i M SOLID GEOMETRY BY FLETCHER DURELL, PH.D. HEAD OF THE MATHEMATICAL DEPARTMENT, THE LAWRENCEVII LE SCHOOL NEW YORK CHARLES E. MERRILL CO. DURELL'S MATHEMATICAL SERIES ARITHMETIC Two BOOK SERIES Elementary Arithmetic, 40 cents Teachers' Edition, 60 cents Advanced Arithmetic, 64 cents THREE BOOK SERIES Book One, 50 cents Book Two, 66 cents Book Three, 60 cents ALGEBRA Two BOOK COURSE Book One, $1.00 Book Two INTRODUCTORY ALGEBRA, 60 cents SCHOOL ALGEBRA, $1.10 GEOMETRY PLANE GEOMETRY, 75 cents SOLID GEOMETRY, 75 cents PLANE AND SOLID GEOMETRY, $1.25 TRIGONOMETRY PLANE TRIGONOMETRY AND TABLES, $1.25 PLANE AND SPHERICAL TRIGONOMETRY AND TABLES, $1.40 PLANE AND SPHERICAL TRIGONOMETRY WITH SURVEYING AND TABLES, $1.50 LOGARITHMIC AND TRIGONOMETRIC TABLES, 60 cents Copyright, 1904, by Charles K Merrill Co. PREFACE ONE of the main purposes in writing this book has been to try to present the subject of Geometry so that the pupil shall understand it not merely as a series of correct deductions, but shall realize the value and meaning of its principles as well. This aspect of the subject has been directly presented in some places, and it is hoped that it per- vades and shapes the presentation in all places. Again, teachers of Geometry generally agree that the most difficult part of their work lies in developing in pupils the power to work original exercises. The second main purpose of the book is to aid in the solution of this difficulty by arranging original exercises in groups, each of the earlier groups to be worked by a distinct method. The pupil is to be kept working at each of these groups till he masters the method involved in it. Later, groups of mixed exercises to be worked by various methods are given. In the current exercises at the bottom of the page, only such exercises are used as can readily be solved in connection with the daily work. All difficult originals are included in the groups of exercises as indicated above. Similarly, in the writer's opinion, many of the nuineri- (iii) 797957 iV PREFACE cal applications of geometry call for special methods of solution, and the thorough treatment of such exercises should be taken up separately and systematically. [See pp. 304-318, etc.] In the daily extempore work only such numerical problems are included as are needed to make clear and definite the meaning and value of the geometric principles considered. Every attempt has been made to create and cultivate the heuristic attitude on the part of the pupil. This has been done by the method of initiating the pupil into original work described above, by queries in the course of proofs, and also at the bottom of different pages, and also by occasional queries in the course of the text where definitions and discussions are presented. In the writer's opinion, the time has not yet come for the purely heuristic study of Geometry in most schools, but it is all-important to use every means to arouse in the pupil the attitude and energy of original investigation in the study of the subject. In other respects, the aim has been to depart as little as possible from the methods most generally used at present in teaching geometry. The Practical Applications (Groups 88-91) have been drawn from many sources, but the author wishes to ex- press his especial indebtedness to the Committee which has collected the Real Applied Problems published from time to time in School Science and Mathematics, and of which Professor J. F. Millis of the Francis W. Parker School of Chicago is the chairman. Page 360 is due almost en- tirely to Professor William Betz of the East High School of Rochester, K Y, FLETCHER DURELL. LAWRENCEVILLE, N. J., Sept. 1, 1904. TO THE TEACHER 1. IN working original exercises, one of the chief dif- ficulties o pupils lies in their inability to construct the figure required and to make the particular enunciation, from it. Many pupils, who are quite unable to do this preliminary work, after it is done can readily discover a proof or a solution. In many exercises in this book the figure is drawn and the particular enunciation made. It is left to the discretion of the teacher to determine for what other exercises it is best to do this for pupils. 2. It is frequently important to give partial aid to tho pupil by eliciting the outline of a proof by questions such as the following: "On this figure (or, in these two tri- angles) what angles are equal, and why?" "What lines are equal, and why?" etc. 3. In many cases it is also helpful to mark in colored crayon pairs of equal lines, or of equal angles. Thus, in the figure on p. 37 lines AB and DE may be drawn with red crayon, AC and DF with blue, and the angles A and D marked by small arcs drawn with green crayon. If colored crayons are not at hand, the homologous equal parts may be denoted by like symbols placed on them, thus : c F or thus: . . P B D P E A B D In solving theorems concerning proportional lines, it is occasionally helpful to denote the lines in a proportion vi TO THE TEACHER (either given or to be proved) by fig- ures denoting the order in which the lines are to be taken. Thus, if OA: OC=OD:OB, the relation may be indicated as on the figure. 4. It is sometimes helpful to vary the symbolism of the book. Thus, in dealing with in- equalities a convenient symbol for "angle" is i t as 2f. A > 3! B. 5. Each pupil need be required to work only so many originals in each group as will give him a mastery of the particular method involved. A large number of exercises is given in order that the teacher may have many to select from and may vary the work with successive classes. 6. It is important to insist that the solutions of exer- cises for the first few weeks be carefully written out; later, for many pupils, oral demonstration will be sufficient and ground can be covered more rapidly by its use. 7. In leading pupils to appreciate the meaning of theorems, it is helpful at times to point out that not every theorem has for its object the demonstration of a new and unexpected truth (i. e., not all are "synthetic"), but that some theorems are analytic, it being their purpose to re- duce an obvious truth to the certain few principles with which we start in Geometry. Their function is, therefore, to simplify and clarify the subject rather than to extend its content. REFERENCES TO PLANE GEOMETRY. DEFINITIONS AND FIRST PRINCIPLES. 41. Parallel lines are lines in the same plane which do not meet, however far they be produced. GENERAL AXIOMS. 1. Things which are equal to the same thing, or to equal things, are equal to each other. 2. If equals be added to equals, the sums are equal. 3. // equals be subtracted from equals, the remainders are equal. 4. Doubles of equals are equal; or, in general, if equals be multiplied by equals the products are equal. 5. Halves of equals are equal; or, in general, if equals be divided by equals the quotients are equal. 6. The whole is equal to the sum of its parts. 7. The whole is greater than any of its parts. 8. A quantity may be substituted for its equal in any process. 9. // equals be added to, or subtracted from, unequals, the results are unequal in the same order; if unequals be added to unequals in the same order, the results are unequal in that order. 10. Doubles, or halves, of unequals are unequal in the same order. 11. // unequals be subtracted from equals, the remainders are unequal in the reverse order, vii Vlll KEFERENCES TO PLANE GEOMETRY. 12. //, of three quantities, the first is greater than the. second, and the second is greater than the third, then the first is greater than the third. GEOMETRIC AXIOMS. 1. Through two given points only one straight line can be passed. 2. A geometric figure may be freely moved in space with- out any change in form or size. 3. Through a given point one straight line and only one can be drawn parallel to another given straight line. Geometric figures which coincide are equal. 71. At a given point in a straight line but one perpen- dicular can be erected to the line. 75. The complements of two equal angles are equal; tlu supplements of two equal angles are equal. 76. The sum of all the angles about a point equals four right angles. BOOK I. 78. // one straight line intersects another straight line, the opposite or vertical angles are equal. 79. //, from a point in a perpendicular to a given line, two oblique lines be drawn cutting off on the given line equal segments from the foot of the perpendicular, the oblique lines are equal and make equal angles with the perpendicular. 81. A triangle is a portion of a plane bounded by three straight lines. 92. The sum of any two sides of a triangle is greater than the third side. 94. The perpendicular is the shortest line that can be drawn from a given point to a given line. 95. If, from a point within a triangle, two lines be drawn to the extremities of one side of the triangle, the sum of the REFERENCES TO BOOK I. IX other two sides of the triangle is greater than the sum of the two lines so drawn. 96. Two triangles are equal if two sides and the included angle of one are equal, respectively, to two sides and the in- cluded angle of the other. 98. Two right triangles are equal if the hypotenuse and an acute angle of one are equal to the hypotenuse and. an acute angle of the other. 99. In an isosceles triangle the angles opposite the equal sides are equal. 100. // two angles of a triangle are equal, the sides oppo- site are equal, and the triangle is isosceles. 101. Two triangles are equal if three sides of one are equal to three sides of the other, respectively. 102. Two right triangles are equal if the hypotenuse and a leg of one are equal to the hypotenuse and a leg of the other. 108. // two sides of a triangle are equal, respectively, to two sides of another triangle, but the third side of the first is greater than the third side of the second, then the angle opposite the third side of the first triangle is greater than the angle oppo- site the third side of the second. 109. Of lines drawn from the same point in a perpen- dicular, and cutting off unequal segments from the foot of the perpendicular, the more remote is the greater. 112. I. Every point in the perpendicular bisector of a line is equally distant from the extremities of the line; and II. Every point not in the perpendicular bisector is un- equally distant from the extremities of the line. 113. Two points each equidistant from the extremities of a line determine the perpendicular bisector of the line. 115. T1i-e perpendicular bisector of a line v'.s tJtc locus of all points equidistant from the extremities of the line. 120. Parallel lines are lines in the same plane which do not meet, however far they be produced. X REFERENCES TO PLANE GEOMETRY. 121. Two straight lines in the same plane, perpendicular to the same straight line, arc parallel. 122. Two straight lines parallel to a third straight line are parallel to each other; Lines parallel to parallel lines are parallel; Lines perpendicular to parallel lines are parallel; Lines perpendicular to non-parallel lines are not parallel. 123. If a straight line is perpendicular to one of two given parallel lines it is perpendicular to the other also. 125. // two straight lines are cut by a transversal, mak- ing the alternate interior angles equal, the two straight lines are parallel. 127. // two straight lines are cut by a transversal, making the exterior interior angles equal,, the two straight lines are parallel. 130. Two angles whose sides are parallel, each to each, are either equal or supplementary. 134. The sum of the angles of a triangle is equal to two right angles. 142. Two right triangles are equal if a leg and an acute angle of one are equal to a leg and the homologous acute angle of the other. 147. A parallelogram is a quadrilateral whose opposite sides are parallel. 149. A rhombus is a rhomboid whose sides are equal. 150. A rectangle is a parallelogram whose angles are right angles. 151. A square is a rectangle whose sides are equal. 155. The opposite sides of a parallelogram are equal, and its opposite angles are also equal. 156. A diagonal divides a parallelogram into two equal triangles. 157. Parallel lines comprehended between parallel lines are equal, REFERENCES TO BOOK II. XI 180. // two sides of a quadrilateral are equal and parallel, the other two sides are equal and parallel and the figure is a parallelogram. 1 62. Two parallelograms are equal if two adjacent sides and the included angle of one are equal, respectively, to two adjacent sides and the included angle of the other. 163. Two rectangles which have equal bases and equal altitudes are equal. 174. In an equiangular polygon of n sides each angle , (n-2)2rf.Z8 2n-4 equals , or rt. Z s. n n 175. The sum of the' exterior angles of a polygon formed by producing its sides in succession equals four right angles. 176. // three or more parallels intercept equal parts on one transversal, they intercept equal parts on every transversal. 177. The line which joins the midpoints of two sides of a triangle is parallel to the third side, and is equal to one-half the third side. 178. The line which bisects one side of a triangle and is parallel to another side bisects the third side. 179. The line which joins the midpoints of the legs of a trapezoid is parallel to the bases and equal to one-half their sum. BOOK II. 197. A circle is a portion of a plane bounded by a curved line, all points of which are equally distant from a point within called the center. 198. A radius of a circle is a straight line drawn from the center to any point on the circumference. 208. Radii of the same circle, or of equal circles, are equal. 216. In the same circle, or in equal circles, equal arcs subtend equal angles at the center. Xli REFERENCES TO PLANE GEOMETRY. 217. In the same circle, or in equal circles, of two unequal central angles the greater angle intercepts the greater arc, and, conversely, of two unequal arcs, the greater arc subtends the greater angle at the center. 218. In the same circle, or in equal circles, equal chords \ subtend equal arcs. 219. In the same circle, or in equal circles, equal arcs are subtended by equal chords. 220. In the same circle, or in equal circles, the greater of two (minor) arcs is subtended by the greater chord; and, CONVERSELY, the greater of two chords subtends the greater (minor) arc. 226. In the same circle, or in equal circles, equal chords are equidistant from the center; and, CONVERSELY, chords which are equidistant from the center are equal. 229. A straight line perpendicular to a radius at its ex- tremity is tangent to the circle. 230. The radius drawn to the point of contact is perpen- dicular to a tangent to a circle. 237. The two tangents drawn to a circle from a point out- side the circle are equal, and make equal angles with a line drawn from the point to the center. 241. // two circles intersect, their line of centers is per- pendicular to their common chord at its middle point. 253. Properties of variables and limits. 1. The limit of the sum of a number of variables equals the sum of the limits of these variables. 2. The limit of a times a variable equals a times the limit of the variable, a being a constant. 3. The limit of th part of variable is th part of the limit 1 a a of the variable, a being a constant. 254. // two variables are always equal, and each ap- proaches a limit, their limits are enual. REFERENCES TO BOOK III. Xlll 257. The number of degrees in a central angle equals the number of degrees in the intercepted arc; that is, a central angle is measured by its intercepted arc. 273. From a given point without a given line to draw a perpendicular to the line. 274. At a given point in a given line erect a perpen- dicular to that line. 279. Through a given point without a given straight line to draw a line parallel to a given line. 286. To circumscribe a circle about a given triangle. BOOK III. 303. The mean proportional between two quantities is equal to the square root of their product. 305. // the antecedents of a proportion are equal, the con- sequents are equal. 307. // Jour quantities are in proportion, they are in pro- portion by alternation ; that is, the first term is to the third as the second is to the fourth. 310. // four quantities are in proportion, they arc in pro- portion by division ; that is, the difference of the first two is to the second as the difference of the last two is to the last. 312. In a series of equal ratios, the sum of all the ante- cedents is to the sum of all the consequents as any one ante- cedent is to its consequent. 314. Like powers, or like roots, of the terms of a propor- tion are in proportion. 317. A line parallel to one side of a triangle and meeting the other two sides, divides these sides proportionally. 318. If a line parallel to the base cut the sides of a triangk, a side is to a segment of that side as the other side is to the cor- responding segment of the second side. 321. Similar polygons are polygons having their ho- XIV REFERENCES TO PLANE GEOMETRY. mologous angles equal and their homologous sides propor- tional. 323. // two triangles are mutually equiangular, they are similar. 326. // two triangles have their homologous sides propor- tional they are similar. 327. // two triangles have an angle of one equal to an angle of the other, and the including sides proportional, the triangles are similar. 328. // two triangles have their sides parallel, or perpen- dicular, each to each, the triangles are similar. 329. // two polygons are similar, they may be separated into the same number of triangles, similar, each to each, and similarly placed. 342. In a right triangle, I. The altitude to f he hypotenuse is a mean proportional between the segments of the hypotenuse; II. Each leg is a mean proportional between the hypote- nuse and the segment of the hypotenuse adjacent to the given leg. 343. The perpendicular to the diameter from any point in the circumference of a circle is a mean proportional between the segments of the diameter; and the chord joining the point to an extremity of the diameter is a mean proportional be- tween the diameter and the segment of the diameter adjacent to the chord. 346. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. 347. In a right triangle, the square of either leg is equal to the square of the hypotenuse minus the square of the other leg. 351. // the square on the side of a triangle equals the sum of the squares on the other two sides, the angle, opposite the first side is a right angle. REFERENCES TO BOOK IV. XV 352. //, in any triangle, a median be drawn to one side, I. The sum of the squares of the other two sides is equal to twice the square of half the given side, increased by twice the square of the median upon that side; and II. The difference of the squares of the other two sides is equal to twice the product of the given side by the projection of the median upon that side. BOOK IV. 383. The area of a rectangle is equal to the product of its base by its altitude. 385. The area of a parallelogram is equal to the product of its base by its altitude. 389. The area of a triangle is equal to one-half the prod- uct of its base by its altitude. 390. Triangles which have equal bases and equal alti- tudes (or which have equal bases and their vertices in a line parallel to the base) are equivalent. 391. Triangles which have equal bases are to each other as their altitudes; Triangles which have equal altitudes are to each other as their bases. 392. Any two triangles are to each other as the products of their bases and altitudes. 397. // two triangles have an angle of one equal to an angle of the other, their areas are to each other as the products of the sides including the equal angles. 398. The areas of any two similar triangles are to each other as the squares of any two homologous sides. 399. The areas of two similar polygons are to each other as the squares of any two homologous sides. XVI REFERENCES TO PLANE GEOMETRY. BOOK V. 444. Formula for the circumference in terms of the radius. 449. The area of a circle is equal to one-half the product of its circumference by its radius. 486. Two points symmetrical with respect to a line or axis are points such that the straight line joining them is bisected by the given line at right angles. 487. A figure symmetrical with respect to an axis is a figure such that each point in the one part of the figure has a point in the other part symmetrical to the given point, with respect to an axis. 489. Two points symmetrical with respect to a point or center are points such that the straight line joining them is bisected by the point or center. 490. A figure symmetrical with respect to a point or center is a figure such that each point in the figure has an- other point in the figure symmetrical to the given point with respect to the center. TABLE OF CONTENTS PAGE BOOK VI. LINES, PLANES AND ANGLES IN SPACE 319 BOOK VII. POLYHEDRONS 360 BOOK VIII. CYLINDERS AND CONES . . . 403 BOOK IX. THE SPHERE 425 NUMERICAL EXERCISES IN SOLID GEOMETRY 4(59 APPENDIX I. MODERN GEOMETRIC CONCEPTS 485 II. HISTORY OF GEOMETRY 490 III. REVIEW EXERCISES. PLANE GEOMETRY .... 491 SOLID GEOMETRY 506 PRACTICAL APPLICATIONS OF SOLID GEOMETRY 510 FORMULAS . 516 (rvii) SYMBOLS AND ABBREVIATIONS H- plus, or increased by. Adj. . . adjacent. minus, or diminished by. Alt. . . alternate. X multiplied by. Art. . . article. -s- divided by. Ax. ... axiom. = equals; is (or are] equal to. Constr. . construction approaches (as a limit). Cor. . . corollary. =c= is (or are) equivalent to. Def. . . definition. > is (or are) greater than. Ex. . . . exercise. < is (or arc) Zess Ma?i. Ext. . . exterior. .'. therefore. Fig. . . figure. J_ perpendicular, perpendicular to, Hyp. . . hypothesis. or, is perpendicular to. Ident. . identity. .k perpendiculars. Int. . . interior. || parallel, or, is parallel to. Post. . . postulate. \\sparallels. Prop. . .proposition. /. , A angle, angles Rt. . . . right. A, A triangle, triangles. Sug. . . suggestion. / 7 , Z7 parallelogram, parallelograms. Sup. . . supplementary. G, cirete, circles. St. ... straight. Q. E. D. quod erat demonstrandum; that is, which was to be proved. Q. E. F. quod erat faciendum; that is, which was to be made. A few other abbreviations and symbols will be introduced and their meaning indicated later on. SOLID GEOMETRY BOOK VI LINES, PLANES AND ANGLES IN SPACE DEFINITIONS AND FIRST PRINCIPLES 497. Solid Geometry treats of the properties of space of three dimensions. Many of the properties of space of three dimensions are determined by use of the plane and of the properties of plane figures already obtained in Plane Geometry. 498. A plane is a surface such that, if any two points in it be joined by a straight line, the line lies wholly in the surface. 499. A plane is determined by given points or lines, if no other plane can pass through the given points or lines without coinciding with the given plane. 500. Fundamental property of a plane in space. A plane is determined by any three points not in a straight line. For, if through a line con- .c necting two given points, A and B, a plane be passed, the plane, if rotated, can pass through a third given point, (7, in but one position. The importance of the above principle is seen from the fact that it reduces an unlimited surface to three points, thus making a vast economy to the attention. It also enables us to connect different planes, and treat of their properties systematically. (319) 320 BOOK VI. SOLID GEOMETRY 501. Other modes of determining a plane. A plane may also be determined by any equivalent of three points not in a straight line, as by a straight line and a point outside the line; or by two intersecting straight lines; or by two parallel straight lines. It is often more convenient to use one of these latter methods of determining a plane than to reduce the data to three points and use Art. 500. \ 502. Representation of a plane in geometric figures. la masoning concerning the plane, it is often an advantage to have the plane represented in all directions. Hence, in drawing a geometric figure, a plane is usually represented to the eye by a small parallelogram. This is virtually a double use of two intersecting lines, or of two parallel lines, to determine a plane (Art. 501). 503. Postulate of Solid Geometry. The principle of Art. 499 may also be stated as a postulate, .thus: Throtigh any three points not in a straight line (or their equivalent) a plane may be passed. 504. The foot of a line is the point in which the line intersects a given plane. 505. A straight line perpendicular to a plane is a line perpendicular to every line in the plane drawn through its foot. A straight line perpendicular to a plane is sometimes called a normal to the plane. LINES AND PLANES 321 506. A parallel straight line and plane are aline and plane which cannot meet, however far they be produced. 507. Rarallel planes are planes which cannot meet, however far they be produced. 508. Properties of planes inferred immediately. 1. A straight line, not in a given plane, can intersect the given plane in but one point. For, if the line intersect the given plane in two or more points, by definition of a plane, the line must lie in the plane. Art. 498. 2. The intersection of two planes is a straight line. For, if two points common to the two planes be joined by a straight line, this line lies in each plane (Art. 498) ; and no other point can be common to the two planes, for, through a straight line and a point outside of 4t only one plane can be passed. Art. 501. Ex. 1. Give an example of a plane surface; of a curved surface; of a surface, part plane and part curved; of a surface composed of different plane surfaces. Ex. 2. Four points, not all in the same plane determine how many different planes ? how many different straight lines ? Ex. 3. Three parallel straight lines, not in the same plane, deter- mine how many different planes ? Ex. 4. Four parallel straight lines can determine how many differ- ent planes ? Ex. 5. Two intersecting straight lines and a point, not in their plane, determine how many different planes f 322 BOOK VI. SOLID GEOMETRY PROPOSITION I. THEOREM 509. If a straight line is perpendicular to each of two other straight lines at their point of intersection, it is per- pendicular to the plane of those lines. Given AB _L lines EC and ED, and the plane MN pass- ing through EG and ED. To prove AB _L plane MN. Proof. Through E draw EG, any other line in the plane MN. Draw any convenient line CD intersecting EC, EG and ED in the points C, G and D, respectively. Produce the line AB to F, making BF=AB. Connect the points C, G, D with A, and also with F. Then, in the A ACD and FCD, CD=CD. Ident. A C= CF, and AD=DF. Art. 112. /. &ACD=AFCD. (Why?) /. Z ACD=Z. FCD. (Why?) Then, in the A ACG and FCG, CG=CG, (Why?) AC= CF, and Z ACG = Z JW#. (Why ?) .'. A J.C#=A ^(7O. (Why?) LINES AND PLANES 323 /. AG=GF. (Why?) .*. B and O are each equidistant from the points A and F. :. BO is J_ AF; that is, AB J_ BO. Art. 113. /. AB _L plane MN, Art. 505. (for it is J_ any line, EG, in the plane MN, through its foot}. Q. E. D. PROPOSITION II. THEOREM 510. All the perpendiculars that can be drawn to a given line at a given point in the line lie in a plane perpen- dicular to the line at the given point. Given the plane MN and the line #<7both J_ line AB at the point B. To prove that BC lies in the plane MN. Proof. Pass a plane AF through the intersecting lines AB and BC. Art. 503. This plane will intersect the plane MN in a straight line BF. Art. 508, 2. But AB J_ plane MN (Hyp.) .-. AB J_ BF. Art. 505. Also AB _L BC. Hyp. .*. in the plane AF, BC and BF _L AB at B. :. BC and BF coincide. Art. 71. But BF is in the plane MN. :. BC must be in the plane MN, (for BC coincides with BF) which lies in the plane MN}. Q. D* 324 BOOK VI. SOLID GEOMETKY 511. COR. 1. At a given point B in the straight line AB, to construct a plane perpendicular to the line AB. Pass a plane AF through AB in any convenient direction, and in the plane AF at the point B construct BF J_ AB (Art. 274). Pass another plane through AB, and in it construct BP AB. Through the lines BF and BP pass the plane MN (Art. 503). MN is the required plane (Art. 509). 512. COR. 2. Through a given external point, P, to pass a plane perpendicular to a given line, AB. Pass a plane through AB and P (Art. 503), and in this plane draw PB _L AB (Art. 273). Pass another plane through AB, as AF, and in AF draw BF JL AB at B (Art. 274). Pass a plane through BP and BF (Art. 503). This will be the plane required (Art. 509). 513. COR. 3. Through a given point but one plane can be passed perpendicular to a given line. Ex. 1. Five points, no four of which are in the same plane, deter- mine how many different planes ? how many different straight lines ? Ex. 2. A straight line and two points, not all of which are in the same plane, determine how many different planes ? Ex. 3. In the figure, p. 322, prove that the triangles GAD and GDF are equal. Ex. 4. In the same figure, if AB=S and BC=Q, find FC. LINES AND PLANES 325 PROPOSITION III. PROBLEM 514. At a given point in a plane, to erect a perpendicu- lar to the plane. p Given the point A in plane MN. To construct a line perpendicular to MN at the point A. Construction. Through the point A draw any line CD in the plane MN. Also through the point A pass the plane PQ JL CD (Art. 511), intersecting the plane MN in the line RS. Art. 508, 2. In the plane PQ draw AK J_ line RS at A. Art. 274. Then AK is the J_ required. Proof. CD _L plane PQ. Constr. /. CD J_ AK. Art. 505. Hence AK J_ CD. But AK _L #. Constr. /. AK JL plane MF. Art. 509. Q. E. F. 515. COR. At a given point in a plane but one perpen- dicular to the plane can be drawn. For, if two Ja. could be drawn at the given point, a plane could be passed through them intersecting the given plane. Then the two Ja would be in the new plane and J_ to the same line (the line of intersection of the two planes, Art, 505), which is im- possible (Art, 71), 326 BOOK VI. SOLID GEOMETBY PROPOSITION IV. PROBLEM 516. From a given point without a plane, to draw a line perpendicular to the plane. A ^ ,B Given the plane MN and the point A external to it. To construct from A a line JL plane MN. Construction. In the plane MN draw any convenient line EC. Pass a plane through EG and A (Art. 503), and in this plane draw AD _L EC. Art. 273. In the plane MN draw LD J_ EC. Art. 274. Pass a plane through AD and LD (Art. 503), and in that plane draw AL J_ LD. Art. 273. Then AL is the _L required. Proof. Take any point C in EC except D, and draw LC and AC. Then A J.DC, AD and LDC are right A. Constr. . Art. 400. + ~D&. Art. 400, Ax. 8. .'