 10, but the altitude of the declining Quadrangular Prisma is the Perpendicular OR, falling from O the top of the Prisma perpendicularly on MR a line drawn in the plain whereon the body re●●eth, 10 therefore multiplied in 6, produceth 60, the solid quantity of that Pris●a: Also the altitude OC being found by mensuration 12. multiplied in 18 the ●ase, bringeth 216, and that is the solid capacity of the declining Prisma IKLMNO.  
Although these common pieces. K.L. are moten as is tofore taught, yet 〈◊〉 may readily thus measure them, multiply the length with the breadth, ●nd the Product in the thickness, so have ye the content or Crassitude.  

Behold the Figure.  
The content of K 216 cubical foot, the Crassitude of L 216 square foot.  
    

The .2. Chapter. How the contents Superficial and solid of a Pyramid may be measured whether it be direct or declining.  
FOr as much as every Pyramid is environed with triangular plains rising fro the base, and concurring or meeting at the top, ye may by the rules given in Planimetra measure the Area of every Triangle▪ whose contents joined together, and the Product to the base, yieldeth the Superficies of that whole Pyramid. But for the content solid whether the Pyramid be direct or inclinate, ye shall first measure the length of the line Perpendicularly falling fro the top to the plain whereon the base resteth, which multiplied in the third part of the bases Area will produce the Crassitude of that Pyramid.  

Example.  
Suppose ABCDE an upright Quadrangular Pyramid, whose base being a square, I measure only the side BC finding it 12, which multiplied in itself, bringeth 144 the Area of the base, then do I measure the length of the line AG 10, for so am I taught in Planimetra, that multiplied in 6, half DE bringeth 60, the Area of the triangle ADE, but seeing the other triangles are equal to the same, considering they have equal bases and altitudes, I need make no more ado, but multiply 60 by 4 so resuiteth 240 the superficies of the four triangles concurring as A▪ and this added to 144, the base bringeth 384 the whole content superficial of that Pyramid. But for the Crassitude I measure the Altitude or line OF falling from A Perpendicularly upon the base, this Perpendiculars length I suppose 8, which multiplied in 144 produceth 1152, and that is the solid con●●nt of that Quadrangular Pyramid. In like manner to attain the Crassitude of 〈◊〉 declining Pentagonal Pyramid IKLMNO, I measure the altitude or Perpendicular IH 20, likewise the Area of the Pentagonal basis, whose side is 14, which by the precepts given in Planimetra I find 337, these multiplied together ●elde 6740, the solid content of the declining Pyramid.    

The .3. Chapter. How Cylinders and Cones are measured.  
THe Cylinders altitude multiplied in the Circumference of his base, or the side of the Cone augmented in half the periphery of his base, & the Productes added to their bases, bring the contents Superficial.  
The solid content of a Cylinder is gotten by augmenting the base in ●is altitude. Likewise the altitude of the Cone multiplied in the third part ●f his base yieldeth his Crassitude.  

Example.  
Admit the Area of the Cylinders ●ase found by the rules in Planime●ra 78 4/7 the circumference being 31 3/7, which multiplied in 21 the Cylinders altitude yieldeth 660, which added ●nto 157 1/7 bringeth 817 1/7, the whole superficies of that Cylinder, but if 〈◊〉 augment 21 in 78 4/7 the bases A●ea, it produceth 1650, and that is ●he Crassitude or solid capacity ●f that Cylinder. Likewise for ●he Superficies of the Cone GHE, I augment 26 the side HG in 31 ●/● half the Circumference GFE, thereof resulteth 817 1/7, whereunto if ye adjoin 314 2/7 that basis Area there amounteth 1131 3/7, the Cones content Superficial, and by multiplying 24 the altitude in 104 16/21, the third part of the Circle GFE. there ariseth 2514 2/7, the solid content of the Cone GHE.  

A note to find the altitude of a solid Cone.  
Multiply the side of the Cone in itself, and likewise the Semidiameter of the Circular base, these deducted the one from the other the quadrate root of the Remainder is the Axis or Altitude of the Cone.     

The .4. Chapter. How excavate or hollow timber. etc. is measured.  
YOu shall by the Rules afore, search the content or Crassitude, as though it were not hollow, then measure the capacity of that hollow, the one subtracted from the other, the Remain uttereth the magnitude of that excavate body.  

Example.  
Admit this round hollow timber log ABC were to be measured the height being 14 foot, the Diameter of the outward Circle 7 foot, the Dimetient of the inward Circle 5 foot, the circumference of the bigger 22 foot, the content superficial or plain of it 38 ½ the Circunference of the less 15 foot and ●/7, his plain 19 ●/14. Now multiply 38 ½, in 14, ● have ye the crassitude of the whole round piece of Timber which is 539 foot. ●hen augment 19 ●/14 into 14 riseth 275 foot, which subtracted from 539, re●ayneth 264, so many foot is in the crassitude about that hollow piece of tym●er. Or thus ye may bring it to pass: Pull 19 9/14 from 38 ½, that is the one super●●ciall content from the other, and the remain multiply in the height 14, So ●aue ye as afore 264 foot. another example of timber hollowed and somewhat fashioned like unto a trough: Suppose it to be DEFG, whose outward breadth DE is 10 foot, the length OF 14 foot, the height FG 8 foot, the inward breadth ●f the hollow HI 5 foot, the length HK 7 foot, the height as before. Now multiply 10 in 14, so have ye 140. This augmented in 8 riseth 1120. Then multiply 7 in 5, that maketh 35, which augmented in 8, surmounteth 280, that ●umme by subtraction taken from 1120, leaveth 840, so many foot is the crassitude of this hollow trough DEFG, even thus of all other. By that which is ●oken ye may gather how to get the capacity of all manner regular vessels, and ●hat is performed in multiplying the plain or content superficial of the inward ●ase or bottom in the profundity or height. Example 35 the inward bases plain augmented in 8, the height produceth 280. So many cubical feet is in the hollow vessel. These 280 augmented in 51, bringeth forth 14280 pints of water, farther divided by 8 showeth 1785 gallons. Thus of all other, and that for liquor contained in any manner vessel.    

The .5. Chapter. How the crassitude and Superficies of a sphere is te be measured.  
FOr the superficies ye shall multiply the diameter in his circumference or get the plain of that circle, as before in Planimetra I have taught, and increase it by 4, so have ye the quantity superficial, and that multiplied in the sixth part of the diameter produceth the crassitude. Likewise the diameter multiplied in his square, and the offcome in 11, this product divided ●y 21 produceth in the quotient also the solid content of the sphere. Many ●o rules could I give you for the like effect. But for better understanding of these, behold the example ensuing.  

Example.  
ABCD the globe whose diameter AC is 14, his circumference being 44, the one multiplied by the other, bringeth 616, and that is the Superficies of the globe which increased by 2 ⅓ bringeth 1437 ⅓ and that is the solid content which is likewise thus produced 14, multiplied in his square yieldeth 2744, this again in 11, maketh 30184, which divided by 21 giveth in the Quotiente 1437 ⅓, the Crassitude agreeable to the former work.    

The .6. Chapter. How fragments or parts of a Globe are measured.  
YE shall augment the whole superficies of the globe by the altitude or thickness of the fragment, and the offcome divide by the dimetient, your quotient is the convex spherical superficies of the fragment, whereunto adding his base or circle ye produce the whole superficies.  
For the crassitude if it be less than half the globe ye shall first detract the altitude of the fragment from the semidiameter of the sphere, the remain ye shall augment by the circular base. This producte shall ye subtract from a number amounting by the multiplication of the semidiameter in the convex spherical superficies of the fragment, the third part of the remain is the crassitude or content solid of that fragment. But if the fragment be greater than half the sphere, then shall ye deduct the spheres semidimetiente from the altitude of the fragment and the remainder multiplied in the circular base the product must be added to the number produced by multiplication of the semidiameter, in the convex spherical superficies of the fragment, the third part of this resulting sum is the desired crassitude.  

Admit ABCD a fragment of the former globe, whose diameter is 14, and ●hat BD the altitude thereof be 4, the spheres whole Superficies being 616, multiplied by 4, bringeth 2464, and that divided by 14, yieldeth in the quotient 176, 〈◊〉 convex Superficies, whereunto if ye adjoin 126, the Area of the base or circled, whose dimetient is AC, there amounteth ●02, the whole Superficies of that portion ABCD, by the like working ye shall find the superficies of EFGH 566. But for their crassitude I work thus, In the lesser portion I find 〈◊〉 altitude 4, which deducted from 7, the spheres semidimetient, leaveth 3, which augmented by 126, the bases Area bringeth 378, and ●his deducted from 1232 (a number produced by multiplication of the spheres semidime●●ent in the convex Superficies of the segment) ●●aueth 854, whose third part is 284 ⅔, so ●uche is the solid content of ABCD. But ●or the crassitude of the other portion, I augment 440 his convex superficies in 7, there ariseth 3080. Likewise I multiply 3, the difference between the semidiameter ●nd the fragments altitude in 126, thereof cometh 378, these added together, ●ake 3458, which divided by 3, yieldeth for a quotient 1152 ⅔, the solid crassitude of the other fragment EGFH. But for as much as the spheres diameter is here supposed known, I shall give you a rule readily by supputation to ●earne the quantity thereof.    

The .7. Chapter. Any spherical segment proponed to attain the spheres dia●● meeter whereof it is the fragment.  
first measure the altitude of the fragment: Secondly, the semidiameter of the circular base, which ye may also attain by knowledge ●f his circumference as was taught in Planimetra, then square the semidiameter and divide the product by the fragments altitude, the quotient added to your divisor yieldeth the spheres diameter. Or thus, add the square of the altitude to the square of the semidimetient, and divide that producte by the altitude, so will your quotient express the diameters quantity.  

Example.  
ABCD the spherical segment his altitude BD 4, the semidiameter of his circular bases AD nigh 6 ⅓ the square thereof is 40, which divided by 4, yieldeth in the quotient 10, which added to 4, bringeth 14 the Diameter of the Sphere. Likewise the square of 4 is 16, added to 40 the square of AD bringeth 56, which divided by 4, giveth for the quotient 14, the sphere 5 dimetient agreeing with the former working.    

The .8. Chapter. How a Pyramid cut off, may be measured.  
IF the Pyramid be unperfect, yea cut off in the top, continued it by rule and line wittyly laid to the two contrary sides▪ and where the joining and common meeting is there that Pyramid is whole and perfect, then measure that whole by the Art afore, and also the Pyramid going from the top of the unperfect to the common meeting. This done ye shall subtract the crassitude of the less from the whole continued Pyramid, so the remain without doubt is the magnitude of the unperfect Pyramid.  

Example.  
Let ABCDE represent an hexagonal imperfect Pyramid encompassed with two equiangle Hexagonal Parallel plains and 6 Isoscheles Trapezia, every of their Bases being the sides of the ●reater Hexagonum 12, every side of the lesser Hexagonum 6. Now, applying rules or lines ●nto the sides BF, GE, I find them to concur at H. Thus have you two Pyramids, ●he greater having his base an equiangle Hexa●onum, whose side is 12: the other is the Pyramid HFG, whose base is the Hexagonum FG, ●uery of his sides being 6, the altitude of the lesser Pyramid is 15 the altitude of the greater ●0. Also by the rules given in Planimetra, ye ●hall find the area of the lesser Hexagonum ●3 ⅗, the area of the greater 294 ⅖, and so consequently by multiplication of them in the third ●arte of their altitudes as was taught before in ●he mensuration of whole Pyramids, the solid content of the lesser will be 468, the crassitude of the greater is 3744, the one deducted for the other, leaveth 3276, the magnitude of that unperfect Pyramid FBCDE.  

A note how by supputation to get the perpendiculares or altitudes of those Pyramids.  
FOr that it were tedious & painful in a solid of great quantity, by rule or line to search out these perpendiculares, ye shall by Arithmetic attain them thus: multiply the greater hexagonal side by the side of the unperfect Pyramid, and the producte divide by the difference or overplus of the sides hexagonal, the quotient showeth the side of the whole great Pyramid, from whose square (if ye deduct the square of the circles semidiameter that encloseth the greater Hexagonum which ye were taught in Planimetra how to find) the root quadrate of the remainder is the altitude of the greater Pyramid, which augmented again by the lesser hexagonal side, and the offcome divided by the greater, bringeth in the quotient, the altitude of the lesser Pyramid.    

Example.  
The unperfect Pyramid side FB being nigh 16 ⅛, I multiply by 12 the grea●er hexagonal side, there ariseth 193 ½, and that divided by 6, the lesser side hexagonal produceth in the Quotiente 32 1/●, the Pyramidal side HB, from the square thereof, if ye deduct the square of BL, the circles semidiameter that encludeth the greater Hexagonum being 144, there will remain 900, whose root is 30, the altitude of the greater Pyramid, which augmented by 6, the lesser hexagonal side, and divided by 12, produceth in the Quotient 15 the altitude of the lesser Pyramid. These lines for that they are some of them incommensurable, can not exactly be expressed, save only in surde numbers: but so nigh as is requisite for any Mechanical mensuration this operation declareth them: such as are expert in Algebra by the former rules with irrational numbers may precisely show their quantity. The very like operation is required in perfect and unperfect Cones, for measuring their Perpendiculars and crassitude. Whereof to the ingenious there need no other Example.    

The .9. Chapter. To cut off from any Cone or Pyramid what part or portion thereof ye will desire, with a plain equedistante to the base, and to find on what part of the solides side the section shall fall.  
THis division of Cones and Pyramids is in two respects to be made, either in consideration of their contents superficial, or in regard of their solid quantities, for either of them I shall give you several rules. First, it behoveth ye to measure the side of the Cone or Pyramid, that side shall ye divide by the quadrate root of the number expressing the part assigned, your quotient is the line or distance from the top of the Cone or Pyramid to the section or place where the plain parallel to the base shall pass, which will divide or cut off from the assigned solides superficies the desired part. The same line or distance is also found with division of the Cones side squared by the number expressing the part to be cut off, for the root quadrat of the quotient shall be your desire. But if ye would from any Cone or Pyramid in like manner make separation, in such sort that their solid contents might retain like proportion, then shall ye multiply the side of the Cone or Pyramid, first in himself, and then again in the offcome, and this last producte shall ye divide by the number expressing the part ye would separate, the root cubical of your Quotiente is the side of the lesser Cone or Pyramid to be cut off from the top of the greater.  

Example.  
Admit ABC the Cone, from whose summitie or top, I would cut off a portion, for example, the Cone AEF whose superficies should be the 9 part of the great Cone ABC, his superficies. I take therefore the root quadrate of 9, that is 3, wherewith I divide 100 the side of the great Cone there ariseth in the quotient 33 ⅓ the line OF or side of the lesser Cone, the same number I produce also by the second rule thus, I divide 10000 the square of AC, the Cones side by 9, the quotient yieldeth 1111 1/9, whose root is likewise 33 ⅓, the side AF. But for operation in solides, I shall give you an other example: Suppose HK the side of the Pyramid HIK 120, from the which Pyramid I would cut off an other portion, for example, HLM that might be an eight part of the great Pyramid HIK, the number of the side 120, I multiply in itself thereof ariseth 14400, which augmented again by 120 maketh 1728000, and this divided by 8 (for that is the number wherewith the part is expressed) I find in the quotient 216000, whose root cubike is 60, and that is the length of HIM or side of the Pyramid HLM, whose solid content is the eight part exactly of the great Pyramid HIK. Even so shall ye do● in all other regulare Pyramids, how many equal sides soever his base have: but if the sides of his base be unequal, and so consequently the sides of the Pyramid also unequal, then shall ye work with every side severally as I have by example showed ye in one: or else work by the rule of three for all the sides are proportionally divided: this to the witty will suffice.  

Here followeth other fashioned bodies somewhat strange in figure, but in effect those that tofore are measured.  
THe first ABCD appeareth to be two round Pyramids joined together, whose Diameter AC is as the round Pyramid afore measured 12 foot, the height or perpendiculare line 20 foot and 5/40, the side 21, wherefore I conclude the crassitude of the one part or half of it to be even as the round Pyramid before measured, that is 759 foot, which doubled, make 1518, the whole content of this Figure, the other Figure ensuing is like to an egg, and not very unlike to the Figure above. Wherefore to measure him, ye shall work as ye did in the other, the crassitude of that so had, and then doubled, bringeth the whole content of him. Also this Figure EFGH is like in content to the other ABCD, according to that form whose crassitude is 1518, this doubled maketh 3036 foot, so much ye may affirm the crassitude or quantity of that figure like unto an egg. Now to get the quantity superficial of him, ye shall multiply the ark FEH in the half circumference of the circle whose Diameter is EG, or the whole circumference in half the ark, so have ye the superficies.     

The .10. Chapter. How wine vessels or barrels are measured.  
SVppose ABCD were the barrel to be moten: first, ye must take the just measure of both the heads severally, the one head AB, I have imagined 7 foot, the other CD as many. Now, take the true height of the mids where the barrel swelleth, which is OF, being here 10 foot. These three diameters keep, them note how many foot, or other measure is contained from the mids of either head to the middle of the swelling G in a straight line moten within the Barrel. Behold from the head H to G the middle swelling, is 6 foot, from G to this letter I as many: now ye must set the three diameters (tofore reserved) upon some charred, paper, or other plain, the one differing from other according to their measures, as ye may see in the figure, then enclose them with an ark on both sides, cutting the middle line, which line crosseth the diameters squirewise in the middle, the one ark is KBECL, the other is KAFDL, the arks drawn as ye see, lo a figure is made like unto an egg: Truly all such fashioned figures I have taught you to measure immediately afore. In him ye may find 1047 13/21, now if ye pull away by any art from this whole sum the number of feet contained in the pieces of the figure superfluous and more than the barrel at either end, than the magnitude or capacity of that vessel must needs remain, as by the example is plainly perceived, although it would seem I had entreated sufficiently, yet I think it worthy of remembrance to tell you how to search the content of these pieces at the ends of the figure, I must suppose that ye know the measuring of a round Pyramid which is entreated before, imagine the diameter  which is 7 foot to be the base, KH the height of the piece here imagined a Pyramid: now if ye (as is before mentioned) multiply the plain of the base in the third part of the height, the content of that figure will come according to that fashioned meeting, which is 51 ⅓, with this number by the precepts of proportion ye shall work as followeth: first know the length of HL that is 16 foot, again GL being 10 foot, both these added make 26. Now say 16 giveth 26, what shall 51 ⅓ bring, so have ye 53 ●/2, so many foot is the piece KBA, and the other CDL as many, which added make 166 ⅚▪ pull this out of the number contained in the whole figure afore found, which is 1047, so have ye left 880 11/14 the capacity of your Barrel. Thus the learned and famous Archimedes hath taught in the 31 proposition of his book de Sph●roibus & Conoidibus.  

another mensuration of vessels more common.  
SOme use when the base plain of the extremes and mids of any vessel are not one, to subtract the less from the bigger Superficies, the medietie that remaineth deduct from the big plain: than whatsoever is left add to the lesser Superficies, the half of the producte showeth the true base to be multiplied in the height, of the which cometh the content.  

Example.  
THe plain of the bigger Superficies containeth 40 foot, the lesser 24, the difference is 16, the half 8 subtracted from 40, leaveth 32, this added to the less superficies 24 make 56, the half is 28, which multiplied in 24 bringeth 67●▪ the whole content. Lo the figure.  
   

Another note of measuring.  
IF the Diameter of the vessel be in the mids 10 foot, but at the ends eight, add them, and take half, so have ye nine, thus nine foot shall be thy diameter, whose plain or base (as before) multiplied in the longitude or height, bringeth a true content. Note well this kind for pillars and other bodies, when they have diameters or bases not agreeable.    

The .11. Chapter. One rule general exactly to measure all kind of wine vessels.  
Forasmuch as there are sundry kinds of wine vessels, as the tun, the pipe, the punshion, hogsheads, butts, barrels, etc. every of them differing from other aswell in quantity as in fashion, to teach several rules for every sort it were over tedious, leving therefore manifold precepts and examples that I might in this case prescribe, I shall for brevity sake set forth one only rule general, whereby ye shall be able with the aid of a small proportional vessel, not only to measure the whole capacity of any manner wine vessel, but also the default and quantity of liquor therein contained, when it is partly empty. And first I will begin with this small proportional vessel, which it behoveth you to procure by some skilful Artificer, so made that their lengths retain ●he same proportion that their circumferences or circles, aswell at the ●ndes and mids, as also at other like and proportional distances taken in the length of either vessel: as by the example following shall more plainly appear.  

Example.  
Admit S and ● the two vessels, ● the greater, S the lesser: now ●o examine whether these two 〈◊〉 proportional, 〈◊〉 search first ●heir lengths, ●●●nding HD 60 ●●ches, and RQ 〈◊〉 inches, likewise 〈◊〉 search the longitude of every depth or diameter, that is to say, BF and LO the diameters or depths of the mids, and AG, CE, KP, MN, the circles of the ends, now if BF, AG and CE retain the same proportion to KP, LO, MN, (every one particularly compared to his like) that HD the length of the greater doth to RQ the longitude of the lesser, then are those vessels likely proportional: but for greater certainty (because error may grow upon small difference) it were requisite also to measure in either of them one other circumference or hoop, as for example in the greater, th● circle passing by TV exactly in the middle between AG and BF, and in the lesser the circle XY situate in like manner between KP and LO. Now if ye found the proportion of these two circumferences also agreeable with the former lengths and circumferences, ye may assuredly gather these vessels to be proportional. But if ye find any discrepance or variance between them, ye shall by the aid of some skilful Artificer reform it in the lesser, till ye have brought it fully agreeing with the proportion of the greater, which is readily proved by the rule of proportion. Whereof I mind not here to use more words, lest in seeking needless plainness, I become over tedious: your small vessel thus prepared, ye shall in the next chapter be taught how to use it.    

The .12. Chapter. How by this small prepared vessel to measure the quantity of the greater.  
Whensoever ye will measure any manner of wine vessels, it behoveth you to consider what sort they are of, and accordingly to prepare your lesser proportional vessel, then shall ye in either of them measure the profundity or greatest diameter, and also the diagonal or cross lines from the bung holes to the opposite or lowest part of either base, these diagonal lines ye shall square, and the productes severally multiply in his correspondent Diameter, of these surmounting sums ye shall divide the greater by the lesser, your quotient openeth how many times the lesser vessel is contained in the greater, or if ye augment the greater of those last surmounting sums by the crassitude of the lesser vessel, and divide by the lesser sum, your quotient will declare the solid content of the greater vessel.  

Example.  
Suppose ABCD the wine vessel, whose capacity I would know EFGH my lesser proportional vessel, first I measure the diameters EG and AC. Admit the first 4, the other 40, again I meet the diagonal lines. Admit I find AB 50 OF 5, the square of 4 is 16, which augmented by 5, bringeth 80, the square of 40 is 1600, & that augmented by 50, yieldeth 8000, which divided by 80, bringeth in the quotient 1000, so many times is the lesser contained in the greater. Now to learn how many pottles or gallons is contained in that great vessel, I try first how much liquoure my little prepared vessel will hold: Admit it contain ●/4 of a pint, this augmented by 80000 produceth 60000, which divided by 80, bringeth 750 so many pints ye may conclude in that great wine vessel which reduced to gallons dividing by 8 yieldeth 93 gallons 3 quarts, the exact quantity of liquor that such a vessel will contain, but if it happen that your vessel be not thoroughly filled, and that ye desire as well to know how much liquor would suffice to fill it, as also what quantity is therein contained, ye shall in the next Chapter receive therein perfect instruction.    

The .13. Chapter. How both the liquor and default or emptiness in wine vessels partly filled is to be moten.  
FOr greater exactness ye shall provide a fine straight rod of 4 or 5 foot in length, exactly divided into 1000 equal portions. Then shall ye move the vessel whose liquor or default ye would measure, till it lie level the bung hole directly upward, this done take your rod divided as is before declared, and let it descend perpendicularly down through the bung, till it come to the bottom of the vessel. This done, note what part▪ the bung or upmoste height of the wine vessel will touch of the rod, and likewise how many parts thereof is wet with the liquor, and last of all how many parts are contained between the liquor and the bung hole. The first I call the diameter or profundity of the vessel, the second the parts of liquor, and the third shall be named (for distinction) the parts of default or emptiness. This done ye shall also measure with your divided rod how many parts the diameter of your lesser vessel containeth, these parts augmented by the parts of liquor tofore found, and the product divided by the diameter of the wine vessel, yieldeth in the quotient the parts of liquor, for your little prepared vessel, which it behoveth ye so to situate that it may also lie level as did the greater vessel, then pour in so much water that by proving with your divided red ye may find the parts of liquor exactly agreeing with your former quotient: This done augment the diameter of the wine vessel cubically, that is to say, by his own square, and the producte in the quantity of liquor now being in your little prepared vessel, the producte divided by the cube of your lesser vessels dimetient yieldeth in the quotient the true quantity of liquor contained in the wine vessel, and that again deducted from the whole content of the vessel found, as was in the former Chapters declared, leaveth the default or emptiness, that is, how many gallons or other measures is requisite to fill up the said vessel.  