. AC 2 = AL 2 + LC 2 . Art. 400, Ax. 8. /. /.ALC is a right Z . Art. 351. But AL JL LD. Constr. /. AL X JOT. Art. 509. Q. E. F. 517r COR. But one perpendicular can be drawn from a given external point to a given plane. LINES AND PLANES 327 PROPOSITION V. THEOREM 518. I. Oblique lines drawn from a point to a plane, meeting the plane at equal distances from the foot of the perpendicular, are equal; II. Of two oblique lines drawn from a point to a plane, but meeting the plane at unequal distances from the foot of the perpendicular, the more remote is the greater. Given AB i. plane MN, BD = BC, and BH > EG. To prove AD = AC, and AH > AC. Proof. I. In the right A ABD and ABC, AB = AB, and BD=BC. /. AABD=AABC. (Why?) (Why?) (Why?) II. On .RET take BF=BG, and draw AF. Then A F= AC (by part of theorem just proved) But AH > AF. :. AH > AC. (Why ?) Ax. 8. Q. E. D. 519. COR. 1. CONVERSELY: Equal oblique lines drawn from a point to a plane meet the plane at equal distances from the foot of the perpendicular drawn from the same point to the plane; and, of two unequal lines so drawn, the greater line meets the plane at the greater distance from the foot of the perpendicular. 328 BOOK VI. SOLID GEOMETRY 520. COR. 2. The locus of a point in space equidistant from all the points in the circumference of a circle is a straight line passing through the center of the circle and perpendicular to its plane. 521. COB. 3. The perpendicular is the shortest line that can l)e drawn from a given point to a given plane. 522. DEF. The distance from a point to a plane is the perpendicular drawn from the point to the plane. PROPOSITION VI. THEOREM 523. If from the foot of a perpendicular to a plane a line l)e drawn at right angles to any line in the plane, the line drawn from the point of intersection so formed to any point in the perpendicular, is perpendicular to the line of the plane. \A-7P Given AB J_ plane MN, and EF _L CD, any line in MN. To prove AF _L CD. Proof. On CD take FP and FQ equal segments. Draw AP, BP, AQ, BQ. Then BP=BQ. Art. 112. Hence AP=AQ. Art. 518. /. in the line AF, the point A is equidistant from P and Q, and F is equidistant from P and Q. :. AF i. CD. (Why?) Q. E. D. Ex. In the above figure, if AS=Q, AF=8, and ^=10, find QF, BF and . LINES AND PLANES 329 PROPOSITION VII. THEOREM 524. Two straight lines perpendicular to the same plane are parallel. A C \ F D Given the lines AB and CD J. plane MN. To prove AB \\ CD. Proof. Draw BD, and through D, in the plane draw FH J_ BD. Draw AD. Then BD _L FIT. Constr. AD J_ FH. Art. 523. CD J_ J^ff. Art. 505. /. BD, AD and CD are all J. FH at the point D. .'. BD, AD and CD all lie in the same plane. Art. 510. .'. AB and CD are in the same plan^. (Why?) But AB and CD are J_ #Z). Art. 505. /. AB and CD are ||. Art. 121. Q. . D. 525. COR. 1. If one of two parallel lines is perpendicu- lar to a plane, the other is perpendicular to the plane also. For, if AB and CD be ||, and AB _L plane PQ, a line drawn from C J_ PQ must be || AB. Art. 524. But CD must coincide with the line so drawn (Art. 47, 3); .-. CD X PQ. 330 BOOK VI. SOLID GEOMETKY 526. COE. 2. If two straight lines are each parallel to a third straight line, they are parallel to each other. For, if a plane be drawn _L to the third line, each of the two other lines must be _L to it (Art. 525), and therefore be || to each other (Art. 524). PROPOSITION VIII. THEOREM 527. If a straight line external to a given plane is paral- lel to a line in the plane, then the first line is parallel to the given plane. Given the straight line AB \\ line CD in the plane MN. To prove AB \\ plane MN. Proof. Pass a plane through the || lines AB and CD. If AB meets MN it must meet it in the line CD. But AB and CD cannot meet, for they are ||. Art. 120. .'. AB and MN cannot meet and are parallel. Art. 506. Q. E. D. 528. COR. 1. If a straight line is parallel to a plane , the intersection of the plane with any plane passing through the given line is parallel to the given line. LINES AND PLANES 331 529. COR. 2. Through a given line (CD) to pass a plane parallel to another given line (AB) . Through P, any point in CD, draw QR || AB (Art. 279) . Through CD and QR pass a plane (Art. 503). This will be the plane required (Art. 527). If AB and CD are not parallel, but one plane can be drawn through CD\\AB. PROPOSITION IX. THEOREM 530. Two planes perpendicular to the same straight line are parallel. M Given the planes MN and PQ _L line AB. To prove MN\\ PQ. Proof. If MN and PQ are not parallel, on being pro- duced they will meet. We shall then have two planes drawn from a point per- pendicular to a given line, which is impossible. Art. 513. .*. MN and PQ are parallel. Art. 507. Q. . D. 332 BOOK VI. SOLID GEOMETRY PROPOSITION X. THEOREM 531. // two parallel planes are cut by a third plane, the intersections are parallel lines. Given MN and PQ two 1 1 planes intersected by the plan* RS in the lines AB and CD. To prove AB \\ CD. Proof. AB and CD lie in the same plane R8. Also AB and CD cannot meet; for if they did meet the planes MN and PQ would meet, which is impossible. Art. 507. .*. AB and CD are parallel. Art. 41. Q. E. D. 532. COR. 1. Parallel lines included between parallel planes are equal. For, if AC and BD are two parallel lines, a plane may be passed through them (Art. 503), intersect- ing MN and PQ in the || lines AB and CD. Art. 531. /. ABDC is a parallelogram. Art. 147. .*. AC=BD. Art. 155. LINES AND PLANES 333 533. COR. 2. Two parallel planes are everywhere equi- distant. For lines _L to one of them are || (Art. 524). Hence the segments of these lines included between the || planes are equal (Art. 532). PROPOSITION XI. * THEOREM 534. If two intersecting lines are each parallel to a given plane, the plane of these lines is parallel to the given plane. D K Given the lines AB and CD, each || plane PQ, and inter- secting in the point F\ and MN a plane through AB and CD. To prove MN\\PQ. Proof. From the point F draw FH _L PQ. Pass a plane through FC and FH, intersecting PQ in HK; also pass a plane through FB and Fit, intersecting PQ in HL. Then HK \\ FC, and HL \\ FB. Art. 528. But FH _L HK and HL. Art. 505. /. FH J_ FC and FB. Art. 123. /. FH _L M^. Art. 509. .-. MN II P#. Art. 530. Q. B. . 334 BOOK VI. SOLID GEOMETRY PROPOSITION XII. THEOREM 535. A straight line perpendicular to one of two parallel planes is perpendicular to the other also. \ T" Given the plane MN \\ plane PQ, and AB JL PQ. To prove AB J_ MN. Proof. Through AB pass a plane intersecting PQ and MN in the lines BO and AF, respectively; also through AB pass another plane intersecting PQ and MN in BJ) and AH, respectively. Then BO 1 1 AF, and J5D || AJ3". Art. 531. But AB _L BG and J9D. Art. 505. .*. AB J_ AJP and AH. Art. 123. .*. A-B _L plane M2V. Art. 509. Q. . D. 536. COR. 1. Through a given point to pass a plane parallel to a given plane. Let the pupil supply the construction. 537. COR. 2. Through a given point but one plane can be passed parallel to a given plane, LINES AND PLANES 335 PROPOSITION XIII. THEOREM 538. If two angles not in the same plane have their corresponding sides parallel and extending in the same direc- tion, the angles are equal and their planes are parallel. \ \ Given the /.BAG in the plane MN, and the B'A'C> in the plane PQ; AB and A'B' \\ and extending in the same direction; and AC and AC' \\ and extending in the same direction. To prove BAC = /.B'A'C', and plane MN \\ plane PQ. Proof. Take AB = AB r , and AC=AC f . Draw AA, BB' t CC', BC, B'C'. Then ABB' A is a ZZ7 , Art. 160. (for AB and A f B' are=and \\ ). /. BB' and A A' are = and IK Art. 155. In like manner CC' and AA' are = and ||. /. BB' and CC' are = and ||. (Why?) .'. BCC'B' is a C3 , and BC=B'C f . (Why ?) /. A ABC= A A'B'C'. (Why ?) /. A = /.A. (Why?) Also AB || A'B', :. AB \\ plane PQ. Art. 527. Similarly AC \\ plane P^. :. plane JOT || plane PQ. Art. 534. o. E. B. 336 BOOK VI. SOLID GEOMETRY A PROPOSITION XIV. THEOREM 539. // two straight lines are intersected by three paral- lel planes^ the corresponding segments of these lines are proportional. Mr wrp. Given the straight lines AB and CD intersected by the || planes MN, P^/and ES in the points A, F, B, and C H, D, respectively. AF GH To prove Proof. Draw the line AD intersecting the plane PQ in G. Draw FG, BD, GH, AC. Then FG \\ BD, and GH \\ AC. Art. 531. Art 317 And AF CH (Why?) " FB HD Q. E. D. Ex. 1. In above figure, if AF=2, FB=5, and CH=3, find CD. Ex 2. If CH=3, HD=, and AB=IQ, find AF and BF. DIHEDKAL ANGLES 337 DIHEDRAL ANGLES 540. A dihedral angle is the opening between two in- tersecting planes. From certain points of view, a dihedral angle may be regarded as a wedge or slice of space cut out by the planes forming the dihedral angle. 541. The faces of a dihedral angle are the planes form- ing the dihedral angle. The edge of a dihedral angle is the straight line in which the faces intersect. 542. Naming dihedral an- gles. A dihedral angle may be named, or denoted, by naming its edge, as the dihedral angle AB; or by naming four points, two on the edge and one on each face, those on the edge coming between the points on the faces, as P-AB-Q. The latter method is necessary in naming two or more dihedral angles which have a common edge. 543. Equal dihedral angles are dihedral angles which can be made to coincide. 544. Adjacent dihedral angles are dihedral angles hav- ing the same edge and a face between them in common. 545. Vertical dihedral angles are two dihedral angles having the same edge, and the faces of one the prolonga- tions of the faces in the other. 546. A right dihedral angle is one of two equal adja- cent dihedral angles formed by two planes. 338 BOOK VI. SOLID GEOMETRY 547. A plane per- pendicular to a given plane is a plane form- ing a right dihedral angle with the given plane. Many of the properties of dihedral angles are obtained most conveniently by using a plane angle to represent the dihedral angle. 548. The plane angle of a dihedral an- gle is the angle formed by two lines drawn one in each face, perpendicular to the edge at the same point. Thus, in the dihedral angle C-AB-F, if PQ is a line in the face AD perpendicu- lar to the edge AB at P, and PR is a line in face AF perpendicular to the edge AB at P, the angle QPR is the plane angle of the dihedral angle C-AB-F. 549. Property of plane angles of a dihedral angle. The magnitude of the plane angle of a dihedral angle is the same at every point of the edge. For let EAC be the plane Z of the dihedral Z E-AB-D at the point A. Then PR \\ AE, and PQ \\ AC (Art. 121.) = /.EAC (Art. 538). D 550. The projection of a point upon a plane is the foot of a perpendicular drawn from the point to the plane. ritt 551. The projection of a line upon a plane is the locus of the pro- jections of all the points of the line on the plane. Thus A'B' is the projection of AB on the plane MN. DIHEDKAL ANGLES PROPOSITION XV. THEOREM 339 552. Two dihedral angles are equal if their plane angle* are equal. Given /.DBF the plane Z of the dihedral C-AB-F, D'B'F' the plane Z of the dihedral /.C'-A'B'-F', and /.DBF = D f B f F f . To prove Z C-AB-F = Z C f -A'B'-F'. Proof. Apply the dihedral Z C f -A'B'-F' to Z C-AB-F so that /.D'B'F' coincides with its equal, /.DBF. Greom. Ax. 2. Then line A'B f must coincide with AB, Art. 515. (/or A'B' and AB are loth plane DBF at the point B). Hence the plane A'B'D f will coincide with plane ABD, Art. 501. (through two intersecting lines only one plane can be passed). Also the plane A f B'F f will coincide with the plane ABF, (same reason). .'. Z G'-A'B'-F' coincides with Z C-AB-F and is equal to it. Art. 47. Q. E. D. 553. COR. The vertical dihedral angles formed by two intersecting planes are equal. In like manner, many other properties of plane angles are true of dihedral angles. 340 ' BOOK VI. SOLID GEOMETRY V PROPOSITION XVI. THEOREM 554. Two dihedral angles have the same ratio as their plane angles. Fig.l Fig,2 Jig. 3 Given the dihedral A C-AB-D and C'-A'B'-D' having the plane A CAD and C'A'D', respectively. To prove Z C'-A'B'-D' /.CAD. Z C-AB-D = Z C'A'D' CASE I. When the plane A C'A'D' and CAD (Figs. 2 and 1), are commensurable. Proof. Find a common measure of the A C'A'D' and CAD, as Z CAR, and let it be contained in Z C'A'D' n times, and in Z CAD m times. Then Z C'A'D' : Z CAD = n : m. Through A'B' and the lines of division of Z C'A'D' pass planes, and through AB and the lines of division of Z CAD pass planes. These planes will divide the dihedral Z C'-A'B'-D' into n, and /.C-AB-D into m parts, all equal. Art. 552. /. /.C'-A'B'-D' : /.C-AB-D = n : m. Hence Z C'-A'B'-D' : Z (J-AB-D = Z C'A'D' : Z CAD. (Why?) DIHEDKAL ANGLES 341 CASE II. When the plane angles C'A'D' and CAD (Figs. 3 and 1) are incommensurable. Proof. Divide the Z CAD into any number of equal parts, and apply one of these parts to the Z C'A'D 1 . It will be contained a certain number of times with a remain- der, as LA'D', less than the unit of measure. Hence the A G'A'L and CAD are commensurable. /. Z C'-A'B'-L\tC-AB-D=. Z C'A'L:Z.CAD. Case I- If now we let the unit of measure be indefinitely dimin- ished, the /.LA'D', which is less than the unit of measure, will be indefinitely diminished. .-. Z C'A'L = Z C'A'D' as a limit, and Z G'A'B'L = Z C'A'B'D' as a limit. Art. 251. Z C' A'B f L Hence becomes a variable, with Z. O~-A-D ~ JJ Z C'-A'B'-D' .. .. ., . as its limit; Art. 253, 3. Z. ^4,-D JL/ , Z Also . ~ . j becomes a variable with as its limit. Art. 253, 3. tO-A'B'-L Z C'A'L But the variable . . = the variable Z \jA.jJ always. Case I. , Z C'-A'B'-D' ., . Z " the hmit =the hmlt Q. E. D. Ex. 1. How many straight lines are necessary to indicate a dihe- dral angle (as /.E-AB-D, p. 338)? How many straight lines are necessary to indicate the plane angle of a dihedral angle ? Hence, what is the advantage of using a plane angle of a dihedral angle instead of the dihedral angle itself ? Ex. 2. Give three additional properties of dihedral angles anal- ogous to properties of plane angles given in Book I, 342 X BOOK VI. SOLID GEOMETKY PROPOSITION XVII. THEOREM 555. If a straight line is perpendicular to a plane, every plane draivn through that line is perpendicular to the plane. Given the line AB _L plane MN, and the plane PQ passing through AB and intersecting MN in RQ. To prove PQ J_ MN. Proof. In the plane MN draw BG JL RQ at B. But AB _L RQ. Art. 505. .*. Z ABC is the plane Z of the dihedral /iP-RQ-M. Art. 548. But Z ABC is a right Z , Art. 505. ( for AB J. MN by hyp . ) . /. PQ -L MN. Art. 547. Q. E. B. 556. COR. A plane perpendicular to the edge of a dihedral angle is perpendicular to each of the two faces form- ing the dihedral angle. DIHEDEAL ANGLES 343 PROPOSITION XVIII. THEOREM 557. If two planes are perpendicular to each other, a straight line drawn in one of them perpendicular to their line of intersection is perpendicular to the other plane. p Given the plane PQ JL plane MN and intersecting it in the line RQ-, and AB a line in PQ _L RQ. To prove AB J_ plane MN. Proof. In the plane MN draw BC JL RQ. :. ZABCis the plane Z of the dihedral ^P-RQ-M. Art. 548. /. Z ABC is a rt. Z , Art. 554. (for P-RQ-Mis a right dihedral Z ). .*. AB JL BC and RQ at their intersection. /. AB _L plane NN. (Why T) Q. . D. 558. COR. 1. If two planes are perpendicular to each other, a perpendicular to one of them at any point of their intersection will lie in the other plane. For, in the above figure, a -L erected at the point B in the plane MN must coincide with AB lying in the plane PQ and _L MN, for at a given point in a plane only one JL can be drawn to that plane (Art. 515). 559. COR. 2. If two planes are perpendicular to each other, a perpendicular to one plane, from a point in the other plane, will lie in tlie other plane. 344 BOOK VI. SOLID GEOMETRY PROPOSITION XIX. THEOREM 560. If two intersecting planes are each perpendicular to a third plane., their line of intersection is perpendicular to the third plane. Given the planes PQ and R8 JL plane MN, and inter- secting in the line AB. To prove AB JL plane MN. Proof. At the point B in which the three planes meet erect a _L to the plane MN. This _L must lie in the plane PQ, and also in the plane R8. Art - 558 - Hence this _L must coincide with AB, the intersection of PQ and RS. Art. 508, 2. /. AB J_ plane MN. Q. . D. 561. COR. If two planes, including a right dihedral angle, are each perpendicular to a third plane, the intersec- tion of any two of the planes is perpendicular to the third plane, and each of the three lines of intersection is perpen' dicular to the other two. Ex. 1. Name all the dihedral angles on the above figure. Ex. 2. If Z CBQ=3Q, find the ratio of each pair of dihedral DIHEDRAL ANGLES 345 PROPOSITION XX. THEOREM 562. Every point in the plane which bisects a given dihedral angle is equidistant from the faces of the dihedral angle. Given plane CB bisecting the dihedral /.A-BR-D, P any point in plane BC, PQ and PT JL faces BA and BD, respectively. To prove PQ = PT. Proof. Through PQ and PT pass a plane intersecting AB in QR, BD in JBT,and BG in PR. Then plane PQT JL planes AB and D. Art. 555. /. plane PQT J_ line RB, the intersection of the planes AB and BD. Art. 560. /. I2 J_ RQ, RP and -RT. Art. 505. /. A QRP and PRT are the plane A of the dihedral ^ A-BR-P and P-BR-D. Art. 548. But these dihedral ^ are equal. Hyp. Art. 554. (Why?) (Why?) Q. . D. rt.A PQR=rt. A PRT. :. PQ=PT. 563. The locus of all points equidistant from the faces of a dihedral angle is the plane bisecting the dihedral angle. 346 BOOK VI. SOLID GEOMETRY PROPOSITION XXI. PROBLEM 564. Through any straight line not perpendicular to a given plane, to pass a plane perpendicular to the given plane. if Given the line AB not J_ plane MN. To construct a plane passing through AB and _L MN. Construction. From a point A in the line AB draw a _L AC to the plane MN. Art. 516. Through the intersecting lines AB and AC pass the plane AD. Art. 503. Then AD is the plane required. Proof. The plane AD passes through AB. Constr. Also plane AD _L plane MN, Art. 555. (for it contains AC, which is _L MN). Q. E. F. 565. COR. 1. Through a straight line not perpendicular to a given plane only one plane can be passed perpendicu- lar to that plane. For, if two planes could be passed through AB _L plane MN, this intersection AB would be J_ MN (Art. 560), which is contrary to the hypothesis. 566. COR. 2. The projection upon a plane of a straight line not perpendicular to that plane is a straight line. For, if a plane be passed through the given line J_ to the given plane, the foot of a _L from any point in the line to the given plane will be in the intersection of the two planes (Art. 559). DIHEDRAL ANGLES 347 PROPOSITION XXII. THEOREM 567. The acute angle which a line makes with its pro- jection on a plane is the least angle which it makes with any line of the plane through its foot. Given line AB meeting the plane MN in the point B, BG the projection of AB on MN, and PB any other line in the plane MN through B. To prove that Z ABC is less than Z ABP. Proof. Lay off PB equal to CB, and draw AC and AP. Then, in the A ABC and ABP, AB=AB. (Why?) BC=BP. (Why!) But AC < AP. Art. 521. /. Z ABC is less than Z ABP Art. 108. Q. . D. 568. DEF. The inclination of a line to a plane is the acute angle which the given line makes with its projection upon the given plane. Ex. 1. A plane has an inclination of 47 to each of the faces of a dihedral angle and is parallel to the edge of the dihedral angle; how many degrees are in the plane angle of the dihedral angle? Ex. 2. In the figure on page 345, if PT=QT,how large is the dihedral L A-BR-Dt if PTRT. how large is it? 348 BOOK VI. SOLID GEOMETKY PROPOSITION XXIII. PROBLEM 569. To draw a common perpendicular to any two lines 90, what kind of trihedral angles are those on the figure ? If ZPJ2$=30, what kind are they? Ex. 2. Are two trirectangular trihedral angles necessarily equal ? Prove this. Ex. 3. Are two lines which are perpendicular to the same plane necessarily parallel ? Are two planes which are perpendicular to the same plane necessarily parallel ? Are two planes which are perpen- dicular to the same line necessarily parallel ? Ex. 4. Let the pupil cut out three pieces of pasteboard of the form indicated in the accompanying figures ; cut them half through where the lines are dotted ; fold them and fasten the edges so as to form three trihedral angles, two of which (Figs. 1 and 2) shall be equal and two (Figs. 1 and 3) symmetrical. By experiment, let the pupil find which pair may be made to coincide, and which not. | Fig. 352 BOOK VI. SOLID GEOMETKY PROPOSITION XXIV. THEOREM 582. The sum of any two face angles of a trihedral angle is greater than the third face angle. Given the trihedral angle S-ABC, with angle A8C its greatest face angle. To prove Z ASB + ^BSC greater than Z A8C. Proof. In the face ASC draw SD, making ^ASD= /.ASB. Take SD = SB. In the face ASC draw the line ADC in any convenient direction, and draw AB and BC. Then, in the A ASB and ASD, SA = SA. SB = SD, and /.ASB= Z.ASD. .'. A ASB= A ASD. /. AB = AD. Also AB + BC > AC. Hence, subtracting the equals AB and BC > DC. Hence, in the A DC and DSC, SC=SC, > DC. /. Z BSC is greater than /.DSC. (Why?) (Why?) (Why?) (Why?) (Why ?) (Why?) = SD, and (Why?) Art. 108. To each of these unequals add the equals /.ASB and :. /.ASB+ Z BSC is greater than /.ASC. (Why?) Q. E. D. Ex. In the above figure, if /.ASC equals one of the other face angles at 5, as Z ASB, how is the theorem proved ? POLYHEDRAL ANGLES 353 PROPOSITION XXV. THEOREM i 583. The sum of the face angles of any convex polyhedral angle is less than four right angles. Given the polyhedral angle S-ABCDE. To prove the sum of the face A at 8 less than 4 rt. A . Proof. Pass a plane cutting the edges of the given poly- hedral angle in the points A, B, C, D, E. From any point O in the polygon ABCDE draw OA OB, OC, OD, OE. Denote the A having the common vertex S as the 8 A, and those having the common vertex as the A . Then the sum of the A of the 8 A = the sum of A of the A . Art. 134. But Z SB A + Z SBC is greater than Z ABC, ZC+Z SCD is greater than ^BCD, etc/ Art .*. the sum of the base A of the S A > the sum of the base A of the A. Ax. 9. . . the sum of the vertex A of the 8 A < the sum of the vertex A of the A, Ax. 11. (if unequals be subtracted from equals, the remainders are unequal in reverse order). But the sum of the A at = 4 rt. A . (Why?) .% the sum of face A at S < 4 rt. A . Ax. 8. Q. E. D. W 354 BOOK VI. SOLID GEOMETRY PROPOSITION XXVI. THEOREM 584. // two trihedral angles have the three face angles of one equal to the three face angles of the other, the tri- hedral angles have their corresponding dihedral angles equal, and are either equal or symmetrical, according as their corresponding face angles are arranged in the same or in reverse order. Given the trihedral A S-ABC and S'-A'B'C', having the face A ASB, A SO and BSC equal to the face A A'S'B', A'S'C' and B'S'C', respectively. To prove that the corresponding dihedral A of 8- ABC and S'-A'B'C' are equal, and that A S-ABC and S'- A'B'C' are either equal or symmetrical. Proof. On the edges of the trihedral A take 8A, SB, SO, S'A', S'B', S'C' all equal. Draw AB, AC, BC, A'B', A'C', B'C'. Then, 1. In the A ASB and A'S'B', SA = S'A f , SB = S'B', and /iASB = /.A'S'B'. (Why?) /. A A8B = A A'8'B'. (Why?) .'. AB=A'B'. (Why?) 2 In like manner AC=A'C f , and BC=B'C'. .'. A ABC=& A' B'C'. (Why?) EXERCISES ON THE LINE AND PLANE 355 3. Take D a convenient point in SA, and draw DE in the face ASB, and DF in the face ASC, each J_ 8A. DE and DF meet AB and A G in points E and F, T"P^T)Pf*f 1 VP 1 V (/or / SAB and SAC are acute). Similarly, take S f D f = SD and construct A D'E'F'. Then, in the rt. A ADE and A'D'E', AD=A'D', and /.DAE^D'A'E'. (Why ?) .-. A ADE=& A'D'E'. (Why?) /. AE=A'E f , and DE=D'E f . (Why?) 4. In like manner it maybe shown that AF=A'F' 1 and DF=D'F'. :. A AEF=& A'E'F'. (Why?) And ^7jP= JEJ'JF' (Why ?) 5. Hence, in the A DEF and D'^J^P, DE=D'E', DF = D f F f and ^7^= E'F'. (Why ?) /. A DEF=& D'E'F'. (Why?) .*. Z EDF= Z ^D^. (Why ?) But these A are the plane A of the dihedral A whose edges are SA and S 1 'A'. :. dihedral Z B-A8~C= dihedral B'-A'8'-C? Art. 552. In like manner it may be shown that the dihedral A at SB and S'B' are equal; and that those at 8C and S'G' are equal. .'. the trihedral A S and S f are either equal or sym- metrical. Arts. 579, 580. Q. E. D. EXERCISES. CROUP 64 THEOREMS CONCERNING THE LINE AND PLANE IN SPACE Ex. 1. A segment of a line not parallel to a plane is longer than its projection in the plane. Ex. 2. Equal straight lines drawn from a point to a plane are equally inclined to the plane, 356 BOOK VI. SOLID GEOMETRY B N Ex. 3. A line and plane perpendicular to the same plane are parallel. Ex. 4. If three planes intersecting in three straight lines are perpendicular to a plane, their lines of intersection are parallel. Ex. 5. If a plane bisects any line at right angles, any point in the plane is equidistant from the ends of the line. Ex. 6. Given AB _L plane MN, and AC J_ plane BS; prove BC _L NE. Ex. 7. Given PQ J_ plane MN, PR _L plane BL, and BS J_ plane .MN; prove QS I. AB. M Ex. 8. If a line is perpendicular to one of two intersecting planes, its projection on the other plane is perpendicular to the line of intersection of the two planes. Ex. 9. Given CE _L DE, AE J_ DE, and / C-AD-E a rt. dihedral Z ; prove CA J_ plane DAE. Ex. 10. The projections of two parallel lines on a plane are parallel. (Is the converse of this theorem also true ?) Ex. 11. If two parallel planes are cut by two non -parallel planes, the two lines of intersection in each of the parallel planes will make equal angles. Ex. 12. If a line is perpendicular to a plane, any plane parallel to the line is perpendicular to the plane. (Is the converse true ?) Ex. 13. In the figure to Prop. VI, given AB J_ MN and AF J. DC; prove BF J_ DC. Ex. 14. Two planes parallel to a third plane are parallel to each other. [Sue. Draw a line -L third plane.] EXERCISES ON THE LINE AND PLANE 357 Ex. 15. The projections upon a plane of two equal and parallel straight lines are equal and parallel. Ex. 16. A line parallel to two planes is parallel to their inter- section. Ex. 17. In the figure to Prop. XXII, if angle GBP is obtuse, prove the angle ABP obtuse. Ex. 18. In a quadrilateral in space (i. e., a quadrilateral whose vertices are not all in the same plane), show that the lines joining the midpoints of the sides form a parallelogram. Ex. 19. The lines joining the midpoints of the opposite sides of a quadrilateral in space bisect each other. Ex. 20. The planes bisecting the dihedral angles of a trihedral angle meet in a line every point of which is equidistant from the three faces. [SuG. See Art. 562.] Ex. 21. Given OQ bisecting PQ JL plane ROS, QR L OR, and QS _L OS', prove PR= PS, PR JL OR, and PS _L OS. Ex. 22. In a plane bisecting a given plane angle, and perpendicu- lar to its plane, every point is equidistant from the sides of the angle. [SuG. See Ex. 21; or through P any point in the bisecting plane pass planes -L to the sides of the Z , etc.] Ex. 23. In a trihedral angle, the three planes bisecting the three face angles at right angles to their respective planes, intersect in a line every point of which is equidistant from the three edges of the tri- hedral angle. Ex. 24. If two face angles of a trihedral angle are equal, the dihe- dral angles opposite them are equal. Ex. 25. In the figure to Prop. XXIV, prove that ASC+ BSC is greater than Z ASD + Z BSD. Ex. 26. The common perpendicular to two lines in space is the shortest line between them. 358 BOOK VI. SOLID GEOMETKY Ex. 27. Given MN \\ ES, and PB = pi ; prove /.ABC=/.abc, and A ABC = A abc. Ex. 28. Two isosceles trihedral angles are equal. symmetrical r\ \ Ex. 29. Any two symmetrical trihe- dral angles are equivalent. [SUG. Take SA, SB, SC, S'A', S'B', S'C', all equal. Pass planes ABC, A'B'C'. Draw SO and S'O' these planes. Then the trihedral A are divided into three pairs of isosceles symmetrical trihedral A, etc.] EXERCISES. CROUP 65 LOCI IN SPACE Find the locus of a point equidistant from Ex. 1. Two parallel planes. Ex. 3. Three given points. Ex. 2. Two given points. Ex. 4. Two intersecting lines. Ex. 5. The three faces of a trihedral angle. Ex. 6. The three edges of a trihedral angle. Find the locus Ex. 7. Of all lines passing through a given point and parallel to a given plane. Ex. 8. Of all lines perpendicular to a given line at a given "point in the line. Ex. 9. Of all points in a given plane equidistant from a given point outside the plane. Ex. 10. Of all points equidistant from two given points and from two parallel planes. Ex. 11. Of all points equidistant from two given points and from two intersecting planes. Ex. 12. Of all points at a given distance from a given plane and equidistant from two intersecting lines. EXERCISES ON THE LINE AND PLANE 359 EXERCISES. CROUP 66 PROBLEMS CONCERNING THE POINT, LINE AND PLANE IN SPACE Ex. 1. Through a given point pass a plane parallel to a given plane. Ex. 2. Through a given point pass a plane perpendicular to a given plane. Ex. 3. Through a given point to construct a plane parallel to two given lines which are not in the same plane. Prove that only one plane can be constructed fulfilling the given conditions. Ex. 4. Bisect a given dihedral angle. Ex. 5. Draw a plane equally inclined to three lines which meet at a point. Ex. 6. Through a given point draw a line parallel to two given intersecting planes. Ex. 7. Find a point in a plane such that lines drawn to it from two given points without the plane make equal angles with the plane. [Sua. See Ex. 23, p. 176.] Ex. 8. Find a point in a given line equidistant from two given points. Ex. 9. Find a point in a plane equidistant from three given points. Ex. 10. Find a point equidistant from four given points not in a plane. Ex. 11. Through a given point draw a line which shall intersect two given lines. [SUG. Pass a plane through the given point and one of the given lines, and pass another plane through the given point and the other given line, etc.] Ex. 12. Through a given point pass a plane cutting the edges of a tetrahedral angle so that the section shall be a parallelogram. [SuG. Produce each pair of opposite faces to intersect in a Straight line, etc.] BOOK VII Polyhedron POLYHEDRONS 585. A polyhedron is a solid bounded by 586. The faces of a polyhe- dron are its bounding planes; the edges of a polyhedron are the lines of intersection of its faces. A diagonal of a polyhedron is a straight line joining two of its vertices which are not in the same face. The vertices of a poly- hedron are the points in which its edges meet or intersect. 