Example.  
Admit ABCD the wine vessel partly filled, ADCE the liquor, ABCE the default or emptiness, FGHI my smaller proportional vessel prepared as I have tofore in the last Chapter declared, LM the strait rod divided into 1000 parts, which being let perpendicularly fall from B till it touch the opposite side at D, I find BD 800 parts, ED that is so much as is w●tte of the rod 600 parts, likewise placing the rod in my lesser vessel at G, admit I find the diameter or profundity GI' 200 parts, this number multiplied by 600, the parts of fullness yieldeth 120000, which divided by 800, bringeth in the quotient 150, the parts of fullness for my smaller vessel: filling therefore the proportional small vessel with liquor till it rise unto K, that is to the 150 part in my divided rod LM, finally by measure I make trial what quantity of liquor is in my little vessel. Admit it 3 pottles: Now to learn the quantity of liquor contained in the great, first I multiply BD the greater diameter being 800 in his square, thereof ariseth 512000000, and that again by 3, so have I 1536000000, which divided 8000000 (the cube of GI' the less dimetient) produceth 192, so many pottles is there in that wine vessel. Now if ye desire to know how many pottles or gallons more will fill it full, ye may by the last Chapter search how many gallons the whole vessel will contain, and from that withdraw 96 gallons, the quantity of liquor already therein contained, the remain is your desire. I think it not necessary in this case to adjoin any farther example, for that this was in the former Chapter plainly both by rule and example already declared. This kind of mensuration serveth not only for wine vessels, whereof there are sundry forms, but also for all manner bodies, what kind of fashion soever they be of, and likewise for their fragments or parts, and it is grounded upon this Theorem. All like solides retain among themselves triple proportion of their like or correspondente sides, that is to say, look what proportion the cube of one side in the lesser solid, retaineth to the cube of his correspondent side in the greater solid, the same proportion doth the lesser body retain to the greater: so that with the aid of the golden rule the first three being known, the fourth is readily found: And as this rule serveth for the whole bodies, so is it also to be applied to all fragments or parts: so that the Superficies wherewith the solides are divided be like, and make equal angles with the sides and Superficies of either solides, for thereupon it must consequently ensue that those fragments correspondently compared are also proportional.  
Thus much I thought to adjoin touching the reason and demonstration of this rule, whereby the ingenious practizer might the better retain in memory the operation, and also be able to apply the same to sundry other uses not here mentioned if occasion be offered. And for as much as this kind of measuring dependeth wholly upon the small proportional Uesselle, whose quantity is supposed known, I think it not amiss to give one rule general for all small vessels how irregular soever they be (their fashion not regarded) exactly to find their solid quantities.    

The .14. Chapter. To measure exactly the solid content of any small body, how disordered or irregular so ever it be, the form or fashion not regarded.  
YE shall prepare a hollow vessel of cubical form so large that it may contain the small irregular body, which being placed therein, ye shall your in so much water that it cover altogether the body, then make a mark where the Superficies or upmoste part of the water toucheth. This done take out the same irregular body, and mark again directly under the former where the brim of the water now toucheth, for the distance of these two marks multiplied by the square of the Cubes side, produceth the crassitude of that irregular body.  

Example.  
Admit A the cubical hollow vessel whose inward side I suppose 20 inches, B is the irregular body, whose crassitude I desire, first therefore I put the solid into my hollow cube, and pouring in water till it be thoroughly covered. Admit the brim of the water reach unto C, then taking out that irregular body again: admit the Superficies of the water fall to D, I measure the distance between C and D, suppose it 7 inches, which multiplied in 400, the square of the cubes side produceth 2800, so many cubical inches are contained in that Irregular body B. Neither is it of great importance whether your vessel A be an exact Cube. For what kind of Prisma soever it be, always rising uniformly, and his Paralelogrammes being Perpendicular upon their base, ye may (changing the form of your Calculation) bring the same to pass, only whereas in the cube ye did multiply his sides square in the distance between the water marks, ye shall now multiply the same distance or difference of water marks in the base of the Prisma, the resulting sum is the Irregulare Solides Crassitude. And thus may you always frame your containing vessel, according to the form and quantity of the Irregulare body that ye desire to measure. marvelous is the appliance of this kinds of mensurations, and strange conclusions may be performed thereby, wherein although I mean not in this treatise to reveal any secrets, reserving them for an other place, yet to give some light to the ingenious to proceed in applying them farther, I shall not think it tedious to show how it may be used to discover the weight of such things, as no way possibly by balance may be found.    

The 15. Chapter. How the weight of any part or portion of a solid body may be known without separation thereof from the body, whereby it might be poised or weighed in Balance.  
FOr as much as neither by common Balance, neither by that kind of balance which the Italians use, called Statera, nor any other hitherto invented, the weight of any fragment or part of a solid body may be known without separating or cutting that Fragment of from the whole body, whereby it may by itself alone be poised in the balance, I thought it not amiss aswell for the rareness of the matter, as for the necessary uses thereof, to set forth this mean of searching weight by water, with the aid of Arithmetic: your vessel therefore being prepared as is tofore declared, whether it be Cube or Rectangular Prisma it forceth not, ye shall first fill it full with water, and throw the solid body thereinto, then softly lift that body out of the water till such time as there remain no more in the water than that portion whose weight ye desire to know, at that instant make a mark on one side of the vessel where the Superficies of the water than toucheth, then take out the body altogether. This done, measure the distance from your mark to the Superficies of the water as it is now after the body is taken quite out. Likewise measure the distance of the waters Superficies from the top of the vessel. This done augment the weight of the whole body by the lesser distance, and divide by the greater, your Quotient will show the true weight of the fragment or portion.  

Example.  
Admit BCD a pillar of an hundred pound in weight, being of Brass, Iron, Silver, or any other Metal, my desire is to know the weight of the foot or portion from B to C, first therefore putting the hole pillar into the vessel A, I fill it full of water, then lifting it softly up till all the pillar be out of the water, save only the foot or fragment BC, I find the Superficies of the water fallen to E, then do I lift out the whole pillar, leaving no part thereof within the vessel, admit now the water fallen unto F, and that by measuring I find OF 8 inches, and GF 20 inches, 8 multiplied in 100 the whole Pillars weight yieldeth 800, which divided by 20 (the greater distance) from the top of the vessel to the lowest water mark, bringeth in the Quotient 40, so many pound weight, I conclude the Pillars foot or portion BC.  
And thus may you by the rising and falling of the water with the aid of Arithmetic know how to cut of from this Pillar or any other body what portion, quantity or weight ye will prescribe, alway it is supposed that it is Corpus Homogeneum, that is to say of matter and kind of substance equally dispersed through out: otherwise if it be Corpus Heterogeneum, that is to say of parts unlike in substance, it requireth more curious calculation, which I reserve for an other place, minding here only to make an Introduction for the witty to proceed farther, and not at large to disclose such secrets as may be showed herein. By this means did Archimedes find the exact quantity of every several metal that was in the King's Crown at Syracuse, without opening or breaking any part thereof: and many more and no less strange conclusions may be done thereby.    
The end of the third Book called Stereometria.   

The Preface.  
ALthough the Rules and Precepts already given abundantly may suffice for the exact mensuration of any manner Solides or vessels that are usually occupied, or may be proponed to be moten, yet for the satisfaction also of such as delighting in matters only new, rare and difficile, seek to reach above the common sort, I have thought good to adjoin this Treatise of the 5 Platonical bodies, meaning not to discourse of their secret or mystical appliances to the Elemental regions and frame of Celestial Spheres, as things remote and far distant from the Method, nature and certainty of Geometrical demonstration, only here I intent Mathematically to confer the Superficial and solid capacities of these Regulare bodies with their Circumscribing or inscribed spheres or Solides, & Geometrically by Algebraycall Calculations to search out the sides, Diameters, Axes, Altitudes, and lines Diagonal, together with the Semidimetients of their Equiangle Bases, containing or contained Circles, and as in the Problems ye shall perceive by Precepts and Examples, the quantities and proportions of all these lines, Superficies and Crassitudes, with numbers Rational and Radical expressed, so have I by Theorems sown the seed of Rules innumerable, wherein the studious may delight himself with infinite variety: Finally I shall in the last Chapter by definitions and Theorems partly set forth the form, nature and proportion of other five uniform Geometrical Solides, created by the transformation of the five bodies Regulare or Platonical: but not in so ample manner, as the novelty or difficulty of the matter requireth, meaning as I see these the first fruits of my studies liked and accepted, to bestow time and travail upon an other peculiar Volume, which shall contain, not only the demonstrations of these Theorems of spherall solides, but also of Conoydall, Parabollical, Hyperbollical and Ellepseycal circumscribed & inscribed bodies, & many more lines & solides produced by the sections of these, and revolution of their sections: but to return to this present Treatise, let no man muse the writing in the English tongue, I have retained the Latin or Greek names of sundry lines and figures, as cords Pentagonal, lines diagonal, Icosaedron, Dodecaedron, or such like, for as the Romans and other Latin writers, notwithstanding the copious and abundante eloquence of their tongue, have not shamed to borrow of the Grecians these and many other terms of art: so surely do I think it no reproach, either to the English tongue, or any English writer, where fit words fail to borrow of them both, but rather should we seem thereby to do them great injury, these being in deed certain testimonies and memorials where such sciences first took their original, and in what languages and countries they chiefly flourished, which names or words how strange soever they seem at the first acquaintance, by use will grow as familiar as these, a triangle, a circle, or such like, which by custom and continuance seem mere English, yet to avoid all obscurity that may grow by the novelty of them, I have adjoined every of their definitions, and so proceeded to Problems and Theorems with such method, as how obscure or hard soever they appear at the first, through the rareness of the matter: I doubt not but by orderly reading, the ingenious Student, having any mean taste of cossicall numbers, shall find them plain and easy.   

Diffynitions.  
PRoportion is a mutual or interchangeable relation of two magnitudes, being of one kind, compared together in respect of their quantities.  

The second dffiinition.  
When the proportion of two magnitudes is such as may be expressed with numbers, then is it certain & apparent and here is called rational: But when the proportion is such as cannot be expressed with numbers, but with their roots only, then is that proportion certain also, but not apparent, and therefore here I name it surde or irrational.   

The third definition.  
When there be three such magnitudes or quantities that the first to the second retain the same proportion that the second doth to the third, those quantities are said to be proportional, and the first to the third retaineth double the proportion of the first to the second, and the second is named mean proportional between the first and the last.   

The fourth Definition  
When four magnitudes are likewise in continual proportion, the first & the fourth are the extremes, and the second and third the means, and the extremes are said to have triple the proportion of the means.   

The fifth definition.  
Any line or number is said to be divided by extreme & mean proportion, when the division or section is such or so placed, that the whole line or number retain the same proportion to the greater part, that the greater doth to the lesser.   

The sixth definition.  
A line is said to be equal in power with two or more lines, when his square is equal to all their squares.   

The seventh definition.  
A line is said to match a superficies in power, when the square of that line is equal to the superficies.   

The eight definition.  
When any equiangle triangle, square, or Pentagonum is in such sort described within a circle, that every of their angles touch the circumference, their sides are called the trigonal, tetragonall and pentagonal cords of that circle.   

The ninth definition.  
About every equilater triangle, square, or Pentagonum, a circle may be described, precisely touching every of those figures angles, and that circle shall be called the circumscribing or containing circle.   

The tenth definition.  
Also within every of these equiangle figures a circle may be drawn, not cutting but only touching every of their sides, this is called the inscribed circle.   

The eleventh definition.  
Any right line drawn from angle to angle in those equiangle figures passing through the superficies, I name the line diagonal.   

The twelfth definition.  
A right line falling from any angle of these superficies perpendicularly to the opposite side shallbe named that plains penpendiculare.   

The .13. Definition.  
TETRAEDRON is a body Geometrical encompassed with four equal equiangle triangles.   

The fourteenth Definition.  
HEXAEDRON or CUBUS is a solid figure, enclosed with six equal squares.   

The fifteenth Definition.  
OCTAEDRON is a body comprehended of eight equal equiangle triangles.   

The .16. Definition.  
ICOSAEDRON is a solid Figure, under twenty equal equiangle triangles conteyne●▪   

The .17. Definition.  
DODECAEDRON is a solid comprehended of twelve equal equiangle pentagonal Superficies.   

The .18. Definition.  
These five bodies are called regular, and about every of them a sphere may be described, that shall with his concave periphery exactly touch every of their solid angles, and it shall be called their comprehending or circumscribing sphere or globe, and these solides shallbe called the inscribed or contained bodies of that sphere.   

The .19. Definition.  
Also within every of these regulare bodies a sphere may be described that shall with his convex superficies precisely touch all the centres of those equiangle figures wherewith these bodies are environed, and such a one I term their inscribed or contained sphere, and those bodies shall be termed the circumscribing solides of that sphere.   

The .20. Definition.  
The semidiameter of this inscribed sphere, forasmuch as it is the very Axis or Kathetus of every Pyramid, having his base one of the equiangle plains, and concurring in the centre, of which Pyramids (intellectually conceived) these bodies seem to be compounded, it shall be named the Axis or Kathetus of that body.   

The 21 definition.  
Every of these body's side, I call any one of those equal right lines wherewith these equiangle Figures are environed that comprehend and include these bodies.   

The 22 definition.  
Any one of the Figures wherewith these solides be environed is called the base of that solid.   

The 23 definition.  
A line falling from any solid angle of these bodies perpendicularly on the opposite plain or base, shall be named that solides Perpendiculare.   

The 24 definition.  
The power of any line given is said to be divided into lines retaining extreme and mean proportion, when two such lines are found as both make their squares joined together, equal to the square of the line given, and also hold such proportion one to another as the two parts of a line divided by extreme and mean proportion.   

The 25 definition.  
One of these regulare solides is said to be described within an other, when all the angles of the internal or inscribed body at once touch the superficies of the comprehending or circumscribing regulare solid.    

The first Problem. To divide any line or number by extreme and mean Proportion.  
ADmit the line divisible AB extend him forth sufficiently, then erect upon B a perpendicular to AB as ye were taught in the first book. Now, place the one foot of your compass in B, and extending the other to A, describe the semicircle ADC, then divide AB in half at F, and placing the one foot of your compass immovable in F (making FD a semidiameter) describe the ark DG. This done, set the one foot of your compass in B & opening the other too G, draw the semicircle GEI, so is your line AB parted by extreme and mean proportion in I, the greater portion BY, the lesser JA. But Arithmetically to find the same portions ye must multiply the square of your given number in 5, and divide by 4, from the square root of the producte, deduct the half of your given number, the remainder is the greater portion, which subtracted from the whole leaveth the lesser part.  

Example.  
THe number given 12, his square 144, multiplied by 5 and divided by 4 yieldeth 180, the half of the given number 6 deducted from the square root of 180, leaveth √² 180— 6, which is the greater portion, and that again deducted rome 12, leaveth this Apotome 18— √² 180 the lesser portion.  
Ye may also by any of the parts known find the quantity of the whole thus: Augment the part given by 12, and divide by his correspondente Portion, the quotient is your desire.  
For example, Suppose the lesser part of a line divided by extreme and mean Proportion, (whose quantity I desire) 10, which augmented by 12, bringeth 120: this divided by 18— √² 180 bringeth 15+ √² 125, so much is the whole line whereof 10 is the lesser part. Again, deducting 10 from 15+ √² 25, there remaineth 5+ √² 125, and that is the greater portion. But forasmuch as both this and the former conclusion may sundry other ways be resolved, here shall ensue Theorems which the Geometer will apply aswell to this purpose as diversly to other necessary conclusions, whereof some shall hereafter ensue.  

The first Theorem.  
ANy number parted by extreme and mean proportion, the square of the greater part is equal to the rectangle or number produced by multiplication of the lease portion in the whole.   

The 2 Theorem.  
If any line or number be parted by extreme and mean Proportion, and to the whole ye adjoin the greater portion, that surmounting number or line is also divided by extreme and mean proportion, his greater portion being the first given line, and the lesser the line or number adjoined.   

The 3 Theorem  
Any line or number parted by extreme and mean proportion, the square of the whole added to the square of the lesser part, is triple to the square of the greater portion.   

The 4 Theorem.  
A line or number divided by extreme and mean proportion, if ye abate the lesser portion out of the greater, the square of the remainder adjoined to the square of the greater portion is triple to the square of the lesser.   

The 5 theorem.  
The proportion of a line parted by extreme and mean proportion to his parts can not be expressed in rational numbers, for if the whole line be rational, the parts are either of them an irrational called Apotome, but in surde numbers the proportion may be thus declared, as √² 5— to 3— √² 5, so is the whole to the greater, and the greater to the less, for they are Proportional.     

The .2. Problem. The dimetiente of any circle given to search out the sides or cords Trigonal, Tetragonal, Pentagonal & Decagonal.  
SUppose the Dimetiente AB of the circle ABCD given, divide this Diameter equally in the point E, whereupon erect the Perpendicular EC and from C to B draw a right line, then fix the one foot of your Compass in A and opening the other to E, draw the ark OF, making intersection with the circle in F, from thence extend right lines to AB, This done, divide the Semidiameter BE, as ye were taught by the first Problem, by extreme and mean Proportion in the point G: Finally, from G to C draw a straight Line, so have ye FB the Cord trigonal, CB the Tetragnall, CG Pentagonal, and EG Decagonall. Now, Arithmetically by the Diameter known to attain these cords, ye must thus work: Triple the square of the Semidiameter the root quadrate of the producte is the side trigonal, double the Semidiameters square and the root quadrate thereof is the Tetragonall Cord: Divide the Semidimetient by extreme and mean Proportion the greater Portion is the side Decagonall, whose square joined to the Semidimetientes square produceth the Square of the Cord Pentagonal.  

Example.  
Admit the Diameter AB 12, the half is 6, whose square tripled, yieldeth 108, this surde number therefore √² 108 is the trigonal Cord. The square of 6 doubled is 72, whose root is the line Tetragonall. The Semidiameter EBB 6 divided by extreme and mean proportion as was taught in the first Problem, yieldeth EG, the greater Portion √² 45— 3, and that is the line Decagonall: whose square joined with the square of 6, produceth 90— √² 1620, whose rote universal is the line CG or Pentagonal Cord.  
The Diameter of the circle 12 

The cord trigonal √² 108  
The cord Tetragonall √² 72  
The cord Pentagonal √² uni. 90— √² 1620  
The cord Decagonall √² 45— 3    
Now because long working with irrational numbers, may breed confusion in such as are not perfect in the rules of Algebra, ye may by the rule of proportion (supposing the Spheres dimetient 1000) reduce furred numbers to integers, although not so exactly as the subtility of geometrical demonstration requireth (considering these cords cannot precisely be expressed in rational numbers) yet for any Mechanical operation or manual mensuration, the difference shall not be sensible.   

Example.  
 
As in these operations by the Diameter ye are brought in knowledge of the cords, so by the knowledge of any cord, ye may also find both the Diameter and all other cords, whereof to the witty there need no farther Example.    

The .3. Problem. The side of any equiangle Triangle given, to find out the Semidiameters of his containing and contained circles, with the true quantity of the Area.  
FOr the Semidiameter of the containing Circle, it behoveth you to square the side given of the Triangle, and divide the Product by 3 the root Quadrate of the Quotient is your desire. The half of that root is the Semidiameter of the certained circle, which multiplied in the Circuit or Perimetry of the Triangle yieldeth a number, whose medietie is the content Superficial or Area of that Figure.  

Example.  
The side given AB 6.  

The containing circle●— √² 12  
Semidiameter BD.— √² 12  
The contained Circles— √² 3  
Semidimetient DE.— √² 3  
The Superficies or Area of the equilater Triangle. √² 243   
Any Equilater triangles Area is also thus gotten, multiply the square of his side by itself, and the product by 3, the offcome divide by 16, the quotients' root Quadrate is the Area.  
Or multiply the square of the side given by 3, and divide the offcome by 4, that Quotient augment again by the square of half the side, the root Quadrate of the Product is the Triangles content.  
Or divide the Triangles side in three equal parts the square of one third part multiply by 3, and divide the offcome by 4, the root Quadrate of your Quotient increased by the triangles Semiperimetrie, produceth his Superficies also.    

The .4. Problem. The side of any square known by supputation to attain the Semidiamiters of his exterior and interior Circles, with the content of his plain or Area.  
THe half of your given side is the Semidiameter of the internal Circle, and if ye double the square of this Semidiamiter, and from the Product extract a Quadrate root, that root is the Semidiameter of the external Circle. Now for the Area either multiply the side first given in itself, or else your lesser Semidiameter in half the Perimetry of your square.  

Example.  
Admit AB the squares side 10, the half being 5 is the Semidimetient of the inward Circle, the square thereof doubled is 50, whose Quadrate root is EG, the containing Circles Semidimetient. Now for the Superficies I augment 10 by itself there ariseth 100 Likewise 5 the internal Circles Semidiameter multiplied by 20 the squares Semiperimetry yieldeth also 100, so much is the Area of the square ABCD.  
AB the side of the square 10.  

EG the containing circles Semidiameter √² 50  
OF the inscribed circles Semidimetient 5  
The Area of the square ABCD 100     

The .5. Problem. The side of any equiangle Pentagonum given by Arithmetic to learn his circumscribing and inscribed circles Semidimetients with the exact quantity of his plain Superficies.  
IT behoveth ye to know that all equiangle Pentagonal figures of what capacity so ever they be, their sides to the Semidiameters of their containing circles always retain one proportion, so that having already by the second Problem (the Semidiameter given) found the side, ye may contrariwise by the rule of proportion in any other, by the side given find the Semidiameter, from whose square if ye deduct half the sides square, the root Quadrate of the Remainder is the Semidimetient of your circle interior, the medietie whereof multiplied by the Perimetrie or circuit of your figure yieldeth the content Superficial or Area thereof.  

Example.  
Suppose the side of your Pentagonum ABCDE 14, for the Semidiameter OF, I work by the rule of proportion thus, √² univers 90— √² 1620 giveth 6, what yieldeth 14, the fourth proportional is √² univers which by Reduction is brought to this number √² univers' 98+ √² 1920 ⅘, and that is the containing circles Semidimetient being very nigh 11 19/21 for exactly neither by integer nor fraction it can be expressed. Now for FG, I must deduct 49 the square of CG fro 98+ √² 1920 ⅘ the square of the Semidiameter tofore found, so there remaineth 49+ √² 1920 ⅘, whose root universal is the line FG or Semidimetient of the intrinsical circle, which augmented by 35 the Semiperimetrie, produceth √² univers' 60025+ √² 2882400500, which surde number resteth between 337 and 338 being very ●igh 337 ¼, so much is the Area of the Pentagonum.  
The side of the Pentagonum. 14 

The containing circles Semidiameter— √² v. 98+ √² 1920 ⅘  
The internal circles Semidimetient— √² v. 49+ √² 1920 ⅘  
The line Diagonal— √² 245+7  
The Pentagonal Area— √² uni. 60025+ √² 2882400500    

Other rules to perform the same.  
Divide the side given by extreme and mean proportion as ye were taught in the 2 problem, and to the whole side, add his greater portion, the sum that thereof resulteth multiply in itself, and from the product subtract the square of half the side given, with the root quadrate of this remainder divide the square of half the side, the quotient add to your divisor, the medietie of this resulting sum is your greater semidiameter, which subducted from your former divisor, leaveth the semidimetient of the intrinsical circle, and this augmented by the pentagonal semiperimetrie produceth his Area.  
Or thus, your given side divided as before by extreme and mean proportion, to the whole add his greater portion, the product square, and thereto add the square of the side pentagonal, the offcoome divide by 5▪ your quotientes quadrate root is the greater semidiameter, which augmented by the pentagonal side, and the sum thereof amounting divided by double his greater portion parted as is before said by extreme and mean proportion, your quotient will express the quantity of the lesser semidimetient: for the Area ye may augment the diameter of the contained circle by 5/4 of the pentagonal side: Or half that lesser semidimetient in the whole circuit of the figure, the surmounting sums declare his superficial quantity.  
Many more rules might in this matter be given, but I think it better to adjoin certain Theorems, whereby the ingenious shall be able not only to conceive the ground and reason of these already taught, but also of himself invent sundry other no less certainly and perchance more spéedyly performing the premises.  

Theorems  

The first Theorem.  
THe square of any equiangle triangles side, is triple to the square of his containing circles semidiameter.   

The second Theorem.  
The square of an equilater triangles side to the inscribed circles semidimetientes square is as 12 to 1.   

The third Theorem.  
The semidiameter of a squares containing circled, is mean proportional between his side and the semidiameter of his contained circle.   

The fourth Theorem.  
In any equiangle pentagonum the square of his side is equal to the square of his containing circles semidiameter, and the square of the cord decagonal joined together.   

The fifth Theorem.  
The Diameter of the circle described within a Pentagonum is equal to the sides hexagonal and decagonall of the comprehending circle.   

The sixth Theorem.  
The square of the side pentagonal together with the square of the line subtending the pentagonal angle, hath proportion to the containing circles semidiameters square, as 5 to 1.   

The seventh Theorem.  
The square of half the side pentagonal being deducted fro the square of the line diagonal subtending one of the pentagonal angles, the root quadrate of the remainder is equal to both semidiameters of the containing and contained circles joined together.   

The eight Theorem.  
In every equiangle pentagonum the containing circles semidimetient retaineth the same proportion to the contained circles diameter that the pentagonal side doth to his line diagonal.   

The ninth Theorem.  
The diagonal line in every equiangle pentagonum doth divide the perpendicular (falling from his subtendent angle to the opposite side) by extreme and mean proportion.   

The tenth Theorem.  
In every equiangle pentagonum the lines diagonal do cut themselves by extreme and mean proportion.   

The eleventh Theorem.  
If within one circled an equilater triangle and a square be described, that square retaineth the same proportion to the square of the triangles inscribed circles diameter, that the semidiameter of the triangles containing circle doth to the semidiameter of the contained.   

The twelfth Theorem.  
Where as with one circle an equiangle Pentagonum and a equilater triangle are described, the pentagonal inscribed circles semidiameter parted by extreme and mean proportion, the greater portion is equal to the triangles contained circles semidimetient.   