587. A convex polyhedron is a polyhedron in which a section made by any plane is a convex polygon. Only convex polyhedrons are to be considered in this book. 588. Classification of polyhedrons. Polyhedrons are sometimes classified according to the number of their faces. Thus, a tetrahedron is a polyhedron of four faces; a hexahedron is a polyhedron of six faces ; an octahedron is one of eight, a dodecahedron one of twelve, and an icosahedron one of twenty faces. Tetrahedron Cube Octahedron Dodecahedron icosahedron (360) POLYHEDEONS 361 The polyhedrons most important in practical life are those deter- mined by their stability, the facility with which they can be made out of common materials, as wood and iron, the readiness with which they can be packed together, etc. Thus, prism means "something sawed off." PRISMS AND PARALLELOPIPEDS 589. A prism is a polyhedron bounded by two parallel planes and a group of planes whose lines of intersection are parallel, 590. The bases o$ a prism are the faces formed by the two parallel planes; Prism the lateral faces are the faces formed by the group of planes whose lines of intersection are parallel. The altitude of a prism is the perpendicular distance between the planes of its bases. The lateral area of a prism is the sum of the areas of the lateral faces. 591. Properties of a prism inferred immediately. 1. The lateral edges of a prism are equal, for they are parallel lines included between parallel planes (Art. 589) and are therefore equal (Art. 532). 2. The lateral faces of a prism are parallelograms (Art. 160), for their sides formed by the lateral edges are equal and parallel. 3. The bases of a prism are equal polygons, for their homologous sides are equal and parallel, each to each, (being opposite sides of a parallelogram), and their homol- ogous angles are equal (Art. 538). 592. A right section of a prism is a section made by a plane perpendicular "to the lateral edges. 362 BOOK VII. SOLID GEOMETRY 593. A triangular prism is a prism whose base is a triangle ; a quadrangular prism is one whose base is a quadrilateral, etc. ObliQiie Prism? Right Prism Regular Prism 594. An oblique prism is a prism whose lateral edges are oblique to the bases. 595. A right prism is a prism whose lateral edges are perpendicular to the bases. 596. A regular prism is a right prism whose bases are regular polygons. 597. A truncated prism is that part of a prism included between a base and a section made by a plane eliliqootQ the base and cutting all the lateral edges. Truncated Prism 598. A parallelepiped is a prism whose bases are paral- lelograms. Hence, all the faces of a parallelepiped are parallelograms. Oblique Parallelepiped Right Rectangular Cube Parallelepiped Parallelepiped 599. A right parallelepiped is a parallelepiped whose lateral edges are perpendicular to the bases, PRISMS AND PAKALLELOPIPEDS 363 600. A rectangular parallelepiped is a right parallele- piped whose bases are rectangles. Hence, all the faces of a rectangular parallelopiped are rectangles. 601. A cube is a rectangular parallelopiped whose edges are all equal. Hence, all the faces of a cube are squares. 602. The unit of volume is a cube whose edge is equal to some linear unit, as a cubic inch, a cubic foot, etc. o '*"*{' 603. The volume of a solid is the number of units of volume which the solid contains. Being a number, a volume may often be determined from other numbers in certain expeditious ways, which it is one of the objects of geometry to determine. 604. Equivalent solids are solids whose volumes are equal. Ex. 1. What is the least number of faces which a polyhedron can have ? Ex. 2. A square right prism is what kind of a parallelopiped ? Ex. 3. Are there more right parallelepipeds or rectangular paral- lelepipeds ? That is, which of these includes the other as a special case ? Ex. 4. Prove that if a given straight line is perpendicular to a given plane, and another straight line is perpendicular to another plane, and the two planes are parallel, then the two given lines are ^parallel. 364 BOOK vn. SOLID GEOMETRY O PROPOSITION I. THEOREM 605. Sections of a prism made by parallel planes cutting all the lateral edges are equal polygons. Given the prism PQ cut by || planes forming the sections AD and A'D' . To prove section AD = section A'D' . Proof. AB, EC, CD, etc., are || A'B', B'C', C'D', etc., respectively. Art. 531. /. AB, BC, <7D ? etc., are equal to A'B', B'C', C'D' r etc., respectively. Art. 157. Also A ABC, BCD, etc., are equal to A. A' B'C', E'UD', etc., respectively. Art. 538. /. ABCDE=A'B'C'D'E', Art. 47. (for the polygons have all their parts equal, each to each, and .'. can be made to coincide). Q. E. D. 606. COR. 1. Every section of a prism made by a plane parallel to the base is equal to the base. 607. COR. 2. All right sections of a prism are equal. PRISMS 365 PROPOSITION II. THEOREM 608. The lateral area of a prism is equal to the product of the perimeter of a right section by a lateral edge. Given the prism RQ, with its lateral area denoted by 8 and lateral edge by E\ and AD a right section of the given prism with its perimeter denoted by P. To prove 8=PXE. Proof. In the prism RQ, each lateral edge = ^7. Art. 591, 1. Also AB JL OR, BG JL IJ, etc. Art. 505. Hence area CU RH=AB X GH=AB X E^\ area C3 GJ= BC X E, \ Art. 385. area EU IQ = CD X E, etc. But S, the lateral area of the prism, equals the sum of the areas of the 7 forming the lateral surface. /. adding, 8= (AB + BG + CD + etc. ) X E. Ax. 2. Or S=PXE. Q. . D. 609. COR. The lateral area of a right prism equals the product of the perimeter of the base by the altitude. Ex. Find the lateral area of a right prism whose altitude is 12 in., and whose base is an equilateral triangle with a side of 6 in. Also find the total area of this figure. 366 BOOK VII. SOLID GEOMETRY O PROPOSITION III. THEOREM i 610. If two prisms have the three faces including a trihedral angle of one equal, respectively, to the three faces including a trihedral angle of the other, and similarly placed, the prisms are equal. Given the prisms AJ and A f J f , having the faces AK, AD, AG equal to the faces A'K', A'D', A'O', respectively, and similarly placed. To prove AJ=A'J'. Proof. The face A EAF, EAB and EAF are equal, respectively, to the face A E'A'F', E f A'B f and B'A'F'. Hyp. /. trihedral ZA = trihedral /.A' ' Art. 584. Apply the prism A'J' to the prism AJ, making each of the faces of the trihedral Z A f coincide with corresponding equal face of the trihedral /.A. Geom. Ax. 2. .*. the plane F'J' will coincide in position with the plane FJ, Art. 500. (for the points G f , F f , K' coincide with G, F, K, respectively). Also the point C r will coincide with the point C. :. G'R' will take the direction of CH. Geom. Ax. 3. .'. R' will coincide with H. Art. 508, 1. In like manner J 1 will coincide with J. Hence the prisms AJ and A'J' coincide in all points. .*. AJ=A'J f . Art. 47. 611. COR. 1. Two truncated prisms are equal if tlie three faces including a trihedral angle of one are equal to the three faces including a trihedral angle of the other. PRISIMS 367 612. COR. 2. Two right prisms are equal if they have equal l)ases and equal altitudes. PROPOSITION IV. THEOREM 613. An oblique prism is equivalent to a right prism whose base is a right section of the oblique prism and whose altitude is equal to a lateral edge of the oblique prism. K Given the oblique prism AD', with the right section FJ; also the right prism FJ f whose lateral edges are each equal to a lateral edge of AD' . To prove AD' =0= FJ'. Proof. AA = FF'. Hyp. Subtracting FA from each of these, AF=A'F'. (Why?) Similarly BO = B'G r . Also AB = AB', and FG=F'G'. Art. 155. And A of face AG = homologous A of face A'O'. Art. 130. /. face A6r = face A'G', Art. 47. (for they have all their parts equal, each to each, and .". can be made to coincide). In like manner face AK=facQ A'K' . But face AD = face AD'. Art. 591, 3. .*. truncated prism AJ= truncated prism AJ'. Art. 611. To each of these equals add the solid FD' . ;.AD'^FJ f . (why?) Q. . P. 368 BOOK VII. SOLID GEOMETRY PROPOSITION V. THEOREM 614. The opposite lateral faces of a parallelopiped are equal and parallel. Given the parallelepiped AH with the base AC. To prove AG = aud \\ EH, and AJ= and || EH. Proof. The base AC is a ZZ7 . Art. 598. /. AB=and \\ EG. (Why?) Also the lateral face AJ is a. 17 , Art. 591, 2. .'. JLF=and || EJ. . (Why ?) /. Z J3J^= Z 0^7. Art. 538. And ~7AG=JEH. Art.162. Also plane AG \\ plane J#. Art. 538. In like manner it may be shown that AJ and BH are equal and parallel. Q. E. D. 615. COR. Any two opposite -faces of a parallelopiped may be taken as the bases. Ex. 1. How many edges has a parallelopiped ? How many faces ? How many dihedral angles ? How many trihedral angles ? Ex. 2. Find the lateral area of a prism whose lateral edge is 10 and whose right section is a triangle whose sides are 6, 7, 8 in. Ex. 3. Find the lateral area of a right prism whose lateral edge is 16 and whose bae is a rhombus with diagonals of 6 arid 8 in. PRISMS 369 PROPOSITION VI. THEOREM 616. A plane passed through two diagonally opposite edges of a parallelepiped divides the parallelepiped into two equivalent triangular prisms. Given the parallelepiped AH with a plane passed through the diagonally opposite edges AF and CH, forming the tri- angular prisms ABC-G and ADC-K. To prove ABC-Go ADC-K. Proof. Construct a plane J_ to one of the edges of the prism forming the right section PQRS, having the diagonal PR formed by the intersection of the plane FHCA. Then PQ \\ SR, and QR \\ PS. (Why ?) /. PQRS is a ZZ7 . (Why ?) ' .*. A PQR= A PSR. (Why ?) But the triangular prism ABC-G =c= a prism whose base is the right section PQR and whose altitude is AF. Art. 613. Also the triangular prism ADC-K o a prism whose base is the right section PSR and whose altitude is AF. (Why?) But the prisms having the equal bases, PQR and PSR, and the same altitude, AF, are equal. Art. 612. /. ABC-G =0= ADC-K. Ax l. Q. . D. 370 BOOK VII. SOLID GEOMETRY PROPOSITION VII. THEOREM 617. If two rectangular parallelepipeds have equal bases, they are to each other as their altitudes. Given the rectangular parallelepipeds P f and P having equal bases and the altitudes A'B' and AB. To prove P f : P=A'B' : AB. CASE I. When the altitudes A'B' and AB are com- mensurable. Proof. Find a common measure of A'B' and AB, as AK, and let it be contained in A'B' n times and in AB m times. Then A'B' : AB = n : m. Through the points of division of A'B' and AB pass planes parallel to the bases. These planes will divide P' into n, and P into m small rectangular parallelepipeds, all equal. Art. 612. /. P : P=n : m. :. P' : P=A'B' : AB. (Why ?) CASE II. When the altitudes A'B' and AB are incom- mensurable. Let the pupil supply the proof, using the method of limits. (See Art. 554). 618. DEF. The dimensions of a rectangular parallele- piped are the three edges which meet at one vertex. PAKALLELOPIPEDS 371 619. COR. If two rectangular parallelopipeds have two dimensions in common, they are to each other as their third, dimensions. PROPOSITION VIII. THEOREM 620. Two rectangular parallelopipeds having equal alti- tudes are to each other as their bases. Given the rectangular parallelopipeds P and P f having the common altitude a, and the dimensions of their bases b, c and b f , c', respectively. P bXc To prove = ' Then Art. 619. Proof. Construct the rectangular parallelepiped, Q, whose altitude is a and the dimensions of whose base are b and c'. P = c_ Q c'' AISO J = | 7 . (Why?) Multiplying the corresponding members of these equalities, = -*.. Ax. 4. f b'Xc' Q. E. D. 621. COR. Two rectangular parallelopipeds having one dimension in common are to each other as the products of the other two dimensions. 372 BOOK VII. SOLID GEOMETRY PROPOSITION IX. THEOREM 622. Any two rectangular parallelepipeds are to each other as the products of their three dimensions. Given the rectangular parallelepipeds P and P f having the dimensions a, 5, c and a f , ~b' ', c', respectively. P aXbXc Proof. Construct the rectangular parallelepiped Q hav- ing the dimensions a, &, c f '. P = c_ Q c'' Q = aX b P 1 a' X I'' Multiplying the corresponding members of these equalities, P n . V 7) V r. Ax. 4. Q. E. B. Then Also Art. 619, Art. 621. P ^ a XbX c P' o'X&'Xc' Ex. 1. Find the ratio of the volumes of two rectangular parallelo- pipeds whose edges are 5, 6, 7 in. and 7, 8, 9 in. Ex.2. Which will hold more, a bin 10x2x7 ft., or one 8x 4 x 5 ft. ? Ex. 8. How many bricks 8 x 4 x 2 in. are necessary to build a wall 80x6 ft, x8 in. f PAKALLELOPIPEDS 373 PROPOSITION X. THEOREM 623. The volume of a rectangular parallelopiped is equal to the product of its three dimensions. u Given the rectangular parallelopiped P having the three dimensions a, 6, c. To prove volume of P=a Xb Xc. Proof. Take as the unit of volume the cube Z7, whose edge is a linear unit. P aX bXc rhen rTxTxV Art ' 622 ' The volume of P is the number of times P contains the p unit of volume U, or Art. 603. /. volume of P=a X b X c. Q. E. D. For significance of this result, see Art. 2. 624. COR. 1. The volume of a cube is the cube of its edge. 625. COR. 2. The volume of a rectangular parallelo- piped is equal to the product of its base by its altitude. Ex. 1. Find the number of cubic inches in the volume of a cube whose edge is 1 ft. 3 in. How many bushels does this box contain, if lbushel = 2150.42cu. in. ? Ex. 2. The measurement of the volume < f a cube reduces to the measurement of the length of what single straight line ? 374 BOOK VII. SOLID GEOMETRY PROPOSITION XI. THEOREM 626. The volume of any parallelopiped is equal to the product of its base ly its altitude. Given the oblique parallelopiped P, with its base denoted by B, and its altitude by H. To prove volume of P=B X H. Proof. In P produce the edge CD and all the edges parallel to CD. On CD produced take FG=CD. Pass planes through F and G J_ the produced edges, forming the parallelopiped Q, with the rectangular base denoted by B f . Similarly produce the edge GI and all the edges || GI. Take IK= GI, and pass planes through 7 and K J_ the edges last produced, forming the rectangular parallelopiped R, with its base denoted by B' f . Then Also But Or [Outline Proof. volume of R = B" X H. volume of P=B"XH. volume of P=BX H. ^R=B" XH=BX H.] Art. 613. Art. 386. Art. 625. Ax. 1. Ax. 8. Q. . P. 'PRISMS 375 PROPOSITION XII. THEOREM 627. The volume of a triangular prism is equal to the product of its base by its altitude. Given the triangular prism PQR-M, with its volume denoted by F, area of base by J5, and altitude by H. To prove V=BXH. Proof. Upon the edges PQ, QR, QM, construct the parallelepiped QK. Hence QK= twice PQR-M. Art. 616. But volume of QK= area PQRT X H. Art. 626. = 2BXH. Ax. 8. .'. twice volume PQR-M=2B X H. Ax. i. .*. volume PQR-M=B X H. Ax. 5. Q. E. D. Ex. 1. If the altitude of a triangular prism is 18 in., and the base is a right triangle whose legs are 6 and 8 in., find the volume. Ex. 2 Find the volume of a triangular prism whose altitude is 24, and the edges of whose base are 7, 8, 9. Also find the total surface. tfc $- -c 376 BOOK VII. SOLID GEOMETRY o PROPOSITION XIII. THEOREM 628. The volume of any prism is equal to the product of its base by its altitude. L Given the prism AK, with its volume denoted by F, area of base by B, and altitude by H. To prove V=BXH. Proof. Through any lateral edge, as AF, and the diag- onals of the base, AC and AD, drawn from its foot, pass 'planes. These planes will divide the prism into triangular prisms. Then F, the volume of the prism AK, equals the sum of the volumes of the triangular prisms. Ax. 6. But the volume of each triangular prism = its base X H. Hence the sum of the volumes of the A prisms = the sum of the bases of the A prisms X H. = BXH. Ax. 8. /. V=BXH. AX. l. Q. E. D. 629. COR. 1. Two prisms are to each other as the pro- ducts of their bases by their altitudes; prisms having equiva- lent bases and equal altitudes are equivalent. 630. COR. 2. Prisms having equivalent bases are to each other as their altitudes; prisms having equal altitudes are to each other as their bases. PYKAMIDS 377 Pyramid PYRAMIDS 631. A pyramid is a polyhedron bounded by a group of planes passing through a common point, and by another plane cutting all the planes of the group. 632. The base of a pyramid is the face formed by the cutting plane; the ^~ lateral faces are the faces formed by the group of planes passing through a com- mon point; the vertex is the common point through which the group of planes passes ; the lateral edges are the inter- sections of the lateral faces. The altitude of a pyramid is the perpendicular from the vertex to the plane of the base. The lateral area is the sum of the areas of the lateral faces. 633. Properties of pyramids inferred immediately. 1. The lateral faces of a pyramid are triangles (Art. 508, 2). 2. The base of a pyramid is a polygon (Art. 508, 2). 634. A triangular pyramid is a pyramid whose base is a triangle; a quadrangular pyramid is a pyramid whose base is a quadrilateral, etc. A triangular pyramid is also called a tetrahedron, for it has four faces. All these faces are triangles, and any one of them may be taken as the base. 635. A regular pyramid is a pyramid whose base is a regular polygon, and the foot of whose altitude coincides with the center of the base. 378 BOOK VII. SOLID GEOMETRY 636. Properties of a regular pyramid inferred immedi- ately. 1. The lateral edges of a regular pyramid are equal, for they are oblique lines drawn from a point to a plane cutting off equal distances from the foot of the per- pendicular from the point to the plane (Art. 518). 2. The lateral faces of a regular pyramid are equal isos- celes triangles. 637. The slant height of a regular pyramid is the alti- tude of any one of its lateral faces. The axis of a regular pyramid is its altitude. 638. A truncated pyramid is the portion of a pyramid included between the base and a section cutting all the lateral edges. 639. A frustum of a pyramid is the part of a pyramid included between the base and a plane parallel to the base. The altitude of a frustum of a pyramid is the perpendicular distance between the planes of its bases. 640. Properties of a frustum of a pyramid inferred im- mediately. 1. The lateral faces of a frustum of a pyramid are trapezoids. 2. The lateral faces of a frustum of a regular pyramid are equal isosceles trapezoids. 641. The slant height of the frustum of a regular pyra- mid is the altitude of one of its lateral faces. Ex. 1. Show that the foot of the altitude of a regular pyramid coincides with the center of the circle circumscribed about the base. Ex. 2. The perimeter of the midsection of the frustum of a pyra- mid equals one -half the sum of the perimeters of the bases, PYKAMIDS 379 PROPOSITION XIV. THEOREM 642. The lateral area of a regular pyramid is equal to half the product of the slant height by the perimeter of the base. o Given 0-ABCDF a regular pyramid with its lateral &rea denoted by 8, slant height by L, and perimeter of its base by P. To prove S=%LXP. Proof. The lateral faces GAB, OBC, etc., are equal isosceles A. Art. 636, 2. Hence each lateral face has the same slant height, L. .'. the area of each lateral face = J L X its base. .'. the sum of all the lateral faces = J L X sum of bases. . 643. COR. The lateral area of Hie frustum of a regular pyramid is equal to one-half the sum of the perimeters of its bases multiplied by its slant height. Ex. Find the lateral area of a regular square pyramid whose slant height is 32, and an edge of whose base is 16. Find the total area also, 5 380 BOOK VII. SOLID GEOMETRY PROPOSITION XV. THEOREM 644. If a pyramid is cut by a plane parallel to the lase, I. The lateral edges and the altitude are divided pro- portionally; II. The section is a polygon similar to the base. Given the pyramid S-ABCDF, with the altitude SO cut by a plane MN, which is parallel to the base and intersects the lateral edges in a, b, c, d, /and the altitude in o. Sa Sb Sc So To prove I. = =-^= = ^Q' II. The section abcdf similar to the base ABCDF. Proof. I. Pass a plane through the vertex 8 II MN. Then SA, SB, SC . . . SO, are lines intersected by three || planes. -^ = _^ = _^L = . . . = ^L. Art. 539. " SA SB SC SO II. ab || AB. (Why?) .*. A Sab and SAB are similar. Art. 328. In like manner the A Sbc, Scd, etc., are similar to the A SBC, SCD, etc., respectively. ab _ / Sb \ _ be _ / Sc \ _ cd _ ' AE SB \ ) BC \SC CD PVKAMIDS 381 That is, the homologous sides of abcdf and ABCDF are proportional. Also Z.dbc = Z ABC, bcd = Z BCD, etc. Art. 538. .*. section obcdf is similar to the base ABCDF. Art. 321. Q. E. D. 645. COR. 1. A section of a pyramid parallel to the base is to the base as the square of its distance from the vertex is to the square of the altitude of the pyramid. abcdf ab For (Why ?> ABCDF ab Sa So ab 2 So But - = - = - /. - (Why!) AB 8A SO AB 2 SO 2 (Whyf) ABCDF 646. COR. 2. // two pyramids having equal altitudes are cut by a plane parallel to their bases at equal distances from the vertices, the sections have the same ratio as the bases. Let S-ABCDF and V-PQR be two pyramids cut as described. Then = (Art . 645); also = . ( } ABCDF so 2 PQR ~VT L But VT=SO, and Vt = So. (Why?) pqr abcdf ABCDF = r - - -- (Why ?) 647. COR. 3. If two pyramids have equal altitudes and equivalent bases, sections made by planes parallel to the bases at equal distances from the vertices are equivalent. 382 BOOK VII. SOLID GEOMETRY PROPOSITION XVI. THEOREM 648. The volume of a triangular pyramid is the limit of the sum of the volumes of a series of inscribed, or of a series of circumscribed prisms of equal altitude, if the number of prisms be indefinitely increased. r o Given the triangular prism O-ABC with a series of in- scribed, and also a series of circumscribed prisms, formed by passing planes which divide the altitude into equal parts, and by making the sections so formed first upper bases, then lower bases, of prisms limited by the next parallel plane. To prove O-ABC the limit of the sum of each series, if the number of prisms in each be indefinitely increased. Proof. Each inscribed prism equals the circumscribed prism immediately above it. Art. 629. .*. (sum of circumscribed prisms) (sum of inscribed prisms) = lowest circumscribed prism, or ABC-K. If the number of prisms be indefinitely increased, the altitude of each approaches zero as a limit. Hence volume ABC-K=0, Art. 253, 2. (for its base, ABC, is constant while its altitude = 0). .*. (sum of circumscribed prisms) (sum of inscribed prisms) = 0. .*. volume O-ABC (either series of prisms) = 0. (for this difference < difference between the two series, which last difference =0). /. O-ABC is the limit of the sum of the volumes of either series of prisms. Q. E. D. PYRAMIDS PROPOSITION XVII. THEOREM 383 649. If two triangular pyramids have equal altitudes and equivalent bases, they are equivalent. o B Given the triangular pyramids O-ABC and O'-A'B'C' having equivalent bases ABC and A'B'O ', and equal altitudes. To prove O-ABC- O'-A'B'C'. Proof. Place the pyramids so that they have the com- mon altitude H, and divide H into any convenient num- ber of equal parts. Through the points of division and parallel to the plane of the bases of the pyramids, pass planes cutting the pyramids. Using the sections so formed as upper bases, inscribe a series of prisms in each pyramid, and denote the volumes of the two series of prisms by V and V. The sections formed by each plane, as KLM and K'L'M', are equivalent. Art. 647. .". each prism in O-ABC =e= corresponding prism in O f -A'B'C f (Art. 629). /. V= V f . Ax. 2. Let the number of parts into which the altitude is divided be increased indefinitely. Then V and V become variables with O-ABC and O'-A'B'C' as their respective limits. Art. 648. But V^V always. (Why?) .-. O-ABC^O'-A'B'C', (Why?) Q. E. D. 384 BOOK VII. SOLID GEOMETRY PROPOSITION XVIII. THEOREM 650. The volume of a triangular pyramid is equal tc one-third the product of its base ly its altitude. D Given the triangular pyramid 0-ABC, having its volume denoted by V, the area of its base by B, and its altitude by 27". To prove V=$BXH. Proof. On ABC as a base, with OB as a lateral edge, construct the prism ABC-DOF. Then this prism will be composed of the original pyra- mid O-ABC and the quadrangular pyramid O-ADFC. Through the edges OD and OC pass a plane intersecting the face ADFC in the line DC, and dividing the quadrangular pyramid into the triangular pyramids 0-ADC and 0-DFC. Then 0-ADC^ 0-DFC. Art. 649. (for they have the common vertex O, and the equal bases ADC and DFC}. But 0-DFC may be regarded as having C as its vertex and DOF as its base. Art. 634. /. 0-DFC^O-ABC. Art. 649.* .*. the prism is made up of three equivalent pyramids. .*. 0-ABC= J the prism. ^ Ax. 5. But volume of prism = B X H. Art. 627. .-. 0-ABC, or F=J B X H. Ax. 5. 0. E. D. Ex. Find the volume of a triangular pyramid whose altitude is 12 ft., and whose base is an equilateral triangle with a side of 15 ft. PYRAMIDS 385 PROPOSITION XIX. THEOREM 651. The volume of any pyramid is equal to one-third the product of its base by its altitude. Given the pyramid 0-ABCDF, having its volume denoted by V, the area of its base by B, and its altitude by H. To prove V=$BXH. Proof. Through any lateral edge, as OD, and the diago- nals of the base drawn from its foot, as AD and BD, pass planes dividing the pyramid into triangular pyramids. Then V, the volume of the pyramid 0-ABCDF, will equal the sura of the volumes of the triangular pyramids. But the volume of each A pyramid = & its base X H. Art. 650. Hence the sum of the volumes of A pyramids = J sum of their bases X H. Ax. 2. = %BXH. Ax. 8. .-. V=$ BX H. Ax. 1. Q. E. D. 652. COR. 1. The volumes of two pyramids are to each other as the products of their bases and altitudes; pyramids having equivalent bases and equal altitudes are equivalent. 653. COR. 2. Pyramids having equivalent bases are to each other as their altitudes; pyramids having equal alti- tudes are to each other as their bases. 654. SCHOLIUM. The volume of any polyhedron may be found by dividing the polyhedron into pyramids, finding the volume of each pyramid separately, and taking their sum. 386 BOOK VII. SOLID GEOMETRY PROPOSITION XX. THEOREM 655. The frustum of a triangular pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean proportional between the two bases of the frustum . Given ABC-DEF the frustum of a triangular pyramid, having the area of its lower base denoted by B, the area of its upper base by b, and its altitude by H. To prove ABC-DEF =c= three pyramids whose bases are B, b and vBb, and whose common altitude is H. Proof. Through E and A C, E and DC, pass planes divid- ing the frustum into three triangular pyramids. Then 1. E-ABC has the base B and the altitude H. 2. E-DFC, that is, C-DEF, has the base b and the altitude H. 3. It remains to show that E-ADC is equivalent to a pyramid having an altitude H, and a base that is a mean proportional between B and b. Denoting the three pyramids by I, II, III, I = A ABE = AB = AC = &ADC = H II A ADE DE DF A DFC III" (Arts. 653, 391, 644, 321. Let the pupil supply the reason for each I II step in detail). or =VlXIII. Art. 303. b. Hence, ABC-DEFosum of three pyramids, as described. Q. . D. PYRAMIDS 387 656. Formula for volume of frustum of a triangular pyramid. y= J H (B + b + PROPOSITION XXI. THEOREM 657. The volume of the frustum of any pyramid is equivalent to the sum of the volumes of three pyramids, whose common altitude is the altitude of the frustum, and ivhose bases are the lower base, the upper base, and a mean propor- tional between the two bases ' hcimj in line \\ plane PACE, in 'which the bases tie). But B-ARC may be regarded as having R for its ver- tex, and ABO for its base, as desired. Art, 634, PKISMATOIDS 389 .'. ABC-PQR =0= sum of three pyramids whose common base is ABC, and whose vertices are P, Q, JR. Q. E. D. Pig. 3 Fig. 4 659. COR. 1. The volume of a truncated right triangu- lar prism (Fig. 3) is equal to the product of its base by one- third the sum of its lateral edges. 660. COR. 2. The volume of any truncated triangular prism (Fig. 4) is equal to the product of the area of its right section by one- third the sum of its lateral edges. PRISMATOIDS 661. A prismatoid is a polyhedron bounded by two polygons in parallel planes, called bases, and by lateral faces which are either triangles, trapezoids or parallelograms. 662. A prismoid is a prismatoid in which the bases have the same number of sides and have their corresponding sides parallel. Erlamoid Ex. The volume of a truncated right parallelepiped equals the area of the lower base multiplied by oue- fourth the sum of the lateral edges (or by a perpendicular from the center of the upper base to tiie Ipwer base), 390 BOOK VII. SOLID GEOMETKY PROPOSITION XXIII. THEOREM 663. The volume of a prismatoid is equal to one-sixth the product of its altitude ~by the sum of its bases and of four times the area of its midsection. G Given the prismatoid ABCD-FGK, with bases B and b, midsection M, volume F, and altitude H. To prove F=i H (B + I + 4 M). Proof. Take any point in the midsection, and through it and each edge of the prismatoid let planes be passed. These planes will divide the figure into parts as follows: 1. A pyramid with vertex 0, base ABCD and altitude J _5T, and whose volume /. =i SX B. Art. 651. 2. A pyramid with vertex 0, base FGK, and altitude J H, and whose volume .*. =i J3"X &. (Why?) 3. Tetrahedrons like 0-ABG whose volume may be determined as follows (see Fig. 2) : AB = 2 PQ. (Why?) .-. A AGB = A PGQ. Art. 398. /. 0-AGB = 4 0-PGQ. Art. 653. But 0-PGQ (or G-PQO) = i PQO X J fi=i HXPQO. :. 0-AGB = i HX4A PQO. (Why ?) .'. the sum of all tetrahedrons like 0-A GB = i jff X 4 M. Or KEGULAK POLYHEDKONS 391 REGULAR POLYHEDRONS 664. DEF. A regular polyhedron is a polyhedron all of whose faces are equal regular polygons, and all of whose polyhedral angles are equal. Thus, the cube is a regular polyhedron. PROPOSITION XXIV. THEOREM 665. But five regular polyhedrons are possible. Given regular polygons of 3, 4, 5, etc., sides. To prove that regular polygons of the same number of sides can be joined to form polyhedral A of a regular polyhedron in but five different ways, and that, conse- quently, but five regular polyhedrons are possible. Proof. The sum of the face A of any polyhedral angle < 360. Art. 583. 1. Each Z of an equilateral triangle is 60. Art. 134. 3 X 60, 4 X 60 and 5 X 60 are each less than 360; but any larger multiple of 60 = or > 360. .*. but three regular polyhedrons can be formed with equilateral A as faces. 2. Each Z of a square contains 90. Art. 151. 3 X 90 is less than 360, but any larger multiple of 90 = or > 360. .*. but one regular polyhedron can be formed with squares as faces. 3. Each Z of a regular pentagon is 108. Art. 174. 3 X 108 is less than 360, but any larger multiple of 108 > 360. 392 BOOK VII. SOLID GEOMETRY /. but one regular polyhedron can be formed with regu* lar pentagons as faces. 4. Each Z of a regular hexagon is 120, and 3 X 120 -360. .*. no regular polyhedron can be formed with hexagons, or with polygons with a greater number of sides as faces. /. but five regular polyhedrons are possible. Q. E. D. 666. The construction of the regular polyhedrons, by the use of cardboard, may be effected as follows: Draw on a piece of cardboard the diagrams given below. Cut the cardboard half through at the dotted lines and en- tirely through at the full lines. Bring the free edges together and keep them in their respective positions by some means, such as pasting strips of paper over them. Octahedron Tetrahedron Hexahedron Dodecahedron Icosahedron POLYHEDRONS 393 POLYHEDRONS IN GENERAL PROPOSITION XXV. THEOREM 667. In any polyhedron, the number of edges increased by two equals the number of vertices increased by the number of faces. p Given the polyhedron AT, with the number of its ver- tices, edges and faces denoted by F, E and F, respectively. To prove E + 2=V+F. Proof. Taking the single face ABCD, the number of edges equals the number of vertices, or E= V. If another face, CRTD, be annexed (Fig. 2), three new edges, CR, RT, TD, are added and two new vertices, R and T. :. the number of edges gains one on the number of ver- tices, or E=V+ 1. If still another face, BQRC, be annexed, two new edges, BQ and QR, are added, and one new vertex, Q. .*. E= F+2. With each new face that is annexed, the number of edges gains one on the number of vertices, till but one face is lacking. The last face increases neither the number of edges nor of vertices. Hence number of edges gains one on number of vertices, for every face except two, the first and the last, or gains F2 in all. .*. for the entire figure, E= F + F 2. That is E + 2= V + F. Ax. 2. Q. B. D. 394 BOOK VII. SOLID GEOMETRY PROPOSITION XXVI. THEOREM 668. The sum of the face angles of any polyhedron equals four right angles taken as many times, less two, as the polyhedron has vertices. Given any polyhedron, with the sum of its face angles denoted by S, and the number of its vertices, edges and faces denoted by F, E, F, respectively. To prove S=(V 2) 4 rt. A. Proof. Each edge of the polyhedron is the intersection of two faces, .'. the number of sides of the faces = 2 E. ^ :. the sum of the interior and exterior A of the faces = 2 E X 2 rt. A , or E X 4 rt. A . Art. 73. But the sum of the exterior A of each face = 4 rt. A . Art. 175 /. the sum of exterior A of the FfsLces = FX 4 rt. A . Ax. 4. Subtracting the sum of the exterior A from the sum of all the A , the sum of the interior A of the F faces = (EXlrt. A)(FX4rt. A). Or S=(EF) 4rt. A. But E + 2=V+F. Art. 667. Hence EF= F 2. Ax. 3. Substituting for E F, S= ( F 2) 4 rt. A . Ax. 8. Q. E. D. Ex. Verify the last two theorems in the case of the cube. COMPARISON OF POLYHEDRONS 395 COMPARISON OF POLYHEDRONS. SIMILAR POLYHEDRONS PROPOSITION XXVII. THEOREM 669. // two tetrahedrons have a trihedral angle of one equal to a trihedral angle of the other, they are to each other as the products of the edges including the equal trihedral angles. c c' Given the tetrahedrons O-ABC and 0'-A'B'C r , with their volumes denoted by Fand V ', respectively, and having the trihedral A and O f equal. V_ = OAXOBX PC V 0'A'XO'B'Y.O'C'' Proof. Apply the tetrahedron O'-A'B'C' to O-ABC so that the trihedral Z 0' shall coincide with its equal, the trihedral Z 0. Draw CP and G'P' JL plane QAB, and draw OP the projection of OC in the plane OAB. Taking OAB and OA'B' as the bases, and CP and C'P' as the altitudes of the pyramids C-OAB and C'-OA'B', respectively. V A OAB X OP &OAB ^ CP Art. 652. Art. 397. (Why ?) But In the V A OA A 'B' X O'P' OA OA X OB C'P' OC ~00'* X 00 A similar rt. / OA X 6 OA'B' OA' X 0' ^7P & OOPandOO'P', 7^7 = X U3 X 00 OA X 05 "7 ' OA' X ' X 00 X C VA> X O'B' X 0'( ^ . AX . o . Q. . D. 396 BOOK vii. SOLID GEOMETRY 670. DEP. Similar polyhedrons are polyhedrons having: the same number of faces, similar, each to each, and simi- larly placed, and having their corresponding polyhedral angles equal. PROPOSITION XXVIII. THEOREM 671. Any two similar polyhedrons may be decomposed into the same number of tetrahedrons, similar, each to each, and similarly placed. c Given P and P f , two similar polyhedrons. To prove that P and P f may be decomposed into the same number of tetrahedrons, similar, each to each. Proof. Take H and H f any two homologous vertices of P and P'. Draw homologous diagonals in all the faces of P and P' except those faces which meet at H and H f , sepa- rating the faces into corresponding similar triangles. Through H and each face diagonal thus formed in P, and through R' and each face diagonal in P', pass planes. Each corresponding pair of tetrahedrons thus formed may be proved similar. Thus, in H-ABC and R'-A'B'C f , the A HBA and R'E'A! are similar. Art. 329. In like manner A REG and R'E'G' are similar; and A ABC and A'B'C' are similar. SIMILAR POLYGONS 397 AC ' Art ' 321 ' /. A ARC and A'H'C f are similar. Art. 326. Hence the corresponding faces of H- ABC and H f -A'B f C f are similar. Also their homologous trihedral A are equal. Art. 584. . tetrahedron H-ABC is similar to H'-A f B'C f . Art. 670. After removing H-ABC from P, and H'-A'B'C' from P', the remaining polyhedrons are similar, for their faces are similar, and the remaining polyhedral A are equal. Ax. 3. By continuing this process, P and P f may be decom- posed into the same number of tetrahedrons, similar, each to each, and similarly placed. Q. . D. 672. COR. 1. The homologous edges of similar polyhe- drons are proportional; Any two homologous lines in two similar polyhedrons have the same ratio as any other two homologous lines. 673. COB. 2. Any two homologous faces of two similar polyhedrons are to each other as the squares of any two homologous edges or lines; The total areas of any two similar polyhedrons are to each other as the squares of any two homologous edges. Ex. 1. In the figure, p. 395, if the edges meeting at OareS, 9, 12 in., and those meeting at 0' are 4, 6, 8 in., find the ratio of the volumes of the tetrahedrons. Ex. 2. If the linear dimensions of one room are twice as great as the corresponding dimensions of another room, how will their surfaces (and .'. cost of papering) compare ? How will their volumes compare ? Ex. 3. How many 2 in. cubes can be cut from a 10 in. cube ? Ex. 4. If the bases of a prismoid are rectangles whose dimensions are a, b and b, a, and altitude is //, find the formula for the volume. Vnfjy 398 BOOK VII. SOLID GEOMETRY PROPOSITION XXIX. THEOREM 674. The volumes of two similar tetrahedrons are to each other as the cubes of any pair of homologous edges. Given the similar tetrahedrons 0-ABC and O'-A'B'C'. V OA To prove ^= F _ , V OA X OB X OC Proof. - = . (Why!) OA x OS ^ 00 O'B' 0>0 OA OB OG But == ' Art ' 672 ' y 0.4 ^ OA ^ OA ~OA 3 = - X - X - = Ax. 8. F Ex. 1. In the above figures, if AB=2 A'B', find the ratio of V to F'. Find the same, if AB=H A'B'. Ex. 2 The measurement of the volume of a regular triangular prism reduces to the measurement of the lengths of how many straight lines ? of a frustum of a regular square pyramid ? Ex. 3. Show how to construct out of pasteboard a regular prism, a parallelepiped, and a truncated square prism. SIMILAK POLYGONS 399 PROPOSITION XXX. THEOREM 675. The volumes of any two similar polyhedrons are to 'each other as the cubes of any two homologous edges, or of any other two homologous lines. BO B 1 O 1 Given the polyhedrons AK and A'K' having their vol- umes denoted by V and V f t and HB and H'B' any pair of homologous edges. V _HB 3 To prove y H'B'* Proof. Let the polyhedrons be decomposed into tetra- hedrons, similar, each to each, and similarly placed. Art. 671. Denote the volumes of the tetrahedrons in P by vi, V2, vs . . . and of those in P f by v'i, 1/2, v's Then Also H f B fi !_!_ 1/1 Art. 674, Ax. 1. (for each of these ratios^. Vl jat= : -2L j that is> *1 = L. Art . 312 . HB Ax. 1. Q. E. D- 400 BOOK VII. SOLID GEOMETRY EXERCISES. CROUP 67 THEOREMS CONCERNING POLYHEDRONS Ex. 1. The lateral faces of a right prism are rectangles. Ex. 2. A diagonal plane of a prism is parallel to every lateral edge of the prism not contained in the plane. Ex. 3. The diagonals of a parallelepiped bisect each other. Ex. 4. The square of a diagonal of a rectangular parallelepiped equals the sum of the squares of the three edges meeting at a vertex. Ex. 5. Each lateral face of a prism is parallel to every lateral edge not contained in the face. Ex. 6. Every section of a prism made by a plane parallel to a lateral edge is a parallelogram. Ex. 7. If any two diagonal planes of a prism which are not par- allel to each other are. perpendicular to the base of the prism, the prism is a right prism. Ex. 8. What part of the volume of a cube is the pyramid whose "base is a face of the cube and whose vertex is the center of the cube ? Ex. 9. Any section of a regular square pyramid made by a plane through the axis is an isosceles triangle. Ex. 10. In any regular tetrahedron, an altitude equals three times the perpendicular from its foot to any face. Ex. 11. In any regular tetrahedron, an altitude equals the sum of the perpendiculars to the faces from any point within the tetrahedron. Ex. 12. Find the simplest formula for the lateral area of a trun- cated regular prism of n sides. Ex. 13. The sum of the squares of the four diagonals of a paral- lelopiped is equal to the sum of the squares of the twelve edges. [Sue. Use Art. 352.] Ex. 14. A parallelepiped is symmetrical with respect to what point? Ex. 15. A rectanglar parallelepiped is symmetrical with respect to how many planes ? (Let the pupil make a definition of a figure sym metrical with respect to a plane. See Arts. 486, 487.) EXERCISES ON POLYHEDRONS 401 Ex. 16. The volume of a pyramid whose lateral edges are the three edges of the parallelepiped meeting at a point is what part of the volume of the parallelepiped ? Ex. 17. If a plane be passed through a vertex of a cube and the diagonal of a face not adjacent to the vertex, what part of the volume of the cube is contained by the pyramid so formed ? Ex. 18. If the angles at the vertex of a triangular pyramid are right angles and each lateral edge equals a, show that the volume of the pyramid is * b Ex. 19. How large is a dihedral angle at the base of a regular pyramid, if the apothem of the base equals the altitude of the pyramid ? Ex. 20. The area of the base of a pyramid is less than the area of the lateral surface. Ex. 21. The section of a triangular pyramid by a plane parallel to two opposite edges is a parallelogram. If the pyramid is regular, what kind of a parallelogram does the section become ? Ex. 22. The altitude of a regular tetrahedron divides an altitude of the base into segments which are as 2 : 1. Ex. 23. If the edge of a regular tetrahedron is a, show that the slant height is -~- ; and hence that the altitude is ^ , and the vol- 2t o Ex. 24. If the midpoints of all the edges of a tetrahedron except two opposite edges be joined, a parallelogram is formed. Ex. 25. Straight lines joining the midpoints of the opposite edges of a tetrahedron meet in a point and bisect each other. Ex. 26. The midpoints of the edges of a regular tetrahedron are, the vertices of a regular octahedron. 402 BOOK VII. SOLID GEOMETRY EXERCISES. GROUP 68 PROBLEMS CONCERNING POLYHEDRONS Ex. 1 . Bisect the volume of a given prism by a plane parallel to the base. Ex. 2. Bisect the lateral surface of a given pyramid by a plane par- allel to the base. Ex. 3. Through a given point pass a plane which shall bisect the volume of a given parallelepiped. Ex. 4. Given an edge, construct a regular tetrahedron. Ex. 5. Given an edge, construct a regular octahedron. Ex. 6. Pass a plane through the axis of a regular tetrahedron so that the section shall be an isosceles triangle. Ex. 7. Pass a plane through a cube so that the section shall be a regular hexagon. Ex. 8. Through three given lines no two of which are parallel pass planes which shall form a parallelepiped. Ex. 9. From cardboard construct a regular square pyramid each of whose faces is an equilateral triangle. EXERCISES. CROUP 69 REVIEW EXERCISES Make a list of the properties of Ex. 1. Straight lines in space. Ex. 9. Right prisms. Ex. 2. One line and one plane. Ex. 10. Parallelopipeds in gen- Ex. 3. Two or more lines and one plane. Ex. 11. Rectangular parallelo- Ex. 4. Two planes and one line. pipeds. Ex.5. Two planes and two lines. Ex 12 Pyramids in general. Ex. 6. Polyhedrons in general. Ex - 13 - Regular pyramids. Ex. 7. Similar polyhedrons. Ex. 14 Frusta of pyramids. Ex.8. Prisms in general. Ex.15. Truncated prisms. BOOK VIII CYLINDERS AND CONES CYLINDERS 676. A cylindrical surface is a curved surface generated .by a straight line which moves so as constantly to touch a given fixed curve and constantly be parallel to a given fixed straight line. Thus, every shadow cast by a point of light at a great distance, as by a star or the sun, approximates the cylindrical form, that is, is bounded Cylindrical surface by a cylindrical surface of light. Hence, in all radiations (as of light, heat, magnetism, etc.) from a point at a great distance, we are concerned with cylindrical surfaces and solids. 677. The generatrix of a cylindrical surface is the mov- ing straight line; the directrix is the given curve, as CDE\ an element of the cylindrical surface is the moving straight line in any one of its positions, as DF. 678. A cylinder is a solid bounded by a cylindrical surface and by two parallel planes. The bases of a cylinder are its parallel plane faces; the lateral surface is the cylindrical surface included between the parallel planes forming its bases; the alti- cylinder tude of a cylinder is the distance between the bases. The elements of a cylinder are the elements of the cylindrical surface bounding it. (403) 404 BOOK VIII. SOLID GEOMETRY 679. Property of a cylinder inferred immediately. All the elements of a cylinder are equal, for they are parallel lines included between parallel planes (Arts. 532, 676). The cylinders most important in practical life are those determined by their stability, the ease with which they can be made from com- mon materials, etc. 680. A right cylinder is a cylinder whose elements are perpendicular to the Oblique circular cylinder 681. An oblique cylinder is one whose elements are oblique to the bases. 682. A circular cylinder is a cylinder whose bases are circles. 683. A cylinder of revolution is a cylin- der generated by the revolution of a rect- angle about one of its sides as an axis. Hence, a cylinder of revolution is a right circular cylinder. Some of the properties of this solid are derived most readily by considering it as generated by a re- volving rectangle ; and others, by regarding it as a particular kind of cylinder derived from the general definition. 684. Similar cylinders of revolution are cylinders gen- erated by similar rectangles revolving about homologous sides. 685. A tangent plane to a cylinder is a plane which contains one element of the cylinder, and which does not cut the cylinder on being produced. Ex. 1. A plane passing through a tangent to the base of a circu- lar cylinder and the element drawn through the point of contact is tangent to the cylinder. (For if it is not, etc.) Ex. 2. If a plane is tangent to a circular cylinder, its intersection the itlane of the base is tangent to the base. CYLINDEKS 405 686. A prism inscribed in a cylinder is a prism whose lateral edges are elements of the cylinder, and whose bases are polygons inscribed in the bases of the cylinder. inscribed prism Circumscribed prism 687. A prism circumscribed about a cylinder is a prism whose lateral faces are tangent to the cylinder, and whose bases are polygons circumscribed about the bases of the cylinder. 688. A section of a cylinder is the figure formed by the intersection* of the cylinder by a plane. A right section of a cylinder is a section formed by a plane perpendicular to the elements of the cylinder. 689. Properties of circular cylinders. By Art. 441 the area of a circle is the limit of the area of an inscribed or circumscribed polygon, and the circumference is the limit of the perimeters of these polygons; hence 1. The volume of a circular cylinder is the limit of the volume of an inscribed or circumscribed prism. 2. The lateral area of a circular cylinder is the limit of the lateral area of an inscribed or circumscribed prism. Also, 3. By methods loo advanced for this book, it may be proved that the perimeter of a right section is the limit of the perimeter of a right sec- tion of an inscribed or circumscribed jpr ism. 406 BOOK VIII. SOLID GEOMETKY PROPOSITION I. THEOREM 690. Every section of a cylinder made by a plane pass- ing through an element is a parallelogram. Given the cylinder AQ cut by a plane passing through the element AB and forming the section ABQP. To prove ABQP a ZZ7 . Proof. AP\\BQ. Art. 531. It remains to prove that v P$ is a straight line || AB. Through P draw a line in the cutting plane || AB. This line will also lie in the cylindrical surface. Art. 676. .*. this line must coincide with PQ, (for the line drawn lies in both the cutting plane and the cylindrical surface, hence, it must be their intersection). :. PQ is a --straight line 1 1 AB. :. ABQP is a ZZ7 . (Why ?) Q. E. D. 691. COR. Every section of a right cylinder made by a plane passing through an element is a rectangle. Ex. 1. A door swinging on its hinges generates what kind of a solid f Ex. 2. Every section of a parallelepiped made by a plane inter- eectiug all its lateral edges is a parallelogram, CYLINDEKS 407 PROPOSITION II. THEOREM 692. The bases of a cylinder are equal. Given the cylinder AQ with the bases APB and CQD. To prove base APJ5 = base CQD. Proof. Let AC and BD be any two fixed elements in the surface of the cylinder AQ. Take P, any point except A and B in the perimeter of the base, and through it draw the element PQ. Draw AB, AP, PB, CD, CQ, QD. Then AC and BD are = and ||. (Why?) .-. AD is a ZZ7. (Why?) Similarly AQ and BQ are CU . :. AB=CD, AP=CQ, and BP=DQ. (Why?) /. A APB = A CQD. (Why ?) Apply the base APB to the base CQD so that AB coin- cides with CD. Then P will coincide with Q, (for A APB = & CQD). But P is any point in the perimeter of the base APB. .*. every point in the perimeter of the lower base will coincide with a corresponding point of the perimeter of the upper base. .*. the bases will coincide and are equal. Art. 47. 9. . D. 408 BOOK VIII. SOLID GEOMETRY 693. COR. 1. The sections of a cylinder made by two parallel planes cutting all the elements arc equal. For the sections thus formed are the bases of the cylinder included between the cutting planes. 694. COR. 2. Any section of a cylinder parallel to the base is equal to the base. PROPOSITION III. THEOREM 695. The lateral area of a circular cylinder is equal to the product of the perimeter of a right section of the cylin- der by an element. Given the circular cylinder AJ, having its lateral area denoted by 8, an element by E, and 'the perimeter of a right section by P. To prove S=PXE. Proof. Let a prism with a regular polygon for its base be inscribed in the cylinder. Denote the lateral area of the inscribed prism by $', and the perimeter of its right section by P f . Then the lateral edge of the inscribed prism is an ele- ment of the cylinder. Constr CYLINDERS 409 Art. 608. If the number of lateral faces of the inscribed prism be indefinitely increased, 8' will approach 8 as a limit. Art. 689, 2. P' will approach P as a limit. Art. 689, 3. And P'XE will approach PXE as a limit. Art. 253, 2. But S' = P'XE always. (Why?) /. S=PXE. (Why?) Q. . D. 696. COR. 1. The lateral area of a cylinder of revolu- tion is equal to the product of the circumference of its base by its altitude. 697. Formulas for lateral area and total area of a cylin- der of revolution. Denoting the lateral area of a cylinder of revolution by 8, the total area by T, the radius by R, and the altitude by H. 8=2 nRH. T=2 nRH + 2 nR 2 :. T=2 nR (H+ R). Ex. 1. If, in a cylinder of revolution, H=W in. and R=7 in., find 8 and T. Ex. 2. If the altitude of a cylinder of revolution equals the radius of the base (H=R), what do the formulas for 8 and T become in terms of -K ? also, in terms of H ? Ex. 3. What do they become, if the altitude equals the diameter of the base ? Ex. 4. In a cylinder of revolution, what is the ratio of the lateral area to the area of the base ? to the total area ? 410 BOOK VIII. SOLID GEOMETRY PROPOSITION IV. THEOREM 698. The volume of a circular cylinder is equal to the product of its base by its altitude. Given the circular cylinder AJ, having its volume denoted by F, its base by B, and its altitude by H. To prove V=BXH. Proof. Let a prism having a regular polygon for its base be inscribed in the cylinder, and denote the volume of the inscribed prism by T 7 , and its base by B' '. The prism will have the same altitude, H, as the cylinder. .-. V' = B'XH. (Why?) If the number of lateral faces of the inscribed prism be indefinitely increased, V will approach V as a limit. Art. 689, 1. B' will approach B as a limit. (Why?) And B' X H will approach B X H as a limit (Why ?) But V' = B'XH always, (Why ?) /. V=BXH. (Why?) Q. E. D. 699. Formula for the volume of a circular cylinder, By use of Art. 450, CYLINDERS PROPOSITION V. THEOREM 411 700. The lateral areas, or the total areas, of two simi- lar cylinders of revolution are to each other as the squares of their radii, or as the squares of their altitudes; and their volumes are to each other as the cubes of their radii, or as the cubes of their altitudes. H' Given two similar cylinders of revolution having their lateral areas denoted by S and $', their total areas by T and T', their volumes by Fand V, their radii by R and R', and their altitudes by H and R' , respectively. To prove 8 : S'=T : T' = R 2 : R' 2 = R 2 : H' 2 -, and V: V f = R* : R f3 = H 3 : H' 3 . g_^_ R+R R R' H' + R 8' 2 RXH R ^R (H+B) = B H+B (R' + R f ) R' R' Also R' = R 2 R = R* ^H 3 nR' 2 R f ~ R' 2 H! R f3 ~ R'* Arts. 321, 3fflL R 2 W 2 R' 2 (Why?) (Why ?) Q. E. D. Ex. If a cylindrical cistern is 12 ft. deep, how much more cement is required to line it than to line a similar cistern 6 ft. deep ? How much more water will the former cistern hold ? 412 BOOK VIII. SOLID GEOMETRY B Conical surface CONES 701. A conical surface is a sur- face generated by a straight line which moves so as constantly to touch a given fixed curve, and constantly pass through a given fixed point. Thus every shadow cast by a near point of light is conical in form, that is, is! bounded by a conical surface of light. Hence, the study of conical surfaces and solids is im- portant from the fact that it concerns all cases of forces radiating from a near point. 702. The generatrix of a conical c \ surface is the moving straight line, as AA f ; the directrix is the given fixed curve, as ABC-, the vertex is the fixed point, as 0; an element is the generatrix in any one of its positions, as BB' '. 703. The upper and lower nappes of a conical surface are the portions above and below the vertex, respectively, as 0-ABC and 0-A'B'C''. Usually it is convenient to limit a conical surface to a single nappe. 704. A cone is a solid bounded by a conical surface and a plane cutting all the elements. 705. The base of a cone is the face formed by the cutting plane; the lateral surface is the bounding conical surface; the vertex of the cone is the vertex of the conical surface; the elements of the cone are the elements of the conical surface; the altitude of a cone is the perpendicular distance from the Vertex to the plane of the base. Oblique circular cone CONES 413 706. A circular cone is a cone whose base is a circle. The axis of & circular cone is the line drawn from the vertex to the center of the base. 707. A right circular cone is a circular cone whose axis is perpendicular to the plane of the base. An oblique circular cone is a circular cone whose axis is oblique to the base. 708. A cone of revolution is a cone gene- rated by the revolution of a right triangle about one of its legs as an axis. Hence a cone of revolution and a right circular cone are the same solid. Cone of revolution 709. Properties of a cone of revolution inferred im- mediately. 1. The altitude, of a cone of revolution is the axis of the cone. 2. All the elements of a cone of revolution are equal. 710. The slant height of a cone of revolution is any one of its elements. 711. Similar cones of revolution are cones generated by similar right triangles revolving about homologous sides. 712. A plane tangent to a cone is a plane which con- tains one element of the cone, but which does not cut the conical surface on being produced. Ex. 1. A plane passing through a tangent to the base of a circular . cone and the element drawn through the point of contact is tangent to the cone. Ex. 2. If a plane is tangent to a circular cone, its intersection with the plane of the base is tangent to the cone. 414 BOOK VTTT. SOLID GEOMETRY 713. A pyramid inscribed in a cone is a pyramid whose lateral edges are elements of the cone and whose base is a polygon in- scribed in the base of the cone. 714. A pyramid circum- scribed about a cone is a pyra- mid whose lateral faces are tan gent to the cone and whose base is a polygon circumscribed about the base of the cone. 715. Properties of circular cones. By Art. 441 the area of a circle is the limit of the area of an inscribed, or of a circumscribed polygon, and the circumference is the limit of the perimeters of these polygons; hence 1. The volume of a circular cone is the limit of the vol- ume of an inscribed or circumscribed pyramid. 2. The lateral area of a circular cone is the limit of the lateral area of an inscribed or circumscribed pyramid. 716. A frustum of a cone is the por- tion of the cone included between the base of the cone and a plane parallel to the base. The lower base of the frustum is the base of the cone, and the upper base of the frustum is the section made by the plane parallel to the base of the cone. Frustum of a cone. What must be the altitude and the lateral surface of a frustum of a cone; also the slant height of the frustum of a cone of revolution ? CONES 415 X PROPOSITION VI. THEOREM 717. Every section of a cone made by a plane passing through its vertex is a triangle ^^ & Y^vv-vr^ *** <^ PROPOSITION XII. THEOREM 729. The volume of Hie frustum of a circular cone is equivalent to the volume of three cones, whose common alti- tude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean proportional between the two bases. Given a frustum of a circular cone having its volume denoted by F, its altitude by H, the area of its lower base by B, and that of its upper base by b. To prove F = 4 H (B + b + l/BXb). Proof. Let the frustum of a pyramid with regular poly- gons for its bases be inscribed in the given frustum. Denote the volume of the inscribed frustum by F', and the areas of its bases by B' and b f . ':. F = J H (B f + b' + VB* X b') . (Why?) If the number of lateral faces of the inscribed frustum be indefinitely increased, V will approach F, B r and b f approach B and b respectively, and B f X b' approach B X 5, as limits. Art. 715. Hence, also, B' + b'+T/B' X b' will approach B + b + V B X 6 as a limit. Art. 253. But 7/= J H (B r + V + *JB' X V) always. (Why t) /. V= J H (B + b + l/BXb). (Why f ) Q. E. J>. 422 BOOK VIII. SOLID GEOMETKY 730. Formula for the volume of the frustum of a circu- lar cone. F=4 H (nR 2 + nr 2 + n& X nr 2 ) . Ex. 1. The measurement of the volume of a frustum of a cone of revolution reduces to the measurement of the lengths of what straight lines? Ex. 2. If a conical oil -can is 12 in. high, how much more tin is required to make it than to make a similar oil -can 6 in. high ? How much more oil will it hold ? Ex. 3. The linear dimensions of a conical funnel are three times those of a similar funnel. How much more tin is required to make the first ? How much more liquid will it hold ? Ex. 4. Make a similar comparison of cylindrical oil -tanks. Of conical canvas tents. EXERCISES. CROUP 7O THEOREMS CONCERNING CYLINDERS AND CONES Ex. 1. Any section of a cylinder of revolution through its axis is a rectangle. Ex. 2. On a cylindrical surface only one straight line can be drawn through a given point. [SuG. For if two straight lines could be drawn, etc.] Ex. 3. The intersection of two planes tangent to a cone is a straight line through the vertex. Ex. 4. If two planes are tangent to a cylinder, their line of inter- section is parallel to an element of the cylinder. [Sue. Pass a plane J_ to the elements of the cylinder.] Ex. 5. If tangent planes be passed through two diametrically opposite elements of a circular cone, these planes intersect in a straight line through the vertex and parallel to the plane of the base. Ex. 6. In a cylinder of revolution the diameter of whose base equals the altitude, the volume equals one-third the product of the total surface by the radius of the base. EXEECISES ON THE CYLINDER AND CONE 423 Ex. 7. A cylinder and a cone of revolution have the same base and the same altitude. Find the ratio of their lateral surfaces, and also of their volumes. Ex. 8. If an equilateral triangle whose side is a be revolved about one of its sides as an axis, find the area generated in terms of a. Ex. 9. If a rectangle whose sides are a and & be revolved first about the side a as an axis, and then about the side &, find the ratio of the lateral areas generated, and also of the volumes. Ex. 10. The bases of a cylinder and of a cone of revolution are concentric. The two solids have the same altitude, and the diameter of the base of the cone is twice the diameter of the base of the cylin- der. What kind of line is the intersection of their lateral surfaces, and how far is it from the base ? Ex. 11. Determine the same when the radius of the cone is three times the radius of the cylinder. Also when r times. Ex. 12. Obtain a formula in terms of r for the volume of the frustum of an equilateral cone, in which the radius of the upper base is r and that of the lower base is 3r. Ex. 13. A regular hexagon whose side is a revolves about a diag- onal through the center as axis. Find, in terms of a, the surface and volume generated. Ex. 14. Find the locus of a point at a given distance from a given straight line. Ex. 15. Find the locus of a point whose distance from a given line is in a given ratio to its distance from a fixed plane perpen- dicular to the line. Ex. 16. Find the locus of all straight lines which make a given angle with a given line at a given point. Ex. 17. Find the locus of all straight lines which make a given angle with a given plane at a given point. Ex. 18. Find the locus of all points at a given dis- tance from the surface of a given cylinder of revolution. Ex. 19. Find the locus of all points at a given dis- tance from the surface of a given cone of revolution. 