The thirteenth Theorem.  
An equilater triangle and an equiangle pentagonum being within one circle described, if two squares be made, the first equal to the square of the pentagonal side, and the square of the pentagonal line diagonal added together, the second equal to the square of the same diagonal and the square of the line, whereby the diagonal exceedeth the side pentagonal, the side of this first square to the side of the second, retaineth the same proportion that the pentagonal diagonal doth to the side of the equilater triangle.   

The fourteenth Theorem.  
The Tetragonall cords square exceedeth the squares of the cords pentagonal and decagonall, by the square of a line mean proportional between the cord decagonal and the excess or difference of the cords hexagonal & decagonal.   

The fifteenth Theorem.  
An equiangle triangles side being rational, his perpendicular and semidiameters as well of the containing as contained circles, are all irrational, but their squares may be expressed in number.   

The sixteenth Theorem.  
The Area of an equilater triangle beareth to the square of his side proportion as his perpendicular to his base doubled.   

The seventeenth Theorem.  
The Area of an equiangle triangle beareth proportion to the square of his side, as √² 3 to 4.   

The eighteenth Theorem.  
Double the Area of an equilater triangle is mean proportional between the square of his side, and the square of the perpendicular falling from one of his angles to the opposite side.   

The nyntenth Theorem.  
The Area of an equilater triangle to the square that is described within his containing circle retaineth the proportion of √² 3 to ●/3.   

The 20 Theorem.  
The side of an equiangle triangle being rational, his Area is irrational, but the square thereof may with number be declared.   

The 21 Theorem.  
The Area of an equiangle Pentagonum, to the Triangles area that is described with in his containing circle, retaineth such proportion as five times the line Diagonal of the Pentagonum to the perimetrie of the Triangle.   

The 22 Theorem.  
If an equiangle Pentagonum and an equilater Triangle be both described in one circle, the rectangle contained of the line Diagonal and the Triangles perpendicular, retaineth proportion to the pentagonal Area, as 6 to 5.   

The 23 Theorem.  
The Pentagonal side being rational, the Area is irrational, proportioned to the square of the side, as √² uniu. 25/16+ √² 125/64 to an unity.   

The 24 Theorem.  
That square whose side is mean proportional between the internal circles semidiameter, and the pentagonal semiperimetrie, is equal to the pentagonal Superficies.   

The 25 Theorem.  
When an equiangle Pentagonum and an equilater Triangle be both within one circle described, a mean proportional between ⅚ of the triangles perpendicular, and the line diagonal is equal in power to the pentagonal Superficies.       

The .6. Problem. The side of any Tetraedron given, to search out the Semidiameters of the circumscribing and inscribed spheres.  
first as is taught in the third problem ye must get the circles semidiameter that containeth the equiangle triangle whose side ye have: Multiply the square of this Semidiameter by 9▪ and divide by 2, the root quadrate of the quotient is the circumscribing spheres dimetient, the medietie whereof if ye augment by itself, and from the product subtract the square of the semidiameter tofore found, the root quadrate of the remain is the Semidiameter of the contained Sphere.  
Or thus more easily & speedily: Multiply the side given in itself, the ofcome divide by 24 the root quadrate of the quotient is the internal Spheres semidiameter, which tripled, yieldeth the semidiameter of the comprehending Sphere.  

Example.  
The trigonal side supposed 10, the containing circles semidiameter is √² 33 ⅓, whose square augmented by 9, and divided by 2, yieldeth in the quotient 150, whose quadrate root is the dimetient of the sphere, the medietie hereof squared, is 37 ½, from which if ye withdraw 33 ⅓, there remaineth 4 ⅙, whose quadrate root is the Axis of this Tetraedron.  
Likewise, if by the second rule ye divide 100 with 24, the quotient is 4 ⅙, whose square root being the Axis of this tetraedron if ye triple it, the resulting number will be √² 37 ½, and that is the semidiameter of the circumscribing sphere, ye may also (by deducting 25 half the sides square from 100 the square of the whole side) find the perpendiculare, for the root of the remainder being √² 75 is the perpendiculare of the base, for the solides perpendiculare I deduct 33 ⅓, the containing circle's semidimetientes square, from 100 the square of the side given, there remaineth 66 ⅔, whose root is the solides perpendiculare.  
The side of Tetraedron 10.  

The containing circles semidiameter— √² 33 ⅓.  
The contained circles semidimetien●— √² 8 ½.  
The perpendicular of the Base— √² 75.  
The comprehending spheres semidiameter √² 37 ½.  
The inscribed spheres semidimetient √² 4 ⅙.  
The Solides perpendiculare— √² 66 ⅔.   
Thus also an other way ye shall most speedily find these spheral semidiameters, square the side given, that squares medietie ye shall triple, the root quadrate of the producte is the comprehending spheres dimetiente, which divided by 3, bringeth the inscribed spheres diameter, the medieties of these are the semidiameters whereof the lesser augmented by 4, produceth the solides perpendiculare.  

Theorems of Tetraedron 1  
THe square of Tetraedrons' side, is equal to the square of his perpendiculare and the square of his containing circles semidiameter added together.   

The 2 theorem.  
The square of Tetraedrons' inscribed circles Semidiameter withdrawn from the square of his bases perpendiculare, leaveth the square of the solides perpendiculare.   

The 3 Theorem.  
This solides perpendiculare is equal to his axis, and containing Spheres Semidimetiente.   

The 4 Theorem.  
The perpendiculare of Tetraedrons' base is equal to his containing and contained circles semidiameter.   

The 5 Theorem.  
Tetraedrons' containing Spheres dimetiente retaineth the same proportion to his Axis, that the square of his side doth to double the square of his contained circles semidimetiente.   

The 6 Theorem.  
Tetraedrons' perpendiculare is ⅔ of his comprehending Spheres dimetiente.   

The 7 Theorem.  
Tetraedrons' Axis is a sixth part of his comprehending Spheres Dimetiente.   

The 8 Theorem.  
Tetraedrons' containing spheres Diameters square retaineth the same proportion to the square of his side, that the perpendiculare of his base doth to the semidiameter of his containing circle.   

The 9 Theorem.  
The square of Tetraedrons' comprehending Spheres dimetiente containeth the square of his inscribed circles semidiameter 18 times.   

The 10 Theorem.  
The square of Tetraedrons' contained circles semidiameter is double to the semidiameters square of his contained Sphere.   

The 11 Theorem.  
Tetraedrons' perpendiculare retaineth the same proportion to his basis containing circles semidiameter, that the contained circles semidimetient doth bear unto his Axis.   

The 12 Theorem.  
The square of the comprehending Spheres dimetiente to the containing Circles semidiameters square, hath proportion as 9 unto 2.   

The 13 Theorem.  
Tetraedrons' side being rational, his containing Spheres diameter is irrational, and their proportion is as √² 2 unto √² 3.   

The 14 Theorem.  
Tetraedrons' side being rational, the Axis is surde, and it beareth proportion to the side as 1. to √² 24.   

The 15 Theorem.  
The Tetraedrons' comprehending Spheres dimetiente, is equal in power to these five, the containing circles semidiameter, the containing Spheres semidimetiente, the inscribed circles semidiameter, the Axis, and the solides Perpendiculare.     

The 7. Problem. The side of any Hexaedron given, to find the semidimetientes of the containing and contained Spheres.  
INcrease the side by itself, the offcome triple, from the producte extract the root quadrate the half thereof is the semidimetient of the circumscribing sphere, from whose square if ye detract the square of that circles semidimetiente that comprehendeth one of the Cubes quadrate plains the root square of the remainder is the semidimetiente of the inscribed sphere: the circles Semidiameter is found by the fourth Problem, admitting the line given the side of a square.  

Example.  
The Cubes side 6, the square thereof tripled 108, half the root thereof is √² 27, and that is the semidiameter of the containing sphere, his containing Circles semidiameter is √² 18, whose square deducted from 27, leaveth 9, his root quadrate being 3 is the Axis or contained spheres semidimetient.  
The Cubes side 6 

The line Diagonal●— √² 72  
The containing Circles Semid. √² 18  
The contained circles Semid. 3  
The containing Spheres Semid. √² 27  
The Axis or inscribed 3  
Spheres Semidimetient 3    

Theorems of Hexaedron. 1  
THE square of the Cubes comprehending Spheres dimetiente is equal to the square of his side, and the square of his basis line diagonal.   

The 2 Theorem.  
The square of his line diagonal containeth the square of his Axis 8 times.   

The 3 Theorem.  
The square of the Cubes comprehending Spheres semidiameter is equal to the square of half the line diagonal of his base, and the square of his Axis added together.   

The 4 Theorem.  
The semidimetiente of the Cubes contained circle is equal to his Axis.   

The 5 Theorem.  
The Hexaedrons' side is mean proportional between his containing Circles semidiameter and his basis line diagonal.   

The 6 Theorem.  
The Cubes line diagonal to his Axis retaineth triple the proportion of his side to his containing circles semidiameter.   

The 7 Theorem.  
The comprehending spheres dimetiente to the Cubes side, observeth the Proportion of √² 3 to 1.   

The 8 Theorem.  
The Cubes side being a line rational, his Axis is also rational.   

The 9 Theorem.  
The Hexaedrons' comprehending Spheres Dimetiente being rational, his Axis is a surde and beareth proportion to the Dimetiente, as 1 to √² 12.   

The 10 Theorem.  
The comprehending spheres diameter hath the same proportion to the line diagonal of his base that the Cubes side hath to the semidimetiente of his containing Circle.     

The 8. Problem. The side of Octaedron given, to search out Arithmetically the containing spheres Diameter and the Axis thereof.  
DOuble the square of the side, and from the offcome extract the root Quadrate, so have ye the comprehending Spheres dimetiente the half (being the semidiameter) if from the square thereof, you abate the square of that circles semidiameter which containeth one of the equilater triangles whereof the body is framed, the root quadrate of the remainder is the Axis.  
Or thus more briefly, square the side given, the producte divide first by 2 then by 3, the root square of the first quotient is the containing spheres semidimetient, the second quotient deducted fro the first, leaveth a remainder, whose root quadrate is the Axis.  
Or thus more readily, from the sixth part of the given sides square, extract the quadrate root, so have ye the axis, whose square augmented by 12, produceth the square of the comprehending spheres diameter.  

Example of the first.  
THe side given 10, his square 100, √² 50 being the quadrate root of half that square, is the comprehending spheres semidimetient, from the square thereof abating 33 ⅓ the square of the containing circles semidiameter, the remainder is 16 ⅔, whose quadrate root is the Octaedrons' axis.   

Example of the second rule.  
100 the square of the side divided first by 2 yieldeth 50, then by 3 bringeth in the quotient 33 ⅓, √² 50 is the containing spheres semidiameter, the one quotient deducted fro the other leaveth 16 ⅔, his root being √² 16 ⅔ is the Axis.   

Example of the third precept.  
The sides square given 100 divided by 6 produceth in the quotient 16 ⅔, the root quadrate thereof is the Axis, the same number augmented by 12 maketh 200, wherefore I conclude √² 200 the comprehending spheres diameter.  
The side of Octaedron 10.  

The containing Circles semidiameter. √² 33 ⅓.  
The contained Circles semidimetient. √² 8 ⅓.  
The containing Spheres semidiameter. √² 50.  
The Axis. √² 16 ⅔.   

Theorems of Octaedron. 1.  
The Axis of Octaedron is mean proportional between the containing and contained circles semidiameters.   

The second Theorem.  
The Octaedrons' side is mean proportional between the diameter and semidiameter of the circumscribing sphere.   

The third Theorem.  
The Axis square increased by 12 yieldeth the compredending spheres diameters square,   

The fourth Theorem.  
The containing spheres diameter hath the same proportion to his Axis, that the Octaedrons' side hath to the contained circles semidiameter.   

The fifth Theorem.  
The side of Octaedron being rational, the comprehending spheres Dimetiente cannot be expressed with number, but his square to the sides square is double.   

The sixth Theorem.  
The side of Octaedron being rational, his Axis is not in number to be expressed, but the square thereof to the square of the side is as 1 to 6.   

The seventh Theorem.  
The Octaedrons' containing spheres semidiameter is equal in power with his Axis and containing circles semidiameter.   

The eight theorem.  
The side of Octaedron is equal in power with the axis, the comprehending spheres semidiametient, and the comprehending circles semidiameter.   

The ninth theorem.  
The diameter of Octaedrons' comprehending sphere being rational, his side is a surde, and retaineth such proportion to the diameter as √² ½ unto 1.   

The tenth theorem.  
The diameter of Octaedron being rational, his containing circles semidiameter is irrational, and hath proportion to his diameter, as √² ⅙ unto 1.   

The eleventh theorem.  
Octaedrons' side being rational his axis is surde, and beareth proportion to the side, as √² ⅙ unto 1.   

The twelfth theorem.  
Octaedrons' side being rational his contained circles semidiameter is irrational, having proportion to the side as √² ●/1● to 1.   

The thirteenth Theorem.  
The axis of Octaedron being rational, his comprehending spheres semidimetient is irrational and their proportion is as an unity unto √² 3.   

The fourteenth Theorem.  
The side of Octaedron being rational, that line which is equal in power to the semidiameters of his contained sphere, and circle, is also rational, and the comprehending spheres diameter to that line, hath triple the proportion of the containing circles semidiameter, to the axis.   

The fifteenth Theorem.  
Octaedrons' contained circles semidimetient his axis, the containing circles semidiameter, and the inscribed spheres diameter, are 4 lines in continual geometrical proportion.     

The .9. Problem. The side of an Icosaedron measured, by supputation to find his axis and containing spheres dimetiente.  
ADmitting the side measured a Cord pentagonal, you shall by the fourth problem search out the containing circles semidiameter, and multiply the square thereof by 5, the quadrate root of the producte is the dimetiente of the comprehending sphere, and so consequently the half thereof shallbe the semidiameter, from whose square if ye subtract his containing circle's semidimetientes square, the root quadrate of the remainder is the axis.  

Example.  
Admit the side of Icosaedron 12, which supposed a Cord pentagonal, the side hexagonal or semidiameter found (as was taught in the fifth problem) is √² v. the square of this semidiameter increased by 5 maketh whose quadrate root is the containing spheres diameter, his medietie being √² v. is the semidiameter. Now to attain the axis I search first (as was taught in the third problem) the circle's semidiameter that containeth the Icosaedrons' triangular base, finding it √² 48, whose square deducted from the square of the containing spheres semidiameter leaveth , whose quadrate root is the axis of this Icosaedron.  
The side of Icosaedron 12 

The circles semidiameter, whereon Icosaedron is framed. √².  
The semidiameter of the containing Circle. √² 48.  
The semidimetient of the contained circle. √² 12:  
The comprehending Spheres semid, √²  
The Axis or Cathetus √² v.    

Theorems of Icosaedron and his parts. 1.  
The diameter of Icosaedrons' comprehending sphere retaineth such proportion to the diameter of the circle whereon the Icosaedron is framed, as √² 5 to .2.   

The second Theorem.  
The semidiameter of this circle whereon the Icosaedron is framed with two cords decagonal of the same circle do make the comprehending spheres dimetient.   

The third Theorem.  
The comprehending spheres semidiameter, doth match in power the Axis and containing circles semidimetient.   

The fourth Theorem.  
If the semidiameter of that circle whereupon the Icosaedron is framed, be divided by extreme and mean proportion, the square of Icosaedrons' side is equal in power to the foresaid semidiameter, and his greater portion.   

The fifth Theorem.  
The diameter of the comprehending sphere being a line rational, the Icosaedrons' side is a line irrational, called of Euclid Minor.   

The sixth Theorem.  
If the side of Icosaedron be a line rational, the dimetient of the comprehending sphere shallbe an irrational line called Mayor.   

The seventh Theorem.  
The diameter of the comprehending sphere being a rational, the semidiameter of the circle whereupon the Icosaedron is made, shall be a line irrational, but his square is rational, and the comprehending spheres diameters square containeth it 5 times.   

The eight theorem.  
The Icosaedrons' side being a line rational, the semidimetiente of that circle whereon the body is framed will be an irrational, called of Euclid Maior, and it retaineth unto this solids side the proportion of √² v. ½+ √² 1/20 unto an unity.   

The ninth Theorem.  
The Icosaedrons' side being rational, his containing circles semidiameter is irrational, and beareth porportion to that side, as 1 to √² 3.   

The tenth Theorem.  
The comprehending spheres diameter being rational, his containing circles semidiameter is an irrational of that kind which Euclid calleth Minor, and it beareth proportion to the diameter as √² univers' ⅙— √² 1/180 unto an unity.   

The eleventh Theorem.  
The Icosaedrons' side admitted rational, the contained circles semidiameter is a surde, and retaineth such proportion to the side, as √² 1/12 unto 1.   

The twelfth Theorem.  
Icosaedrons' containing spheres dimetient being rational, his contained circles semidimetient is an irrational called Minor, and the proportion between them is as 1 unto √² universalss 1/24— √² 1/2880.   

The 13 Theorem.  
Icosaedrons' side supposed rational, his axis is an irrational called Binomium and beareth proportion to the side, as √² uni. 7/24+ √² 5/64 unto 1.   

The 14 theorem.  
The diameter of Icosaedrons' containing sphere admitted rational, his axis is irrational, and retaineth such proportion to his dimetient as √² uni. 1/120+ √² 1/180 unto 1.   

The 15 theorem.  
The semidiameter of the containing and contained circles, although the proportion of either to the spheres dimetient be surde, yet the proportion between themselves is rational, and as 2 to 1.   

The 16 theorem.  
Icosaedrons' side being parted by extreme and mean proportion, if ye adjoin to the side his greater portion, the square of that whole line added to the square of the side, produceth the square of the comprehending spheres diameter.   

The 17 theorem.  
The diameter of that circle whereon Icosaedron is framed, retaineth such proportion to the containing circles semidimetient of the body, that the Icosaedrons' side doth to the cord pentagonal of the contained circle.   

The 18 theorem.  
If from the square of the comprehending spheres dimetient ye abate the Icosaedrons' sides square, that line which matcheth in power the remain, being divided by extreme and mean proportion, will make his greater segment the Icosaedrons' side.   

The 19 Theorem.  
The pentagonal line diagonal of that circle, whereon Icosaedron is framed, exceedeth in power the contained spheres diameter, by a line matching in power the excess of Icosaedrons' basis containing circles diameters square, about the square of his side.   

The 20 Theorem.  
Icosaedrons' side is mean proportional between his diameter and the decacagonall cord of that circle whereon Icosaedron is framed.     

The .10. Problem. The side of Dodecaedron given, by calculation to find his axis and containing spheres diameter.  
Using the aid of the fift problem if you search out the triangular or trigonal cord of that circle that containeth this Dodecaedrons' pentagonal base, admitting the same an Icosaedrons' side, ye may by the last chapter search out the axis and containing spheres dimetient, which are in all points equal and agreeable in proportion with the axis and spherical Diameter of this Dodecaedron.  
Or thus more speedily and with less confusion in working, by the first problem divide this dodecaedrons side given by extreme and mean proportion, adding thereto his greater portion, then triple the square of this whole line, the root quadrate of the product is the containing spheres Dimetiente. Again it behoovethe you by the fifth problem to learn the semidiameter of the circle that containeth one of the dodecaedrons pentagonal superficies, and deduct the square thereof from the square of the comprehending spheres semidimetient, the root quadrate of the remainder is the axis of the body.  

Example.  
The side of Dodecaedron 10, divided by extreme and mean proportion by the first problem maketh his greater portion √² 125— 5, which added to 10, produceth √² 125+5, the square hereof tripled is 450+ √² 112500 the root square hereof is the comprehending spheres dimetient, whose medietie being √² v. 112 ½+ √² 112500/16 is that spheres semidiameter. Again, the semidiameter of this Icosaedrons' containing circle I find by the fifth problem √² v. 50+ √² 500, whose square taken from the square of the containing spheres semidiameter leaveth 62 ½+ √² 112500/16— √² 500, whose root quadrate is the axis or inscribed spheres semidimetiente.  
The side of Dodecaedron 10 

The containing circles semid. √² uni. 50+ √² 500  
The contained circles semidimetient √² uni 25+ √² 500  
The diameter of the containing sphere √² v. 450+ √² 112500  
The axis √² uni. 62 ½+ √² 1●2500/16— √² 500·  
The pentagonal diagonal line √² 125+5.    
The axis of this body is also thus found, I add the line diagonal being √² 125+5 to 10 the dodecaedrons side, thereof ariseth √² 125+15, this medietie squared maketh 350/4+ √² 28125/4 from whence I deduct 25+ √² 500 the square of the contained circles semidimetiente, there remaineth 250/●+ √² 28125/4— √² 500, whose root quadrate is the dodecaedrons axis exactly agreeing with the former operation.  

Theorems of Dodecaedron.  
THe square of the diameter of Dodecaedrons' comprehending sphere is triple to the square of his basis line diagonal.   

The second theorem.  
The Dodecaedrons' side being added to the diagonal line of his basis maketh a line, whose square joined to the square of the side▪ produceth the square of the containing globes dimetient.   

The third Theorem.  
The diameter of Dodecaedrons' comprehending sphere being rational, the side of that body shall be an irrational called Apotome, bearing proportion to the diameter as √² uni. ½— √² 5/36 unto 1.   

The fourth Theorem.  
The side of dodecaedron being rational, his comprehending spheres dimetient is an irrational Binomium, having the squares of his names or compounding lines in proportion as 5 to 1, and this irrational diameter to the Dodecaedrons' side, retaineth proportion, as √² univers 9/2+ √² 45/4 unto an unity.   

The fifth theorem.  
If dodecaedrons side be adjoined to his diagonal line, the medietie thereof squared is equal to the square of the axis and the square of the contained circles semidiameter.   

The 6 theorem.  
Dodecaedrons' side being rational, the containing circles semidimetient is an irrational called Mayor, and beareth such proportion to the side, as √² uni. ½+ √² 1/20 unto 1.   

The 7 theorem.  
The square of Dodecaedrons' side added to the square of his basis diagonal produceth a quantity, whose fifth part is the containing circles semidiameters square.   

The 8 theorem.  
Dodecaedrons' side being rational, the diagonal of his basis shall be an irrational called Binomium, retaining proportion to the side, as √² 5/4+½ unto 1.   

The 9 theorem.  
Dodecaedrons' axis being divided in extreme and mean proportion, maketh his greater part the less semidimetient of his basis.   

The 10 Theorem.  
The side of Dodecaedron being rational, his basis contained circles diameter is an irrational Mayor, proportioned to the side, as √² v. 1+ √² frac45; to an unity.     

The .11. Problem. The side of Tetraedron known to find his superficial and solid content.  
FOr the superficies ye shall augment the squared square of the side by 3, the productes quadrate root is your desire, or multiply the containing circles semidimetient in the perimetrie of the triangle, or augment the contained circles semidiameter in the perimetrie of the triangle and double the product, so have you by every of these operations his content superficial. For his crassitude thus, Multiply the semidimetient of the containing circle in the axis of the body, and the product in the side, Or the semidimetient of the contained circled in the diameter of the contained sphere, and the offcome in the side: Or the axis in the perimetrie of the traingle, and the product in the semidiameter of the containing circle, the offcome divided by 3, yieldeth in the quotient the solid quantity.  

Examples of the Superficial quantity.  
The side of Tetraedron 10, his squared square 10000 multiplied in 3, bringeth 30000, I conclude √² 30000 the Superficial content. Likewise the side being 10, I find by the third Problem his containing circles Semidiameter √² 33 ⅓, which increased by 30 the triangular Perimetrye yieldeth √² 90000/3, Also the semidiameter of the intrinsical circle being √² 8 ⅓ multiplied in 30 bringeth √² 90000/12 which doubled amounteth to √² 30000 the whole Superficies of that body agreeing with both the former operations.   

Examples of the solid capacity.  
The side admitted as before 10, by the third Problem the Semidiameter of the containing circle is √² 33 ⅓, the Axis ye shall by the sixth Problem find √² 4 ⅙, these multiplied together make √² 2500/18, and this again in 10, bringeth √² 250000/18. Likewise the Semidiameter of the contained Circle found by the third Problem √² 8 ⅓, augmented by √² 50/3, the contained Spheres Dimetient, there ariseth √² 1250/9, and that again in 10 the side triangular bringeth √² 125000/9. Also the Axis √² 4 ⅙ multiplied by 30 the triangles Perimetrie, produceth √² 22500/6, and this again in √² 33 ⅓ the containing Circles Semidiameter, maketh √² 2250000/18, whose third part is √² 250000/18 the solid capacity of that body exactly agreeing with the former workings.  
The side of Tetraedron 10 

The containing circles Semidiameter √² 33 ⅓  
The contained circles Semidimetient √² 8 ⅓  
The containing Spheres Semidimetient √² 37 ½  
The Axis or Kathetus √² 4 ⅙  
The content Superficial √² 30000  
The content solid √² 13888 8/9    

Theorems of Tetraedrons' superficial and solid contents.  
THe side of Tetraedron being Rational, the line that matcheth in power his superficies is an irrational, retaining proportion to the side as √²² 3 unto 1.   

The second theorem.  
The Tetraedrons' superficies to the square of his side beareth proportion as √² 3 unto 1.   

The third theorem.  
The triangle whose basis is equal to the Perimetrie of this solides basis, and his altitude or Perpendicular to the Diameter of the containing circle is equal to the superficial quantity of this body.   

The fourth theorem.  
The right angled Paralelogramme, whose altitude is the contained circles diameter, and his base the triangular Perimetrie, is equal to the Tetraedrons' superficies.   

The fifth theorem.  
That equilater triangle is equal to the Tetraedrons' superficies, whose side beareth proportion to the solides side as 2 to 1.   