424 BOOK VIII. SOLID GEOMETRY EXERCISES. CROUP 71 PROBLEMS CONCERNING THE CYLINDER AND CONE Ex. 1. Through a given element of a circular cylinder, pass a plane tangent to the cylinder. Ex. 2. Through a given element of a circular cone, pass a plane tangc pt to the cone. Ex. 3. About a given circular cylinder circumscribe a prism, with a regular polygon for its base. Ex. 4. Through a given point outside a circular cylinder, pass a plane tangent to the cylinder. Ex. 5. Through a given point outside a given circular cone, pass a plane tangent to the cone. [Suo. Through the vertex of the cone and the given point pass a line, and produce it to meet the plane of the base.] Ex. 6. Into what segments must the altitude of a cone of revolution be divided by a plane parallel to the base, in order that the volume of the cone be bisected ? Ex. 7. Divide the lateral surface of a given cone of revolution into two equivalent parts by a plane parallel to the base. Ex. 8. If the lateral surface of a cylinder of revolution be cut along one element and unrolled, what sort of a plane figure is formed ? Hence, out of cardboard construct a cylinder of revolution with given altitude and given circumference. Ex. 9. If the lateral surface of a cone of . revolution be cut along one element and unrolled, what sort of a plane figure is formed ? Hence, out of cardboard construct a cone of revolution of given slant height. Ex. 10. Construct an equilateral cone out of pasteboard. Ex. 11. Construct a frustum of a cone of revolution out of paste- board. BOOK IX THE SPHERE 731. A sphere is a solid bounded by a surface all points of which are equally distant from a point within called the center. 732. A sphere may also be defined as a solid generated by the revolution of a semicircle about its diameter as an axis. . Some of the properties of a sphere may be obtained more readily from one of the two definitions given, and some from the other. A sphere is named by naming the point at its center, or by naming three or more points on its surface. 733. A radius of a sphere is a line drawn from the center to any point on the surface. A diameter of a sphere is a line drawn through the center and terminated at each end by the surface of the sphere. 734. A line tangent to a sphere is a line having but one point in common with the surface of the sphere, however far the line be produced. (425) 426 BObK IX. SOLID GEOMETKY 735. A plane tangent to a sphere is a plane having but one point in common with the surface of the sphere, how- ever far the plane be produced. 736. Two spheres tangent to each other are spheres whose surfaces have one point, and only one, in common. 737. Properties of a sphere inferred immediately. 1. All radii of a sphere, or of equal spheres, are equal. 2. All diameters of a sphere, or of equal spheres, are equal. 3. Two spheres are equal if their radii or their diame- ters are equal. PROPOSITION I. THEOREM 738. A section of a sphere made by a plane is a .circle. Given the sphere 0, and PCD a section made by a plane cutting the sphere. To prove that PCD is a circle. Proof. From the center 0, draw OA J_ the plane of the section. THE SPHERE 427 Let be a fixed point on the perimeter of the section, and P any other point on this perimeter. Draw A C, AP, OC, OP. Then the A GAP and OAC are rt. A. Art. 505. OP= OC. (Why ?) OA=OA. (Why?) A0AP=AOJLO. (Why?) .'. AP=AC. (Why?) But P is any point on the perimeter of the section PCD. .'. every point on this perimeter is at the distance AC from A. :. PCD is a circle with center A. Art. 197. Q. E. D. 739. COR. 1. Circles which are sections of a sphere made by planes equidistant from the center are equal; and conversely. 740. COR. 2. Of two circles on a sphere, the one made by a plane more remote from the center is smaller; and conversely. 741. DEF. A great circle of a sphere is a circle whose plane passes through the center of the sphere. 742. DEF. A small circle of a sphere is a circle whose plane does not pass through the center of the sphere. 743. DEF. The axis of a circle of a sphere is the diameter of a sphere which is perpendicular to the plane of the circle. Thus, on figure p. 426, BB' is the axis of PCD. 744. DEF. The poles of a circle of a sphere are the extremities of the axis of the circle. Thus, B and B', of figure p. 426, are poles of the circle PCD. 428 BOOK IX. SOLID GEOMETRY 745. Properties of circles of a sphere inferred imme- diately. 1. The axis of a circle of a sphere passes through the center of the circle; and conversely. 2. Parallel circles have the same axis and the same poles. 3. All great circles of a sphere are equal. 1 great circle on a sphere bisects the sphere and its surface. 5. Two great circles on a sphere bisect each other. For the line of intersection of the two planes of the circles passes through the center, and hence is a diameter of each circle. 6. Through two points (not the extremities of a diameter) on the surface of a sphere, one, and only one, great circle can be passed. For the plane of the great circle must also pass through the center of the sphere (Art. 741), and through three points not in a straight line only one plane can be passed (Art. 500). 7. Through any three points on the surface of a sphere, not in the same plane with the center, one small circle, and only one, can be passed. 746. DEF. The distance between two points on the sur- face of a sphere is the length of the minor arc of a great circle joining the points. Ex. I. If the radius of a sphere is 13 in., find the radius of a circle on the sphere made by a plane at a distance of 1 ft. from the center. Ex. 2. What geographical circles on the earth's surface are great, and what small circles ? Ex. 3. What is the largest number of points in which two circles on the surface of a sphere can intersect ? Why ? THE SPHERE 429 PROPOSITION II. THEOREM 747. All points in the circumference of a circle of a sphere are equally distant from each pole of the circle. Given ABC a circle of a sphere, and P and P f its poles. To prove the arcs PA, PB, PC equal, and arcs P'A, P'B, P'G equal. Proof. Draw the chords PA, PB, PC. The chords PA, PB and PC are equal. Art. 518. /. arcs PA, PB and PC are equal. Art. 218. In like manner, the arcs P'A, P'B and P'C may be proved equal. Q. . D. 748. DEF. The polar distance of a small circle on a sphere is the distance of any point on the circumference of the circle from the nearer pole. The polar distance of a great circle on a sphere is the dis- tance of any point on the circumference of the great circle from either pole. 749. COR. The polar distance of a great circle is the quadrant of a great circle. 430 BOOK IX. SOLID GEOMETRY PROPOSITION III. THEOREM 750. If a point on the surface of a sphere is at a quad- rant's distance from two other points on the surface, it is the pole of the great circle through those points. Given PB and PC quadrants on the surface of the sphere 0, and ABC a great circle through B and C. To prove that P is the pole of ABC. Proof. From the center draw the radii OB, OC, OP. The arcs PB and PC are quadrants. (Why ?) .'. A POB and POC are rt. A . (Why ?) /. PO I. plane ABC. (Why?) .'. P is the pole of the great circle ABC. (Why ?) Q. . D. 751. COR. Through two given points on the surface of a sphere to describe a great circle. Let A and B be the given points. From A and B as centers, with a quadrant as radius, describe arcs on the surface of the sphere inter- secting at P. With P as a center and a quadrant as a radius, describe a great circle. THE SPHERE 431 PROPOSITION IV. THEOREM 752. A plane perpendicular to a radius at its extremity is tangent to the sphere. Given the sphere 0, and the plane MN J. the radius OA of the sphere at its extremity A. To prove MN tangent to the sphere. Proof. Take P any point in plane MN except A. Draw OP. Then OP > OA. (Why?) .*. the point P is outside the surface of the sphere. But P is any point in the plane MN except A. .'.plane MN is tangent to the sphere at the point A, (for every paint in the plane, except A, is outside the surface of the sphere). Art. 735. Q. E. D. 753. COR. 1. A plane, or a line, which is tangent to a sphere, is perpendicular to the radius drawn to the point of contact. Also, if a plane is tangent to a sphere, a perpendicu- lar to the plane at its point of contact passes through the center of the sphere. 754. COR. 2. A straight line perpendicular to a radius of a sphere at its extremity is tangent to the sphere. 755. COR. 3. A straight line tangent to a circle of a sphere lies in the plane tangent to the sphere at the point of contact. 432 BOOK IX. SOLID GEOMETKY 756. COR. 4. A straight line drawn in a tangent plane, and through the point of contact is tangent to the sphere at that point. 757. COR. 5. Two straight lines tangent to a sphere at a given point determine the tangent plane at that point. 758. DEP. A sphere circumscribed about a polyhedron is a sphere in whose surface lie all the vertices of the polyhedron. 759. DEF. A sphere inscribed in a polyhedron is a sphere to which all the faces of the polyhedron are tangent. PROPOSITION V. PROBLEM 760. To circumscribe a sphere about a given tetra- hedron. Given the tetrahedron ABCD. To circumscribe a sphere about ABCD. Construction and Proof. Construct E and F the centers of circles circumscribed about the A ABC and BCD, re- spectively. Art. 286. Draw EH _L plane ABC and FK _L plane BCD. Art, 514. Draw EQ and FG to - jjs \ s^\ | / 2*~ ''' . P' Fig-. 2 ^8' 3 Given the material sphere 0. To construct the radius of the sphere. Construction. With any point P (Fig. 1) of the surface of the sphere as a pole, describe any convenient circum- ference on the surface. On this circumference take any three points A, B and C. Construct the A ABC (Fig. 2) having as sides the three chords AB, BC, AC, obtained from Fig. 1, by use of the compasses. Art. 283. Circumscribe a circle about the A ABC. Art. 286. Let KB be the radius of this circle. Construct (Fig. 3) the right A Jcpb, having for hypot- enuse the chord pb (Fig. 1) and the base kb. Art. 284. Draw bp' J_ bp and meeting pk produced at p' . Bisect pp f at 0. Then op is the radius of the given sphere. Proof. Let the pupil supply the proof, Q. E. F. 436 BOOK IX. SOLID GEOMETRY PROPOSITION VIII. THEOREM 769. The intersection of two spherical surfaces is the circumference of a circle whose plane is perpendicular to the line joining the centers of the spheres, and whose center is w that line. Given two intersecting and 0' which, by rotation about the line 00' as an axis, generate two intersecting spherical surfaces. To prove that the intersection of the spherical surfaces is a O, whose plane J_ 00', and whose center lies in 00'. Proof. Let the two circles intersect in the points P and Q, and draw the common chord PQ, Then, as the two given rotate about 00' as an axis, the point P will generate the line of intersection of the two, spherical surfaces that are formed. But PR is constantly J_ 00'. Art. 241, /. PR generates a plane A. 00 f Art. 510. Also PR remains constant in length. .*. P describes a circumference in that plane. Art. 197. Hence the intersection of two spherical surfaces is a O , whose plane J_ the line of centers, and whose center is in the line of centers. Q. E. D. The above demonstration is an illustration of the use of the second definition of a sphere (Art. 732). THE SPHERE 437 PROPOSITION IX. THEOREM 770. A spherical angle is measured by the arc of a great circle described from the vertex of the angle as a pole, and included between its sides, produced, if necessary. Given /.BAG a spherical angle formed by the intersec- tion of the arcs of the great circles BA and CA, and BO an arc of a great circle whose pole is A. To prove /.BAG measured by arc BC. Proof. Draw AD tangent to AB, and AF tangent to AC. Also draw the radii OB and OG. Then AD _L AO. Art. 230. Also OB _L AO ( for AB is a quadrant) . :. OB \\AD. (WhyT) Similarly OG\\AF. :. Z BOG= Z DAP Art. 538. But /.BOG is measured by arc BG. Art. 257. /. /.DAF, that is, /.BAG, is measured by arc BG. Us- ( Wh y ? ) Q. E. D. 771. COR. A spherical angle is equal to the plane angle of the dihedral angle formed by the planes of its sides. 438 BOOK IX. SOLID GEOMETRY SPHERICAL TRIANGLES AND POLYGONS 772. A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles, as ABCD. The sides of the spherical polygon are the bounding arcs; the vertices are the points in which the sides in- tersect; the angles are the spherical angles formed by the sides. The sides of a spherical polygon are usually limited to arcs less than a semicircumference. 773. A spherical triangle is a spherical polygon of three sides. Spherical triangles are classified in the same way as plane triangles; viz., as isosceles, equilateral, scalene, right, obtuse and acute. 774. Relation of spherical polygons to polyhedral angles. If radii be drawn from the center of a sphere to the ver- tices of a spherical polygon on its surface (as OA, OB, etc., in the above figure), a polyhedral angle is formed at O, which has an important relation to the spherical poly- gon ABCD Each face angle of the polyhedral angle equals (in num- ber of degrees contained) the corresponding side of the spheri- cal polygon; Each dihedral angle of the polyhedral angle equals the corresponding angle of the spherical polygon. Hence, corresponding to each property of a polyhedral angle, there exists a property of a spherical polygon, and conversely. THE SPHERE 439 Hence, also, a trihedral angle and its parts correspond to a spherical triangle and its parts. Of the common properties of a polyhedral angle and a spherical polygon, some are discovered more readily from the one figure and some from the other. In general, the spherical polygon is simpler to deal with than a polyhedral angle. For instance, if a trihedral angle were drawn with the plane angles of its dihedral angles, nine lines would be used, forming a complicated figure in solid space; whereas, the same magnitudes are represented in a spherical triangle by three lines in an approximately plane figure. On the other hand, the spherical polygon, because of its lack of detailed parts, is often not so suggestive of properties as the poly- hedral angle. PROPOSITION X. THEOREM 775. The sum of two sides of a spherical triangle is greater than the third side. Given the spherical triangle ABC, of which no side is larger than AB. To prove AC+ BC > AB. Proof. From the center of the sphere, 0, draw the radii OA, OB, OC. Then, in the trihedral angle 0-ABC, AOC+BOC > AOB. Art. 582. /. AC-}-BG > AB. Art. 774. Q. . D. 440 BOOK IX. SOLID GEOMETRY 776. COR. 1. Any side of a spherical triangle is greater than the difference between the other two sides. 111. COR. 2. The shortest path between two points on the surface of a sphere is the arc of the great circle joining those points. For any other path between the two points may be made the limit of a series of arcs of great circles connect- ing successive points on the path, and the sum of this series of arcs of great circles connecting the two points is greater than the single arc of a great circle connecting them. PROPOSITION XI. THEOREM 778. The sum of the sides of a spherical polygon is less than 360. Given the spherical polygon ABCD. To prove the sum of the sides of ABCD < 360. Proof. From 0, the center of the sphere, draw the radii OA, OB, OC, OD. Then Z AOB + Z.BOC + Z COD-}- ^DOA < 300. (Why ?) /. AB + BG + GD + DA < 360 . Art. 774. Q. . D. SPHEKICAL TKIANGLES 441 D A' 779. DEF. The polar p' D ' triangle of a given triangle is the triangle formed by taking the vertices of the given triangle as poles, and describing arcs of great cir- cles. (Hence, if each pole be regarded as a center, the radius used in describing each arc is a quadrant.) Thus A'B'C 'is the polar triangle of ABC-, also D'E'F' is the polar triangle of DEF. PROPOSITION XII. THEOREM 780. // one spherical triangle is the polar of another, then the second triangle is the polar of the first. Given A'B'C' the polar triangle of ABC. To prove ABC the polar triangle of A'B'C'. Proof. B is the pole of the arc A'C'. Art. 779. .*. arc A'B is a quadrant. (Why ?) Also C is the pole of the arc A'B'. (Why ?) .*. arc A'C is a quadrant. (Why ?) /. A' is at a quadrant's distance from both B and C. /. A f is the pole of the arc BC. Art. 750. In like manner it may be shown that B' is the pole of AC, and C' the pole of AB. q. E. j>. 442 BOOK IX. SOLID GEOMETRY PROPOSITION XIII. THEOREM 781. In a spherical triangle and its polar, each angle of one triangle is the supplement of the side opposite in the other triangle. A' Given the polar A ABC and A'B'C' with the sides of ABC denoted by a, &, c, and the sides of A'B'C' denoted by a', &', c', respectively. To prove A + a'= 180, B + &' = 180, C + c' = 180 f A' + a = 180, JB' + 6 = 180, (7 + c = 180. Proof. Produce the sides AB and A G till they meet B'O in the points D and .F, respectively. Then B' is the pole of AF :. arc 5^=90. Art. 780. Also C' is the pole of AD :. arc / D = 90. (Why?) Adding, '^ + (71)= 180. (Why?) Or J3'.F+ ^O 7 + D^-180 . Ax. 6. Or JB / C / + D^ T -180. But B'C' = a', and D^is the measure of the Z A. Art. 770. /. A + a' = 180. In like manner the other supplemental relations may be proved as specified. Q. E. D. 782. DEF. Supplemental triangles are two spherical triangles each of which is the polar triangle of the other. This new name for two polar triangles is due to the property proved in Art. 781. SPHERICAL TRIANGLES 443 PROPOSITION XIV. THEOREM 783. The sum of the angles of a spherical triangle is greater than 180, and less than 540. Given the spherical triangle ABO. To prove A + B -f G > 180 and < 540. Proof. Draw A'B'C', the polar triangle of ABC, and denote its sides by ', b' ', c' . Then ^ + a' = 180 ) J? _j- ' 180 f Art. 781. (7 _j_ c ' = 180 ) .-. A + J5 + (7 + ' + ^ + c' = 540. . . (1) AX. 2. But C a' + ' -f c r < 360 Art. 778. { a' + J' + c r > Subtracting each of these .in turn from (1), ^ _j_ B + C > 180 and < 540. Ax. 11. Q. E. D. 784. COR. A spherical triangle may have one, two or three right angles; or it may have one, two or three obtuse angles. 785. 'DEF. A birectangular spherical triangle is a spher- ical triangle containing two right angles. 786. DEF. A trirectangular spherical triangle is a spherical triangle containing three right angles. 444 BOOK IX. SOLID GEOMETRY 787. COR. The surface of a sphere may be divided into eight trirectangular spherical triangles. For let three planes J_ to each other be passed through the center of a sphere, etc. 788. DBF. The spherical excess of a spherical triangle is the excess of the sum of its angles over 180. 789. DEF. Symmetrical spherical triangles are triangles which have their parts equal, but arranged in reverse order. B' Fig. 2 Fig-. 3 Fig.l ' Three planes passing through the center of a sphere form a pair of symmetrical spherical triangles on opposite sides of the sphere (see Art. 580), as A ABC and A'B'C* of Fig. 1. 790. Equivalence of symmetrical spherical triangles. Two plane triangles which have their parts equal, but ar- ranged in reverse order, may be made to coincide by lifting up one triangle, turning it over in space, and placing it upon the other triangle. But two symmetrical spherical triangles cannot-be made to coincide in this way, because of the curvature of a spherical surface. Hence the equivalence of two sym- metrical spherical triangles must be demonstrated in some indirect way. SPHE1UCAL TRIANGLES 445 791. Property of symmetrical spherical triangles. Two isosceles symmetrical spherical triangles are equal, for they can be made to coincide. PROPOSITION XV. THEOREM 792. Two symmetrical spherical triangles are equivalent. Given the symmetrical spherical A ABC and A'B'C', formed by planes passing through 0, the center of a sphere. (See Art. 789.) To prove A ABC=a= A A'B'C'. Proof. Let Pbe the pole of a small circle passing through the points A, J3, C. Draw the diameter POP'. Also draw PA, PB, PC, P'A', P'B', P'C', all arcs of great . Then PA = PB = PC. Art. 747. Also P'A' = PA, P'B' = PB, P'C' = PC. Arts. 78, 215. ; .% P'A' = P'B' =P'C'. Ax. 1. Hence PAB and P'A'B' are symmetrical isosceles A. Similarly A PA C= A P'A'G'. Art. 791. And A PBC= A P'E'G'} Adding A PAB+ A PAC+ A PEG ^ A P*A'B'+ A PA'O+ A P'B' a. Ax. 2. Or A ABC- A A'B'C'. Ax. 6. In case the poles P and P' fall outside the A A.B<7 and A'B'C 1 ', let the puuil suoply the demonstration. Q. E. D. 446 BOOK IX. SOLID GEOMETRY PROPOSITION XVI. THEOREM 793. On the same sphere, or on equal spheres, two tri- angles are equal, I. // two sides and the included angle of one are equal to two sides and the included angle of the other; or II. If two angles and the included side of one are equal to two angles and the included side of the other, the corresponding equal parts being arranged in the same order in each case. I. Given the spherical A ABC and DEF, in which AC=DF, CB=FE, and Z<7= /.F. To prove A ABC= A DEF. Proof. Let the pupil supply the proof (see Book I, Prop. VI). II. Given the spherical A ABC and DEF, in which To prove A ABC= A DEF. Proof. Let the pupil supply the proof (see Book I, Prop. VII) . _ Ex. 1. If the line of centers of two spheres is 10 in., and the radii are 12 in. and 3 in., how are the spheres situated with reference to each other ? Ex. 2. The tank on a motor car is a cylinder 35 inches long and 15 inches in diameter. How many gallons of gasolene will it hold ? Ex. 3. In an equilateral cone, find the ratio of the lateral area to the area of the base. SPHERICAL TRIANGLES 447 PROPOSITION XVII. THEOREM 794. On the same sphere, or on equal spheres, two tri- angles are symmetrical and equivalent, I. If two sides and the included angle of one are equal to two sides and the included angle of the other; or II. // two angles and the included side of one are equal to two angles and the included side of the other, the corresponding equal parts being arranged in reverse order. A I. Given the spherical A ABC and DEF, in which AB = DE, AC = DF, and /.A = /.D, the corresponding parts being arranged in reverse order. To prove A ABC symmetrical with A DEF. Proof. Construct the &D'E'F' symmetrical with ADEF. >t Then A ABC may be made to coincide with A D'E'F', Art. 793. (having two sides and the included Z equal and arranged in the same order). But A D'E'F' is symmetrical with the A DEF. :. A ABC, which coincides with A D'E'F', is symmet- rical with A DEF. II. The second part of the theorem is proved in the same way. Q. . D. 448 BOOK IX. SOLID GEOMETRY PROPOSITION XVIII. THEOREM 795. If two triangles on the same sphere, or equal spheres, are mutually equilateral, they are also mutually equiangular, and therefore equal or symmetrical. Given two mutually equilateral spherical A ABC and A'B'C' on the same or on equal spheres. To prove A ABC and A'B'C' equal or symmetrical. Proof. From and 0', the centers of the spheres to which the given triangles belong, draw the radii OA, OB, OC, O'A', 0'B f , O'C'. Then the face A at = corresponding face A at 0' '. Art. 774. Hence dihedral A at 0= corresponding dihedral A at O f . Art. 584. .*. A of spherical A A BC= homologous A of spherical & A'B'C'. Art. 774. .*. the A ABC and A'B'C' are equal or symmetrical, according as their homologous parts are arranged in the same or in reverse order. Art. 789. 796. NOTE. The conditions in Props. XVI and XVIII which make two spherical triangles equal are the same as those which make two plane triangles equal. Hence many other propositions occur in spherical geometry which are identical with corresponding proposi- tions in plane geometry. Thus, many of the construction problems of spherical geometry are solved in the same way as the corresponding construction problems in plane geometry j as, to bisect a given angle, etc, SPHERICAL TRIANGLES 449 PROPOSITION XIX. THEOREM 797. // two triangles on the same sphere are mutually equiangular, they are also mutually equilateral, and there- fore equal or symmetrical. Given the mutually equiangular spherical A Q and Q* on the same sphere or on equal spheres. To prove that Q and Q f are mutually equilateral, and therefore equal or symmetrical. Proof. Construct P and P f the polar A of Q and #', respectively. Then A P and P f are mutually equilateral. Art. 781. /. A P and P f are mutually equiangular. Art. 795. But Q is the polar A of P, and Q' of P' '. Art. 780. /. A Q and Q' are mutually equilateral. Art. 781. Hence Q and Q' are equal or symmetrical, according as their homologous parts are arranged in the same or in reverse order. Art. 789. Q. E. D. 798. COR. If two mutually equiangular triangles are on unequal spheres, their corresponding sides have the same ratio as the radii of their respective spheres. CC 450 BOOK IX. SOLID GEOMETRY PROPOSITION XX. THEOREM 799. In an isosceles spherical triangle the angles oppo< site the equal sides are equal. Given the spherical A ABC in which AB=AC. To prove Z=ZO. Proof. Draw an arc from the vertex A to D, the mid- point of the base. Let the pupil supply the remainder of the proof. PROPOSITION XXI. THEOREM (CONY. OF PROP. XX) 800. If two angles of a spherical triangle are equal, the sides opposite these angles are equal, and the triangle is isosceles. A. Given the spherical A ABC in which B= Z G. To prove AB = AC. Proof. Construct A A'B'C' the polar A of ABC. Then A'C' = A'B'. Art. 781. /. Z C'= Z-B'. Art. 799. .'. AB = AG. Art. 781. Q. . D. SPHEKICAL TRIANGLES 451 PROPOSITION XXII. THEOREM 801. In any spherical triangle, if two angles are un- equal, the sides opposite these angles are unequal, and the greater side is opposite the greater angle, and CONVERSELY^, Given the spherical A ABC in which Z BAG is greater than Z.C. To prove BC > BA. Proof. Draw the arc AD making /.D AC equal to Z C. Then DA = DC. Art. 800. To each of these equals add the arc BD. :. BD + DA = BD + DC, or EG. (Why f) But in A BDA, BD + DA > BA. (Why ?) /. BC > BA. . Ax. 8. Let the pupil prove the converse by the indirect method (see Art. 106). Q. E. D. Ex. 1. Bisect a given spherical angle. Ex. 2. Bisect a given arc of a great circle on a sphere. Ex. 3. At a given point in an arc on a sphere, construct an angle equal to a given spherical angle on the same sphere. Ex. 4. Find the locus of the centers of the circles of a sphere formed by planes perpendicular to a given diameter of the given sphere. 452 BOOK IX. SOLID GEOMETRY SPHERICAL AREAS 802. Units of spherical surface. A spherical surface may be measured in terms of, either 1. The customary units of area, as a square inch, a square foot, etc., or 2. Spherical degrees, or spherids. 803. A spherical degree, or spherid, is one -ninetieth part of one of the eight trirectangular triangles into which the surface of a sphere may be divided (Art. 787), or TTO part of the surface of the entire sphere. A solid degree is one-ninetieth part of a trirectangular angle (see Art. 774). 804. A lune is a portion of the surface of a sphere bounded by two semicircumferences of great circles, as PBP'C of Fig. 1, The angle of a lune is the angle formed by the semicircumferences which bound it, as the angle BPC. 805. A zone is the portion of the surface of the sphere bounded by two parallel planes. A zone may also be defined as the sur- face generated by an arc of a revolving semicircumference. Thus, if QFQ' (Fig. 2) generates a sphere by rotating about QQ f , its diameter, any arc of QFQ', as EF, generates a zone. 806. A zone of one base is a zone one of whose bounding planes is tangent to the sphere, as the zone generated by the arc QE of Fig. 2. 807. The altitude of a zone is the perpendicular dis- tance between the bounding planes of the zone. The bases of a zone are the circumferences of the circles of the sphere formed by the bounding planes of the zone. SPHERICAL AREAS 453 PROPOSITION XXIII. THEOREM 808. The area generated ~by a straight line revolving about an axis in its plane is equal to the projection of the line upon the axis, multiplied by the circumference of a circle ivhose radius is the perpendicular erected at the midpoint of the line and terminated by the axis. Given AB and XYin the same plane, CD the projection of AB on JTY, P# the -L bisector of A; and a surface generated by the revolution of AB about XT, denoted as To prove area AB=CDX2 nPQ. Proof. 1. In general, the surface generated by AB is the surface of a frustum of a cone (Fig. 1). /. area AB = AB X 2 nPR. Draw AF LED, then A ABF and PQR are similar. /. A: AF=PQ: PR. :. AB X PR = AFX PQ, or CD X PQ. Substituting, area AB= CD X 2 nPQ. 2. If AB\\XY (Fig. 2), the surface generated by AB is the lateral surface of a cylinder. /. area AB=CDX2 nPQ. Art. 697. 3. If the point A lies in the axis XY (Fig. 3), let the pupil show that the same result is obtained. p. E. p, Art. 728. Art. 328. (Why?) (Why?) Ax. 8. 