The sixth theorem.  
A Quadrangular Prisma having his base a square, whose side is mean proportional between the Axis and the containing circles semidimetient, and his altitude the Tetraedrons' side, is equal to the crassitude of Tetraedron.   

The seventh theorem.  
The solid of Tetraedron may by imagination be parted into 4 equal Trigonal Pyramids, having for their bases the triangles of that body, and meeting with their tops or Vertices in the centre of the Sphere.   

The eight theorem.  
A triangular Prisma, whose altitude is equal to the third part of the Axis, and his base one of the equilater triangles wherewith the body is encompassed, is equal to one of those Pyramids that meet with their tops or Vertices at the centre of the body.   

The ninth theorem.  
A right angled Quadrangular direct Prisma, having for his longitude, latitude and profundity these three lines, Tetraedrons' side, his Axis, and containing circles Semidimetient, is equal to the Tetraedron.   

The tenth theorem.  
If four right lines be found in continual proportion, so as the first being equal to the Tetraedrons' side, bear proportion to the last, as √² 72 unto 1, the Cube of the second is equal to the Tetraedron in crassitude.     

The .12. Problem. The side of a Cube measured, to find his Superficial and solid quantity.  
Multiply the side first by 2, then by 3, these Productes multiplied together, declare the Superficial content. Or double the comprehending Spheres Diameters square. Or triple the diagonal lines square. Or multiply the Cubes base by 6, every of these ways ye have the superficies of this body.  
For the Crassitude thus, Augment the side in the base, or the Axis in the square of the line diagonal, or the superficies in the sixth part of his altitude, any of the Productes is your desire.  

Example.  
Admit Hexaedrons' side 10, by the seventh Problem ye shall find the Diameter of his containing Sphere √² 300, his Axis 5, his line diagonal √² 200. The side first increased by 2 maketh 20, then by 3 bringeth 30, these together multiplied make 600. Likewise the square of the comprehending spheres Diameter 300 doubled maketh 600, and the square of the line diagonal being 200 tripled maketh also 600, I conclude therefore 600 the content Superficial.   

Example of the solid content.  
10 the side augmented in 100, the base maketh 1000 Also 5 the Axis in 200 the square of the line diagonal produceth 1000 Likewise 600 the superficies in 5/3 the sixth part of the side, yieldeth also 1000, thus I find by all these rules 1000 the Cubes solid quantity.  
The side of the Cube 10 

The containing circles Semidiameter— √² 50  
The contained circles Semidiameter— 5  
The containing Spheres Dimetient— √² 300  
The Axis or Kathetus— 5  
The line Diagonal— √² 200  
The cubes Superficial quantity— 600  
The solid capacity— 1000    

Theorems of Hexaedrons' superficies and Crassitude. Theorem first.  
THe comprehending spheres Dimetients square doubled is equal to the Cubes superficial content.   

The second theorem.  
A Paralelogramme whose altitude is double to the Cubes side, and his base triple, is equal to the superficies of that solid.   

The third theorem.  
That square is equal to the Cubes superficies, whose side is mean proportional between the Cubes side and a line 6 times his length.   

The fourth theorem.  
Any Prisma whose base is equal to the square of the Cubes basis line diagonal, and his altitude the Cubes Axis, is equal to his solid quantity.   

The fifth theorem.  
The Cube may intellectually be divided into 6 Quadrate Pyramids, every of them having to his base one of the Cubes square basis, and concurring or meeting at the centre of the sphere, which is a common Vertex to them al.   

The sixth theorem.  
Any Pyramid having his Base equal to the square of the comprehending spheres Diameter, and his altitude the side of the Cube, is equal to his crassitude.   

The seventh theorem.  
A Tetraedron whose side is proportioned to the Cubes side, as √²² 12 unto an unity, have equal superficial contents.   

The eight theorem.  
An Octaedron whose side is equal to a line mean proportional between the Cubes side and his Diameter, hath his superficies equal to the Cubes.   

The ninth theorem.  
If 7 lines be continually proportional, the last retaining proportion to the first as 72 unto 1, the Tetraedron of the second is equal to the Cube of the first.   

The tenth theorem.  
If 4 lines be continually proportional the last bearing proportion to the first, as 1 to √² 2/9 the cube of the first is equal to the Octaedron of the second.     

The .13. Problem. Octaedrons' side given to search his superficial and solid content.  
Multiply the containing circles diameter in the circuit or perimetrie of the Triangle, or increase the squared square of the side given by 12, the quadrate root of the producte is the superficies: for the solid content work thus, multiply the diameter of the interior circle in the diameter of the internal sphere, and the producte in the side. Or the dimetiente of the containing circle in the Axis of the body, and the product in the side. Or the Axis in the perimetrie of the triangle and the resulting sum in the containing circles dimetient, the product divided by 3 yieldeth the solid content.  

Example.  
THe side of Octaedron supposed 10, the contained circles diameter by the third problem I find √² 400/3, which augmented by 30, the triangles perimetrie produceth √² 360000/3  
The squared square of the triangular side is 10000 which augmented by 12 bringeth 120000, whose quadrate root is this bodies superficial quantity.   

The example of the solid content.  
THe side being 10 by the third problem I find the containing circles semidimetient √² 33 ⅓, the contained spheres diameter by the eight problem is √² 66 ⅔, these multiplied together make 20000/9, which increased by 10 the triangular side produceth √² 2000000/9  
The diameter of the containing circle √² 400/3 augmedted by √² 50/3 the axis, bringeth √² 20000/9, and this again in 10 maketh as before √² 2000000/9  
The axis √² 50/3 multiplied by 30 the triangular perimetry, bringeth √² 45000/3 and this again in √² 400/3 the containing circles dimetiente, produceth √² 18000000/9 whose third part is √² 222222 2/9 the solid capacity of that Octaedron.  
The side of Octaedron 10 

The containing Spheres dimetient √² 200.  
The containing circles diameter √² 400/3.  
The Axis or Cathetus √² 16 ⅔.  
The internal circles diameter √² 33 ⅓.  
The content superficial √² 120000.  
The solid capacity √² 222222 2/9.    

Theorems of Octaedrons' content superficial and solid. 1.  
THe side of Octaedron admitted rational, his content superficial is irrational, and the line that matcheth it in power is a surde called of Euclid a line mediall, retaining such proportion to the side, as √²² 12 to 1.   

The 2 theorem.  
The superficial quantity of Octaedron beareth proportion to the square of his side as √² 12 to an unity.   

The 3 theorem.  
A rectangular parallelogramme having the one of his containing sides the perimetrie of Octaedrons' triangular basis, and the other equal to the dimetient of his containing circle, is equal to the whole superficies of this body.   

The 4 theorem.  
An equilater triangle, whose side beareth proportion to the side of Octaedron, as √² 8 to 1 is equal to the capacity superficial of that body.   

The 5 theorem.  
The square whose side is mean proportional between the containing circles diameter, and the perimetrie of this body's basis, is equal to his superficies.   

The 6 theorem.  
A quadrangulare Prisma having to his base a square (whose side is mean proportional between the contained circles diameter and the contained spheres diameter, and his altitude the solides side is equal to this Octaedrons' crassitude·   

The 7 theorem.  
This body may be divided into 8 equal trigonal Pyramids, whose bases are the equilater triangles wherewith this solid is environed, every of these Pyramids meet at the centre of the containing and contained sphere, which is the common vertex to them all.   

The 8 theorem.  
To every of these Pyramids that Prisma is equal, whose altitude is the contained spheres semidiameter, and his base ⅓ of one of those equilater triangles, area, wherewith this body is environed.   

The 9 Theorem.  
A quadrangular direct Pyramid having his base equal to the square of a line in mean proportion between the axis and the perimetrie of the triangle, and his altitude the containing circles dimetiente, is equal to this regulare solid.   

The 10 Theorem.  
If four lines in continual proportion have the first and greatest equal to Octaedrons' side, retaining proportion to the last, as 1 to √² 2/9 the Cube of the second is equal to this solid.     

The .14. Problem. The side of Icosaedron known, by supputation to learn the contents superficial and solid of that body.  
COnsidering this body is environed with 20 equilater triangles, ye shall by the third problem get one of those triangles Area, which increased by 20 yieldeth your desire. Or augment the squared square of the given side by 75, and from the producte extract the root quadrate. Likewise if you add the semidiameter of the inscribed circle to the diameter of the circumscribed circle, the ofcoome multiplied by double the triangles perimetrie, produceth the Icosaedrons' content superficial. For the solid capacity add the containing and contained circles semidimetientes together, the producte increase by double the side known, the resulting sum augmented by the Icosaedrons' axis (which ye were in the ninth problem taught to find) yieldeth the gross capacity. Also if ye augment the squared square of the side by 8 ⅓ the quadrate root of the producte increased by the axis of this body bringeth the desired crassitude. Or if ye augment the sides squared square by 2 1/12 and the resulting sum again in the square of the inscribed spheres dimetiente, the root quadrate of the producte is likewise the Icosaedrons' solid capacity.  

Examples of the former rules.  
THe Icosaedrons' side supposed 12, by the third problem ye shall find the area of the equilater triangle √² 3888, which increased by 20 produceth √² 1555200. Likewise the squared square of 12 being 20736 augmented by 75, yieldeth, 1555200, whose quadrate root is your desired superficies. In like manner if ye add √² 12, the semidimetiente of the inscribed circle, to √² 192 the dimetiente of the circumscribed circle (for so shall ye find them by the third problem the triangles side being 12) the product will be √² 300, which multiplied by 72 double the triangles perimetrie, yieldeth again √² 1555200 the superficies of that whole Icosaedron.  
Now for the solid content if you add √² 192 (for so is the dimetiente of the containing circle) to √² 12, the semidimetiente of the contained circle, there will amount √² 300, which augmented by 24 the triangles side doubled bringeth √² 17280, and this by √² v. 720/10— R. 20— 48, (for so is the axis as ye may by the ninth problem perceive) multiplied, produceth √² uni. 7257600+ √² 48372940800000, and that is the solid capacity of this body. The same sum is also produced by multiplication of 20736, the squared square of 12, in 8 1/●, and the producte thereof in √² v. 42+ √² 1620, the axis. Also if ye multiply 20736 the sides squared square by 2 1/12 the ofcoome will be 43200, which augmented again by 168+ √² 25920 the square of the inscribed spheres dimetiente produceth 7257600+ √² 48372940800000, whose universal quadrate root is the solid capacity of this Icosaedron, and by reduction to numbers rational it falleth out between 3769 and 3770.  
The side of Icosaedron. 12.  

The containing circles diameter. √² 192.  
The contained circles semid. √² 12.  
The comprehending Spheres semid. √² v  
The Axis. √² v.  
The superficial content √² 1555200.  
The solid capa. √² v. 7257●00+ √² 48372940800000.   

Theorems of Icosaedrons' contents superficial and solid. 1.  
THe superficies of Icosaedron to the square of his side, beareth the proportion of √² 75 to 1.   

The second Theorem.  
Any parallelogramme whose base is equal to double the perimetrie of the Icosaedrons' triangular base, and his altitude to the circumscribing Circles Diameter and the inscribed circles semidiameter added together, is equal to the Icosaedrons' 20 triangles.   

The 3 Theorem.  
When Tetraedrons' side retaineth such proportion to Icosaedrons' side, as Icosaedrons' diameter doth to the semidiameter of the circle whereon it is framed, that Tetraedrons' superficies is equal to Icosaedrons'.   

The 4 Theorem.  
When the square of Octaedrons' side containeth the square of Icosaedrons' sides medietie 10 times, then is that Octaedrons' superfiiciall quantity equal to Icosaedrons'.   

The 5 Theorem.  
A direct quadrangular rightangled Prisma that hath for his three dimensions, these three right lines, double the side, the Axis, and a line componed of the Diameter of the containing circle, and the semidiameter of the contained circle, is equal to the solid contents of Icosaedron.     

The .15. Problem. The side of Dodecaedron given, to search out by Arithmetical calculation the superficial and solid content.  
THis body as was before declared among the definitions is encompassed with 12 equal equiangle Pentagonal Superficies, so that, if ye search the Area of any one (as was taught in the fift Problem) and increase the same by 12, the Dodecaedrons' superficies resulteth, Or by the forenamed fift Problem, ye shall learn the quantity of the containing circles semidiameter, and also of the line diagonal, the semidiameter ye shall augment by 3, the line diagonal by 5, these productes multiplied together, bring the desired Superficies also.  
In like manner for the crassitude if ye multiply the semidiameter of the containing circle in the line diagonal and the offcome in the Axis of this body (which ye were in the tenth Problem taught to find) the producte augmented by 5, yieldeth your desire. Or the Semidiameter of the contained circle in the Dimetient of the contained sphere, and the product in the pentagonal perimetrie bringeth the solid content also.  

Example.  
Admit the Dodecaedrons' side 1, by the 5 and 10 Problem, I find his containing spheres semidimetient, √² uni. 9/8+ √² 180/256, the Axis √² v. ⅝+ √² 180/256— √² 1/20, the containing circles semidimetient √² v. ½+ √² 1/20, the line diagonal of the base √² univers 3/2+ √² 5/4. Now, for the Dodecaedrons' superficies, I search first the area of one Pentagonum, which I find by the fifth problem √² uni. 25/16+ √² 625/320, this augmented by 12, produceth √² v. 900/4+ √² 810000/20. Likewise the containing circles semidimetiente tripled, maketh √² uni. 9/2+ √² 81/20, and the line diagonal of the base increased by 5, yieldeth √² uni. 150/4+ √² 12500/16, these multiplied together, produce √² uni. √² 253125/16+ √² 253125/80+ √² 455625/16+ √² 455625/80, which contracted, maketh √² v. 225+ √² 40500, so much is the Dodecaedrons' superficial quantity, and being reduced to rational numbers, it falleth out between 20 and 21 very nigh 20 13/20. Now, to attain the solid quantity I augment √² uni. ½+ √² 1/20 the containing circles semidimetiente, in √² v. 3/2+ √² 5/4 the line diagonal, there ariseth √² uni. 1+ √² 5/16+ √ 9/80, and this again multiplied in √² uni. ⅝+ √² 45/64— √² 1/20 thee Dodecaedrons' Axis, the producte augmented by 5, maketh √² uni. 10 ⅝+ √ 28125/1024+ √ 78125/1024+ √ 28125/64+ √² 50625/1024+ √ 140625/1024— √ 625/20, so much is the Dodecaedrons' crassitude. Again, by the other rule I augment √² v. ¼+ √ 1/20 the semidimetiente of the contained circle, in √² v. 5/2+ √ 45/4— √ ⅘ the Diameter of the inscribed sphere, thereof resulteth √² v. 1 7/40+ √ 95/64+ √ 5/16— √ 1/20, which multiplied again in 5, the Pentagonal perimetrie produceth √² uni. 29 ⅜+ √ 28125/64+ √ 3125/16— √ 625/20 this is also the Dodecaedrons' solid capacity exactly agreeing with that former, operation, and by reduction is found to rest between √² 58 and √² 59, being very nigh 7 9/14.  
The side of Dodecaedron 1 

The containing circles √² v. ½+ √² 1/20  
Semidimetiente √² v. ½+ √² 1/20  
The contained circles semidiameter, √² v. ¼+ √² 1/●  
The line Diagonal, √² uni. 3/2+ √² 5/4  
The containing spheres semidimetiente, √² uni. 9/8+ √² 180/256  
The Axis or Kathetus, √² v. ⅝+ √² 180/256— √² 1/20  
The content superficial, √² uni. 225+ √² 40500  
The crassitude or solid capacity, √² v. 29 ⅜+ √² 55125/64    

Theorems of Dodecaedrons' Superficial and solid quantities. 1.  
A Pentagonal equiangle superficies (whose side to the side of Dodecaedrons' retaineth the proportion of the comprehending Spheres Diameter to the medietie of the line diagonal) is equal to the Dodecadrons content superficial.   

The 2 theorem.  
A rightangled Parallelogramme contained of the Pentagonal perimetrie, and triple the Diameter of the contained Circle, is equal to Dodecaedrons' superficies.   

The 3 theorem.  
That Cube whose side is mean proportional between the semidiameter of the internal Circle, and the perimetrie of the Pentagonal base, hath a superficial content equal to Dodecaedrons'.   

The 4 theorem.  
When an Icosaedrons' side is mean proportional between the Pentagonal bases diagonal and his circumscribing circles trigonal cord, the Superficies of that Icosaedron is equal to the superficies of Dodecaedron.   

The 5 theorem.  
If two mean proportional lines be found between the Pentagonal diagonal and his circumscribing circles cord trigonal, that Icosaedron whose side is the mean proportional nighest to the cord trigonal, is equal to the Dodecaedron.     

The .16. Problem. The diameter of any sphere known, to search out the sides, Axes and containing or contained circles semidiameters of all such bodies regulare as are therein to be described, both Arithmetically and Geometrically.  
COnsidering that in any Sphere all the regulare bodies may be described as it is by Euclid sufficiently demonstrated, and that both the proportion of their sides Axes and perpendiculars to the spheres dimetient, and also the manner of working in every of them is different, I think it best to adjoin several rules for every of their operations, & first for the invention of their sides peruse the precepts following: square the spheres diameter, and from the producte deduct ⅓, the root quad●●●● of the remainder is the Tetraedrons' side: Double the square of the Spheres semidiameter, so have ye the square of Octaedrons' side, divide the Diameters square by 3, the root quadrate of your quotient is the Cubes side, For the Icosaedrons' side ye shall divide the Diameters square by 5, the rote of the quotient note, and supposing the same a circles semidiameter, by the second Problem search out the circles pentagonal Cord, for that is the Icosaedrons' side. Finally, for the side of Dodecaedron, ye shall divide the square of the spheres Diameter by 3, and from the producte extract the quadrate root, this root divided by extreme and mean proportion, (as ye were taught in the first Problem) yieldeth for his greater part the Dodecaedrons' side. Now the sides of every regulare body thus found, for their perpendiculares, Axes, and Semidiameters of their containing and contained circles, ye shall resort for Tetraedron to the sixth Problem, for Octaedron to the seventh, for Hexaedron to the eight, and so in order for the rest, where ye are taught by the side to search out all those forenamed lines, and so shall ye find the exact quantity of all perpendiculares Axes and semidiameters of any regulare body in that sphere contained.  

Example.  
It would be over tedious to show the calculation for finding of every particular line, wherefore I shall only give examples of the sides, for that the practise of their rules hath not been yet set forth in any former Problem. Admit the spheres Diameter that shall comprehend these regulare bodies 10, the square thereof divided by 3, yieldeth 33 ⅓ which deducted from 100, leaveth 66 ⅔, the quadrate rote thereof is the inscribed Tetraedrons' side. The square of the spheres semid. is 25, which doubled maketh 50, the Zenzike rote thereof, is the Octaedrons' side. 100 divided again by 3, yieldeth as before 33 ⅓, the root square thereof is the Cubes side, which divided by extreme and mean proportion (as was taught in the first Problem) maketh the greater segment √² 41 ⅔— √² 8 ⅓, so much I affirm the contained Dodecaedrons' side. Now, for the side of Icosaedron I divide 100 by 5, thereof ariseth 20, whose root quadrate admitted a side hexagonal, his correspondente Cord Pentagonal by the second Problem ye shall find √² v 50— √² 500, the sides thus known for the Axis, Semidiameters, and other lines, 〈…〉 use the same supputation that you did in those 5 Problems past, where ye were taught by the sides known to attain all the other lines, and for more plainness I shall at the end of this chapter adjoin a Table containing the true quantity of all the rest, which ye may use in stead of an Example to direct you, if happily you err in your supputations, and for the farther satisfaction of such as seek to reach beyond the common sort, and will not content themselves with bore rules and precepts, unless they may also conceive some ground and reason of their workings, I have thought good to every of these Problems ensuing, to adjoin his peculiar figure, with means Geometrical (no regard had to Irrational numbers without aid of Arithmetical supputation) to search out the sides, Diameters, and Axis, of all the regular bodies inscribed or circumscribed of spheres, by knowledge of their Diameters, or mutually conferred together by knowledge of some side, according to the tenure of the Chapter wherein they are placed. And although brevity (which in this trifling treatise I have chiefly affected) compel me not to stay in making demonstration of every rule and Theorem, yet the very construction of the figures well weighed and conferred with Euclides 5 last books of Solides, will give sufficient light to the ingenious both to understand the cause of these, and to invent many more whereof there is no end.  
Geometrically without aid of Arithmetical calculation, to attain the quantity of all these forenamed lines ye shall thus work.  
Admit the Diameter of the comprehending sphere given AB, which ye shall as was taught in the first book divide in two equal parts at C, and in three at E, A being a third part, upon either of those sections errear Perpendiculars, and (describing a Semicircle upon the Diameter) note their intersections with FD, drawing lines from either of them to AB, so is OF the Cubes side, AD Octaedrons' side, FB the side of Tatraedron, OF divided by extreme and mean proportion (as ye were taught in the first Problem at G) maketh AG the Dodecaedrons' side, which extended out to H, ye shall make FH equal to FG, drawing the right line HB, and from F extend a parallel to HB, till it cross the Diameter in I, erecting thereupon the Perpendicular IK, so is the Cord KB the inscribed Icosaedrons' side, IL is a third part of IB, ML a parallel to IK, MB is the Diameter of Icosaedrons' basis containing circle, whose medietie MN is the Diameter of the contained circle, the half thereof MS parted by extreme and mean proportion in V, so as SV be the greater segment, un will be the Semidiameter of Dodecaedrons' contained circle, and NB the Semidimetient of his containing circle, NC the Axis both of Icosaedron and Dodecaedron, VB is the Perpendicular of Dodecaedrons' basis, and MA his Solides altitude, BS the perpendicular of Icosaedrons' basis, and MA likewise his Solides altitude, OF (the inscribed cubes side) is also Dodecaedrons' basis line Diagonal, OF is the greater Semidiameter of Tetraedrons' base, and EP his medietie the lesser Semidiameter, EC Tetraedrons' Axis, EBB his Perpendicular or altitude, FB the cubes line diagonal, OB his medietie the greater Semidiameter of the cubes base, OC the less Semidiameter, and Hexaedrons' axis Octaedrons' containing circles Semidimetient, OF the Sediameter of his contained circle RO, his Axis CO, and OF his altitude. Thus have ye Geometrically in one figure the exact quantities and proportions of all the regular bodies sides, Diameters, Axes, Perpendiculars, and lines diagonal, whereby ye may also be able both to conceive some reason of such rules as are past, or Theorems that shall ensue. And also invent divers means to abbreviate such painful calculation as by the former rules ye shallbe forced to enter into, while ye labour with irrational number to search out the hidden proportions of these unknown lines, as by proof the industrious will soon perceive.  
The comprehending Spheres Di+metient given AB 10 

Tetraedrons' 

side, BF √² 66 ⅔  
Basis greater semid, FE, √² 22 2/●  
Basis less semid. PE. √² 5 5/●  
Axis, EC, 1 ⅔  
Altitude, EBB, 6 ⅔    
Hexaedrons' 

side, FAVORINA, √² 33 ⅓  
Basis greater Semid. OF, √² 16 2/●  
Basis less semid. OC, √² 8 ⅓  
Axis CO, √² 8 ⅓  
Altitude OF, √² 33 ⅓    
Octaedrons' 

side AD, √² 50  
Basis greater semid. OF, √² 16 ⅔  
Basis less semid. OR √² 4 ⅓  
Axis QF, √² 8 ⅓  
Altitude OF √² 33 ⅓    
Icosaedrons' 

side KB, √² v. 50— √² 500  
Basis greater sem. MN, √² v. 16 ⅔— √² 55 5/●  
Basis less semid. N. S √² v. 4 ⅙— √² 3 17/36  
Axis, CN, √² v. 8 ⅓+ √² 55 5/9  
Altitude MA, √² v. 33 ⅓+ √² 888 8/●    
dodecaedrons 

side AG, √² 41 ⅔+ √² 8 ⅓  
Basis greater semi. NB, √² v. 16 ⅔— √² 55 5/9  
Basis less semid. NV, √² v. 4 ⅙+ √² 3 17/36  
Axis CN, √² v. ⅓+ √² 55 5/9  
Altitude MA, √² v. 33 ⅓+ √² 888 8/9      

Theorems of the Regular bodies in one containing sphere described. Theorem first.  
THe containing spheres diameters square to the square of the inscribed Tetraedrons' side is as 3 to 2.   

The second theorem.  
The spheres Dimetient is in power double to Octaedrons' side.   

The third theorem.  
The spheres Diameter is in power triple to the Cubes side.   

The fourth theorem.  
The spheres Dimetient being rational, Icosaedrons' side is such an irrarationall as Euclid calleth Minor, and beareth proportion to the Diameter, 〈◊〉 √² uni. ½— √² 1/20 to 1.   

The fifth theorem.  
The spheres Dimetient rational, Dodecaedrons' side is an irrational Apotome, retaining proportion to the dimetient, as √² 5/12— √² 1/12 unto 1.   

The sixth theorem.  
Tetraedrons' Axis, is a sixth part of his spheres dimetient.   

The seventh theorem.  
Octaedron and the Cubes Axis are equal, and triple in power to the Axis of Tetraedron.   

The eight theorem.  
The Axis of Icosaedron and Dodecaedron are equal, either of them being such an irrational as Euclid calleth Mayor, the spheres dimetient admitted rational.   

The ninth Theorem.  
Tetraedrons' side, is the mean proportional between his Perpendicular and his diameter.   

The tenth Theorem.  
Octaedrons' containing circles semidimetient is mean in proportion between his inscribed circles semidiameter and Tetraedrons' side.   

The eleventh Theorem.  
The Hexaedrons' side is a mean proportional between Tetraedrons' side, and Octaedrons' basis containing circles semidiameter.   

The twelfth theorem.  
The cubes side being rational, the dodecaedrons side is an irrational Apotome, bearing proportion to the side of the Cube as √² 1 ¼— ½ unto 1.   

The thirteenth theorem.  
Octaedrons' side being rational, the side of Icosaedron is such an irrational as Euclid nameth Minor, and retaineth proportion to Octaedrons' side as √² un. 1— √² ⅕ to an unity.   

The fourteenth theorem.  
Octaedrons' side being rational, the Dodecaedrons' side is an irrational Apotome, retaining proportion thereunto as √² v. 1— √² 5/9 unto an unity.   

The fifteenth theorem.  
The semidiameter of Dodecaedrons' bases containing circle, hath the same proportion to the Cubes side, that Icosaedrons' side hath to the Diameter of his containing sphere.   

The 16 theorem.  
Hexaedrons' side being rational, the contained Dodecaedrons' circles semidiameter is an irrational named by Euclid Maior, having proportion to the Cubes side, as √² v. ⅛+ √² ●/320 unto 1.   

The 17 theorem.  
Tetraedrons' side admitted rational, the contained circles semidiameter of Icosaedron is an irrational Minor, bearing proportion to the Tetraedrons' side as √² univers ⅛— √² 1/320 unto √² 2.   

The 18 Theorem.  
Octaedrons' contained circles diameter to the dimetient of dodecaedrons contained circle, hath proportion as 1 to √² v. 1+ √² ⅕.   

The 19 theorem.  
Dodecaedrons' contained circles semidiameter being an irrational Mayor, Icosaedrons' contained circles semidimetient shall be an irrational Minor, bearing proportion as √² v. 1+ √² ⅕ to √² uni. 1— √² ⅕.   

The 20 Theorem.  
Dodecaedrons' side, to octaedrons axis, retaynethe such proportion, as the greater part of a line divided by extreme and mean proportion, to the medietye of the whole.   

The 21 Theorem.  
If from the square of dodecaedrons dimetient, ye subtract the square of his side, the root of the remainder (divided by extreme and mean proportion) maketh his greater part the Cubes side, and his less the dodecaedrons side.   

The 22 theorem.  
A right line equal in power to the diameter and semidiameter of dodecaedrons basis containing circle, retaineth the same proportion to their containing spheres diameter, that dodecaedrons side doth to the side of Icosaedron.   

The 23 theorem.  
Dodecaedrons' side retaineth the same proportion to the side of the cube, that the semidiameter of dodecaedrons containing circle doth to the diameter of his contained circle.   

The 24 theorem.  
Dodecaedrons' basis internal circles semidiameter divided by extreme and mean proportion, maketh his greater part the Icosaedrons' inscribed circles semidimetient.   

The 25 theorem.  
If two lines equal in power to the spheres diameter retain the proportion of a line (divided by extreme and mean proportion) to his greater part, the lesser of those lines is Icosaedrons' side: but if their proportion be as the whole to the lesser part, then is that less part the side of dodecaedron.     

The .17. Problem. Arithmetically and geometrically to search out all the sides, diameters, perpendiculars, and lines diagonal, with the bases semidiameters, of all such regular bodies as shall circumscribe or comprehend any sphere whose dimetient is known.  
seeing these bodies as it is demonstrated by Euclid, ar● of such uniform composition that they will both receive an inward sphere touching with his convex superficies every of their bases centres, and also an outward sphere enclosing and with his concave peripherye touching every of their angles, for resolution of this Problem it shall be requisite to show how the diameter of the containing sphere may be found by knowledge of the contained spheres dimetient: which done, by the proposition last passed, the sides, and semidiameters, may consequently be known. But because these 5 bodies being described without one sphere, are not also environed of one, but have three several different comprehending spheres, the largest environing Tetraedron, the next encompassing the cube and Octaedron, and the last enclosing Icosaedron and dodecaedron (for this as it is by demonstration approved of Euclid, so is it also apparent by these former problems) it seemeth therefore requisite to prescribe three several rules for the enquiring out of their Dimetientes. And first for Tetraedrons' Diameter, ye shall only increase the diameter given by 3, the producte is your desire. And for the spheres dimetiente that containeth the cube and Octaedron, ye shall divide the diameter given by √² ●/3 the quotient or resulting sum is likewise your demand. Or divide the same diameter by √² univers ⅓— √² 8/90, So have ye the spheres dimetiente that comprehendeth Dodecaedron and Icosaedron.  

Example.  
Admit the diameter of the sphere whereon these bodies shall be described 10, this augmented by 3, maketh 30, the diameter of tetraedron: Again the square of 10 augmented by 3, bringeth 300, so is √² 300, the diameter of octaedron and the cube: I divide 100 by ⅓— √² 4/45 thereof ariseth 1500,— √² 1800000, I conclude therefore √² v. 1500— √² 1800000, the diameter of the sphere that shall include the dodecaedron & Icosaedron, whose inscribed spheres diameter is 10, and proceeding by the last problem ye shall find the dodecaedrons side √² v. 1250.— √² 1512500, Icosaedrons' side √² univers 1050— √² 1012500, and so forth of all the other solides sides, and semidiameters: for considering their operation is nothing different from that was showed in the 16 problem, it were in vain here again to make thereof a superfluous recital.   

Geometrically to perform the same.  
Upon AB the diameter given of the sphere (which all these regular bodies shall circumscribe) describe the semicircle, AKB, and upon the same centre C, describe the semicircle DNE, having his diameter DE triple to AB the diameter given, and from A, errere the perpendicular OF, which divided in two equal parts at ω, maketh A ω the less semidiameter of Tetraedrons' base, and OF the greater: from F extend cords to ED, OF is the tetraedrons side, and EA his perpendiculare. Now if ye fix one foot of the compass in C (and opening the other to ω) describe the semicircle HLI, it will touch the medietie of FE at G, whereby ye have HI the diameter of Hexaedron and octaedron, HL, the side of Octaedron, AB the side of the cube, A ω the semidimetient of Octaedrons' basis containing circle, AK, the semidiameter of the Cubes containing circle, AC the semidimetient of the cubes inscribed circle, and AM the medietie of AK is the octaedrons basis lesser semidimetiente. Thus have you found the diameters, sides, and circular semidiameters, of these first 3 bodies: for the other two ye shall thus proceed, from N as ye were taught in the first book, draw NO parallel to CE equal to the medietie thereof, and (coupling CO together with a straight line) from P where it cutteth the greatest semicircles circumference let fall the perpendicular PQ, and upon S (leaving SQ equal to 23 part of QE) erect the perpendicular SIR, then from R and P, extend strait lines to C & E, and to the middle of RE extend the right line CT, cutting the left semicircle in φ, from thence draw a parallel to RE, cutting RC, CE in π, x, and (making Cx a semidiameter) describe the semicircle β π x crossing CP in ρ, from thence to x draw the right line x ρ, for that is the circumscribing Icosaedrons' side, and φ x is the greater semidimetient of his base, wx the lesser, β x the diameter of Icosaedrons' comprehending sphere: This done from R draw a parallel to the semicircles diameter, crossing the greatest circumference in Y, fro thence to the end of the greatest dimetient, draw the line DIE, from whose medietie θ, draw a strait line to the centre C, then as was taught in the first book, cut D θ in half at μ, and divide D μ in extreme & mean proportion at λ by the first problem, from these points λ, FIE, to the centre C extend right lines, cutting the circumference of the last described semicircle in the points α z, from them to β, draw lines again, and divide the cord α β in extreme and mean proportion at γ: Likewise β z is divided in half at δ, being both the contact of the lest circumference, and also the intersection made with θ C, but where λ C conreth with β z place this letter ε, so is β γ the circumscribing dodecaedrons side β x his dimetient, β δ the semidiameter of his containing circle, ε δ the lesser semidiameter of the dodecaedrons pentagonal basis, and α β his line diagonal. Thus have ye in one figure all the sides and diameters both circular and spherall of all such regulare solides as comprehend or circumscribe the assigned sphere.  
The diameter of the internal given Sphere 10 

Tetraedrons' 

Diameter DE, 30  
side OF, √² 600  
Containing circles sedimiameter, OF, √² 200  
Inscribed circles semidimetient A ω √² 50  
Perpendiculare or Altitude, A 20    
Hexaedrons' 

Diameter HI √² 300  
side AB, 10  
Basis greater semidiameter, AK, √² 50  
Basis lesser Semidimetient, AC 5  
Basis line diagonal, √² 200  
Altitude AB 10    
Octaedrons' 

Diameter HI √² 300  
side HL, √² 150  
Basis greater semid. AK √² 50  
Basis less semidiameter, AM, √² 25/2  
Perpendicular AB, 10    
Icosaedrons' 

Diameter β x √² v. 1500— √² 1800000  
side ρ x. √² viii. 1050— √² 1012500  
Basis greater semidiamet. ω φ √² v. 350— √² 112500  
Basis less semidiameter φ w, √² v. 87 ½,— √² 7031 ¼  
Altitude 10    
Dodecaedrons' 

Diameter β x √² v. 1500— √² 1800000  
side γ β √² uni. 1250— √² 1512500  
Basis greater semid. δ β— √² v. 350— √² 112500  
Basis less semidiameter ε δ √² v. 37 ½— √² 781 ¼  
Altitude AB, 10      

Theorems of the internal Spheres circumscribing Solides. 1  
TEtraedrons diameter is rational and triple to the contained Spheres diameter, the internal Spheres diameter supposed rational.   

The 2 Theorem.  
Tetraedrons' sides square retaineth the same proportion to the square of the internal spheres Diameter, that his containing spheres Diameter doth to his Axis.   

The 3 Theorem.  
Tetraedrons' Basis containing circles semidiameter, is double in power to the contained spheres Diameter.   

The 4 Theorem.  
The square of Tetraedrons' inscribed circles semidiameter to his Axis, holdeth the same proportion in power, that his perpendiculare doth to his inscribed Spheres diameter.   

The 5 Theorem.  
Tetraedrons' side, is double to Octaedrons' side, one Sphere being contained of them both.   

The 6 Theorem.  
The square of Tetraedrons' side, to the square of the Cubes side, retaineth the same proportion, that Tetraedrons' comprehending spheres Diameter doth to his Axis.   

The 7 Theorem.  
The diameter of Tetraedrons' inscribed circle, is equal to the Diameter of Octaedrons' containing circle.   

The 8 Theorem.  
Tetraedrons' diameter to the diameter of the sphere that comprehendeth Octaedron and the Cube, hath the proportion of 3 to his quadrate root.   

The 9 Theorem.  
Octaedron and the Cubes comprehending spheres diameter being rational, the diameter of Icosaedron and Dodecaedrons' containing sphere is an irrational Minor, bearing proportion to the forenamed diameter, as √² v. 5— √² 20 unto 1.   

The 10 Theorem.  
Icosaedron and Dodecaedrons' axis being rational, their comprehending spheres dimetiente, shall be an irrational Minor, bearing proportion thereunto, as √² uni. 60— √² 2880 unto an unity.   

The 11 theorem.  
Icosaedrons' axis being rational, his side is an irrational Apotome proportioned thereunto, as √² v. 42— √² 1620 unto an unity.   

The 12 theorem.  
Dodecaedrons' Axis rational, his side is a surde Minor proportioned to the Axis, as √² v. 50— √² 2420 unto an unity.   

The 13 Theorem.  
Icosaedrons' side being an irrational Apotome of the sixth order, Dodecaedrons' side is an irrational Minor, retaining such Proportion thereto, as √² v. 50— √² 2420 unto √² 27— √² 15.   

The 14 Theorem.  
Icosaedrons' axis being rational, his basis containing circles semidiameter is an Apotome of the first order bearing proportion to the Axis, as 3— √² 5 unto an unity.   

The 15 Theorem.  
Dodecaedrons' axis being rational, his Pentagonal basis greater semidiameter is an Apotome of the first order proportional to his side, as √² v 7— √² 45 to √² v. 25— √² 605.     

The .18. Problem. The side of any Tetraedron given, to find the sides, Diameters, and Axes, of all such regulare bodies as may therein be described.  
Having heretofore at large set forth by Problems sundry ways (the sides of these bodies given) to find the semidiameters of their containing and contained circles, the diameters of their comprehending and comprehended spheres, with their contents superficial and solid: having also by Theorems showed manifold diversity of proportions rational and surde of these bodies, their Superficies and lines compared with theirs comprehending and contained spheres, there remaineth only now to confer these bodies mutually inscribed or circumscribed one with an other, and to search out by the side of any one known, the sides and diameters both circular and spherall, with the capacities superficial and solid of all such bodies as may within or without the same body be described, I shall therefore first begin with Tetraedron, and so proceed with the rest. Tetraedron receiveth only Octaedron and Icosaedron, for the Cube and Dodecaedron cannot possibly therein be so placed, that all their angles at one instant might exactly touch his superficies, the Tetraedrons' side therefore given parted in two equal portions, either medietie is the inscribed Octaedrons' side: Likewise the medietie of Tetraedrons' sides square, is the square of Octaedrons' diameter which divided by 12, produceth a number, whose quadrate root is the Octaedrons' Axis. For the inscribed Icosaedron ye shall divide the medietie of Tetraedrons' side by extreme and mean proportion, and double the square of the less portion, the quadrate root of the producte is the side. Or deduct one of these portions from the other, & add the square of the remains medietie to the square of Octaedrons' sides medietie, the rote quadrate of the producte doubled, is the Icosaedrons' Diameter: Or if from the last producte, ye abate the third part of Icosaedrons' sides square, the root quadrate of the remain is the Icosaedrons' Axis.  

Example.  
Tetraedrons' side supposed, Octaedrons' side is ½, √² ½ his Diameter, whose square divided by 12 bringeth 1/24, the root being √² 1/24, is the Octaedrons' Axis. Likewise for Icosaedron the medietie of Tetraedrons' side divided in extreme and mean proportion by the first Problem, maketh the lesser portion ●/4— √² 5/16, the square hereof doubled, hath for his root √² uni. 14/8— √² 180/64, so much is the inscribed Icosaedrons' side. Again, the difference of Tetraedrons' sides medieties Portions divided by extreme and mean proportion is √² 20/16— 1, the square of half this difference is 9/16— √² 80/256, which added to the square of Octaedrons' sides medietie, produceth ⅝— √² 5/16, the root thereof doubled, is √² v. 5/2— √² 5 the true quantity of Icosaedrons' dimetient. Now by subtracting 7/12— √² 45/144 the third part of Icosaedrons' sides square, from ⅝— √² 5/16, the last producte whose root ye doubled to make the Diameter your remainder will be this number 1/24+ √² 45/144— √² 5/16, whose root universal is the inscribed Icosaedrons' Axis.  
The containing Tetraedrons' side 1 

Tetraedrons' 

Diameter. √² 3/2  
Axis, √² 1/24    
The inscribed Octaedrons' 

side ½  
Diameter √² ½  
Axis, √² 1/24    
The contained Icosaedrons' 

side √² uni. 7/4— √² 45/16  
Diameter √² uni. 5/2— √² 5  
Axis, √² 1/24.       

Or thus geometrically without respect of number.  
Admit AB the Tetraedrons' side given, thereon I describe the semicircle ACB, erecting the perpendiculare DC, and drawing the line AC, that Cord AC is the inscribed Octaedrons' dimetiente, and AD his side. Now divide DC by extreme and mean proportion as was taught in the first problem at E, and from E erect the perpendiculare OF, cutting AC in F, conclude FC the inscribed Icosaedrons' side: again part AC in half at H, and from H to E extend a strait line, for that shall be the Icosaedrons' semidiameter. For the Axis ye shall cut of from DB his twelfth part, as ye were taught in the first book: suppose it BY, upon I erect a perpendiculare, cutting the circumference in K, so is the Cord KB the inscribed Icosaedron and Octaedrons' Axis. As for their bases circular semidiameters they are found as was taught in the fifth and sixteenth problems, forasmuch as both their sides and dimetientes are known.  
 

Theorems of Tetraedrons' inscribed solides. 1.  
Tetraedrons' side is double to the side of his contained Octaedron.   

The second Theorem.  
Octaedrons' diameter is a mean proportional between his side and the side of his containing Tetraedron.   

The third Theorem.  
If any equilater triangle be so described and situate within the Tetraedrons' equiangle triangular base, that the angles of the inscribed triangle divide the sides of the base by extreme and mean proportion, the medietie of this inscribed triangles side, is the side of this Icosaedron.   

The fourth Theorem.  
Tetraedrons' side divided by extreme and mean proportion, his lesser part is double in power to this Icosaedrons' side.   

The fifth Theorem.  
The dimetient of Octaedron divided by extreme and mean proportion maketh his lesser portion the Icosaedrons' side.   

The sixth Theorem.  
Icosaedrons' diameter is equal in power with Octaedrons' side, and the excess or difference of his two parts, the one deducted fro the other.   

The seventh Theorem.  
The side of Tetraedron being rational, his inscribed Icosaedrons' side is an Apotome, bearing proportion to the Tetraedrons' sides, as √² v. 1 ¾— √² ●●/16 unto 1.   

The eight theorem.  
The Axes of these inscribed bodies Octaedron and Icosaedron are equal.   

The ninth theorem.  
The side of Tetraedron being rational, the contained bodies axes are rational, in power commensurable, and bear proportion to the side as √² 24 to 1.   

The tenth theorem.  
Octaedrons' diameter being rational, Icosaedrons' dimetient is an irrational Minor, proportioned to the dimetiente of Octaedron as √² v. 5— √² 20 unto an unity.     

The .19. Problem. The side of a Cube given, to find the sides, diameters and axes of all such regulare bodies as may therein be described.  
THe Cube is capable of three bodies, Tetraedron, Octaedron, and Icosaedron, for the greatest Dodecaedron that may within a Cube be imagined, will only with 12 of his angles touch the Cubes superficies, his other eight angles remaining within the body under every of the Cubes angles, dividing his semidimetients by extreme and mean proportion: this body therefore omitted because his inscription is unperfect, I shall give rules for the other three, and first of Tetraedron, double the square of the Cubes given side, the product root quadrate is the Tetraedrons' side, the same given sides square tripled bringeth his comprehending spheres diameters square, which divided by 36 yieldeth in the quotient a number, whose quadrate root is the Tetraedrons' axis. The Cubes sides squares medieties root quadrate, is the inscribed Octaedrons' side.  
The Octaedrons' diameter is equal to the Cubes side,  
The Cubes sides square divided by 12, bringeth the contained Octaedrons' Axis square. For the contained Icosaedrons' side, ye shall part the containing Cubes side given by extreme and mean proportion, as ye were taught in the first problem, the greater part is your desire. And if you add the square thereof to the square of Hexaedrons' side, the root quadrate of the resulting sum is the Icosaedrons' dimetient. And if ye square the medietie of this dimetient, and from it deduct a third part of the square of this inscribed Icosaedrons' side, the root quadrate of the remain is the Icosaedrons' Axis.  

Example.  
Hexaedrons' side given 1, his square doubled is 2, √² 2 is the contained Tetraedrons' side, the given sides square tripled is 3, his quadrate root is tetraedrons diameter, that square divided by 36 yieldeth 1/12, √² 1/12 is the axis. Likewise √² ½ being the quadrate root of half the Cubes side is the inscribed Octaedrons' side, the Octaedrons' diameter being equal to the Cubes side is 1 also, and the Cubes sides square divided by 12 yieldeth 1/12, I conclude √² 1/12 the axis. Again for Icosaedron I divide 1 by extreme and mean proportion, the greater part is √² 5/4— ½, so much is the Icosaedrons' side, the square of this side added to the square of the Cubes side, maketh 2 ½— √² 5/4, whose root universal is the diameter: from the square of this roots medietie being ⅝— √² 1/64, if ye deduct ½— √² 5/36 the third part of the sides square, there will remain ⅛+ √² 5/36— √² 5/64 the root universal thereof is Icosaedrons' axis.  
The Cubes side 1. His inscribed.  

Tetraedrons' 

side √² 2.  
Diameter √² 3  
Axis √² 1/12.    
Octaedrons' 

Side √² ½.  
Diameter. 1.  
Axis √² 1/12.    
Icosaedrons' 

Side √² 5/4— ½.  
Diameter √² v. 2 ½— √² ●/●.  
Axis √² v. ⅛+ √² 5/36— √² ●/64.     
Ye may also with the compass find out all these sides, diameters and axis, no regard had to any number, so the containing Cubes side be known. Admit the Cubes side given AB, extend the same line out to D, and upon A as a centre, making AB the semidiameter, describe the semicircle BDC, and upon A erect the perpendiculare AC, as ye were taught in the first book: concurring with the circumference in C, again as ye were taught in the first book cut of from AB 1/24 part. Admit it FB, upon F erect an other perpendiculare crossing the circumference in G, divide also CB in two equal parts in I, then open your compass to the length of AB, and fixing one foot in B, with the other cross the circumference in E. last of all by the first problem divide AB by extreme and mean proportion in H, so as AH may be his greater portion, and extend right lines from H to C, from B to C, from E to BD, and from G to B. Thus have ye the diameters sides and axes of those inscribed bodies, for CB is the inscribed Tetraedrons' side, DE the Tetraedrons' dimetiente, GB the Tetraedrons' axis, BY is the inscribed Octaedrons' side, AC the Octaedrons' dimetiente, GB also Octaedrons' axis, AH is the side of the inscribed Icosaedron, and CH his diameter, as for the axis ye may geometrically find it by the sixteenth problem, the comprehending spheres dimetiente being known, whereof I mind not here to make any new recital, considering it is sufficient plainly declared before.  

Theorems of Hexaedrons' inscirbed regulare Solides.  
THe inscribed Tetraedrons' side is double in power to the containing cubes side.   

The 2 theorem.  
The containing Cube and his contained tetraedron have one comprehending sphere, but their axes are different, and their proportion in power triple.   

The 3 theorem.  
The containing cubes axis is equal to his inscribed Octaedrons' semidiameter, and their axes in power retain triple proportion.   

The 4 Theorem.  
Hexaedrons' side divided by extreme and mean proportion, maketh his greater part the Icosaedrons' side.   

The 5 Theorem.  
The inscribed Icosaedrons' dimetient is in power equal to his containing cubes side, and the greater portion thereof, it being divided by extreme and mean proportion.   

The 6 Theorem.  
The Cubes side being rational, his inscribed Icosaedrons' side is an apotome bearing proportion to the cubes side, as √² 1 ¼— ½ unto an unity.   

The 7 Theorem.  
Hexaedrons' side is mean proportional between his inscribed Tetraedron and Octaedrons' side.   

The 8 theorem.  
The Cubes side is equal in power to his axis, and the contained tetraedrons semidiameter.   

The 9 theorem.  
Hexaedrons' diameter to his inscribed Tetraedrons' perpendicular, retaineth the proportion of 3 to 2.   

The 10 Theorem.  
The comprehending cubes side being rational, his contained Icosaedrons' side is an irrational Binomye, bearing proportion unto the cubes side as √² universalis ⅛+ √² 5/576 to 1.     

The .20. Problem. Octaedrons' side given, to search out all his contained bodies, sides, diameters and axes.  
DOdecaedron within this body may not aptly be described, for as much as only eight of his solid angles touch the superficies of this body being situate in the centres of octaedrons bases, the other 12 falling quite within this body, not touching his superficies on any part. But Tetraedron may be inscribed and all his solid angles placed in the centres of octaedrons triangles. Likewise the cubes solid angles take their place in the centres of his bases, only Icosaedron hath his situation somewhat more strange and intricate, every of his 12 angles resting in a side of octaedron, and divide his 12 sides by extreme and mean proportion. Now to attain the diameters axes etc. Peruse these rules ensuing.  

For Tetraedron.  
Augment octaedrons side by 2, and divide by 3, the quotient is tetraedrons side, the square thereof augmented by 3, and parted by 2, yieldeth his diameter, which divided by 6▪ declareth the axis.  
The square of octaedrons side multiplied by 2 and divided by 3 produceth this Cubes diameters square, which again divided by 3, showeth his sides square, half the side is the axis.   

For Icosaedron.  
Divide the side given by extreme and mean proportion, the square of the lesser part double, and from the product extract the root quadrate, so have ye the inscribed Icosaedrons' side, deduct one of those former found parts of the given side from the other, and square the difference, for that added to the square of Icosaedrons' side bringeth the square of his dimetient. Now for the axis ye may deduct the third part of Icosaedrons' sides square from the square of his semidimetiente, the root quadrate of the remainder is his cathetus.  

Examples of Tetraedron.  
Octaedrons' side given 1▪ increased by 2, and divided by 3, maketh ⅔, the Tetraedrons' side, the square thereof being ⅘ augmented by 3 and parted by 2, yieldeth ⅔ √² ⅔ is his diameter, which divided by 6, bringeth √² 1/54 the axis.    

Of the Cube.  
The square of Octaedrons' side 1, augmented by 2, and divided by 3, produceth ⅔ the diameters square, that again divided by 3, yieldeth 2/9, √² 2/9 is the cubes side, whose half being √² 1/18 is the axis.   