454 BOOK IX. SOLID GEOMETKY PROPOSITION XXIV. THEOREM 809. The area of the surface of a sphere is equal to the product of the diameter of me sphere by the circumference of a great circle. A Given a sphere generated by the revolution of the semi- circle ACE about the diameter AE, with the surface of the sphere denoted by S, and its radius by R. To prove 8=AE X 2 nR. Proof. Inscribe in the given semicircle the half of a regular polygon of an even number of sides, as ABCDE. Draw the apothem to each side of the semipolygon, and denote it by a. From the vertices B, C, D draw Ja to AE. Then area AB = AF X 2 na. area BC= FOX 2 na. area CD=OKX2 na. area DE=KEX 2 na. Adding, area ABCDE=AE X 2 na. If, now, the number of sides of the polygon be indefi- nitely increased, area ABCDE approaches 8 as a limit. Art. 441. And a approaches R as a limit. (Why?) .*. AE X 2 na approaches AE X 2 nR as a limit, f Why ?) J3ut area ABCDE '=AE X 2 na always. ,-, $=AEX2 nil. (WhyT) . E. 9. Art. 808. SPHERICAL AREAS 455 810. Formulas for area of surface of a sphere. Substituting for AE its equal 2 R, #=4 nR 2 . Also denoting the diameter of the sphere by D, R= J D. 811. Con. 1. T/ie surface of a sphere is equivalent to four times the area of a great circle of the sphere. 812. COR. 2. The areas of the surfaces of tivo spheres are to each other as the squares of their radii, or of their diameters. For, if S and $' denote the surfaces, R and R' the radii, and D and D f the diameters of two spheres, 8 = 4KR 2 = R 2 8 = KD 2 = D 2 S' InR' 2 R' 2 ' ! 8' TiD' 2 D' 2 ' 813. Property of the sphere. The following property of the sphere is used in the proof of Art. 809 : //, in the generat- ing arc of any zone, a broken line be inscribed, whose vertices divide the arc into equal parts, then, as the number of these parts is increased indefinitely, the area generated by the broken line approaches the area of the zone as a limit. Hence COR. 3. The area of a zone is equal to the circumference of a great circle multiplied by the altitude of the zone. Thus the area generated by the arc BC = FOX2 rtR. 814. COR. 4. On the same sphere, or on equal spheres, the areas of two zones are to each other as the altitudes of the zones. _ Ex. 1. Find the area of a sphere whose diameter is 10 in. Ex. 2. Find the area of a zone of altitude 3 in., on a sphere whose radius is 10 in. 456 BOOK IX. SOLID GEOMETKY , PROPOSITION XXV. THEOREM 815. The area of a lune is to the area of the sur- face of the sphere as the angle of the lune is to four right is. Given a sphere having its area denoted by 8, and on the sphere the lune ABCD of Z A with its area denoted by L. To prove L:S=A : 360. Proof. Draw FBH, the great O whose pole is A, inter- secting the bounding arcs of the lune in B and D. CASE I. When the arc BD and the circumference FBH are commensurable. Find a common measure of BD and .FBH, and let it be contained in the arc BD m times, and in the circumference FBHn times. Then arc BD : circumference FBH=m : n. Through the diameter AC, and the points of division of the circumference FBH pass planes of great . The arcs of these great will divide the surface of the sphere in n small equal lunes, m of them being contained in the lune ABCD. :. L : S=m : n. :. L : $=arc BD : circumference FBH. (Why ?) Or L : S = A : 360. Art. 257. CASE II. When the arc BD and the circumference FBH are incommensurable. Let the pupil supply the proof. Q. E. D. SPHERICAL AREAS 457 816. Formula for the area of a lune in spherical degrees, or spherids. The surface of a sphere contains 720 spherids (Art. 803). Hence, by Art. 815, L L spherids A m L_ '' fl r 720 spherids" 360' * 2 " that is, the area of a lune in spherical degrees is equal to twice the number of angular degrees in the angle of the lune. 817. Formula for area of a lune in square units o S=4 TtR* (Art. 810) /.r^ 2 -^, or(L = ^g- 818. COE. 1. On the same sphere, or on equal spheres, two lunes are to each other as their angles. 819. COR. 2. Two lunes ivith equal angles, but, on un- equal spheres, are to each other as the squares of the radii of their spheres. ' Ex. 1. Find the area in spherical degrees of a lune of 27. "Ex. 2. Find the number of square inches in the area of a lune of t 27, on a sphere whose radius is 10 in. A solid symmetrical with respect to a plane is a solid in which a line drawn from any point in its surface JL the given plane and produced its own length ends in a point on the surface ; hence Ex. 3. How many planes of symmetry has a circular cylinder ? A cylinder of revolution ? Ex. 4. Has either of these solids a center of symmetry f Ex. 5. Answer the same questions for a circular cone. Ex. 6. For a cone of revolution. For a sphere. ^ Ex. 7. For a regular square pyramid. For a regular pentagonal pyramid. 458 BOOK IX. SOLID GEOMETKY PROPOSITION XXVI. THEOREM 820. If two great circles intersect on a hemisphere, the sum of two vertical triangles thus formed is equivalent to a lune ivhose angle is that angle in the triangles which is formed by the intersection of the two great circles. Given the hemisphere ADBF, and on it the great circles AFB and DFC, intersecting at F. To prove A AFC+ A BFD o lune whose Z is BFD. ' Proof. Complete the sphere and produce the given arcs of the great circles to intersect at F' on the other hemisphere. Then, in the A AFG and BF'D, arc AF=arc BF' , (each being the supplement of the arc BF). In like manner arc CF= 'arc DF'. And arc AC = arc DB. . A AFC^ A BF'D. Art. 795. Add the A BFD to each of these equals ; /. A AFC + A BFD<* A BF'D + A BFD. Ax. 3. Or .-. A AFC + A ^D^lune FBF'D, Ax. 6 Q. E. D, SPHERICAL AREAS 459 PROPOSITION XXVII. THEOREM 821. The number of spherical degrees, or spherids, in the area of a spherical triangle is equal to the number of angular degrees in the spherical excess of the triangle. Given the spherical A ABO whose A are denoted by A , B, C, and whose spherical excess is denoted by E. To prove area of A ABC E spherids. Proof. Produce the sides AC and BC to meet AB pro- duced in the points D and F, respectively. A ABC + & CDB = luuQ ABDC=2A spherids. ) V Art. 816. A ABC + A ACF=\unQ BCFA = 2 B spherids. j A ABC + & GFD=lune of /_BCA = 2 C spherids. Art. 820- Adding, and observing that A ABC+A CDB+&ACF -f A CFD = hemisphere ABDFC, 2 A ABC-}- hemisphere = 2 (A + B + C) spherids. Or, 2 A ABC+ 360 spherids = 2 (A + B + C) spherids. Ax. 8. /. A ABC + 180 spherids = (A +B'+ C) spherids. Ax. 5. /. A ABC= ( A + B + C180) spherids. Ax. 3. Or area A ABC=E spherids. Art. 788. Q. . D. 460 BOOK IX. SOLID GEOMETRY 822, Formula for area of a spherical triangle in square units of area. Comparing the area of a spherical A with the area of the entire sphere . area A : 4 nR 2 = E spherids : 720 spherids. 4 KR 2 X E KR 2 E :. area A = -705, or area A-- 823. The spherical excess of a spherical polygon is the sum of the angles of the polygon diminished by (n 2) 180 ; that is, it is the sum of the spherical excesses of the tri- angles into which the polygon may be divided. PROPOSITION XXVIII. THEOREM 824. The area of a spherical polygon, in spherical de- grees or spherids^ is equal to the spherical excess of the polygon. A Given a spherical polygon ABCDF of n sides, with its spherical excess denoted by E. To prove area of ABCDF=E spherical degrees. Proof. Draw diagonals from A, any vertex of the poly- gon, and thus divide the polygon into n 2 spherical A. The area of each A = (sum of its A 180) spherids. Art. 821. /. sum of the areas of the A = [sum of A of the A (n 2) 180] spherids. Ax. 2. /. area of polygon = E spherids, Art. 823. (for the sum of 4 of the A (n 2) 180=^). SPHERICAL VOLUMES 461 SPHERICAL VOLUMES 825. A spherical pyramid is a por- tion of a sphere bounded by a spheri- cal polygon and the planes of the great circles forming the sides of the polygon. The base of a spherical pyramid is the spherical polygon bounding it, and the vertex of the spherical pyramid is the center of the sphere. Thus, in the spherical pyramid O-ABCD, the base is A BCD and the vertex is 0. 826. A spherical wedge (or ungula) is the portion of a sphere bounded by a lune and the planes of the sides of the lune. 827. A spherical sector is the portion of a sphere gene- rated by a sector of that semicircle whose rotation generates the given sphere. 828. The base of a spherical sector is the zone gene- rated by the revolution of the arc of the plane sector which generates the spherical sector. Let the pupil draw a spherical sector in which the base is a zone of one base. 462 BOOK IX. SOLID GEOMETRY 829. A spherical segment is a portion of a sphere included between two parallel planes. The bases of a spherical segment are the sections of the sphere made by the parallel planes which bound the given segment ; the altitude is the perpendicular distance between the bases. 830. A spherical segment of one base is a spherical seg- ment one of whose bounding planes is tangent to the sphere. PROPOSITION XXIX. THEOREM 831. The volume of a sphere is equal to one -third the product of the area of its surface by its radius. Given a sphere having its volume denoted by V, sur- face by $, and radius by R. To prove V=%$XR. Proof. Let any polyhedron be circumscribed about the sphere. Pass a plane through each edge of the polyhedron and the center of the sphere. These planes will divide the polyhedron into as many pyramids as the polyhedron has faces, each pyramid hav- ing a face of the polyhedron for its base, the center of the SPHERICAL VOLUMES 4G3 sphere for its vertex, and the radius of the sphere for its altitude. .*. volume of each pyramid = J base X R. (Why ?) .*. volume of polyhedron = J (surface of polyhedron) XR. If the number of faces of the polyhedron be increased indefinitely, the volume of the polyhedron approaches the , volume of the sphere as a limit, and the surface of the polyhedron approaches the surface of the sphere as a limit. Hence the volume of the polyhedron and J (surface of the polyhedron )X R, are two variables always equal. Hence their limits are equal, Or V=bSXR. (Why?) Q. . D. 832. Formulas for volume of a sphere. Substituting 8=4: TtR 2 , or &=7tD 2 in the result of Art. 831, also F= D 833. COR. 1. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diameters. 834. COR. 2. The volume of a spherical pyramid is equal to one -third the product of its base by the radius of the sphere. 835. COR. 3. The volume of spherical sector is equal to one -third the product of its base (the bounding zone) by the radius of the sphere, 464 BOOK IX. SOLID GEOMETRY 836. Formula for the volume of a spherical sector. De- noting the altitude of the sector by H and the volume by V, F=4 (area of zone) X J2, = j(2 nEH) E. Art. 813. PROPOSITION XXX. THEOREM 837. The volume of a spherical segment is equal to one- half the product of its altitude by the sum of the areas of its bases, plus the volume of a sphere whose diameter is the altitude of the segment. Given the semicircle ABC A' which generates a sphere by its rotation about the diameter A A '; ED and CF semi- chords _L AA', and denoted by r and r'; and DF denoted by tf. To prove volume of spherical segment generated by BCFD, or F=J(7ir 2 + 7ir /2 ) H+i nH 8 . Proof. Draw the radii OB and 0(7. Denote OF by h, and OD by k. Then F=vol. OBC + vol. OCF vol. OBD. .'. F= 7l# 2 #-f 4 7tr' 2 h J 7tr 2 A'. Arts. 836, 723. But H=h-k, h*=lP-r* 9 and /t 2 =^ 2 -r 2 , (Why?) SPHERICAL VOLUMES 465 .-. F=4 n [2 R 2 (h-k)-i (R 2 -h 2 ) h(R 2 -k 2 ) k] . Ax. 8. = 4 7t [2 R 2 '(h-k) + R 2 'U-AO (A 3 -* 3 )]. = i 7t JT [3 2 U 2 + M + k 2 )] . x But 7i 2 2 M -{- k 2 = H 2 . s As. 4. ,<* Subtract each member from 3 ft 2 -f 3 & 2 and divide by 2. *s. 3, 5. Then A 2 + M + ^ 2 = l (/i 2 + k 2 )- /. F= i 7t# [f (r 2 + r' 2 ) + Y ] Ax. 8. Q. E. D. 838. Formula for volume of a spherical segment of one base. In a spherical segment of one base r f = o, and r 2 = (2 R-H) H (Art. 343). Substituting for r and r' these values in the result of Art. 837, 839. Advantage of measurement formulas. The student should observe carefully that, by the results obtained in Book IX, the measurement of the areas of certain curved surfaces is reduced to the far simpler work of the measure- ment of the lengths of one or more straight lines; in like manner the measurement of certain volumes bounded by a curved surface is reduced to the simpler work of linear measurements. A similar remark applies to the results of Book VIII. Ex. 1. Find the volume of a sphere whose radius is 7 in. Ex. 2. Find the volume of a sphere whose diameter is 7 in. Ex. 3. In a sphere whose radius is 8 in., find the volume of a spherical segment of one base whose altitude is 3. DD 466 BOOK IX. SOLID GEOMETRY EXERCISES. GROUP. 72 THEOREMS CONCERNING THE SPHERE Ex. 1. Of circles of a sphere whose planes pass through a given point within a sphere, the smallest is that circle whose plaiie is perpendicular to the diameter through the given point. Ex. 2. If a point on the surface of a given sphere is equidistant from three points on a given small circle of the sphere, it is the pole of the small circle. Ex. 3. If two sides of a spherical triangle are quadrants, the third side measures the angle opposite that side in the triangle. Ex. 4. If a spherical triangle has one right angle, the sum of its other two angles is greater than one right angle. Ex. 5. If a spherical triangle is isosceles, its polar triangle is isosceles. Ex. 6. The polar triangle of a birectangular triangle is birectan- gular. Ex. 7. The polar triangle of a trirectangular triangle is identical with the original triangle. Ex. 8. Prove that the sum of the angles of a spherical quadri- lateral is greater than 4 right angles, and less than 8 right angles. What, also, are the limits of the sum of the angles of a spherical hexagon ? Of the sum of the angles of a spherical w-gon ? Ex. 9. On the same sphere, or on equal spheres, two birectan- gular triangles are equal if their oblique angles are equal. Ex. 10. Two zones on the same sphere, or on equal spheres, are to each other as their altitudes. Ex. 11. If one of the legs of a right spherical triangle is greater^ than a quadrant, another side is also greater than a quadrant. [SuG. Of the leg which is greater than a quadrant, take the end remote from the right angle as a pole, and describe an arc.] Ex. 12. If ABC and A'B'C' are polar triangles, the radius OA is perpendicular to the plane OB'C'. EXEKCISES ON THE SPHEKE 467 Ex. 13. On the same sphere, or on equal spheres, spherical tri- angles whose polar triangles have equal perimeters are equivalent. Ex. 14. Given OAO' ', OBO> ', and AB arcs of great circles, intersecting so that Z OAB= Z CfBA; prove that AOA=A(yAB. A [Sue. Show that /.OBA = /.WAS.] Ex. 15. Find the ratio of the volume of a sphere to the volume of a circumscribed cube. Ex. 16. Find the ratio of the surface of a sphere to the lateral surface of a circumscribed cylinder of revolution; also find the ratio of their volumes. Ex. 17. If the edge of a regular tetrahedron is denoted by a, find the ratio of the volumes of the inscribed and circumscribed sphez'js. Ex. 18. Find the ratio of the two segments into which a hemi- sphere is divided by a plane parallel to the base of the hemisphere and at the distance $ R from the base. EXERCISES. GROUP 73 SPHERICAL LOCI Ex. 1. Find the locus of a point at a given distance a from the surface of a given sphere. Ex. 2. Find the locus of a point on the surface of a sphere that; is equidistant from two given points on the surface. Ex. 3. If, through a given point outside a given sphere, tangent planes to the sphere are passed, find the locus of the points of tangency. Ex. 4. If straight lines be passed through a given fixed point in space, and through another given point other straight lines be passed perpendicular to the first set, njicl the locus of the feet of the perpendiculars. 468 BOOK IX. SOLID GEOMETRY EXERCISES. GROUP 74 PROBLEMS CONCERNING THE SPHERE Ex. 1. At a given point on a sphere, construct a plane tangent to the sphere. Ex. 2. Through a given point on the surface of a sphere, draw an arc of a great circle perpendicular to a given arc. Ex. 3. Inscribe a circle in a given spherical triangle. Ex. 4. Construct a spherical triangle, given its polar triangle. Given the radius, r, construct a spherical surface which shall pass through Ex. 5. - Three given points. Ex. 6. Two given points and be tangent to a given plane. Ex. 7. Two given points and be tangent to a given sphere Ex. 8. One given point and be tangent to two given planes. Ex. 9. One given point and be tangent to two given spheres. Given the radius, r, construct a spherical surface which shall be tangent to Ex. 10. Three given planes. Ex. 11. Two given planes and one given sphere. Ex. 12. Construet a spherical surface which shall pass through three given points and be tangent to a given plane. Ex. 13. Through a given straight line pass a plane tangent to a given sphere. [SuG. Through the center of the sphere pass a plane X given line, etc.] When is the solution impossible ? Ex. 14. Through a given point on a sphere, construct an arc of a great circle tangent to a given small circle of the sphere. [Sua. Draw a straight line from the center of the sphere to the given point, and produce it to intei-sect the plane of the small circle, NUMERICAL EXERCISES IN SOLID GEOMETRY For methods of facilitating numerical computations, see Arts. 493-6. EXERCISES. CROUP 75 LINES AND SURFACES OF POLYHEDRONS Find the lateral area and total area of a right prism whose / Ex. 1. Base is an equilateral triangle of edge 4 in., and whose altitude is 15 in. Ex. 2. Base is a triangle of sides 17, 12, 25, and whose altitude is 20. f Ex. 3. Base is an isosceles trapezoid, the parallel bases being 10 and 15 and leg 8, and whose altitude is 24. * Ex. 4. Base is a rhombus whose diagonals are 12 and 16, and whose altitude is 12. Ex. 5. Base is a regular hexagon with side 8 ft., and whose alti- tude is 20 ft. Ex. 6. Find the entire surface of a rectangular parallelepiped 8 X 12 X 16 in. ; of one p X q X r ft. Ex. 7. Of a cube whose edge is 1 ft. 3 in. Ex. 8. The lateral area of a regular hexagonal prism is 120 sq. ft. and an edge of the base is 10 ft. Find the altitude. Ex. 9. How many square feet of tin are necessary to line a box 20X6X4 in. ? / Ex. 10. If the surface of a cube is 1 sq. yd., find an edge in inches, Ex. 11. Find the diagonal of a cube whose edge is 5 in. Ex. 13. If the diagonal of a cube is 12 ft., find the surface, 470 SOLID GEOMETKY Ex. 13. If the surface of a rectangular parallelepiped is 208 sq. in.," and the edges are as 2 : 3 : 4, find the edges. In a regular square pyramid Ex. 14. If an edge of the base is 16 and slant height is 17, find the altitude. j Ex. 15. If the altitude is 15 and a lateral edge is 17, find an edge of the base. Ex. 16. If a lateral edge is 25 and an edge of the base is 14, find the altitude. In a regular triangular pyramid Ex. 17. If an edge of the base is 8 and the altitude is 10, find the slant height. Ex. 18. Find the altitude of a regular tetrahedron whose edge is 6. Find the lateral surface and the total surface of Ex. 19. A regular square pyramid an edge of whose base is 16, and whose altitude is 15. Ex. 20. A regular triangular pyramid an edge of whose base is 10, and whose altitude is 12. ^ Ex. 21. A regular hexagonal pyramid an edge of whose base is 4, and whose altitude is 21. Ex. 22. A regular square pyramid whose slant height is 24, and whose lateral edge is 25. / Ex. 23. A regular tetrahedron whose edge is 4. Ex. 24. A regular tetrahedron whose altitude is 9. Ex. 25. A regular hexagonal pyramid each edge of whose base is a, and whose altitude is 6. Ex. 26. In the frustum of a regular square pyramid the edges of the bases are 6 and 18, and the altitude is 8. Find the slant height. Hence find the lateral area. Ex. 27. In the frustum of a regular triangular pyramid the edges of the bases are 4 and 6, and the altitudi is 5, Find the slant height. Hence and the lateral area. NUMERICAL EXERCISES IN SOLID GEOMETRY 471 Ex. 28. In the frustum of a regular tetrahedron, if the edge of the lower base is &i, the edge of the upper base is 6 2 , and the altitude is a, show that =iT/H&i W -f-4a 2 . Ex. 29. In the frustum of a regular square pyramid the edges of the bases are 20 and 60, and a lateral edge is 101. Find the lateral surface. EXERCISES. GROUP 76 LINES AND SURFACES OP CONES AND CYLINDERS Ex. 1 . How many square feet of lateral surface has a tunnel 100 yds. long and 7 ft. in diameter. Ex. 2. The lateral area of a cylinder of revolution is 1 sq. yd., and the altitude is 1 ft. Find the radius of the base: Ex. 3. The entire surface of a cylinder of revolution is 900 sq. ft. and the radius of the base is 10 ft. Find the altitude. In a cylinder of revolution Ex. 4. Find R in terms of 8 and H. Ex. 5. Find H in terms of E and T. Ex. 6. Find T in terms of S and H. Ex. 7. How many sq. yds- of canvas are required to make a coni- cal tent 20 ft. in diameter and 12 ft. high ? Jt Ex. 8. A man has 400 sq. yds. of canvas and wants to make a conical tent 20 yds. in diameter. What will be its altitude ? ^ Ex. 9. The altitude of a cone of revolution is 10 ft. and the lat- eral area is 11 times the area of the base. Find the radius of the base. In a cone of revolution Ex. 10. Find T in terms of S and L. Ex. 11. Find 12 in terms of T and L. J* Ex. 12. How many square feet of tin are necessary to make a funnel the diameters of whose ends are 2 in. and 8 in., and whose altitude is 7 in. ? It Ex. 13. If the slant height of a frustum of a cone of revolution makes an angle of 45 with the base, show that the lateral area of the frustum is (n 2 r 2 2 ) TT i/2. 472 SOLID GEOMETRY EXERCISES. CROUP 77 SPHERICAL LINES AND SURFACES Ex. 1. Find in square feet the area of the surface of a sphere whose radius is 1 ft. 2 in. f> Ex. 2. How many square inches of leather will it take to cover a baseball whose diameter is 3i in.? Ex. 3. How many sq. ft. of tin are required to cover a dome in the shape of a hemisphere 6 yds. in diameter ? K Ex. 4. What is the radius of a sphere whose surface is 616 sq. in. ? Ex. 5. Find the diameter of a globe whose surface is 1 sq. yd. Ex^jS. If the circumference of a great circle on a sphere is 1 ft., find the area of the surface of the sphere. Ex. 7. If a hemispherical dome is to contain 100 sq. yds. of sur- face, what must its diameter be ? * Ex. 8. Find the radius of a sphere in* which the area of the sur- face equals the number of linear units in the circumference of a great circle. Find the area of a lune in which X Ex. 9. The angle of the lune is 36, and the radius of the sphere is 14 in. Ex. 10. The angle of the lune is 18 20', and the diameter of the sphere is 20 in. ft Ex. 11. The angle of the lune is 24, and the surface of the sphere is 4 sq. ft. Find the area of a spherical triangle in which * Ex. 12. The angles are 80, 90, 120, and the diameter of the sphere is 14 ft. + Ex. 13. The angles are 74 24', 83 16', 92 20', and the radius of the sphere is 10. % Ex. 14. The angles are 85, 95, 135, and the surface of the sphere is 10 sq. ft. NUMERICAL EXERCISES IN SOLID GEOMETRY 473 Ex. 15. If the sides of a spherical triangle are 100, 110, 120 and the radius of the sphere is 16, find the area of the polar triangle. Ex. 16. If the angles of a spherical triangle are 90, 100, 120* and its area is 3900, find the radius of the sphere. Ex. 17. If the area of an equilateral spherical triangle is one- third the surface of the sphere, find an angle of the triangle. Ex. 18. In a trihedral angle the plane angles of the dihedral angles are 80, 90, 100; find the number of solid degrees in the trihedral angle. S Ex. 19. Find the area of a spherical hexagon each of whose angles is 150, on a sphere whose radius is 20 in. Ex. 20. If each dihedral angle of a given pentahedral angle is 120, how many solid degrees does the pentahedral angle contain ? Ex. 21. In a sphere whose radius is 14 in., find the area of a zone 3 in. high. Ex. 22. What is the area of the north temperate zone, if the earth is taken to be a sphere with a radius of 4,000 miles, and the distance between the plane of the arctic circle and that of the tropic of Cancer is 1,800 miles ? Ex. 23. If Cairo, Egypt, is in latitude 30, show that its parallel of latitude bisects the surface of the northern hemisphere. Ex. 24. How high must a person be above the earth's surface to see one -third of the surface ? S/ Ex. 25. How much of the earth's surface will a man see who is 2,000 miles above the surface, if the diameter is taken as 8,000 miles? Ex. 26. If the area of a zone equals the area of a great circle, find the altitude of the zone in terms of the radius of the sphere. Ex. 27. If sounds from the Krakatoa explosion were heard at a distance of 3,000 miles (taken as a chord) on the surface of the earth, over what fraction of the earth's surface were they heard ? Ex. 28. The radii of two spheres are 5 and 12 in. and their cen- s are 13 in. apart. Find the area of the circle of intersection and also of that part of the surface of each sphere not included by the other sphere. 474 SOLID GEOMETKY EXERCISES. CROUP 78 VOLUMES OF POLYHEDRONS Find the volume of a prism XEx. 1. Whose base is an equilateral triangle with side 5 in., and rhose altitude is 16 in. )/ Ex. 2. Whose base is a triangle with sides 12, 13, 15, and whose Altitude is 20. Ex. 3. Whose base is an isosceles right triangle with a leg equal to 2 yds., and whose altitude is 25 ft. Ex. 4. Whose base is a regular hexagon with a side of 8 ft., and hose altitude is 10 yds. S Ex. 5. Whose base is a rhombus one of whose sides is 25, and one of whose diagonals is 14, and whose altitude is 11. Ex. 6. Whose base contains 84 sq. yds., and whose lateral faces are three rectangles with areas of 100, 170, 210 sq. yds., respectively. XEx. 7. How many bushels of wheat are held by a bin 30 x 10 x 6 ft. , if a bushel is taken as li cu. ft.? Ex. 8. How many cart-loads of earth are in a cellar 30 x 20x6 ft., if a cart-load is a cubic yard ? / Ex. 9. 'If a cubical block of marble costs $2, what is the cost of a cube whose edge is a diagonal of the first block ? Ex. 10. Find the edge of a cube whose volume equals the sum of the volumes of two cubes whose edges are 3 and 5 ft. Ex. 11. Find the edge of a cube whose volume equals the area of its surface. Ex. 12. If the top of a cistern is a rectangle 12x8 ft., how deep ust the cistern be to hold 10,000 gallons ? Ex. 13. Find the inner edge of a peck measure which is in the shape of a cube. Ex. 14. A peck measure is to be a rectangular parallelepiped with square base and altitude equal to twice the edge of the base, Find its dimensions, NUMERICAL EXERCISES IN SOLID GEOMETRY 475 Ex. 15. Find the volume of a cube whose diagonal is a. Find the volume of a pyramid Ex. 16. Whose base is an equilateral triangle with side 8 in. r and whose altitude is 12 in. Ex. 17. Whose base is a right triangle with hypotenuse 29 and one leg 21, and whose altitude is 20. Ex. 18. Whose base is a square with side 6, and each of whose lateral edges is 5. Ex. 19. Whose base is a square with side 10, and each of whose lateral faces makes an angle of 45 with the base. Ex. 20. If the pyramid of Memphis has an altitude of 146 yds. and a square base of side 232 yds., how many cubic yards of stone does it contain ? What is this worth at $1 a cu. yd.? Ex. 21. A church spire 150 ft. high is hexagonal in shape and each side of the base is 10 ft. The spire has a hollow hexagonal interior, each side of whose base is 6 ft., and whose altitude is 45 ft. How many cubic yards of stone does the spire contain ? Ex. 22. If a pyramid contains 4 cu. yds. and its base is a square with one side 2 ft., find the altitude. Ex. 23. A heap of candy in the shape of a frustum of a regular square pyramid has the edges of its bases 25 and 9 in. and its altitude 12 in. Find the number of pounds in the heap if a pound is a rectan- gular parallelepiped 4x3x2 in. in size. Ex. 24. Find the volume of a frustum of a regular triangular pyramid, the edges of the bases being 2 and 8, and the slant height 12. Ex. 25. The edges of the bases of the frustum of a regular square pyramid are 24 and 6, and each lateral edge is 15; find the volume. Ex. 26. If a stick of timber is in the shape of a frustum of a regular square pyramid with the edges of its ends 9 and 15 in., and with a length of 14 ft., find the number of feet of lumber in the stick. What is the difference between this volume and that of a stick of the same length having the shape of a prism with a base equal to the area of a midsection of the first stick ? 