Of Icosaedron.  
Octaedrons' side 1 divided by extreme and mean proportion, the greater part ● √² 5/4— ½, the lesser 3/2— √² 5/●▪ this latter parts square doubled is 7,— √ 45, the root quadrate universal thereof is Icosaedrons' side. Again by subtraction of the parts ye shall found the difference √² 5— 2, the square thereof added to the square of Octaedrons' side given, maketh 10,— √² 80. I conclude √² uni. 10,— √² 80 the Icosaedrons' diameter. Finally for the axis I deduct 7/3— √² 5 the third part of Icosaedrons' sides square, from 5/2— √² 5, the semidimetient of Icosaedrons' containing sphere, there remaineth ⅙ √² ⅙ is the axis.  
The comprehending Octaedrons' side 1 

Tedraedrons' 

side ⅔  
Diameter √² ⅔  
Axis, √² 1/54    
Hexaedrons' 

side √² 2/9.  
Diameter √² ⅔.  
Axis √² 1/18.    
Icosaedrons' 

side √² uni. 7— √² 45.  
Diameter √² uni. 10— √² 80.  
Axis √² ⅙.      
Or thus without aid of numbers, admit AB the Octaedrons' side given, thereon (making C the medietie a centre) I describe the semicircle ADB, and from C, I erect the perpendicular CD, drawing lines from D to AB, the semidiameter CB, I divide in 3 equal parts at E and I, upon I, I errere the perpendicular IK, concurring with the circumference in K, from E, I draw EM paralele to DB, cutting AD in M, as was taught in the first book of this treatise: again by the first problem I divide AB by extreme and mean proportion in F, and thereupon I raise the perpendicular FG, crossing AD in G, and from F to D, I extend a straight line▪ finally I draw the right line KB, cutting of a third part at H, and thus is the figure fully finished, containing all these bodies, sides, Diameters, and Axes: For A is Tetraedrons' side, KB his Semidiameter, HB his axis, AM the cubes side, KB his semidiameter, MD his axes, AG Icosaedrons' side, F D his semidiameter, and KB his axes.  

Theorems of Octaedrons' inscribed regular bodies. 1.  
THe containing Octaedrons' side beareth proportion to the side of his contained tetraedron, as 3 to 2.   

The second theorem.  
Octaedrons' side, to his inscribed cubes side, beareth the proportion of √² 4 ½ unto 1.   

The third Theorem.  
The side of tetraedron, to the side of the cube, is as 2 unto √² 2.   

The fourth Theorem.  
Octaedrons' containing dimetient, is triple in power to his inscribed Icosaedrons' axis.   

The fifth theorem.  
The conteyaing octaedrons diameter is triple to the contained cubes side.   

The sixth Theorem.  
The axes of these 3 inscribed bodies are proportional, and the cubes axes it the mean proportion between the other two.   

The seventh Theorem.  
The side of Octaedron divided by extreme and mean proportion, that quantity which is double in power to the lesser part, is equal to the inscribed Icosaedrons' side.   

The eight theorem.  
Icosaedrons' diameter is equal in power to the side of the containing Octaedron, and the d●fference or excess of Octaedrons' sides parts being divided in extreme and mean proportion.   

The ninth Theorem.  
Octaedrons' dimetient divided by extreme and mean proportion maketh his lesser segment the inscribed Icosaedrons' side.   

The tenth theorem.  
Octaedrons' side being rational, Icosaedrons' side is an Apotome proportioned to the Octaedrons' side, as √² univers 7— √² 45 unto an unity.     

The .21. Problem. Icosaedrons' side given, in line or number to set forth all the sides, Diameters and Axes of his contained regulare bodies.  
ICosaedron is a body of such uniform capacity, that he receiveth all the other four regulare bodies, whereof Tetraedron, the Cube, and Dodecaedron, have their solid angles all resting in the centres of his bases, and Octaedrons' angles are situate in the medieties or middle sections of Icosaedrons' opposite sides, and his three diameters cross themselves at right angles upon the centre of his spheres. Now for the invention of their sides, Diameters, and Axes, note the rules ensuing.  

Of Tetraedron.  
Divide the given side by extreme and mean proportion, adjoining thereunto his greater part the producte square, and from it subtract the third part of the square of the given side, the root Quadrate of the remainder is the Tetraedrons' containing spheres Diameter, and the third part of his square doubled is the square of the Tetraedrons' side, whose third part deducted fro the fourth part of the Diameters square, leaveth the square of the Axis, or divide the Dimetient by six, so have ye the Axis also.   

Of Octaedron.  
THe side of Icosaedron given, being parted by extreme and mean proportion, having his greater part to him adjoined, yieldeth the Octaedrons' dimetient, whose squares medieties root is the side, and the third part of his sides square abated from the square of the Semidimetient, leaveth the square of his Axis.   

Of Hexaedron.  
THe Cubes Diameter is equal to the Diameter of Tetraedron, which found as before, ye shall divide the square thereof by 3 the quotientes root Quadrate is the Cubes side, whose medietie is his Axis.  
The inscribed regular Solides lines.  

AB The comprehending Icosaedrons' side given 1  
EBB The comprehending Icos. Diameter √² v 5/2+ √² 5/4  
SK The comprehending Icosaedrons' Axis, √² v. 7/24+ √² 5/64  
CP Tetraedrons' side, √² v. 7/9+ √² 5/9  
CK Tetraedrons' Diameter, √² v. 1 ⅙+ √² ¼  
SN Tetraedrons' Axis, √² v. 7/216+ √² 5/5184  
CR Hexaedrons' side, √² v. 7/18+ √² 5/36  
CK Hexaedrons' Dimetient, √² v. 1 ⅙+ √² 1 ¼  
CT Hexaedrons' Axis √² v. 7/72+ √² 5/576  
CF Octaedrons' side, √² v. ¾+ √² 5/16  
CB Octaedrons' Dimetient, √² 5/4+½  
CT Octaedrons' Axis √² v. ⅛— √² 25/2880  
CM Dodecaedrons' side √² 5/36+⅙  
CK Dodecaedrons' Diameter √² v. ⅙+ √² 5/4  
VZ dodecaedrons axis, √² v. ⅛+ √² 1/320+ √² V.D. 1/18 √² 1/102●   

Theorems of Icosaedrons' inscribed bodies. Theorem first.  
THe containing Icosaedrons' dimetient is in power equal with his side and the diameter of his inscribed Octaedron.   

The second theorem.  
The inscribed Octaedrons' dimetient, is in power equal to the dimetient of the contained Cube and the diameter of Icosaedrons' contained circle.   

The third theorem.  
Octaedrons' dimetient, (divided by extreme and mean proportion) his greater segment is the comprehending Icosaedrons' side.   

The fourth theorem.  
Icosaedrons' dimetient in power is equal with the diameter of his containing circle, and the dimetient of his contained cube.   

The fifth theorem.  
Tetraedrons' side, is double in power to the side of the contained cube.   

The sixth theorem.  
Tetraedrons' diameter, is triple in power to the inscribed Cubes side.   

The seventh theorem.  
Dodecaedrons' inscribed side is the third part of such a line as divided by extreme and mean proportion maketh his greater part the containing Icosaedrons' side.   

The eight Theorem.  
Octaedrons' diameter retaineth the same proportion to the side of Dodecaedron that the diameter of Tetraedron doth in power to the side of the Cube.   

The ninth Theorem.  
Icosaedrons' diameter retaineth the same proportion to the dimetient of Octadron, that the side of Icosaedron beareth to the semidiameter of that circle, whereon Icosaedron is framed.   

The tenth Theorem.  
Icosaedrons' axis is triple in power to the axis of the contained Cube.     

The .22. Problem. The side of any Dodecaedron given, both Arithmeticallye and Geometrically to search out the sides diameters and axis of all the regulare bodies therein described.  
SUch is the resemblance and mutual conformity between Icosaedron and this body, that the 12 solid angles of Icosaedron will precisely rest in the centres of Dodecaedrons' pentagonal bases, the angles of the inscribed Octaedron have their place in the medietie of the six opposite sides of Dodecaedron, which coupled together with right lines make 12 sides containing his 8 triangles, and 3 diameters, crossing themselves at right angles on the centre of his spheres. But the solid angles of the internal Cube and Tetraedron are situate in the angles of the containing Dodecaedron, and one sphere comprehendeth them all three. As for the longitude of their axes, sides and diameters, peruse these precepts following.  

Of Tetraedron.  
Divide the side given by the first problem into extreme and mean proportion, and to the whole side adjoin his greater part, the square of the resulting quantity tripled maketh the Tetraedrons' dimetientes square, which augmented by 2 and divided by 3 bringeth the square of Tetraedrons' side, which divided by 24 yieldeth in the quotient the square of Tetraedrons' Axis: extract the roots quadrate of these squares, so have ye the desired lines.   

Of Hexaedron.  
Divide Tetraedrons' sides square by 2, so have ye the square of the Cubes side, which augmented by 3 showeth the diameters square, and that divided by 12 produceth the axis square, extract the Zenzike roots of these numbers, so have ye the lines.   

Of Octaedron.  
DOdecaedrons side given added to the side of Hexaedron late found, maketh the diameter of Octaedron, the root quadrate of his squares medietie is the side, and the root quadrate of the dimetientes twelfth part is the axis.   

Of Icosaedron.  
THe side of Dodecaedron being given, first search the diameter of his containing circle by the fifth & tenth problems, and deduct the square thereof from the square of the Cubes dimetiente found as is before declared, the root quadrate of the remain is the Icosaedrons' dimetient: Now for his side ye shall add the squares of the Cubes side and Dodecaedrons' sides together, the root universal of the producte ye shall reserve for a divisor, then multiply Icosaedrons' dimetiente by Dodecaedrons' side, and the product divide by your reserved divisor, the quotient is the Icosaedrons' side, the square of this side deduct from triple the square of Icosaedrons' semidimetiente, and fro the third part of the remainder extract the Zenzike root, for that is the Icosaedrons' Axis.  
Because these rules of themselves are apparent enough, I shall only adjoin an example for the last, with a table containing the numbers ready calculate of all the rest, which shall supply the place of examples for the other.  

An example of Icosaedron.  
THe Dodecaedrons' side given 20 divided by extreme and mean proportion maketh his greater part √² 500— 10, which added unto 20, is √² 500+10, the square thereof tripled is 1800+ √² 1800000, so is the square of the Cubes dimetient, and from it deducting 800+ √² 128000 Dodecaedrons' basis containing circles diameters square, there will remain 1000+ √² 968000, the root quadrate universal thereof is the diameter of the contained Icosaedron, which multiplied by the Dodecaedrons' side createth √² v. 400000+ √² 154880000000, and that divided by √² v. 1000+ √² 200000 (for so much ariseth for a divisor by addition of the Cubes sides square to the square of Dodecaedrons' side) there will result of that division √² v. 280+ √² 72000, so much is the contained Icosaedrons' side, whose square deducted fro triple the square of Icosaedrons' semidiameter, 17600/80+ √² 70560000/320, whose third parts root is √² v. 1880/12+ √² 102000/16 so much is the contained Icosaedrons' axis, and that doubled showeth the quantity of his contained spheres dimetiente.  
The containing Dodecaedrons' side 20.  

Dodecaedrons' 

Diameter √² v. 1800+ √² 1800000  
Axis √² v. 250+ √² 60500.    
The contained Tetraedrons' 

side √² v. 1200+ √² 800000  
Diameter √² v. 1800+ √² 1800000  
Axis √² v. 50+ √² 1388 8/9    
The inscribed Hexaedrons' 

side √² v. 600+ √² 200000  
Diameter √² v. 1800+ √² 1800000  
Axis √² v. ●50+ √² 12500    
The included Octaedrons' 

side √² v. 700+ √² 450000  
Diameter √² v. 1400+ √² 1800000  
Axis √² v. 116 ⅔+ √² 12500    
The internal Icosaedrons' 

side √² v. 280+ √² 72000  
Diameter √² v. 1000+ √² 968000  
Axis √² v. 156 ⅔+ √² 24500     
For the geometrical searching out of all these lines, behold the figure following, where AB is the Dodecaedrons' side given, BC such a line as being divided into extreme and mean proportion maketh his greater part equal to , and therefore the inscribed Cubes side AC is the inscribed Octaedrons' diameter, and AFC a semicircle described thereon; the centre being E, DC is equal to AB, and perpendiculare to AC, AD is the diameter of Dodecaedron, the contained Cube, and Tetraedron, AGD a semicircle described upon his medietie the centre N, BG is an ark described upon the centre D, GH is a perpendiculare let fall on AD, HK is the fifth part of HD, KI an other perpendiculare ereared on that point cutting the semicircle in I, AI is the diameter of the inscribed Icosaedron, and AVI a semicircle thereon described, BL equal to AI, LM a parallel to DC, falling perpendicularly upon AC extended to M, AV is equal to LM, it is the contained Icosaedrons' side, VX a perpendiculare from V to AI, XY is a third part of AXE, YZ a perpendiculare to AI, cutting the semicircle at Z, ZI is the contained Icosaedrons' inscribed spheres diameter, WI his medietie is the Icosaedrons' axis, AO is a third part of AD, OPEN a perpendiculare upon AD, PD the inscribed Tetraedrons' side, PA the Cubes side, OR a parallel to DC, RC the diameter of the inscribed Octaedrons' containing circle, S his middle point, ES Octaedrons' Axis, OF a perpendiculare ereared on the diameter, AC cutting his semicircles circumference in F, OF Octaedrons' side, T W is the semidiameter of Icosaedrons' containing circle, TN the semidiameter of Dodecaedrons' containing circle, OPEN the semidimetient of Tetraedrons' base, and DQ the circle's semidiameter that containeth the Cubes square. Thus have ye in one figure all the sides, diameters, and axes, both of the containing Dodecaedron, and his inscribed Tetraedron, Octaedron, Hexaedron, and Icosaedron, with the semidiameters also of the circles environing their bases, but jest the intricate crossing and concurring of lines and arks might bread confusion to such as have not been trained in geometrical demonstration. I shall adjoin a table containing the sides, axes, and diameters, both circular and spherall of every body particularly by itself, expressed also exactly with numbers and Algebraicall characters, the containing Dodecaedrons' side admitted an unity wherein the skilful Arithmetrician shall find matter abundante to exercise or delight himself.  
The comprehending Dodecaedrons' 

side AB, 1  
Basis line diagonal, √² 1 ½+ √² 5/2 BC.  
Containing spheres diameter, √² v. 4 ½+ √² 11 ●/● AD.  
Basis containing circles Semid. √² v. ½+ √² 1/20 TN.  
Axis, √² uni. ⅝+ √² 121/320, AT.    
Dodecaedrons' inscribed regulare 

Tetraedrons' 

side √² uni. 3+ √² 5 DP.  
Diameter, √² uni. 4 ●/2+ √² 11 ●/4, AD.  
Basis greater semid. √² v. 1+ √² 5/9 OP.  
Axis, √² uni. ⅛+ √² 5/●76, ON.  
Perpendiculare √² uni. 2+ √² 2 2/9 ODD.    
Octaedrons' 

side, √² v. 1 ¾+ √² ●5/16, AF.  
Diameter √² uni. 7/2+ √² 45/4, AC.  
Basis greater semid. √² v. 7/12+ √² 45/144, RS.  
Axis, √² v. 7/24+ √² 5/64, ES.    
Hexaedrons' 

side √² uni. 3/2+ √² 5/4, AP.  
Diameter, √² uni. 4 ½+ √² 11 ¼, AD.  
Basis diagonal, √² uni. 3+ √² 5. PD.  
Containing circles semid √² v. ¾+ √² 5/16, DQ.  
Axis, √² uni. ⅜+ √² 5/64 NQ.    
Icosaedrons' 

side, √² uni. 7/10+ √² 9/20, LM.  
Dimetient, √² v. 2 ½+ √² 121/20, AI.  
Basis cont. circles semid. √² v. 7/30+ √² 1/20, Tw.  
Axis, √² v. 47/120+ √² 49/320, wI.  
The Circles Semidiameter whereon Icos. is framed▪ √² v. ½+ √ 121/500 VX.      

Theorems of Dodecaedrons' inscribed regulare Solides.  
TEtraedrons inscribed sides square joined to the Cubes sides square maketh the diameters square of Hexaedron, all being described within one Dodecaedron.   

The 2 Theorem.  
Tetraedrons' side is double in power to the cubes side, and equal to his line diagonal.   

The 3 Theorem.  
Dodecaedrons' Diameter is equal in power to the dimetiente of Octaedron and Dodecaedrons' side.   

The 4 Theorem.  
Icosaedrons' dimetiente is equal in power to his own side, and such an other line also (as being by extreme and mean proportion divided) maketh his greater part equal to Icosaedrons' side.   

The 5 theorem.  
A line equal in power to Dodecaedron and his inscribed Cubes side, hath such proportion to Icosaedrons' diameter, as Dodecaedrons' side, hath to the side of his contained Icosaedron.   

The 6 theorem.  
Octaedrons' diameter divided by extreme and mean proportion maketh his greater part the Cubes side, and his lesser the side of Dodecaedron.   

The 7 theorem.  
Hexaedrons' sides square is equal to the square of Dodecaedrons' side, and a rightangled Parallelogramme contained of the Cubes side and Dodecaedrons' side.   

The 8 theorem.  
Octaedrons' diameters square, is equal to the square of Tetraedrons' side, and a right angled Parallelogramme contained of Dodecaedrons' side and his inscribed Cubes side.   

The 9 theorem.  
Octaedrons' diameter being by extreme and mean proportion divided, the excess or difference of these parts squares deducted fro the square of the diameter, leaveth the square of the inscribed Tetraedrons' side.   

The 10 theorem.  
Hexaedrons' diameter is equal in power to Dodecaedrons' Basis containing circles diameter of the inscribed Icosaedron.   

The 11 theorem.  
The excess of Dodecaedrons' basis containing circles diameters square above the square of his line diagonal, added unto his contained Spheres diameter, createth the square of his inscribed Tetraedrons' side.   

The 12 theorem.  
Dodecaedrons' dimetient is equal in power with these 3 lines, the greater diameter of his own basis, the diameter of Icosaedrons' contained sphere, and the greater diameter of Icosaedrons' basis.   

The 13 theorem.  
Octaedrons' dimetiente retaineth the same proportion to the diameter of his basis, that Dodecaedrons' diameter doth to the side of his inscribed Tetraedron.   

The 14 theorem.  
Octaedrons' containing circles diameter retaineth the same proportion to his contained spheres dimetiente, that Tetraedrons' side doth to the side of the Cube.   

The 15 theorem.  
Dodecaedrons' side being rational, the side of his contained Icosaedron is an irrational Binomie of the 3 order, bearing proportion to the containing dodecaedrons side, as √² 9/20+ √² 5/20 to an unity.   

The 16 theorem.  
Octaedrons' diameter, to the side of Icosaedron, is proportioned in power, as 5 to 1.   

The 17 theorem.  
Octaedrons' diameter beareth such proportion to the dimetient of Icosaedron. as a Cord Pentagonal doth to a Cord hexagonal of the same circle.   

The 18 theorem.  
Icosaedrons' diameter divided by extreme and mean proportion maketh his greater part the diameter of his comprehending dodecaedrons Basis internal Circle.   

The 19 theorem.  
Octaedrons' dimetient exceedeth in power the dimetiente of Icosaedron by the diameter of Dodecaedrons' basis internal circle.   

The 20 theorem.  
A line in power equal to the sides of Dodecaedron and the Cube to the diameter of Icosaedron, retaineth such proportion as the less segment of Dodecaedrons' basis perpendiculare (divided in extreme and mean proportion) doth to the less semidiameter of the same basis.   

The 21 Theorem.  
Icosaedrons' side holdeth the same proportion to Octaedrons' dimetiente, that the semidiameter of Dodecaedrons' contained circle doth to the perpendiculare of his base.   

The 22 Theorem.  
If any line be divided in extreme and mean proportion, and from the whole ye abate the medietie of the greater part, the remain being by extreme and mean proportion again divided, the lesser segment of this last divided line, to the medietie first abated, retaineth such proportion, as Dodecaedrons' side, to the side of his inscribed Icosaedron.   

The 23 Theorem.  
If Icosaedrons' side doubled be divided in extreme and mean Proportion, and to the greater part the Icosaedrons' side adjoined, the whole line thereof resulting, is the diameter of the inscribed Octaedron.   

The 24 theorem.  
The contained Icosaedrons' semidiameter to the less Semidimetiente of the comprehending Dodecaedrons' base, retaineth such proportion as Dodecaedrons' basis line diagonal to the side.   

The 25 theorem.  
Icosaedrons' inscribed diameter to the dimetiente of Dodecaedrons' containing circle, beareth such proportion, as the medietie of Dodecaedrons' basis diagonal, to the greater portion of his side by extreme and mean Proportion divided.      

The .23. Problem. The side, Diameter or Axis of any regulare body known, to search out all those forenamed lines in any regulare body that shall include or circumscribe that proponed solid.  
ALthough this question of circumscribing bodies might be divided into 5 Chapters, and in every one as many different rules, precepts, and Theorems taught, as was in the former of bodies inscribed, yet for brevity sake, I think best to remit the more ample handling hereof to the ingenious studente, who comparing the rules, and well weighing the Theorems already given, shall easily apply them to this purpose, and invent many m●e perhaps of greater facility and no less certainty: and in this Chapter I will only open one way, leaving a large field for others to invent and exercise themselves in at pleasure. It shall therefore be requisite when the side, Diameter or Axis of any regulare body is proponed, to consider by the fift Problems last past what bodies may contain or circumscribe the same, and resorting to their peculiar chapters, search out the numbers appropriate to the containing and contained solides sides, axes and diameters, which found by three quantities known, using the rule of proportion, ye may readily find the fourth, as by the example shall more plainly appear.  

Example.  
Suppose the side of a tetraedron given 10, for the sides, diameters and axes I repair first to the 19 problem, where I found the containing cubes side, being 1, the contained tetraedrons side √² 2, saying therefore by the rule of proportion √² 2, the contained tetraedrons side found heretofore in the 19 problem, giveth 1 for his containing cubes side, what yieldeth 10 the side given your fourth proportional number will be √² 50, the containing cubes side. Likewise in the 22 problem, I found the containing dodecaedrons side being 1, the contained tetraedrons side √² v. 3+ √² 5. augmenting therefore 10, the side given by tetraedrons side in that problem found, and dividing by √² uni. 3+ √ 5, your quotient will be √² 62 ½— √² 12 ½, so much conclude the side of a dodecaedron that shall contain or comprehend this tetraedron, whose side is 10. In like manner may ye search out the other sides, diameters, and axes, of all the comprehending bodies, whereof I leave to give any farther examples, these two being sufficient to the ingenious to proceed with like order in the rest.  
But for such as not contented with one kind of working will delight themselves in the diversity of rules and kinds of calculation, I have thought good to adjoin these Theorems ensuing, which well weighed and compared with such as are already past, shall yield matter abundantly for the invention of many more conclusions and strange operations, than hitherto hath been used or published by any.  

Theorems of these bodies mutually circumscribed and conferred with their inscribed regular bodies. 1.  
TEtraedron may be contained or circumscribed of all the other four regular bodies, and his side being rational, his containing Octaedrons' side is also rational, proportioned thereunto, as 3 to 2.   

The 2 theorem.  
Tetraedrons' comprehending cubes side is equal to the dimetient of his inscribed Octaedron.   

The 3 theorem.  
Tetraedrons' containing Octaedrons' side to the side of his inscribed Octaedron is triple.   

The 4 Theorem.  
Tetradrons side being rational, his encompassing Icosaedrons' side is an Apotome, and triple to tetraedrons inscribed Icosaedrons' side.   

The 5 theorem.  
Tetraedrons' side being rational, his circumscribing dodecaedrons side is an Apotome of the 6 order, proportioned to the side as √² uni. ¾— √² 5/16 unto an unity.   

The 6 Theorem.  
Hexaedron hath only 3 circumscribing regular bodies, for no tetraedron may be so placed about a cube but that his superficies shall either cut or not touch some of the Hexaedrons' angles.   

The 7 Theorem.  
Hexaedrons' containing Octaedrons' side is triple to his contained Octaedrons' side, and to the Hexaedrons' side it beareth such proportion, as tetraedrons diameter to his containing circles semidimetient.   

The 8 theorem.  
Hexaedrons' side being rational, his comprehending Icosaedrons' side, is an irrational Minor, proportioned to the cubes side, as √² v. 31 ½ √² 911 ¼ unto 1.   

The 9 theorem.  
Hexaedrons' side being rational, his comprehending Dodecaedrons' side is an Apotome of the sixth he order proportioned to the side as √² v. 1 ½— √² ¼ to an unity.   

The 10 Theorem.  
Hexaedrons' Dimetient is mean proportional between his side and the Dimetiente of his containing Octaedron.   

The 11 theorem.  
Octaedron may be comprehended of all the other regular bodies, and his side is the medietie of his encompassing tetraedrons side.   

The 12 Theorem.  
Octaedrons' external cubes side is equal to his diameter, and double in power to his side.   

The 13 Theorem.  
Octaedrons' side being rational, his including Icosaedrons' side is an Apotome of the 6 order, having proportion to Octaedrons' side, as √² univer 3— √ 5 unto 1.   

The 14 theorem.  
Octaedrons' side being rational, his environing dodecaedrons side is an Apotome, proportioned to the Octaedrons' side, as √² uni. 7— √² 45 unto 1.   

The 15 theorem.  
Octaedrons' comprehending Dodecaedrons' side is equal to his contained Icosaedrons' side.   

The 16 Theorem.  
Icosaedron may be comprehended perfectly of all the other regulare solides, and his containing tetraedrons side is triple to the side of his contained tetraedron.   

The 17 theorem.  
Icosaedrons' side being rational, his comprehending cubes side is a Binomy, and the Icosaedrons' Diameter in power equal to them both.   

The 18 theorem.  
Icosaedrons' side rational his encompassing Octaedrons' side is a Binomye and equal to the medietie of his external Tetraedrons' side.   

The 19 theorem.  
Icosaedrons' side rational, his comprehending dodecaedrons side is an irrational Apotome, bearing proportion to the side as √² v 17 ½— √² 281 ¼ unto 1.   

The 20 theorem.  
Icosaedrons' comprehending cubes side is double in power to his comprehended Octaedrons' side.   

The 21 theorem.  
Dodecaedron can perfectly be comprehended of no regulare body save only of Icosaedron, whose side retaineth such proportion to the side of Dodecaedron, as √² uni. 13 ½— √ 101 ¼ unto an unity.   

The 22 theorem.  
Dodecaedron and Icosaedron having equal sides, the cubes side that containeth Icosaedron is equal to the cubes side contained of dodecaedron.   

The 23 theorem.  
Dodecaedron and octaedron having equal and rational sides, the side of octadrons comprehending Icosaedron is an Apotome of the same order, that Dodecaedrons' inscribed Tetraedrons' side is a Binomye, and their names or componing quantities equal.   

The 24 theorem.  
Dodecaedron and the cube having equal sides, whether they be rational or furred, the dodecaedrons containing Icosaedrons' side is triple to the cubes comprehending dodecaedrons side.   

The 25 theorem.  
Dodecaedron and Tetraedron having equal Sides, whether they be rational or irrational, the dodecaedrons containing Icosaedrons' side to the Tetraedrons' comprehending Dodecaedrons' side, retaineth such proportion, as Tetraedrons' diameter doth to the less semidiameter of his basis.     

The .24. Problem. The side, diameter, Axis, or altitude, of any regular body, or any semidiameter, perpendicular or line diagonal of their base given, to search out the content Superficial and solid, not only of that body, but also of any other regular solid that shall inscribe or contain that body or any of his spheres.  
FOr resolution of this Problem I might prescribe rules innumerable of most strange and intricate operation, such is the variety of proportions between these body's sides, dimetientes, axes, and other their superficial and solid lines, that if I should set forth for every peculiar question that might herein be proponed, but the tenth part of such precepts as offer themselves upon consideration of these solides forms, proportions & nature, this only chapter should grow to an huge treatise. But to conclude it in shortest form and fewest words, the whole variety may be reduced to these principal points, what manner line is proponed or given, and what content or capacity is demanded. If the line given be none of the bodies sides, but some diameter, axis, perpendicular or other line before named, ye have in the 6 problem and those other four that immediately ensue, the proportion of all such lines to the solides sides in rational and irrational numbers expressed, so that by conversion of the proportions there found, using the rule of proportion (as hath been partly before declared, and shall in this Chapter by example be somewhat more plainly showed) ye may find out the correspondent side of that solid, whose diameter, axis or other line is given. Then if it be the content superficial or solid of the same body that is required, ye shall resort to the 10 problem, and the other four following, where ye shall find in the correspondente Chapter by the side tofore known, the means how to search his forenamed contents.  
But if those contents be not demanded of the same body, they are either of bodies inscribed or circumscribed, and that either of the proponed regulare solid, or some of his spheres, for bodies described within his spheres, it behoveth ye first by the side known to learn the Diameter of that Sphere, and then repair unto the sixteenth Problem, where ye shall be taught to find all the inscribed Solides sides, but if it be an inscribed body of your proponed solid, then have ye precepts for the invention of his side in the eighteenth Problem, or some other of the four immediately ensuing. Likewise if it be the content of a circumscribing body that is required, that body doth either circumscribe some Sphere, and then have ye Precepts in the seventeenth Problem for the invention of his side, or else it containeth the solid, whose side is known, and then may ye by the rules given in the last Problem search out his side, and the side of every solid so found (resorting to the sixth, seventh, eight, ninth and tenth Problems) ye shall there in his answerable Chapter, receive rules for the invention of his capacity superficial and solid, or else work for every several body according to the rules ensuing.  

Rules for the content Superficial and solid of the five Regular bodies.  

For Tetraedron.  
AVgment the square of the side found by √² 3, and the Cube of the side by √² 1/●2, the first producte is the Superficies, the second is the solid content of Tetraedron.   

For Octaedron.  
MVltiplie the sides square by √² 12, and the sides cube by √² 2/9, so have ye two productes equal to the two contents solid and Superficial of Octaedron.   

For Hexaedron.  
THe square of the side augmented by 6, giveth his Superficies, which augmented by ⅙ of the side, produceth the crassitude.   

For Icosaedron.  
INcrease the sides squared square by 75, the root Zenzike of the producte is the Superficies, and the Cube of the side augmented by √² 7.2 31/●2+ √² 5 245/576 yieldeth the crassitude.   

For Dodecaedron.  
MVltiplie the sides square by √² v. 225+ √ 40500, so have ye the superficies, or augment the sides Cube by √² v. 29 ⅜ + √² 861 21/64, the product is the crassitude.  
For more plainness I shall adjoin v. questions and examples of these principal varieties, whereunto all other peculiar questions may be referred.  

The first question. I have a Dodecaedron, whose comprehending spheres diameter I know to be five, I demand his capacity superficial and solid.  
COnsidering the Dimetient of this solid is known unto me, I repair unto the tenth Problem, where I find the Dodecaedrons' side being 10, the comprehending spheres diameter √² v. 450+ √² 112500, converting therefore the limits or bounds of this proportion, I say thus √² v. 450+ √² 112500 giveth 10, what yieldeth 5, working by the rule of proportion, ye shall find the fourth proportional number √² 10 5/12— √² 2 1/12, so much is the side of the proponed Dodecaedron, which known I resort to the fifteenth Problem, there am I willed to search out the Semidiameter and line diagonal of his basis, and the Axis or Semidimetient of his inscribed sphere, his Axis is √² v. 2 1/12+ √² 3 17/36 his containing circles semidimetient √² v. 4 ⅙— √² 3 17/36, his basis line diagonal is √² 8 ⅓, so that by multiplication and division of these numbers according to the Precepts there given, I find the Dodecaedrons' superficies √² v. 7812 ½— √ 1220703● ¼ and his crassitude √² un 1808 97/216+ √² 654097 28403/46656— √² 523278 1033/11664. In like manner if ye augment 12 ½— √² 86 29/3● the square of the Dodecaedrons' found side, in 225+ √² 40500. Or √² v. 3038 7/36+ √ 4709502 1033/1296 √² 8477105 5/144— √² 5128125 2125/11664 being the Cube of the same side, in √² v. 29 ⅜+ √² v 861 1/64, for so are ye taught to do by the rule of this Chapter, the first Producte will be the Dodecaedrons' superficies, and the last his crassitude, exactly agreeing with your former operations.   

The second question. A cube is proponed, whose Diameter is the Zenzike root of 108▪ I would know the superficial and solid contents of such a Tetraedron as this Cubes contained sphere should circumscribe.  
Because the diameter of this Cube is given, I resort to the sixteenth Problem, by the rules there prescribed, I find the Axis 3, which doubled maketh 6, the Diameter of this cubes contained sphere, and considering the Tetraedron, whose capacity is required, must be circumscribed of this sphere, I search out again by the same chapter this spheres inscribed Tetraedrons' side, finding it the zenzike root of 24 ye may also thereby learn the Axis and containing circles semidiameter, and so consequently the capacities of this Tetraedron, whereof ye have examples in the eleventh Problem: Or by the first rule of this Problem if ye augment 24 the square of Tetraedrons' side by √² 3 there ariseth √² 1728 for his superficies, and by multiplication of √² 13824 the Cube of this inscribed Tetraedrons' side by √² 12/72, there amounteth √² 192, so much conclude the Tetraedrons' crassitude,   

The third question. I demand the superficial and solid capacity of a Dodecaedron, circumscribing such an Octaedrons' containing sphere, as hath for his side this irrational Mayor √² v. 20+ √² 387 ⅕.  
BY the eight Problem I find this Octaedrons' comprehending spheres dimetient √² v. 46+ √² 7744/5 the medietie hereof being √² uniu. 10+ √² 484/5 is the Axis of the Dodecaedron, whose contents are required, which known by the rules of the seventeenth Problem, I find the side √² v. 500— √² 234256, and by extraction of this irrational numbers root, it appeareth 4, whose square augmented by the root Ȝenzike universal of 2●5+ √² 40500 produceth √² v. 57600+ √² 2654208000, so much is the Dodecaedrons' superficies. Likewise by multiplication of √² uniu. 275/8 + √² ●●125/64 in the Cube of 4 there ariseth √² v. 962560/8+ √² 92484403200●/64 for the crassitude of this Dodecaedron.   

The fourth question. There is a Dodecaedron, whose side is this irrational Apotome √² v. 1120— √² 1152000, my desire is to know his inscribed Icosaedrons' superficies and crassitude.  
FOrasmuch as this Icosaedron whose contents are required, is described within the Dodecaedrens, whose side is known, I search out by the rules given in the xxij. Chapter, the side of his contained Icosaedron. Or with more speed ye may thus work, saying a containing Dodecaedrons' side being an unity, giveth for the side of his contained Icosaedron √² v 7/10+ √² 9/20, what shall √² universalis 11200— √² 1152000, by multiplication of these latter numbers ye shall produce √² v. 784— √² 518400, which ye shall find (by extraction of the universal ●enzike root) 8, the side thus found, ye may search out his contents by the fifteenth Problem, or by the fourth rule tofore prescribed thus, I augment 64 by √² 75 there amounteth √² 307200 the inscribed Icosaedrons' superficies. Or increasing 512 the side cubically multiplied by √² v. 2 21/72+ √² 5 245/576 there amounteth √² v. 637155 5/9 + √² 372827022222 2/9 the desired cassitude.   

The fifth question. An Icosaedron is offered, whose basis containing circles semidimetient is this surde Binomye √² uni. 14 14/15 + √² 204 ⅘ the superficies and crassitude of his containing Dodecaedron is required.  
COnsidering the semidiameter of this Icosaedrons' basis is given, I repair to the 9 Problem, wherein I receive rules for the invention of the comprehending Dodecaedrons' side, or more particularly to set forth the whole operation, I am referred to the 9 Problem, where I perceive by the 9 Theorem, that the square of this known semidiamiter tripled, maketh the square of the Icosaedrons' side: I augment therefore the given Binomyes' square by 3, there ariseth 44 ⅘ + √² 36864/20, the root zenzike universal thereof is the Icosaedrons' side. And for his containing Dodecaedrons' side I work by the precepts of the last Chapter, saying: √² v. 2●0+ √² 72000 giveth 20, what yieldeth √² uni. 44 ⅕+ √² 36864/20 your fourth Proportional number will be 8. Or thus by the fifteenth Theorem of the 22 Problem: Divide √² uni. 44 ⅘+ √² 36●6●/20 by √² 9/20+ √² 5/●●●, so will the quotient be 8 also, and that is the comprehending Dodecaedrons' side. Now for his contents solid and superficial, I repair to the fifteenth Problem, where I am referred to the 10 Problem, and working by the precepts there given, I find the Dodecaedrons' basis line diagonal √² v. 96+ √² 5120, his basis greater Semidiameter √² uni. 32+ √² 4006/20▪ his Axis √² uni 40+ √² 495616/320, triple the Semidiameter is √² uni. 288+ √² ●2●776/20, the line diagonal increased by 5, maketh √² uni. 2400+ √² 3200000, these multiplied together, created this furred rote √² v. 691200+ √² 53084160000+ √² 265420800000+ √² 95552488000, resting between the Ȝenzike roots of 1745904, and 1745905, being very nigh 1321 ●63/2642 so much is the desired content Superficial. Again by multiplication of √² uni. 96+ √² 5120 the line diagonal, in √² uni. 3●+ √² 4●●6/20, there amounteth √² v. 4096+ √² 5242880+ √² 1887436 ⅘, which increased by √² uni. 40+ √² 495616/320, the Axis produceth √² univers 163840+ √² ●0305289240576/320+ √² 66520453480448/3200, and this multiplied by 5, createth √² uni 7700480+ √² 59190018048000 the contained Dodecaedrons' content solid. The like is brought to pass by the last rules, for 64 the Dodecaedrons' sides square augmented by √² uni ●25+ √² 40500, maketh √² v. 92160●+ √² 679477248000 the superficies, and 512 the Cube of 8 multiplied in √² uni. 29 ●/8+ √² ●●125/64, createth √² v. 7700480+ √ 59190018048000 the solid capacity of a Dodecaedron described within the Icosaedron, whose side was given exactly agreeing with the former calculations.  
Theorems of the superficial and solid quantities of these regulare bodies conferred by mutual inscription and circumscription, and first:  

Of Tetraedron.  

TEtraedrons superficies is double to the superficies of his inscribed Octaedron.   

The 2 theorem.  
Tetraedrons' superficies to the superficies of his containing Cube beareth proportion, as 110 √² 3.   

The 3 theorem.  
Tetraedrons' circumscribing Cubes content superficial, to the content superficial of his contained Octaedron, retaineth such proportion as the same Octaedrons' superficies doth to the square of his side.   

The 4 theorem.  
Tetraedrons' encompassing Octaedrons' superficies containeth the superficies of his inscribed Octaedrons' 9 times.   

The 5 theorem.  
One base of Tetraedrons' containing Icosaedron, is equal unto 9 of his inscribed Icosaedrons' triangular bases.   

The 6 theorem.  
Tetraedron to his contained Octaedron retaineth such proportion as their sides.   

The 7 theorem.  
Tetraedrons' containing Octaedron to the Cube described without the same Tetraedron, retaineth such proportion as the perpendiculare of the containing Octaedrons' basis to the less semidiameter of the Tetraedrons' base.   

The 8 theorem.  
Tetraedrons' circumscribing Octaedron containeth his inscribed Octaedron 27 times.   

The 9 theorem.  
Tetraedrons' containing Cube holdeth the same proportion to Tetraedrons' inscribed Octaedron, that the Cubes diameter doth to the Octaedrons' Axis.   

The 10 theorem.  
Tetraedrons' comprehending Icosaedron, containeth his inscribed Icosaedron 27 times.    

Of Hexaedron.  

The 11 theorem.  
Hexaedrons' superficies to the superficies of his inscribed Tetraedron, holdeth such proportion as Tetraedrons' dimetiente doth to the Cubes side.   

The 12 theorem▪  
Hexaedrons' superficial content to the con●ente superficial of his inscribed Octaedron, retaineth the proportion of the Cubes diameter to Octaedrons' Semidimetiente.   

The 13 theorem.  
Hexaedrons' containing Octaedrons' content superficial hath such Proportion to the superficial quantity of his inscribed Tetraedron, as the Octaedrons' contained spheres dimetientes square to the greater semidiameters square of Tetraedrons' base.   

The 14 theorem.  
Hexaedrons' comprehending Octaedrons' superficies containeth his inscribed Octaedrons' superficies ● times.   

The 15 theorem.  
The proportion of Hexaedrons' comprehending Dodecaedrons' superficies▪ to the superficies of his inscribed Icosaedron, is compounded of double, the proportion between a Cord trigonal and a cord Pentagonal of one circled, adjoined to the proportion of the Pentagonal, diagonal to the same Cord trigonal.   

The 16 theorem.  
Hexaedrons' solid capacity to his inscribed Tetraedron, beareth such proportion as the Hexaedrons' comprehending spheres semidimetiente doth unto the Tetraedrons' Axis.   

The 17 theorem.  
Hexaedron to his contained Octaedron, retaineth the Proportion of his Diameter to the Octaedrons' axis.   

The 18 theorem.  
Hexaedrons' comprehending Octaedron to his contained Tetraedron, beareth the proportion of 27 unto 2.   

The 19 theorem.  
Hexaedrons' external Octaedron containeth his internal Octaedron 27 times.   

The 20 theorem.  
Hexaedrons' comprehending Dodecaedrons' solid capacity to the content solid of his inscribed Icosaedron, retaineth the same Proportion that the line diagonal of the Dodecaedrons' Basis doth unto that part of his side whereunto the whole side is proportioned as the square of a Cord trigonal to the square of a Cord Pentagonal.    

Of Octaedron.  

The 21 theorem.  
Octaedrons' base doubled, is equal unto 9 of his contained Tetraedrons' triangular bases.   

The 22 theorem.  
Octaedrons' superficies to the content superficial of his inscribed Cube, beareth such proportion as the Octaedrons' axis to the Cubes sides third part.   

The 23 theorem.  
Octaedrons' circumscribed tetraedrons superficies containeth the superficies of his inscribed tetraedron 9 times.   

The 24 theorem.  
Octaedrons' comprehending Cubes Diagonal lines square, is triple to the superficies of his comprehended Cube.   

The 25 theorem.  
Octaedrons' external tetraedrons superficies to the superficial quantity of his internal Cube, is proportioned, as √² 27 unto an unity.   

The 26 theorem.  
Octaedron is proportioned to his inscribed tetraedron, as 13 ½ unto 1.   

The 27 theorem.  
Octaedron to his inscribed Cube retaineth double the Proportion of their sides.   

The 28 theorem.  
Octaedrons' external tetraedron to his internal tetraedron, retaineth triple the proportion of their sides that is 27 unto 1   

The 29 theorem.  
Octaedrons' circumscribing Cube to his inscribed Cube, retaineth double the proportion of the comprehending Cubes Diameter to the internal Cubes side.   

The 30 theorem.  
Octaedrons' comprehending Dodecaedron to his included Icosaedron, beareth such Proportion as the square of a cord Trigonal doth to the greater part of a Pentagonal cords square (of one Circle) divided by extreme and mean Proportion.    

Of Icosaedron.  

The 31 Theorem.  
Icosaedrons' comprehending Tetraedrons' superficies containeth the superficies of his internal Tetraedron 9 times.   

The 32 Theorem.  
Icosaedrons' containing Hexaedrons' superficies to the superficies of his contained Octaedron is proportioned as the Cubes diameter to his axis.   

The .33. Theorem.  
Icosaedrons' circumscribing Octaedrons' superficies to the superficies of his internal Cube, beareth such proportion as 9 unto √² 12.   

The .34. Theorem.  
Icosaedrons' comprehending Tetraedrons' superficies is double to the superficies of his comprehending Octaedron.   

The .35. Theorem.  
Icosaedrons' comprehending Octaedrons' superficies hath the same proportion to his inscribed Tetraedrons' superficies that Octaedrons' lesser spheres dimetientes square hath to the square of the greater semidiameter of Tetraedrons' base.   

The .36. Theorem.  
Icosaedrons' comprehending Tetraedron containeth his enclosed Tetraedron 27 times.   

The 37 Theorem.  
Icosaedrons' contained tetraedron is a third part of his contained Cube.   

The .38. Theorem.  
Icosaedrons' encompassing tetraedron to hi● enclosed Cube, foldeth the proportion of their diameters squares.   

The 39 Theorem.  
Icosaedrons' external Hexaedron containeth his internal Octaedrons' 6 times.   

The 40 Theorem.  
The proportion of Icosaedrons' solid capacity to his comprehended Dodecaedron, is equal to the proportion of the Icosaedrons' semidiameter to the Dodecaedrons' axis, and the difference (of a pentagonal diagonals proportion to his Cord trigonal of one self same circle deducted from the proportion of these solides diameters) joined together.    

Of Dodecaedron.  

The .41. Theorem.  
THe proportion of Dodecaedrons' superficies to his inscribed Icosaedrons' superficies, is compounded of the proportion of Dodecaedrons' basis pentagonal diagonal, to the Cord trigonal of his containing circle, and the proportion of dodecaedrons greater spheres semidiameter to his inscribed Icosaedroms axis.   

The .42. Theorem.  
If from the proportion of Dodecaedrons' external Icosaedrons' greater spheres semidiameter to Dodecaedrons' axis, ye deduct the proportion between the Dodecaedrons' pentagonal basis diagonal and his containing circles cord trigonal, there remaineth the proportion of their contents superficial.   

The .43. Theorem.  
The superficies of Dodecaedrons' containing Icosaedron to the superficies of his contained Icosaedron, retaineth quadruple the proportion of Dodecaedrons' comprehending spheres semidiameter to his axis.   

The .44. Theorem.  
Dodecaedrons' contained Cubes superficies and side, and his contained Tetraedrons' superficies and diameter, are reciprocally proportional, that is to say, the cubes superficies retaineth the same proportion to the Tetraedrons' superficies that the Tetraedrons' diameter doth to the Cubes side.   

The .45. Theorem.  
Dodecaedrons' contained Icosaedron and Octaedron have equal superficial quantities.   

The 46 Theorem.  
The proportion of Dodecaedron to his inscribed Icosaedron is equal to the proportion of Dodecaedrons' greater spheres semidiameter to Icosaedrons' axis, and the proportion of Dodecaedrons' basis diagonal line to the inscribed Icosaedrons' side.   

The 47 Theorem.  
The proportion between dodecaedron and his comprehending Icosaedron, is componed of the proportion between Icosaedrons' containing spheres dimetient and dodecaedrons internal spheres diameter, added to the proportion between Icosaedrons' side, and dodecaedrons basis line diagonal.   

The 48 Theorem.  
Dodecaedrons' inscribed Icosaedron to his inscribed Octaedron, retaineth suc●e proportion, as Icosaedrons' axis doubled to the diameter of Octaedrons' contained sphere.   

The 49 Theorem.  
Dodecaedrons' comprehended Tetraedron to his internal Hexaedron, beareth the proportion of tetraedrons contained spheres dimetient to the Cubes comprehending spheres diameter.   

The 50 Theorem.  
Dodecaedrons' comprehending Icosaedron to his contained Icosaedron holdeth sixefolde the proportion of his containing and contained spheres diameters.        

The .25. Problem. A Metamorphosis or transformation of the five regulare bodies.  
HItherto have I only entreated of the five regulare bodies, Theorically and practically opening sundry means to search out the proportion and quantities of their sides, diameters, axes, perpendiculares, altitudes, and contents both superficial and solid, and that not only in these solides considered by themselves, but also conferred with other, aswell by inscription as circumscription, both of themselves and their spheres: so as I suppose hardly any question may be proponed concerning these bodies, which by the form●● problems or theorems may not be resolved. Yet before I finish this Treatise I thought good to adjoin one Chapter of the transformation of these Solides into such bodies (as though they may not be termed regulare, for that Euclid hath demonstrate only five, and no more possibly to be found or imagined) yet have they such uniform composition and convenience with th●se solides, that they are not only environed with equilater and equiangle superficies as they be, but also have all their sides equal, and one comprehending sphere exactly and at once touching all their solid angles, they are also capable of the regulare bodies, and only herein different, that whereas the regulare solides be environed with one kind of plainness and receive one internal sphere, and so consequently one a●is. Thei● Transformed bodies are encompassed with several kind● of plainness, and have several axes and contained spheres, of which their diversity ariseth manifold more strange, rare, and different kinds of proportions, than may in a very large volume (I will not say be demonstrate, but only by Theorems) be declared. Transformed regulare solydes I call them, both to avoid the forging of new names, which I should be enforced to use for distinction sake, and also because they seem to be created by the uniform section of the regulare bodies, and may by addition of certain proportional equal Pyramids, be reduced to the regulare solides, and are in sundry proportions and proprieties so agreeable and resemblante to those regulare solides, whose names they bear, that they seem only to loose the form, and yet still to retain the nature of them. I mean not here in so ample manner as the novelty of the matter requireth to entreat of them, but only by definitions and Theorems open so much as may be sufficient to explain the composition, form, nature, and proportion of these, and also give light to the ingenious infinitely to proceed for invention of the like, whose use and appliance may be manifold to conclusions no less strange than necessary: but thereof in due place for the nature of these transformed solides peruse their several definitions and Theorems immediately ensuing.  

Of Tetraedron transformed. The first Definition.  
TEtraedron transformed is a solid encompassed with four equal equilater triangles, and four equal equiangle plains hexagonal, having equal sides with the triangles.  
 

The first Theorem.  
This solid hath 18 equal sides, 36 plain angles, and 12 solid angles, and may be circumscribed of a sphere as exactly as any of the forenamed regulare bodies: the right line drawn from this spheres centre to one of the solid angles I call the transformed tetraedrons semidiameter.   

The 2 theorem.  
This solid also receiveth two intrinsical spheres, the lesser touching all the centres of his hexagonal plains, the greater touching the centres of the trigonal plains, and making circular sections concentrical with the hexagonal bases, the semidiameters of these spheres I call the solides Axes.   

The 3 theorem.  
Within this solid may Octaedron be described, and his six angles shall rest in the medieties of those his sides that are peculiar to the hexagonal plains and not common both to the hexagonal and trigonal bases.   

The 4 theorem.  
Within this solid may two Tetraedrons' of different quantity be described, the greater having his solid angles resting in the centres of the trigonal plains, and the lesser hath his angles situate in the hexagonal bases centres.   

The 5 theorem.  
Icosaedron may within this transfigured Tetraedron be described, having his solid angles placed in the hexagonal plains, and four of his trigonal bases concentrical, with those hexagonal bases wherein they are situate.   

The 6 theorem.  
This solides semidiameter is equal in power to the semidiameter of his triangular plains containing circle, and his greater Axis.   

The 7 theorem.  
The square of this solides side deducted from the square of his semidiameter, leaveth the square of his less axis.   

The 8 theorem.  
The squares of the axes deducted one from an other, the Zenzike root of the remain is the semidiameter of the hexagonal basis concentrical circle tofore mentioned in the second theorem.   

The 9 theorem.  
The altitude of this transformed solid is equal to his Axes both joined together, and beareth proportion to his side, as four unto the root Zenzike of six.   

The 10 theorem.  
This solid semidiameter beareth to his side the proportion of √² 11 to √² 8.   

The 11 theorem  
The greater axis to the lesser is proportioned as 5 unto 3.   

The 12 theorem.  
The medietie of this solides altitude is equal to the semidiameter of the concentrical circle made in the hexagonal bases by the periphery of the solidus greater internal sphere.   