476 SOLID GEOMETRY Ex. 27. Find the volume of a prismoid whose bases are rectangles 5 x 2 ft. and 7 x 4 ft., and whose altitude is 12 ft. Ex. 28. How many cart-loads of earth are there in a railroad cut 12 ft. deep, whose base is a rectangle 100x8 ft., and whose top is a rectangle 30x50 ft.? Ex. 29. Find the volume of a prismatoid whose base is an equi- lateral triangle with side 12 ft., and whose top is a line 12 ft. long parallel to one side of the base, and whose altitude is 1& ft. Ex. 30. If the base of a prismatoid is a rectangle with dimensions a and &, the top is a line c parallel to the side b of the base, and the altitude is h, find the volume. EXERCISES. CROUP 79 VOLUMES OF CONES AND CYLINDERS Ex. 1. How many barrels of oil are contained in a cylindrical tank 20 ft. long and 6 ft. in diameter, if a barrel contains 4 cu. ft. f )^ Ex. 2. How many cu. yds. of earth must be removed in making a tunnel 450 ft. long, if a cross-section of the tunnel is a semicircle of 15 ft. radius ? V Ex. 3. A cylindrical glass 3 in. in diameter holds half a pint. Find its height in inches. y Ex. 4. If a cubic foot of brass be drawn out into wire ^ inch in ^ diameter, how long will the wire be ? Ex. 5. A gallon measure is a cylinder whose altitude equals the diameter of the base. Find the altitude. * Ex. 6. Show that the volumes of two cylinders, having the altitude of each equal to the radius of the other, are to each other as E : E' . Ex. 7. In a cylinder, find E in terms of V and H ; also V in terms of 8 and E. A Ex. 8. A conical heap of potatoes is 44 ft. in circumference and 6 ft. high. How many bushels does it contain, if a bushel is lieu, ft.f y Ex. 9. What fraction of a pint will a conical wine-glass hold, if its altitude is 3 in. and the diameter of the top ia 2 in. ? NUMERICAL EXERCISES IN SOLID GEOMETRY 477 Ex. 10. Find the ratio of the volumes of the two cones inscribed in, and circumscribed about, a regular tetrahedron. Ex. 11. If an equilateral cone contains 1 quart, find its dimen- sions in inches. Ex. 12. In a cone of revolution find V in terms of R and L ; also find V in terms of R and S. Ex. 13. Find the volume of a frustum of a cone of revolution, whose radii are 14 and 7 ft., and whose altitude is 2 yds. ' Ex. 14. What is the cost, at 50 cts. a cu. ft., of a piece of marble in the shape of a frustum of a cone of revolution, whose radii are 6 and 9 ft., and whose slant height is 5 ft.? Ex. 15. In a frustum of a cone of revolution, the volume is 88 cu. ft., the altitude is 9 ft., and R = 2r. Find r. EXERCISES. CROUP 8O SPHERICAL VOLUMES Ex. 1. Find the volume of a sphere whose radius is 1 ft. 9 in. Ex. 2. If the earth is a sphere 7,920 miles in diameter, find its ***+***.. volume in cubic miles. T' / Ex. 3. Find the diameter of a sphere whose volume is 1 cu. ft. Ex. 4. What is the volume of a sphere whose surface is 616 sq. in. ? ML //^rty Ex. 5. Find the radius of a sphere equivalent to the sum of two spheres, whose radii are 2 and 4 in. Ex. 6. Find the radius of a sphere whose volume equals the area of its surface. Ex. 7. Find the volume of a sphere circumscribed about a cube whose edge is 6. " Ex. 8. Find the volume of a spherical shell whose inner and outer diameters are 14 and 21 in. Ex. 9. Find the volume of a spherical shell whose inner and outer Surfaces are 20 TT and 12 TT. 478 SOLID GEOMETRY Find the volume of Ex. 10. A spherical wedge whose angle is 24, the radius of the sphere being 10 in. - Ex. 11. A spherical sector whose base is a zone 2 in. high, the radius of the sphere being 10 in. V "icallsc Ex. 12. A spherical ^segment of two bases whose radii are 4 and 7 and altitude 5 in. ' Ex. 13. A wash-basin in the shape of a segment of a sphere is 6 in. deep and 24 in. in diameter. How many quarts of water will the basin hold ? / Ex. 14. A plane parallel to the base of a hemisphere and bisect- ing the altitude divides its volume in what ratio ? / Ex. 15. A spherical segment 4 in. high contains 200 cu. in. ; find the radius of the sphere. f Ex. 16. If a heavy sphere whose diameter is 4 in. be placed in a conical wine-glass full of water, whose diameter is 5 in. and altitude 6 in., find how much water will run over. EXERCISES. GROUP 81 EQUIVALENT SOLIDS I Ex. 1. If a cubical block of putty, each edge of which is 8 inches, be molded into a cylinder of revolution whose radius is 3 inches, find the altitude of the cylinder. Ex. 2. Find the radius of a sphere equivalent to a cube whose edge is 10 in. -*, Ex. 3. Find the radius of a sphere equivalent to a cone of revolu- tion, whose radius is 3 in. and altitude 6 in. Ex. 4. Find the edge of a cube equivalent to a frustum of a cone of revolution, whose radii are 4 and 9 ft. and altitude 2 yds. Ex. 5. Find the altitude of a rectangular parallelepiped, whose base is 3 x5 in. and whose volume is equivalent to a sphere of radius 7 in. NUMEKICAL EXERCISES IN SOLID GEOMETEY 479 V Ex. 6. Find the base of a square rectangular parallelepiped, whose altitude is 8 in. and whose volume equals the volume of a cone of revolution with a radius of 6 and an altitude of 12 in. Ex. 7. Find the radius of a cone of revolution, whose altitude is 15 and whose volume is equal to that of a cylinder of revolution with radius 6 and altitude 20. Ex. 8. Find the altitude of a cone of revolution, whose radius is 15 and whose volume equals the volume of a cone of revolution with radius 9 and altitude 24. X. Ex. 9. On a sphere whose diameter is 14 the altitude of a zone of one base is 2. Find the altitude of a cylinder of revolution, whose base equals the base of the zone and whose lateral surface equals the surface of the zone. EXERCISES. GROUP 82 SIMILAR SOLIDS Ex. 1. If on two similar solids L, L' and I, V are pairs of homol- ogous lines; Af A 1 and a, a' pairs of homologous areas, V, V and v, v f pairs of homologous volumes, CL :L' = l : l' = VA : i/A'=&v : tfv'. show that A : A' = L Z : L n = a : a'= vl : F'i. Ex. 2. If the edge of a cube is 10 in., find the edge of a cube having 5 times the surface. Ex. 3. If the radius of a sphere is 10 in., find the radius of a sphere having 5 times the surface. Ex. 4. If the altitude of a cone of revolution is 10 in., find the altitude of a similar cone of revolution having 5 times the surface. Ex. 5. In the last three exercises, find the required dimension if the volume is to be 5 times the volume of the original solid. Ex. 6. The linear dimensions of one trunk are twice as great as those of another trunk. How much greater is the faunae ? Tbe surface? 480 SOLID GEOMETRY Ex. 7. How far from the vertex is the cross -section which bisects the volume of a cone of revolution ? Which bisects the lateral surface ? Ex. 8. If the altitude of a pyramid is bisected by a plane parallel to the base, how does the area of the cross-section compare with the area of the base ? How does the volume cut off compare with the volume of the entire pyramid ? Ex. 9. Planes parallel to the base of a cone divide the altitude into three equal parts ; compare the lateral surfaces cut off. Also the volumes. Ex. 10. A sphere 10 in. in diameter is divided into three equiva- lent parts by concentric spherical surfaces. Find the diameters of these surfaces. Ex. 11. If the strength of a muscle is as the area of its cross- section, and Goliath of Gath was three times as large in each linear dimension as Tom Thumb, how much greater was his strength ? His weight ? How, then, does the activity of the one man compare with that of the other ? Ex. 12. If the rate at which heat radiates, from a body is in pro- portion to the amount of surface, and the planet Jupiter has a diame- ter 11 times that of the earth, how many times longer will Jupiter be in cooling off ? [Suo. How many times greater is the volume, and therefore .the original amount of heat in Jupiter ? How many times greater is its surface ? What will be the combined effect of these factors ?] CROUP 83 MISCELLANEOUS NUMERICAL EXERCISES IN SOLID GEOMETRY Find S, T and V of Ex. 1. A right triangular prism whose altitude is 1 ft., and the sides of whose base are 26, 28, 30 in. Ex. 2. A cone .of revolution the radius of whose base is 1 ft. 2 in., and whose altitude is 35 in. Ex. 3. A frustum of a square pyramid the areas of whose bases are 1 sq, ft. and 36 sq, in., and whose altitude is 9 in. NUMEKICAL EXEECISES IN SOLID GEOMETEY 481 Ex. 4. A pyramid whose slant height is 10 in., and whose base is an equilateral triangle whose side is 8 in. k Ex. 5. A cube whose diagonal is 1 yd. % Ex. 6. A frustum of a cone of revolution whose radii are 6 and 11 in. and slant height 13 in. /^ Ex. 7. A rectangular parallelepiped whose diagonal is 2^29, and whose dimensions are in the ratio 2 : 3 : 4. * Ex. 8. Find the volume of a sphere inscribed in a cube whose 'edge is 6 ; also find the area of a triangle on that sphere whose angles are 80, 90, 150. X Ex. 9. Find the volume of the spherical pyramid whose base is the above triangle. X Ex. 10. Find the angle of a lune on the same sphere, equivalent to that triangle. Ex. 11. On a cube whose edge is 4, planes through the midpoints of the edges cut off the corners. Find the volume of the solid re- maining. Ex. 12. How is V changed if H of a cone of revolution is doubled and R remains unchanged ? If R is doubled and H remains unchanged? If both H and B are doubled ? jL Ex, 13. In an equilateral cone, find S and V in terms of E. Ex. 14. A piece of lead 20x8 x2 in. will make how many spher- ical bullets, each f in. in diameter ? Ex. 15. How many bricks are necessary to make a chimney in the shape of a frustum of a cone, whose altitude is 90 ft., whose outer diameters are 3 and 8 ft., and whose inner diameters are 2 and 4 ft., counting 12 bricks to the cubic ft.? Ex. 16. If the area of a zone is 300 and its altitude 6, find the radius of the sphere. Ex. 17. If the section of a cylinder of revolution through its axis is a square, find S, T, V in terms of E. Ex. 18. If every edge of a square pyramid is fy find b in terms of T. DD 452 SOLID GEOMETRY ^ Ex.19. A regular square pyramid has a for its altitude and also for each side of the base. Find the area of a section made by a plane parallel to the base and bisecting the altitude. Find also the volumes of the two parts into which the pyramid is divided. / Ex. 20. If the earth is a sphere of 8,000 miles diameter and its atmosphere extends 50 miles from the earth, find the volume of the atmosphere. Ex. 21. On a sphere, find the ratio of the area of an equilateral spherical triangle, each of whose angles is 95, to the area of a lune whose angle is 80. Ex. 22. A square right prism has an altitude 6a and an edge of the base 2a. Find the volume of the largest cylinder, sphere, pyramid and cone which can be cut from it. I/ Ex. 23. Obtain a formula for the area of that part of a sphere 'illuminated by a point of light at a distance a from tne sphere whose radius is E. */ Ex. 24. On a sphere whose radius is 6 in., find an angle of an / equilateral triangle whose area is 12 sq. in. Ex. 25. Find the volume of a prismatoid, whose altitude is 24 and whose bases are equilateral triangles, each side 10, so placed that the mid-section of the prismatoid is a regular hexagon. / Ex. 26. On a sphere whose radius is 16, the bases of a zone are equal and are together equal to the area of the zone. Find the alti- tude of the zone. /" Ex. 27. Find the volume of a square pyramid, the edge of whose base is 10 and each of whose lateral edges is inclined 60 to the base. ^ Ex. 28. An irregular piece of ore, if placed in a cylinder partly filled with water, causes the water to rise 6 in. If the radius of the cylinder is 8 in., what is the volume of the ore ? Ex. 29. Find the volume of a truncated right triangular prism, if the edges of the base are 8, 9, 11, and the lateral edges are 12, 13, 14. - Ex. 30. In a sphere whose radius is 5, a section is taken at tho distance 3 from the center. On this section as a base a cone is formed whose lateral elements are tangent to the sphere, Find tho lateral surface and volume of the cone. NUMEKICAL EXERCISES IN SOLID GEOMETRY 483 Ex. 31 . The volume of a sphere is l,437i cu. in. Find the surface. Ex. 32. A square whose side is 6 is revolved about a diagonal as an axis; find the surface and volume generated. Ex. 33. Find the edge of a cubical cistern that will hold 10 tons of water, if 1 cu. ft. of water weighs 62.28 Ibs. Ex. 34. A water trough has equilateral triangles, each side 3 ft., for ends, and is 18 ft. long. How many buckets of water will it hold, if a bucket is a cylinder 1 ft. in diameter and li ft. high ? Ex. 35. The lateral area of a cylinder of revolution is 440 sq. in., and the volume is 1,540 cu. in. Find the radius and altitude. Ex. 36. The angles of a spherical quadrilateral are 80, 100, 120, 120. Find the angle of an equivalent equilateral triangle. if Ex. 37. A cone and a cylinder have equal lateral surfaces, and their axis sections are equilateral. Find the ratio of their volumes. Ex. 38. A water-pipe in. in diameter rises 13 f.t. from the ground. How many quarts of water must be drawn from it before the water from under the ground comes out ? If a quart runs out in 5 seconds, how long must the water run ? Ex. 39. A cube immersed in a cylinder partly filled with water causes the water to rise 4 in. If the radius of the cylinder is 6 in., what is an edge of the cube ? Ex. 40. An auger hole whose diameter is 3 in. is bored through the center of a sphere whose diameter is 8 inches. Find the volume remaining. Ex. 41. Show that the volumes of a cone, hemisphere, and cylin- der of the same base and altitude are as 1 '. 2 : 3. Ex. 42. The volumes of two similar cylinders of revolution are as 8 : 125 ; find the ratio of their radii. If the radius of the smaller is 10 in., what is the radius of the larger ? Ex. 43. An iron shell is 2 in. thick and the diameter of its outer surface is 28 in. Find its volume. Ex. 44. The legs of an isosceles spherical triangle each make an angle of 75 with the base. The legs produced form a lune whose area is four times the area of the triangle. Find the angle of the lune. 484 SOLID GEGMETEY CROUP 84 EXERCISES INVOLVING THE METRIC SYSTEM Find S, T, V of Ex. 1. A right prism the edges of whose base are 6m., 70 dm., 900 cm., and whose altitude is 90 dm. Ex. 2. A regular square pyramid an edge of whose base is 30 dm., and whose altitude is 1.7 m P Ex. 3. A sphere whose radius is 0.02 m. Ex. 4. A frustum of a cone of revolution whose radii are 10 dm. and 6 dm., and whose slant height is 50 cm. Ex. 5. A cube whose diagonal is 12 cm. Ex. 6. A cylinder of revolution whose radius equals 2 dm., and whose altitude equals the diameter of the base. Ex. 7. Find the area of a spherical triangle on a sphere whose radius is 0.02 m., if its angles are 110, 120, 130. Ex. 8. Find the number of square meters in the surface of a sphere, a great circle of which is 50 dm. long. Ex. 9. How many liters will a cylindrical vessel hold that is 10 dm. in diameter and 0.25 m. high ? How many liquid quarts ? Ex. 10. A liter measure is a cylinder whose diameter is half the altitude. Find its dimensions in centimeters. Ex. 11. Find the surface of a sphere whose volume is 1 cu. m,, APPENDIX J. MODERN GEOMETRIC CONCEPTS 840. Modern Geometry. In recent times many new geometric ideas have been invented, and some of them developed into important new branches of geometry. Thus, the idea of symmetry (see Art. 484, etc.) is a modern geometric concept. A few other of these modern concepts and methods will be briefly mentioned, but their thorough consideration lies beyond the scope of this book. 841. Projective Geometry. The idea of projections (see Art. 345) has been developed in comparatively recent times into an important branch of mathematics with many practical applications, as in engineering, architec- ture, construction of maps, etc. 842. Principle of Continuity. By this principle two or more theorems are made special cases of a single more general theorem. An important aid in obtaining continuity among geometric principles is the application of the con- cept of negative quantity to geometric magnitudes. Thus, a negative line is a line opposite in direction to a given line taken as positive. For example, if OA is +, OB is , (485) 486 GEOMETKY. APPENDIX Similarly, a negative angle is an angle formed by rotating a line in a plane in a direction opposite from a direction of rotation taken as positive. Thus, if the line OA rotating from the position OA forms the positive angle AOB, the same line rotating in the opposite direction forms the negative angle AOB'. Similarly, positive and negative arcs are formed. In like manner, if P and P f are on opposite sides of the line AB and the area PAB is taken as positive, the area P'AB will be negative. As an illustration of the law of continuity, we may take the theorem that the sum of the triangles formed by drawing lines from a point to the vertices of a polygon equals the area of the polygon. Applying this to the quadrilateral ABCD, if the point P falls within the quadrilateral, APA+APC + APOD + APAD=ABCD (Ax. 6). Also, if the point falls without the quadrilateral at P', A P'AB + A P'BC + A P'CD + A P'AD = ABCD, since AP'AD is a negative area, and hence is to be subtracted from the sum of the other three triangles. 843. The Principle of Reciprocity, or Duality, is a principle of relation between two theorems by which each theorem is convertible into the other by causing the words for the same two geometric objects in each theorem to ex- change places. MODERN GEOMETRIC CONCEPTS 487 Thus, of theorems VI and VII, Book I, either may be converted into the other by replacing the word "sides" by "angles," and "angles" by "sides." Hence these are termed reciprocal theorems. The following are other instances of reciprocal geometric properties : 1. Two points determine a straight line. 2. Three points not in the same straight line determine a plane. 3. A straight line and a point determine a plane. 1. Two lines determine a point. 2. Tliree planes not through the same straight line determine a point. 3. A straight line and a plane determine a point. The reciprocal of a theorem is not necessarily true. Thus, two parallel straight lines determine a*plane, but two parallel planes do not determine a line. However, by the use of the principle of reciprocity, geometrical properties, not otherwise obvious, are fre- quently suggested. 844. Principle of Homology. Just as the law of reci- procity indicates relations between one set of geometric concepts (as lines) and another set of geometric concepts (as points), so the law of homology indicates relations between a set of geometric concepts and a set of concepts outside of geometry: as a set of algebraic concepts, for instance. Thus, if a and 6 are numbers, by algebra (a + 5) (a 5) = a 2 6 2 . Also, if a and b are segments of a line, the rectangle (a-j-ft)X(a &) is equivalent to the difference between the squares a 2 and 6 2 . By means of this principle, truths which would be over- looked or difficult to prove ill one department of thought 488 GEOMETRY. APPENDIX are made obvious by observing the corresponding truth in another department of thought. Thus, if a and I are line segments, the theorem (a+ 6) 2 + (a &) 2 =2(a 2 -f-6 2 ) is not immediately obvious in geo- metry, but becomes so by observing the like relation between the algebraic numbers a and &. i 845. Non-Euclidean Geometry. Hyperspace. By vary- ing the properties of space, as these are ordinarily stated, different kinds of space may be conceived of, each having its own geometric laws and properties. Thus, space, as we ordinarily conceive it, has three dimensions, but it is possible to conceive of space as having four or more dimensions. To mention a single property of four dimen- sional space, in such a space it would be possible, by simple pressure, to turn a sphere, as an orange, inside out without breaking its surface. As an aid toward conceiving how this is possible, consider a plane in which one circle lies inside another. No matter how these circles are moved about in the plane, it is impossible to shift the inner circle so as to place it outside the other without breaking the circumference of the outer circle. But, if we are allowed to use the third dimension of space, it is a simple matter to lift the inner circle up out of the plane and set it down outside the larger circle. Similarly if, in space of three dimensions, we have one spherical shell inside a larger shell, it is impossible to place the smaller shell outside the larger without breaking the larger. But if the use of a fourth dimension be allowed, that is, the use of another dimension of freedom of motion, it is possible to place the inner shell outside the larger without breaking the latter. 846. Curved Spaces. By varying the geometric axioms of space (see Art. '47), different kinds of space may be conceived of. Thus, we may conceive of space such that through a given point one line may be drawn parallel to a given line (that is ordinary, or Euclidean space); or such MODEKN GEOMETRIC CONCEPTS 489 that through a given point no line can be drawn parallel to a given line (spherical space) ; or such that through a given point more than one line can be drawn parallel to a given line (pseudo- spherical space). These different kinds of space differ in many of their properties. For example, in the first of them the sum of the angles of a triangle equals two right angles; in the second, it is greater; in the third, it is less. These different kinds of space, however, have many properties in common. Thus, in all of them every point in the perpendicular bisector of a line is equidistant from the extremities of the line. EXERCISES. GROUP 85 Ex. 1. Show by the use of zero and negative arcs that the princi- ples of Arts. 257, 263, 258, 264, 265, are particular cases of the general theorem that the angle included between two lines which cut or touch a circle is measured by one-half the sum of the intercepted arcs. Ex. 2. Show that the principles of Arts. 354 and 358 are particular cases of the theorem that, if two lines are drawn from or through a point to meet a circumference, the product of the segments of one line equals the product of the segments of the other line. Ex. 3. Show by the use of negative angles that theorem XXXVIII, Book I, is true for a quadrilateral of the form ABCD. [BCD is a negative angle ; the angle at the vertex D is the reflex angle Ex. 4. What is the reciprocal of the state- ment that two intersecting straight lines deter- mine a plane ? Ex. 5. What is the reciprocal of the statement that three planes perpendicular to each other determine three straight lines perpen' dicular to each other ? H. HISTORY OF GEOMETRY 847. Origin of Geometry as a Science. The beginnings of geometry as a science are found in Egypt, dating back at least three thousand years before Christ. Herodotus says that geometry, as known in Egypt, grew out of the need of remeasuring pieces of land parts of which had been washed away by the Nile floods, in order to make an equi- table readjustment of the taxes on the same. The substance of the Egyptian geometry is found in an old papyrus roll, now in the British museum. This roll is, in effect, a mathematical treatise written by a scribe named Ahmes at least 1700 B.C., and is, the writer states, a copy of a more ancient work, dating, say, 3000 B. C. 848. Epochs in the Development of Geometry. From Egypt a knowledge of geometry was transferred to Greece, whence it spread to other countries. Hence we have the following principal epochs in the development of geometry; 1. Egyptian : 3000 B. C. 1500 B. C. 2. Greek : 600 B. C. 100 B. C. 3. Hindoo : 500 A. D. 1100 A. D. 4. Arab : 800 A. D. 1200 A. D. 5. European : 1200 A. D. In the year 1120 A. D., Athelard, an English monk, visited Cordova, in Spam, in the disguise of a Mohamme- dan student, and procured a copy of Euclid m the Arabic language. This book he brought back to central Europe, where it was translated into Latin and became the basis of all geometric study m Europe till the year 1633, when, (490) HISTORY OF GEOMETRY 491 owing to the capture of Constantinople by the Turks, copies of the works of the Greek mathematicians in the original Greek were scattered through Europe. HISTORY OF GEOMETRICAL METHODS -*. f- 849. Rhetorical Methods. By rhetorical methods in the presentation of geometric truths, is meant the use of definitions, axioms, theorems, geometric figures, the rep- resentation of geometric magnitudes by the use of letters, the arrangement of material in Books, etc. The Egyptians had none of these, their geometric knowledge being re- corded only in the shape of the solutions of certain numeri- cal examples, from which the rules used must be inferred. Thales (Greece 600 B.C.) first made an enunciation of an abstract property of a geometric figure. He had a rude Idea of the geometric theorem. Pythagoras (Italy 525, B.C.) introduced formal defini- tions into geometry, though some of those used by him were not very accurate. F,or instance, his definition of a point is "unity having position." Pythagoras also arranged the leading propositions known to him in something like logical order. Hippocrates (Athens, 420 B. C.) was the first systemati- cally to denote a point by a capital letter, and a segment of a line by two capital letters, as the line AB, as is done at present. He also wrote the first text -book on geometry. Plato (Athens, 380 B. C.) made definitions, axioms and postulates the beginning and basis of geometry. To Euclid (Alexandria, 280 B. C.) is due the division of geometry into Books, the formal enunciation of theo- rems, the particular enunciation, the formal construction, 492 GEOMETKY. APPENDIX proof, and conclusion, in presenting a proposition. He also introduced the use of the corollary and scholium. Using these methods of presenting geometric truths, Euclid wrote a text -book of geometry in thirteen books, which was the standard text -book on this subject for nearly two thousand years. The use of the symbols A, ZZ7 , || , etc., in geometric proofs originated in the United States in recent years. 850. Logical Methods. The Egyptians used no formal methods of proof. They probably obtained their few crude geometric processes as the result of experiment. The Hindoos also used no formal proof. One of their writers on geometry merely states a theorem, draws a figure, and says "Behold ! " The use of logical methods of geometric proof is due to the Greeks. The early Greek geometricians used experi- mental methods at times, in order to obtain geometric truths. For instance, they determined that the angles at the base of an isosceles triangle are equal, by folding half of the triangle over on the altitude as an axis and observ- ing that the angles mentioned coincided as a fact, but without showing that they must coincide. Pythagoras (525 B. C.) was the first to establish geo- metric truths by systematic deduction, but his methods were sometimes faulty. For instance, he believed that the converse of a proposition is necessarily true. Hippocrates (420 B.C.) used correct and rigorous de- duction in geometric proofs. He also introduced specific varieties of such deduction, such as the method of reduc- ing one proposition to another (Art. 296), and the reductio ad absurdum. HISTORY OF GEOMETRY 493 The methods of deduction used by the Greeks, however, were de- fective in their lack of generality. For instance, it was often thought necessary to have a separate proof of a theorem for each different kind of figure to which the theorem applied. Thus, the theorem that the sum of the an- gles of a triangle equals two right angles was proved, (1) for the equilateral triangle by use of the regular hexagon ; (2) for the right triangle by the use of a rectangle ; (3) for a scalene triangle by dividing the scalene triangle into two right triangles. The Greeks appeared to fear that a general proof might be vitiated if it were applied to a figure in any way special or peculiar. In other words, they had no conception of the principle of continuity (Art. 842). Plato (380 B. C.) introduced the method of proof by analysis, that is, by taking a proposition as true and work- ing from it back to known truths (see Art. 196). To Eudoxus (360 B. C.) is virtually due proof by the method of limits; though his method, known as the method of exhaustions, is crude and cumbersome. Apollonius (Alexandria, 225 B. C.) used projections, transversals, etc., which, in modern times, have developed into the subject of projective geometry. 851. Mechanical Methods. The Greeks, in demonstra- ting a geometrical theorem, usually drew the figure em- ployed in a bed of sand. This method had certain advan- tages, but was not adapted to the use of a large audience. At the time when geometry was being developed in Greece, the interest in the subject was very general. There was scarcely a town but had its lectures on the subject. The news of the discovery of a 494 GEOMETKY. APPENDIX new theorem spread from town to town, and the theorem was redemon- strated in the sand of each market place. The Greek treatises, however, were written on vellum or papyrus by the use of the reed, or calamus, and ink. In Roman times, and in the middle ages, geometrical figures were drawn in wax smeared on wooden boards, called tablets. They were drawn by the use of the stylus, a metal stick, pointed at one end for making marks, and broad at the other for erasing marks. These wax tablets were still in use in Shakespeare's time (see Hamlet Act I, Sc. 5, 1. 107). The blackboard and crayon are modern inventions, their use having developed within the last one hundred years. The Greeks invented many kinds of drawing instruments f6r tracing various curves. It was due to the influence of Plato (380 B. C.) that, in constructing geometric figures, the use of only the ruler and compasses is permitted. HISTORY OF GEOMETRIC TRUTHS. PLANE GEOMETRY 852. Rectilinear Figures. The Egyp- tians measured the area of any four- sided field by multiplying half the sum of one pair of opposite sides by half the sum of the other pair; which was equivalent to using the a~rc^.b-\-d formula, area = X _ This, of course, gives a correct result for the rectangle and square, but gives too great a result for other quadrilaterals, as the trapezoid, etc. Hence Joseph, of the Book of Genesis, in buying the fields of the Egyptians for Pharoah in time of famine by the use of this formula, in many cases paid for a larger field than he obtained. The Egyptians had a special fondness for geometrical constructions, probably growing out of their work as temple HISTORY OF GEOMETRY 495 builders. A class of workers existed among them called "rope -stretchers," whose business was the marking out of the foundations of buildings. These men knew how to bisect an angle and also to construct a right angle. The latter was probably done by a method essentially the same as forming a right triangle whose sides are three, four and five units of length. Ahmes, in his treatise, has various constructions of the isosceles trapezoid from different data. Thales (600 B. C.) enunciated the following theorems: If two straight lines intersect, the opposite or vertical angles are equal; The angles at the base of an isosceles triangle are equal; Two triangles are equal if two sides and the included angle of one are equal to two sides and the included angle of the other; The sum of the angles of a triangle equals two right angles ; Two mutually equiangular triangles are similar. Thales used the last of these theorems to measure the height of the great pyramid by measuring the length of the shadow cast by the pyramid and also measuring the length of the shadow of a post of known height at the same time and making a proportion between these quantities. Pythagoras (525 B. C.) and his followers discovered correct formulas for the areas of the principal rectilinear figures, and also discovered the theorems that the areas of similar polygons are as the squares of their homologous sides, and that the square on the hypotenuse of a right triangle equals the sum of the squares on the other two sides. The latter is called the Pythagorean theorem. They also discovered how to construct a square equivalent to a given parallelogram, and to divide a given line in mean and extreme ratio, 496 GEOMETRY. APPENDIX To Eudoxus (380 B. C.) we owe the general theory of proportion in geometry, and the treatment of incommen- surable quantities by the method of Exhaustions. By the use of these he obtained such theorems as that the areas of two circles are to each other as the squares of their radii, or of their diameters. In the writings of Hero (Alexandria, 125 B. C.) we first find the formula for the area of a triangle in terms of its sides, K=V s(s a) (s b) (s c) - Hero also was the first to place land-surveying on a scientific basis,, It is a curious fact that Hero at the same time gives an incorrect formula for the area of a triangle, viz., lT=ia(&-f-c), this formula being apparently derived from Egyptian sources. Xenodorus (150 B. C.) investigated isoperemetrical figures. The Romans, though they excelled in engineering, ap- parently did not appreciate the value of the Greek geom- etry. Even after they became acquainted with it, they continued to use antiquated and inaccurate formulas for areas, some of them of obscure origin. Thus, they used the Egyptian formula for the area of a quadrilateral, K=^- X . They determined the area of an equilat- JJ eral triangle whose side is a, by different formulas, all incorrect, as K=-r , T=i(a 2 +a), and j=Ja 2 . 