The 13 theorem.  
The less semidiameters of this solides hexagonoll and trigonal bases are proportioned one to an other, as the squares of their greater semidiameters.   

The 14 theorem.  
The greater diameter of this solides hexagonal basis, to the greater dimetient of his trigonal base, is in power triple.   

The 15 theorem.  
The side of this solid is equal to the side of his lesser inscribed tetraedron, and to his greater internal tetraedrons side it retaineth such proportion as 3 to 5.   

The 16 theorem.  
This solides inscribed Octaedrons' side is triple to the medietie of his containing transfigured Tetraedrons' side.   

The 17 theorem.  
The side of this solides inscribed Octaedron (being by extreme and mean proportion divided) the lesser segments square is half the square of the inscribed Icosaedrons' side.   

The 18 theorem.  
The superficies of this transformed Tetraedron is equal to both the superficies of an Octaedron and Icosaedron, having equal sides with this solid joined together.   

The 19 Theorem.  
This solides crassitude is equal unto two Pyramids, whereof the one hath for his base a triangle whose side is double to the side of this solid, and his axis equal to the greater axis of this body, the other is an hexagonal Pyramid, whose base is an equiangle hexagonum, having his side double to the side of this solid, and his altitude equal to this transformed tetraedrons lesser axis.   

The 20 Theorem.  
This transfigured solid may be resolved into 23 equal tetraedrons, every of them of equal sides unto this figure, and all joined together make his exact crassitude, which is proportioned to the Cube of his side, as 23 to the quadrate root of 72.    

Of the transfigured Cube. The second Definition.  
A transformed Cube is a figure geometrical environed with 6 equiangle Octogonall and 8 equilater triangular plains or bases, whose sides are all equal.  
 

The 21 Theorem.  
This figure hath 36 equal sides, 72 plain angles, and 24 solid angles, it may be encompassed of a sphere, exactly touching with his concave periphery every of their solid angles, and the semidimetiente of that comprehending sphere is called also this solides semidimetiente.   

The 22 theorem.  
This transfygured body is also capable of two internal spheres, the less touching only the centres of the Octogonall plains, the greater both touching exactly with his convex circumference the triangular bases centres, and also cutting the plains Octagonall describing therein concentrical circular sections, the semidimetientes of these spheres are the axis of the solid.   

The 23 theorem.  
Within this solid may a Tetraedron be described, whose angles shall rest in the centres of the trigonal bases.   

The 24 theorem.  
Octaedron may also herein be placed, his angles resting in the centres of the Octogonall plains.   

The 25 theorem.  
Hexaedron may likewise be inscribed by drawing straight lines conjoining all the trigonal bases centres.   

The 26 theorem.  
This transformed figure receiveth also Icosaedron, whose 12 solid angles are placed in his 6 Octagonall bases, every of them receiving two of his angles.   

The 27 theorome.  
Dodecaedron may not exactly be within this figure described, for 12 of his angles being situate in the 6 Octagonall bases, dividing their less semidiameters by extreme and mean proportion, the other 8 angles will not touch his superficies, but rest precisely in the greater Axes of this solid, directly under the centres of the 8 trigonal plains, uniformly and proportionally cutting every of these forenamed greater Axes.   

The 28 theorem.  
The side of this transfigured body is triple in power to the greater semidiameter of his triangular base, and to the less semidiametiente of that base it is proportioned as 1 unto √² 1/12.   

The 29 theorem.  
The side of this transformed body being rational, the greater semidiameter of his Octogonall bases is an irrational Mayor, bearing proportion to the solides side, as √² uni. 1+ √² ½ unto an unity.   

The 30 theorem.  
This solides side being rational, his Octogonall bases less semidimetiente is an irrational Binomie, retaining such proportion to the side, as √² v. ¼+ √² ½   

The 31 theorem.  
This solid side being rational, his greater axis is an irrational Binomie proportioned to the side, as √² ⅔+ √² ¾ unto an unity.   

The 32 theorem.  
The less Axis of this Figure is equal to the loss semidimetiente of his Octogonall basis, and retaineth unto his side the proportion of √² ½+●/2 to an unity.   

The 33 theorem.  
This transformed Figures side being rational, his comprehending spheres dimetiente is an irrational Mayor, proportioned to the side, as √² v. 7+ √² 32 to 1.   

The 34 theorem.  
This solides contained Tetraedrons' side is proportioned to the greater Axis, as √² 8/3 to an unity.   

The 35 theorem.  
The side of the inscribed tetraedron is double in power to the side of the inscribed Cube.   

The 36 theorem.  
This transfigured Cubes less altitude is double in power to the side of his inscribed Octaedron.   

The 37 theorem.  
The Octogonall bases less Diameter divided by extreme and mean Proportion, maketh his greater segment, the inscribed Icosaedrons' side.   

The 38 theorem.  
The side of this transformed solid, being by extreme and mean Proportion divided, if to the less portion ye adjoin a line in power double, that whole line thereof resulting, is the side of the tofore mentioned inscribed Dodecaedron.   

The 39 theorem.  
The Cube of the transfigured Solides Octogonall bases lesser Diameter, hath a superficies exceeding the superficial content of all this Solides Octogonall bases, by a superficies proportionate to the superficial quantity of his trigonal bases, as the Cubes dimetiente to his side.   

The 40 theorem.  
The Cube whose side is equal to the less altitude of the transfigured hexaedron, exceedeth in solid capacity this transformed figure, by the solid content of an Octaedron described upon the same side with this transfygured hexaedron.    

Of Octaedron transformed. The 3 Definition.  
A Transfigured Octaedron is a Geometrical Figure▪ encompassed with 14 bases, whereof 8 are equal equiangle hexagonal plains, and the other 6 are equal squares.  
 

The 41 theorem.  
A transformed Octaedron hath 36 equal sides, 72 plain angles, and 24 solid angles, and may be enclosed with a containing sphere exactly touching all his Angles.   

The 42 theorem.  
This solid also receiveth two internal spheres, the one only touching all the hexagonal bases centres, the other both touching all the centres of the squares▪ and also with his periphery cutting the hexagonal plains, delineating in them concentrical circles, the semidiameters of these spheres are called the Axes.   

The 43 theorem.  
Within this solid may Tetraedron be described, and his angles situate in the centres of the hexagonal bases.   

The 44 theorem.  
This Figure also receiveth the Cube, with his 8 solid angles resident in the ● centres of his hexagonal plains.   

The 45 Theorem.  
The inscribed Octaedrons' 6 solid angles rest in the 6 centres of the Quadrate plains.   

The 46 theorem.  
This solides inscribed Icosaedron hath all his Angles situate in those his 12 sides which are peculiar to the hexagonal bases, not perticipating with the quadrate plains.   

The 47 theorem.  
This solides side is equal to the greater semidiameter of his Hexagonal basis, and double in power to the greater semidimetientes of his quadrate plains.   

The 48 theorem.  
The greater dimetient of the Hexagonal basis, is equal in power to his less diameter and side.   

The 49 theorem.  
The less diameter of the Hexagonal basis, is triple in power to his less diameter of the quadrate basis.   

The 50 theorem.  
The greater Axis of this solid, is double in power to his side.   

The 51 theorem.  
This solides semidimetiente, is equal in power to his less Axis, and side.   

The 52 Theorem.  
The semidimetiente of the concentrical circle, is equal to the greater semidiameter of the quadrate plains, and the square thereof added to the square of the less Axis produceth the square of the greater.   

The 53 Theorem.  
The inscribed tetraedrons side is double to the side of this transfigured solid.   

The 54 theorem.  
The side of the inscribed Cube, is mean proportional between the side of the inscribed tetraedron, and the side of this solid.   

The 55 theorem.  
The diameter of this body is fivefolde in power greater than the side of his inscribed Cube.   

The 56 theorem.  
This solidus inscribed Octaedrons' side, is equal to the side of his inscribed tetraedron, and double in power to the inscribed Cubes side, or the greater Axis, his equal.   

The 57 theorem.  
If the greater diameter of this transformed Figures quadrate bases be divided by extreme and mean proportion, triple the less part is the inscribed Icosaedrons' side.   

The 58 Theorem.  
The side of an Octaedron transformed being rational, the right line that matcheth in power his superficies, is an irrational called of Euclid, Potens rationale & mediale, proportioned to the square of the side, as √² v. √² 432+8 unto an unity.   

The 59 Theorem.  
This transfigured Octaedron may intellectually be divided into 41 Pyramids rising fro the superficies, and concurring with their tops or vertices in the centre of this solid, whereof 8 are hexagonal, and 6 Tetragonall, according to the bas●s from whence they ascend, and the hexagonal Pyramids joined together, are triple to the solid capacity of the tetragonall Pyramids.   

The 60 theorem.  
Octaedron transformed to the Cube of his side retaineth the Proportion, of √² 28 unto an unity, and may be resolved into 24 equal Octaedrons', having every of their sides equal to this transformed Figure.    

Of the transfigured Icosaedron. The fourth Definition.  
ICosaedron transfigured, is a solid body encompassed with 32 Equiangle and equilater bases, whereof 20 are hexagonal, and the other 12 Pentagonal plains.  
 

The 61 theorem.  
This transformed figure hath 90 equal sides, 180 plain angles, and 60 solides, one comprehending sphere environeth all his angles, and the semidiameter of that sphere, is the semidimetient of this solid.   

The 62 theorem.  
This solid also conceiveth two internal spheres, the one only touching all the Hexagonal bases centres, the other both touching all the centres of the Pentagonal bases, and also cutting the plains Hexagonal, deliniating in them concentrical circles, the spheres semidimetients are this solides Axes.   

The 63 theorem.  
Within this body may Tetraedron be described, and his angles situate in 4 of the Hexagonal bases centres.   

The 64 theorem.  
This solides inscribed cube hath his solid angles also placed in the centres of his Hexagonal plains.   

The 65 Theorem.  
Icosaedron transfigured, receiveth an internal Octaedron, whose solid angles rest in the medieties of 6 such sides of his hexagonal bases, as communicate not with his Pentagonal plains, but are situate Parallel, Perpendicular, and opposite one to an other.   

The 66 Theorem.  
Icosaedron may also herein be inscribed, his 12 solid angles resident in the centres of his 12 Pentagonal bases.   

The 67 theorem.  
Dod●caedron may also perfectly be described within this solid, all his solid angles being placed in the centres of all the hexagonal bases.   

The 68 theorem.  
The side of this transformed figure being rational, his semidiameter is an irrational Mayor, proportionate to his side, as √² univers 3 ⅝+ √² 6 2●/6● to an unity.   

The 69 theorem.  
The greater semidimetient of the Hexagonal bases, is equal to the side of this transfigured body, and to the greater semidiameter of the Pentagonal plains, it beareth proportion as 1 unto √² v. ½+ √² 1/20.   

The 70 theorem.  
The less semidimetients of these Pentagonal and Hexagonal bases, are proportionate, as √² v. ⅓+ √² 4/45 unto an unity.   

The 71 Theorem.  
This transfigured solides side being rational, his greater Axis is a Binomye proportioned to the side, as √² v. 3 ⅛+ √² 1681/320 unto an unity.   

The 72 theorem.  
The side of this solid being rational, his lesser axis is also a Binomye, & to the side proportionate as √² v. 21/8+ √² 405/64 unto an unity.   

The 73 theorem.  
The greater axis in power exceedeth the lesser, by the semidimetient of the concentrical circle, which retaineth the same proportion to the side of this transformed solid, that √² v. ½— √² 1/26 doth unto an unity, and it is the greater part of this solides Pentagonal bases semidiameter, divided by extreme and mean proportion.   

The 74 theorem.  
This transformed figures inscribed Tetraedrons' side, is double in power to the side of the inscribed cube, and to the side of this solid, it beareth the proportion of √² v. 7+ √² 45 unto 1.   

The 75 theorem.  
The side of Octaedron described within a transformed Icosaedron, whose side is rational, shall be a Binomye proportional to the rational side, as √² uniu. 6 ¾+ √² 405/16 to 1.   

The 76 theorem.  
This transformed figures side being rational, his inscribed Icosaedrons' side is an irrational Binomye, retaining such proportion to this transformed solides side, as triple his greater altitude to the diameter of his comprehending Icosaedron, that is as √² v. 4 ⅕+ √² 3 ⅕ to an unity.   

The 77 theorem.  
The side of a Dodecaedron described within this transfigured Icosaedron, being divided by extreme and mean proportion, maketh his greater part this transformed solides side, and is proportioned thereunto, as √² v. 3/2+ √² ●/4 unto an unity.   

The 78 theorem.  
The superficies of this transformed Icosaedron, is equal to the superficial capacities of one Dodecaedron, and 6 Icosaedrons', having their sides all equal to the side of this transformed figure.   

The 79 theorem.  
The superficies of a transfigured Icosaedron, is equal to the superficial capacity of a dodecaedron and an Octraedron of equal sides to this solid, and the superficies of such a transformed Tetraedron, as hath a side equal to both the sides of those two bodies joined together.   

The 80 theorem.  
A transfigured Icosaedron may be resolved into 12 Pentagonal and 20 hexagonal Pyramids, concurring with their tops or vertices all in the centres of this transformed body, and their bases the Pentagonal and Hexagonal equiangle superficies tofore mentioned in the definition of this solid. The 12 Pentagonal Pyramids crassitude, to a Dodecaedron created upon this transfigured solides side, is proportioned, as his greater axis, to that line, whose greater segment is this Pentagonal basis less semidimetient. And the crassitude of the other 20 Hexagonal Pyramids, is ⅔ of this transfigured solides containing Icosaedron.    

Of Dodecaedron transformed. The fifth Definition.  
A Transformed Dodecaedrens is a massy or solid figure, comprehended of 12 equiangle decagonal, and 20 equilater triangular bases.  
 

The 81 Theorem.  
A transfigured Dodecaedron hath 90 equal sides, 180 plain Angles, and ●0 solid Angles, comprehended of one sphere, with his concave periphery touching all those angles, whose semidimetient is the semidiameter of this transformed solid.   

The 82 Theorem.  
This transformed figure hath two internal sphere●, the greater with his convex superficies touching all the trigonal bases centres, and cutting the bases or plains Decagonall, describing in them concentrical circles, the less sphere only toucheth all the Decagonall bases centres resting wholly within this solid, and the semidimetientes of these spheres are called the Axes of this transformed figure.   

The 83 theorem.  
This transfigurate body receiveth an internal Tetraedron, whose solid angles rest in the centres of his trigonal bases.   

The 84 theorem.  
Hexaedron may also within this body be described, his angles likewise remaining in the trigonal bases centres.   

The 85 theorem.  
Within a transfigured Dodecaedron may an Octaedron be framed, and his angles will be resident in the medieties of those his sides that be peculiar to the Decagonall, not communicating with the trigonal bases, but opposite, parallel, and Orthogonal, one to an other.   

The 86 theorem.  
Icosaedron within this transfigurate body may be described, and all his 12 angles in the centres of his 12 Decagonal bases situate.   

The 87 theorem.  
Dodecaedron may also within this transfigurate body be described, all his 20 solid angles placed in the 20 trigonal bases centres.   

The 88 theorem.  
This solides semidiameter is equal in power to his less Axis, and the greater semidiameter of his Decagonal bases.   

The 89 theorem.  
The side of this solid being rational, his Decagonall bases less semidiameter is an irrational Mayor, proportioned to the greater semidimetient of the same bases, as √² v. 5/4+ √² 5/4 unto √² v. 1 ½+ √² 1 ¼.   

The 90 theorem.  
The side of this transfigurate body, is triple in power to the greater semidiameter of his trigonal bases, and to the less semidimetient of the same bases, it is proportioned, as the Diameter of a Cube to his Axis.   

The 91 theorem.  
The dimetient of this solid, is equal in power to these three, his side, his less altitude, and the lesser diameter of his decagonall bases.   

The 92 theorem.  
The square of this solides greater axis, added to the third part of his sides square, maketh the square of his semidimetient.   

The 93 theorem.  
The square of the less altitude of this solid, deducted from the square of the greater, leaveth the square decagonall basis concentrical circles diameter.   

The 94 theorem.  
Twenty direct or upright equilater triangular Pyramids may upon these solides 20 triangular bases of s●tch altitude be erected, and to this solid in such sort adjoined, that the wh●●e body thereof resulting, shall be a dodecaedron: whose side containeth in power fiu●fold the side of this transformed figure.   

The 95 theorem.  
Every of these adjoined Pyramids sides, are equal to the greater portion of this transfigured dodecaedrons side, divided by extreme and mean proportion.   

The 96 theorem.  
The square of this transformed figures trigonal bases greater semidimetient, deducted from the square of his sides greater portion divided by extreme and mean proportion, leaveth the square of his adjoined Pyramids altitude.   

The 97 theorem.  
The greater altitude of this solid, is equal in power to the sides of his inscribed cube, and tetraedron, and if the power thereof be divided in two lines retaining the proportion of a line divided by extreme and mean proportion to his less part, the lesser of these lines is this transfigured solides inscribed Dodecaedrons' side.   

The 98 theorem.  
This transformed Dodecaedrons' inscribed Octaedrons' dimetient, is equal in power to the inscribed Icosaedrons' diameter, and the less dimetient of his decagonall bases, and the power of the less altitude of this solid, (being divided in two right lines retaining extreme and mean proportion) the lesser of those lines is the inscribed Icosaedrons' side.   

The 99 theorem.  
The superficies of a transfigured Dodecaedron doth exceed the superficies of an Icosaedron made upon this transformed solides side, by the superficies of ●n equiangle Decagonum, whose side retaineth the same proportion to the side of this solid, that the side doth to his Triangular bases lesser semidiameter, and the superficies of all this solides Decagonal bases, to the superficies of all his Trigonal bases, retaineth the proportion of his decagonal basis lesser diameter, to the semidimetient of his trigonal bases contained circle.   

The 100 Theorem.  
A transformed Dodecaedron may be resolved into 32 Pyramids, rising from his equilater bases, and concurring with their tops or Vertices at his Centre, whereof 12 are Decagonal, and 20 Trigonal, according to the bases from whence they rise, and the proportion of all the decagonal Pyramids crassitude, to the solid content of all the trigonal, is compounded of the proportion between the Decagonal basis less diameter, & the trigonal basis semidiameter, added to the proportion of the less altitude to the greater.  

Theorems of these transformed bodies conserred both with their circumscribing regular bodies, and also between themselves.  

The 1 theorem.  
THe square of a transformed tetraedrons diameter, added to 8 times the square of his side, produceth the square of his comprehending tetraedrons diameter.   

The 2 theorem.  
The superficies of a transfigured Tetraedron, is proportioned to the superficies of his comprehending Tetraedron, as 7 to 9   

The 3 Theorem.  
If the side of a transformed Octaedron, to the side of a transfigured Tetraedron, retain the proportion of √² 7 unto √² 12, then is the superficial capacity of the transformed Octaedrons' Hexagonal plains, equal to the whole superficies of Tetraedron transformate.   

The 4 Theorem.  
The solid capacity of a transformed Tetraedron, to his circumscribing Tetraedron, holdeth the proportion of 23 unto 27.   

The 5 Theorem.  
The solid quantity of a transfigured Tetraedron, to the content solid of a transfigurate Octaedron framed on the same side, is proportioned as 23 to 96.   

The 6 Theorem.  
Hexaedron transformed, may of a cube be circumscribed, and his side deducted from the cubes side, leaveth a line double in power to the transformed Hexaed●ōs side.   

The 7 Theorem.  
If ye adjoin √² ⅙ of a transfigured Hexaedrons' side unto his greater Axis, the product is the semidimetient of his containing cube.   

The 8 Theorem.  
The containing cubes superficies, exceedeth the content superficial of his inscribed transformed Hexaedrons' Octogonal bases, by the square of a line retaining such proportion to the transfigured solides side, as a Cubes diameter, to the greater semidimetient of his quadrate bases.   

The 9 Theorem.  
A transfigured cube, with four Tetraedrons' made on his side, are equal to the crassitude of his comprehending Hexaedron.   

The 10 theorem.  
A transformed Tetraedron, and transfigured Hexaedron of equal sides, have the differenees between their crassitudes and the solid capacities of their comprehending regular bodies equal.   

The 11 theorem.  
A transfigured Octaedron, may within an Octaedron be described, and the dimetient of that containing Octaedron to the side of the transformed body, retaineth the same proportion, that the side of a transfigured cube doth to the default or difference between his greater Axis, and the semidimetient of his comprehending Hexaedron.   

The 12 theorem.  
A transformed Octaedron, and a transformed Tetraedron having equal sides, their comprehending regular solides sides are equal also, and their superficial quantities in double proportion.   

The 13 theorem.  
Octaedrons' superficies, doth surmeunt the content superficial of his inscribed Octaedron transfigured by the difference of an equiangle Hexagonum, (whose side is double to the transformed solides side) from the square of the same transfigured solides less altitude.   

The 14 theorem.  
If two mean proportional lines be found between the sides of Octaedron transformed, and his circumscribed regular body the Octaedron having to side the less of those mean proportional lines, deducted from the crassitude of the containing Octaedron, leaveth the solid quantity of the transfigured body.   

The 15 theorem.  
If two mean proportional lines be found between the side of a comprehending Octaedron, and the greater Axis of his inscribed transfigured Octaedron, and likewise two other mean proportionals between the inscribed transformed solides side and his quadrate bases greater diameter: A cube h●●yng for ●his side the greater of the former two mean proportional lines, is equal to the circumscribing Octaedron. And if from that cube ye deduct a cube, having his side the lesser of the latter mean proportionals, the residue or remaining quantity, is the crassitude of the inscribed transfigured octaedron.   

The 16 theorem.  
Icosaedron transformed may be enclosed with an Icosaedron, and all the 180 plain angles of the transfigured body, concurring and resting in the 30 sides of the containing Icosaedron, whose side is triple to the side of the contained transformed body.   

The 17 theorem.  
The semidimetient of the comprehending Icosaedron is equal in power to the lesser axis, and a mean proportional between the transfigured body, and his containing Icosaedrons' sides.   

The 18 theorem.  
The greater diameter of a transfigured Icosaedrons' pentagonal basis, divided by extreme and mean proportion, and the greater part adjoined to his greater altitude, produceth his comprehending Icosaedrons' dimetient.   

The 19 theorem.  
The superficies of an Icosaedron surmounteth the superficial quantity of all his inscribed transfigurate Icosaedrons' hexagonal plains, by a superficies retaining such proportion to the superficial capacity of all his pentagonal plains, as the less dimetient of the comprehending Icosaedrons' basis doth to the less diameter of the transformed solides pentagonal basis.   

The 20 theorem.  
Two such lines being found, that the first be to a transfigured Icosaedrons' side double, and the second line equal to the greater segment of this transformed Icosaedrons' pentagonal basis greater semidiameter parted by extreme and mean proportion, a pentilater Prisma, having for his altitude this second ly●e, and for the side of his pentagonal equiangle basis the first line, shall be equal to the excess or difference of the containing Icosaedron above the crassitude of his inscribed transfigured Icosaedron.   

The 21 theorem.  
Dodecaedron transfigured, may of a regular Dodecaedron be circumscribed, whose side being divided in two lines retaining double extreme and mean proportion, if from the greater ye deduct the lesser, the remain is the side of the inscribed transfigured Dodecaedron.   

The 22 theorem.  
The side of Dodecaedron transformed, being subtracted from the side of his comprehending regular Dodecaedron, and from the square of that remains medietie again deducting one third part of the square of the transfigured Solides side, there remaineth a superficies, whose quadrate root or line matching it in power, added to the greater Axis of the transfigured body, maketh the semidimetient of his comprehending regular Dodecaedron.   

The 23 theorem.  
The superficial content of all transformed Dodecaedrons' Decagonall bases, to the superficial capacity of all his trigonal plains, retaineth such proportion as the less Diameter of the Decagonal bases, to the less semidimetient of his Triangular bases, and if unto the Perpendicular of the Triangular bases ye adjoin triple the Decagonall bases lesser dimetient. The square whose side is mean proportional between that resulting line, and the circuit or Perimetrie of one Decagonall base, is equal to the universal superficies of the transfigured Dodecaedron.   

The 24 theorem.  
The superficies of a comprehending regular Dodecaedron, surmounteth the superficial quantity of all his contained transfigurate Dodecaedrons' decagonal bases, by a superficies, retaining the same proportion to the superficial content of all his trigonal plains, that the lesser of two such right lines retaining double extreme and mean proportion (as joined together make the lesser part of the comprehending Dodecaedrons' Pentagonal Perpendicular divided by extreme and mean proportion) doth unto the less semidiameter of the transformate Dodecaedrons' trigonal bases.   

The 25 Theorem.  
A transformed Dodecaedrons' comprehending Dodecaedrons' diameter, being into two right lines retaining quadruple extreme and mean proportion divided, and the lesser of them again into two lines retaining double extreme and mean proportion parted: A trigonal Pyramid, having for his equilater triangular bases side, a line double to the transfigured Dodecaedrons' side, and his altitude fivefoulde the less line produced by the latter section of the Dodecaedrons' diameters lesser part, shallbe equal to the excess or difference, whereby the containing regular Dodecaedron, surmounteth the solid capacity of his inscribed transfigurate Dodecaedron.      
The end of this Mathematical discourse of the Platonical regular Solides, and their Metamorphosis compiled and invented by Thomas Digges Gentleman. ANNO AET. 25.  
DIFFICILIA QVAE PULCHRA.   


HIS DEDIT FUNDAMINA VIRTUS.    

❧ Imprinted at London by Henry Bynneman, dwelling in Knightriders street, at the sign of the Mermaid.  
ANNO DOMINI. 1571.