853. The Circle. Thales enunciated the theorem that every diameter bisects a circle, and proved the theorem that an angle inscribed in a semicircle is a right angle. To Hippocrates (420 B. C.) is due the discovery of nearly all the other principal properties of the circle given in this book. HISTORY OF GEOMETRY 497 The Egyptians regarded the area of the circle as equiva- lent to ff of the diameter squared, which would make The Jews and Babylonians treated n as equal to 3. Archimedes, by the use of inscribed and circumscribed regular polygons, showed that the true value of n lies between 3| and 3yf; that is, between 3.14285 and 3.1408. The Hindoo writers assign various values to 7t, as 3, 3i, 1/10, and Aryabhatta (530 A. D.) gives the correct ap- proximation, 3.1416. The Hindoos used the formula 1/2 _ 1/4 AB 2 (^ ee Art. 468) in computing the numeri- cal value of 71. Within recent times, the value of n has been computed to 707 decimal places. The use of the symbol n for the ratio of the circum- ference of a circle to the diameter was established in mathematics by Euler (Germany, 1750). HISTORY OF GEOMETRIC TRUTHS. SOLID GEOMETRY 854. Polyhedrons. The Egyptians computed the vol- umes of solid figures from the linear dimensions of such figures. Thus, Ahmes computes the contents of an Egyp- tian barn by methods which are equivalent to the use 3c of the formula V=aXbX . As the shape of these barns is not known, it is not possible to say whether this formula is correct or not. Pythagoras discovered, or knew, all the regular poly- hedrons except the dodecahedron. These polyhedrons were supposed to have various magical or mystical properties. Hence the study of them was made very prominent. FF 498 GEOMETEY. APPENDIX Hippasus (470 B.C.) discovered the dodecahedron, but lie was drowned by the other Pythagoreans for boasting of the discovery. Eudoxus (380 B. C.) showed that the volume of a pyra mid is equivalent to one -third the product of its base by its altitude. E. F. August (Germany, 1849) introduced the prisma- toid formula into geometry and showed its importance. 855. The Three Round Bodies. Eudoxus showed that the volume of a cone is equivalent to one -third the area of its base by its altitude. Archimedes discovered the formulas for the surface and volume of the sphere. Menelaus (100 A. D.) treated of the properties of spherical triangles. Gerard (Holland, 1620) invented polar triangles and found the formulas for the area of a spherical triangle and of a spherical polygon. 856. Non-Euclidean Geometry. The idea that a space might exist having different properties from those which we regard as belonging to the space in which we live, has occurred to different thinkers at different times, but Lobatchewsky (Russia, 1793-1856) was the first to make systematic use of this principle. He found that if, instead of taking Geom. Ax. 2 as true, we suppose that through a given point in a plane several straight lines may be drawn parallel to a given line, the result is not a series of absur- dities or a general reductio ad absurdum; but, on the con- trary, a consistent series of theorems is obtained giving the properties of a space. III. REVIEW EXERCISES EXERCISES. GROUP 86 REVIEW EXERCISES IN PLANE GEOMETRY Ex. 1. If the bisectors of two adjacent angles are perpendiculai to each other, the angles are supplementary. Ex. 2. If a diagonal of a quadrilateral bisects two of its angles, the diagonal bisects the quadrilateral. Ex. 3. Through a given point draw a secant at a given distance from the center of a given circle. Ex. 4. The bisector of one angle of a triangle and of an exterior angle at another vertex form an angle which is equal to one -half the third angle of the triangle. Ex. 5. The side of a square is 18 in. Find the circumference of the inscribed and circumscribed circles. Ex. 6. The quadrilateral ADBC is inscribed in a circle. The diag- onals AB and DC intersect in the point F. Arc AD = 112, arc A C = 108, LAFC= 74. Find all the other angles of the figure. Ex. 7. Find the locus of the center of a circle which touches two given equal circles. Ex. 8. Find the area of a triangle whose sides are 1 m., 17 dm., 210 cm. Ex. 9. The line joining the midpoints of two radii is perpendicular to the line bisecting their angle. Ex. 10. If a quadrilateral be inscribed in a circle and its diag- onals drawn, how many pairs of similar triangles are formed ? Ex. 11. Prove that the sum of the exterior angles of a polygon (Art. 172) equals four right angles, by the use of a figure formed by drawing lines from a point within a polygon to the vertices of the polygon. (499) 500 GEOMETRY. APPENDIX Ex. 12. In a circle whose radius is 12 cm., find the length of the tangent drawn from a point at a distance 240 mm. from the center. Ex. 13. If two sides of a regular pentagon be produced, find the angle of their intersection. Ex.|14. In the parallelogram ABCD, points are taken on the diagonals such that AP=BQ=CR=DS. Show that PQRS is a parallelogram. Ex. 15. A chord 6 in. long is at the distance 4 in. from the center of a circle. Find the distance from the center of a chord 8 in. long. Ex. Ii6. If B is a point in the circumference of a circle whose* center is \O, PA a tangent at any point P, meeting OB produced at A, and PD perpendicular to OB, then PB bisects the angle APD. Ex. 17. Construct a parallelogram, given a side, an angle, and a diagonal. Ex. 18. Find in inches the sides of an isosceles right triangle whose area is 1 sq. yd. Ex. 19. Given the line a, construct q vi/2 1)^ o Ex. 20. If two lines intersect so that the product of the segments of one line equals the product of the segments of the other, a cir- cumference may be passed through the extremities of the two lines. Ex. 21. Find the locus of the vertices of all triangles on a given base and having a given area. Ex. 22. On the figure p. 206, prove thatl^ 2 +Z^=ZB 2 +JW 2 . Ex. 23. The area of a rectangle is 108 and the base is three times th'i altitude. Find the dimensions. Ex. 24. If, on the sides AC and BC of the triangle ABC, the squares, AD and BF, are constructed, AF and DB are equal. Ex. 25. If the angle included between a tangent and a secant is half a right angle, and the tangent equals the radius, the secant passes through the center of the circle. REVIEW EXERCISES IN PLANE GEOMETRY 501 Ex. 26. The sum of the areas of two circles is 20 sq. yds., and the difference of their areas is 15 sq. yds. Find their radii. Ex. 27. Construct an isosceles trapezoid, given the bases and a leg. Ex. 28. Show that, if the alternate sides of a regular pentagon be produced to meet, the points of intersection formed are the vertices of another regular pentagon. Ex. 29. If a post 2 ft. 6 in. high casts a shadow 1 ft. 9 in. long, how tall is a tree which, at the same time, casts a shadow 66 ft. long ? Ex. 30. If two intersecting chords make equal angles with the diameter through their point of intersection, the chords are equal. Ex. 31. From a given point draw a secant to a circle so that the external segment is half the secant. Ex. 32. Find the locus of the center of a circle which touches a given circle at a given point. Ex. 33. If one diagonal of a quadrilateral bisects the other diagonal, the first diagonal divides the quadrilateral into two equi- valent triangles. Ex. 34. In a given square inscribe # square having a given side. Ex. 35. A field in the shape of an equilateral triangle contains one acre. How many feet does one side contain ? Ex. 36. If perpendiculars are drawn to a given line from the ver- tices of a parallelogram, the sum of the perpendiculars from two opposite vertices equals the sum of the other two perpendiculars. Ex. 37. Any two altitudes of a triangle are reciprocally propor- tional to the bases on which they stand. Ex. 38. Construct a triangle equivalent to a given triangle and having two given sides. Ex. 39. The apothem of a regular hexagon is 20. Find the area of the inscribed and circumscribed circles. Ex. 40. M is the midpoint of the hypotenuse AB of e, right tri- angle ABC. Prove 8 MC 2 =A 2 +BC 2 -^AC 2 , 502 GEOMETRY. APPENDIX Ex. 41. Transform a given triangle into an equivalent right tri- angle containing a given acute angle. Ex. 42. The area of a square inscribed in a semicircle is to the area of the square inscribed in the circle as 2 : 5. Ex. 43. If, on a diameter of the circle 0, OA = OB and AC is par- allel to BD, the chord CD is perpendicular to AC. Ex. 44. Find the radius of a circle whose area is equal to one- third the area of the circle whose radius is 7 in. Ex. 45. State and prove the converse of Prop. XXI, Book III. Ex. 46. If, in a given trapezoid, one base is three times the other base, the segments of each diagonal are as 1 : 3. Ex. 47. If two sides of a triangle are 6 and 12 and the angle included by them is 60, fiad the length of the other side. Also find this when the included angle is 45; also, when 120. Ex. 48. How many sides has a polygon in which the sum of the interior angles exceeds the sum of the exterior angles by 540? Ex. 49. If the four sides of a quadrilateral are the diameter of a circle, the two tangents at its extremities, and a tangent at any other point, the area of the quadrilateral equals one-half the product of the diameter by the side opposite it in the quadrilateral. Ex. 50. An equilateral triangle and a regular hexagon have the same perimeter; find the ratio of their areas. Ex. 51. To a circle whose radius is 30 cm. a tangent is drawn from a point 21 dm. from the center. Find the length of the tangent. Ex. 52. If two opposite sides of a quadrilateral are equal, and the angles which they make with a third side are equal, the quad- rilateral is a trapeaoid. Ex. 53. If two circles are tangent externally and two parallel diameters are drawn, one in each circle, a pair of opposite extremities of the two diameters and the point of contact are collinear. Ex. 54. If, in the triangle ABC, the line AD is perpendicular to BD, the bisector of the angle B, a line through D parallel to BC bisects AC. REVIEW EXERCISES IN PLANE GEOMETRY 503 Ex. 55. Bisect a given triangle by a line parallel to a given line. Ex. 56. If two parallelograms have an angle of one equal to the supplement of an angle of the other, their areas are to each other as the products of the sides including the angles. Ex. 57. The sum of the medians of a triangle is less than the perimeter, and greater than half the perimeter. Ex. 58. If PARE is a secant to a circle through the center O, PT a tangent, and TE perpendicular to PB, then PA : PR=PO : PB. Ex. 59. Two concentric circles have radii of 17 and 15. Find the length of the chord of the larger which is tangent to the smaller. Ex. 60. On the figure, p. 244, (a) Find two pairs of similar triangles; (1} Find two dotted lines which are perpendicular to each other; (c) Discover a theorem concerning points, not connected by lines on the figure, which are collinear; (d) Discover a theorem concerning squares on given lines. Ex. 61. One of the legs, AC, of an isosceles triangle is produced through the vertex, C, to the point F, and F is joined with D, the mid- point of the base AB. DF intersects BC in E, Prove that CF is greater than CE. Ex. 62. The line of centers of two circles intersects their common external tangent at P. PABCD is a secant intersecting one of the two circles at A and B and the other at C and D. Prove Ex. 63. Trisect a given parallelogram by lines drawn through a given vertex. Ex. 64. Find the area of a triangle the sides of which are the chord of an arc of 120 in a circle whose radius is 1 ; the chord of an arc of 90 in a circle whose radius is 2 ; and the chord of an arc of 60 in a circle whose radius is 3. Ex. 65. Construct a triangle, given the median to one side and a median and altitude on the other side. Ex. 66. Two circles intersect at P and Q. The chord CQ is tan- gent to the circle QPB at Q. APB is any chord through P. Prove that AC is parallel to BQ. 504 GEOMETRY. APPENDIX Ex. 67. In the triangle ABC, from D, the midpoint of BC, DE and DF are drawn, bisecting the angles ADB and ADC, and meeting AB at E and AC at ^. Prove EF \\ BC. Ex. 68. Produce the side BC of the triangle ABC to a point P, so that PBXPC=RA 2 . Ex. 69. In a given circle inscribe a rectangle similar to a given rectangle. Ex. 70. In a given semicircle inscribe a rectangle similar to a given rectangle. Ex. 71. The area of an isosceles trapezoid is 140 sq. ft., one base is 26 ft., and the legs make an angle of 45 with the other base. Find the other base. Ex. 72. Cut off one-third the area of a given triangle by a Hue perpendicular to one side. Ex. 73. Find the sides of a triangle whose area is 1 sq. ft., if the sides are in the ratio 2:3:4. Ex. 74. Divide a given line into two parts such that the sum of the squares of the two parts shall be a minimum. Ex. 75. If, from any point in the base of a triangle, lines are drawn parallel to the sides, find the locus of the center of the paral- lelogram so formed. Ex. 76. Three sides of "a quadrilateral are 845, 613, 810, and the fourth side is perpendicular to the sicies 845 and 810. Find the area. Ex. 77. If BP bisects the angle ABC, and DP bisects the angle CDA, prove that angle P=i sum of angles A and C. Ex. 78. Two circles intersect at P and Q. Through a point A in one circumference lines APC and AQD are drawn, meeting the other in C and D. Prove the tan- gent at A parallel to CD. Ex. 79. In a given triangle, draw a line parallel to the base and terminated by the sides so that it shall be a mean proportional be- tween thy segments of one side. KEYIEW EXERCISES IN PLANE GEOMETRY 505 Ex. 80. Find the angle inscribed in a semicircle the sum of whose Rides is a maximum. Ex. 81. The bases of a trapezoid are 160 and 120, and the alti- tude 140. Find the dimensions of two equivalent trapezoids into which the given trapezoid is divided by a line parallel to the base. Ex. 82. If the diameter of a given circle be divided into any two segments, and a semicircumference be described on the two segments on opposite sides of the diameter, the area of the circle will be di- vided by the semicircumferences thus drawn into two parts having the same ratio as the segments of the diameter. Ex. 83. On a given straight line, AB, two segments of circles are drawn, APB and AQB. The angles QAP and QBP are bisected by lines meeting in R. Prove that the angle JR is a constant, wherever P and Q may be on their arcs. Ex. 84. On the side AB of the triangle ABC, as diameter, a cir- cle is described. EF is a diameter parallel to EC. Show that EB bisects the angle ABC. Ex. 85. Construct a trapezoid, given the bases, one diagonal, and an angle included by the diagonals. Ex. 86. If, through any point in the common chord of two inter- secting circles, two chords be drawn, one in each circle, through the four extremities of the two chords a circumference may be passed. Ex. 87. From a given point as center describe a circle cutting a given straight line in two points, so that the product of the distances of the points from a given point in the line may equal the square of a given line segment. Ex. 88. AB is any chord in a given circle, P any point on the circumference, PM is perpendicular to AB and is produced to meet the circle at Q ; AN is drawn perpendicular to the tangent at P. Prove the triangles NAM and PAQ similar. Ex. 89. If two circles ABCD and EBCF intersect in B and C and have common exterior tangents AE and DF cut by EC produced at G and H, then GH 2 =C 2 -\-'A! 2 , 506 GEOMETRY. APPENDIX EXERCISES. CROUP 87 REVIEW EXERCISES IN SOLID GEOMETRY Ex. 1. A segment of a straight line oblique to a plane is greater than its projection on the plane. Ex. 2. Two tetrahedrons are similar if a dihedral angle of one equals a dihedral angle of the other, and the faces forming these dihedral angles are similar each to each. Ex. 3. A plane and a straight line, both of which are parallel to the same line, are parallel to each other. Ex. 4. If the diagonal of one face of a cube is 10 inches, find the volume of the cube. Ex. 5. Construct a spherical triangle on a given sphere, given the poles of the sides of the triangle. Ex. 6. Given AB _L MN, AE and BF _L ME; prove EF J. PM. Ex. 7. The diagonals of a rectangular paral- lelepiped are equal. Ex. 8. What portion of the surface of a sphere is a triangle each of whose angles is 140? Ex. 9. Through a given point pass a plane parallel to two given straight lines. Ex. 10. Show that the lateral area of a cylinder of revolution is equivalent to a circle whose radius is a mean proportional between the altitude of the cylinder and the diameter of its base. Ex. 11. The volumes of polyhedrons circumscribed about equal spheres are to each other as the surfaces of the polyhedrons. Ex. 12. Find /Sand T of a regular square pyramid an edge of whose base is 14 dm., and whose lateral edge is 250 cm. Ex. 13. If two lines are parallel and a plane be passed through each line, the intersection of these plane? is parallel to the given lines. REVIEW EXEECISES IN SOLID GEOMETRY 507 If Ex. 14. Given PH J_ plane AD, /.PEH= /.PFH; prove Z PEF= Z PFE. Ex. 15. If a plane be passed through the midpoints of three and the altitude by H; prove that i=li(6 1 6 2 ) 2 -f4 J H" 2 . Ex. 24. If the opposite sides of a spherical quadrilateral are equal the opposite angles are equal. 508 GEOMETRY. APPENDIX Ex. 25. Obtain the simplest formula for the lateral surface of a truncated triangular right prism, each edge of whose base is a, and whose lateral edges are p, q, and r. Ex. 26. The area of a zone of one base is a mean proportional between the remaining surface of the sphere and its entire surface. Find the altitude of the zone. Ex. 27. The lateral edges of two similar frusta are as 1 : a. How do their areas compare ? Their volumes ? Ex. 28. Construct a spherical surface with a given radius, r, which shall be tangent to a given plane, and to a given sphere, and also pass through a given point. Ex. 29. The volume of a right circular cylinder equals the area of the generating rectangle multiplied by the circumference generated by the point of intersection of its diagonals. Ex. 30. On a sphere whose radius is 8i inches, find the area of a zone generated by a pair of compasses whose points are 5 inches apart. Ex. 31. The perpendicular to a given plane from the point where the altitudes of a regular tetrahedron intersect equals one -fourth the sum of the perpendiculars from the vertices of the tetrahedron to the same plane. Ex. 32. Two trihedral angles are equal or symmetrical if their corresponding dihedral angles are equal. Ex. 33. On a sphere whose radius is a, a zone has equal bases and the sum of the bases equals the area of the zone. Find the alti- tude of the zone. Ex. 34. A plane which bisects two opposite edges of a tetrahedron bisects the volume of the tetrahedron. Ex. 35. Find the locus of all points in space which have their distances from two given parallel lines in a given ratio. Ex. 36. If a, b, c are the sides of a spherical triangle, a', b', cf the sides of its polar triangle, and a>&>c, then a'S=side of a regular polygon li altitude of a triangle. of n sides. 61 and & 2 =bases of a trapezoid. LENGTHS OF LINES 1. In a right triangle, C being the right angle, c 2 =a 2 + 6 2 . Art. 346. 2. In a right triangle, I and m being the projections of a and b on c, and h, the altitude on c, a 2 = ZXc,WmXc. Art. 342. 3. In an oblique triangle, tn being the projection of & on c, if a is opposite an acute Z , a' 2 = 6 2 -j- c 2 2 cX m > Art. 349. if a is opposite an obtuse L , a 2 = & 2 + c 2 + 2 c X w. Art. 350. 4. fc c =tl/s(s a) (s 6) (s c). Art. 393. 5. w c =4-|/2 (a 2 + 6 2 ) c 2 . Art. 353. 6. f c = f/a&s (s c). Art. 363. 7. If I and m are the segments of c made by the bisector of the angle opposite, a:b = l:m. Arts. 332, 336. (514} FORMULAS OF PLANE GEOMETKY 515 8. If I and m are the segments of a line, a, divided in extreme and mean ratio, and I > m, a :l=l :m. Art. 370. (P:P f =:'. Art. 341. 9. In similar polygons, ( ^ ; ^ ; &/ ^ ^ 10. In circles, C : C' = R : R' ; also C : C'=>D : D 1 . Art. 442. 11. C=2 irR, or C=irD. Art. 444. 12. An arc = eB * r ** le X*B. Art. 445. J-oU 13. In inscribed regular polygons, S 2n =V R ( 2 R 1/4 R> Sn 2 ) . Art. 467 . AREAS OF PLANE FIGURES 1. In a triangle, K=l Ih. Art. 389. 2. In a triangle, K=\/s (sa) (s b) (s c). Art. 393. 3. In an equilateral triangle, K= C ?^. Ex. 4, p. 257. 4. In a parallelogram, K=by^li. Art. 385. 5. . In a trapezoid, K=i h (61 + & 2 ). Art. 394. 6. In a regular polygon, K=$ rXP. Art. 446. 7. In a circle, K^I? or JT=i vD\ Art. 442. 8. In a sector of a circle, Z"=i i?X arc, Art. 453. 9. In a segment of a circle, jK^= sector A formed by the chord and radii of the segment. 10. In a circular ring, E=Tr (R 2 R n ). Art. 449. 11. In any two similar plane figures, K:K'=a* :a /2 ; Art. 399, also a : a' = /K : /! 7 . Art. 314. 12. In two circles, K : K'=R 2 : .R /2 = Z> 2 : Z> /2 =C 2 : C /2 ; Art. 452. also R : R'=D : D> = C : C'=/K : /tf'. Art. 314. FOEMULAS OF SOLID GEOMETRY SYMBOLS B, &=areas of the lower and up- P=perimeter of right section. per bases of a frustum. P, p = perimeters of lower and up- l=lateral edge (or element); per bases of a frustum. or r, r f = radii of bases. = spherical excess. = area of lateral surface; or H= altitude. =area of surface of sphere, I, b, 7& = length, breadth, height. etc. i=slant height. T=area of total surface. Art. 608. 2. In a regular pyramid, S=i LX?. Art. 641. 3. In a frustum of a regular pyramid, S=i(P-\-p) L. Art. 643. 4. In a cylinder of revolution, S=2 ^BH. Art. 697. T=27rB(R + H). Art. 697. 5. In a cone of revolution, S=TRL. Art. 721. T=TTK (L + E). Art. 721. 6. In a frustum of a cone of revolution, S=^L (B-\- r). Art. 727. 7. In a sphere, S=4 irg 2 , O r S=irIP. Art. 810. 8. In a zone, =2 ^EH. Art. 813. 7J-J?2 A 9. In a lune, 8=-~~ . Art. 817. 10. In a spherical triangle, S^-. Art. 822. loO 11. In a spherical polygon, s=-~ Art. 824. 180 (516) VJi-' ' * 2 FOKMULAS OF SOLID GEOMETRY 517 FORMULAS FOR VOLUMES 1. In a prism, V=B^H. Art. 628. 2. In a parallelepiped, F=lX^Xh. Art. 626. 3. In a pyramid, F=i BX H. Art. 651. 4. In a frustum of a pyramid, F=i H (B + 1 -\-\/ Bb). Art. 656. 5. In a prismatoid, V=\ H (B + 6 +4 M). Art. 663. 6. In a cylinder, V=BXH- Art. 698. 7. In a cylinder of revolution, V=irR' i H. Art. 699. 8. In a cone, F=i BXH. Art. 722. 9. In a circular cone, F=t TJ2 2 ^. Art. 723. 10. In a frustum of a cone, V=% H (B-\-l -\-\/ Bb). Art. 729. 11. In a frustum of a cone of revolution, V=\TTR( U 2 + r 2 + Rr) . Art . 730 . 12. In a sphere, F= ^.B 3 , or F=|7rD s . Art. 832. 13. In a spherical sector, F=t ^-B 2 ^. Art. 836. 14. In a spherical segment of two bases, F=i (Trr 2 + 7T/ 2 ) Jff + i 7T# 3 . Art. 837. 15. In a spherical segment of one base, F=T J H 2 (R ^H). Art. 838. CONSTANTS 1 aere = 43,560 sq. ft. ^2 = 1.2599 + 1 bushel = 2150.42 cu. in. ^3 = 1.4422 + 1 gallon = 231 cu. in, ^=- 3183 + l /2 = 1.4142+ v/7r=1.7725- !/3 = 1.7321 - ^=0.5642 + GG 513 GEOMETRY. APPENDIX SUMMARY OF METRIC SYSTEM TABLE FOR LEHGTH 10 millimeters (mm.) =1 centimeter (cm.) 10 cm. =1 decimeter (dm.) 10 dm. =1 meter (m.). 10 m. =1 Dekameter (Dm.) 10 Dm. =1 Hektometer (Hm.) 10 Hm. =1 Kilometer (Km.) 10 Km. = 1 Myriameter (Mm.) Similar tables are used for the unit of weight, the gram ; for the unit of capacity, the liter ; for the unit of land measure, the are; and for the unit of wood measure, the stere. TABLE FOR SQUARE MEASURE 100 sq. mm. = l sq. cm. 100 sq. cm. =1 sq. dm., etc. TABLE FOR CUBIC MEASURE 1000 cu. mm. = l cu. cm. 1000 cu. cm. =1 cu. dm., etc. A liter =1 cu. dm. A gram weight of 1 cu. cm. of water at 39.2 Fahrenheit. An are =100 sq. m. A stere =1 cu. m. EQUIVALENTS 1 meter =39.37 inches. 1 liter =1.057 liquid quarts, or .9581 dry quarts. 1 kilogram = 2. 2046 Ibs. av. 1 hektare =2.471 acres. 1 sq. m. 1550 sq. in. INDEX OF DEFINITIONS AND FORMULAS PAGE Altitude of cone . . . .412 of cylinder 403 of frustum of cone . .414 of frustum of pyramid . 378 of prism 3d of pyramid .... 377 of spherical segment . . 402 of zone 452 Angle, dihedral . . . .337 formed by two curves . 433 of lune ...... 452 polyhedral 349 spherical 433 tetrahedral 349 trihedral 349 Angles of spherical polygon 438 Appolonius 493 Arabs 490 Archimedes . . . 497,498 Aryabhatta 497 August, E. F 498 Axis of circle of sphere . 427 of circular cone . . .413 of regular pyramid . . 378 of sphere 427 Babylonians 497 Base of cone . 412 of pyramid 377 of spherical pyramid . .461 of spherical sector . . .461 PAGE Bases of cylinder . . .403 of frustum of cone . .414 of frustum of pyramid . 378 of prism 361 of spherical segment . . 462 of zone 452 Bodies, The Three Round . 492 Center of sphere .... 425 Circle 496 great 427 small .427 Cone 412 altitude of 412 axis of 413 base of 412 circular 413 circular, formula for vol- ume of 418 circular, formulas for lat- eral, a total area of .417 elements of 412 lateral surface of . . .412 oblique circular . . .413 of revolution . . . .413 right circular . . . .413 vertex of ..... 412 Cones, similar . . . .413 Conical surface . . . .412 directrix of . . . .412 element of .... 412 519 520 INDEX OF DEFINITIONS AND FORMULAS PAGE Conical surface, generatrix of 412 nappes of 412 vertex of 412 Constants 513 Continuity, principle of . 485 Cube 363 Curved spaces .... 488 Cylinder 403 altitude of .... 403 bases of 403 circular 404 circular, properties of . 405 elements of 403 lateral surface of ... 403 oblique 404 of revolution .... 404 of revolution, formulas for lateral and total areas of 409 right 404 right circular .... 404 right section of ... 405 section of 405 Cylinders of revolution, similar 404 Cylindrical surface . . . 403 directrix of 403 element of 403 generatrix of .... 403 Degree, spherical .... 452 Determined plane , 319,320- Diagonal of polyhedron . 360 Diameter of sphere . . . 425 Dihedral angle . . . .337 edge of 337 faces of 337 plane angle of . . . .338 right 337 PAGE Dihedral angles, adjacent . 337 equal 337 vertical 337 Directrix of conical surface 412 of cylindrical surface . 403 Distance from point to plane 328 on surface of sphere . . 428 polar, of great circle . . 429 polar, of small circle . . 429 Dodecahedron . . . .360 Duality, principle of . . 486 Edge of dihedral angle . . 337 Edges of polyhedral angle . 349 of polyhedron .... 360 Egyptians . 490, 491, 492,494, 497 Elements of conical surface 412 of cylindrical surface . 403 Equivalent solids . . . 363 Euclid 491 Eudoxus . . . 493,496,498 Eulor 497 European 490 Faces of dihedral angle . . 337 of polyhedral angle . . 349 of polyhedron .... 360 Figures, rectilinear . . . 494 Formulas of plane geometry for lengths of lines 514, 515 for plane geometry, for areas of plane figures . 515 of solid geometry, for areas 516 of solid geometry, for vol- umes 517 Foot of line 320 Frustum of cone . . .414 altitude of 414 bases of . . 414 INDEX OF DEFINITIONS AND FORMULAS 521 PAGE Frustum of cone, formula for lateral area of ... 420 formula for volume of . 422 lateral surface of . . .414 slant height of . . .414 Frustum of pyramid . .378 altitude of ..... 378 bases of ...... 378 slant height of . . .378 Generatrix of conical surface 412 of cylindrical surface . 403 Geometry, epochs in develop- ment of . . . . .490 history of ... 490-498 modern ...... 485 Non-Euclidean . . 488,496 origin of ..... 490 projecti ve ..... 485 solid ...... 319 Gerard Greeks . 498 490,492,493,494 Hero ....... 496 Hexahedron ..... 360 Hindoos . . . 490,492,497 Hippasius ...... 498 Hippocrates . . 491,492,496 History of geometry . 490-498 Homology, principle of . 487 Hyperspace ..... 488 Icosahedron ..... 360 Inclination of line to plane 347 Jews 497 Lateral area of frustum of pyramid, formula for . 379 Lateral area of frustum prism of pyramid .... Lateral edges of -prism . of pyramid .... Lateral faces of prism . of pyramid .... Lateral surface of cone of cylinder .... of frustum of cone Line parallel to plane perpendicular to plane Lobatchewsky Logical methods . Lune angle of formula for area ofj spherical degrees formula for area of, square units of area PAGE of . 361 . 377 . 361 . 377 . 361 . 377 . 412 . 403 . 414 . 321 . 320 . 498 . 492 . 452 . 452 in . 457 in . 457 Mechanical methods . . . 493 Menelaus 498 Methods, logical . . . .492 mechanical 493 rhetorical . . . . .491 Metric system, summary of 514 Modern geometry . . .485 Nappes of cone . . . .412 Negative quantities . 485, 486 Non-Euclidean geometry 488, 498 Octahedron Origin of geometry . . . 360 . . . 490 Parallelepiped right . . . 362 362 rectangular . . . 363 522 INDEX OF DEFINITIONS AND FORMULAS PAGE Parallel planes . . . .321 Perpendicular planes . . 338 Plane 319 determination of . 319,320 Planes, parallel . . . .321 Plato 491,493 Polar distance of circle . 429 triangle 441 Pole 427 Polygon, spherical . . . 438 Polyhedral angle . . .349 convex 349 edges of 349 face angles of .... 349 faces of 349 vertex of 349 Polyhedral angles, equal 350 symmetrical .... 350 vertical . . . . . .350 Polyhedron 360 convex 360 diagonal of 360 edges of 360 faces of 360 regular 391 section )f 360 vertices of 360 Polyhedrons 497 classification of ... 360 similar 396 Postulate of solid geometry 320 Prism 361 altitude of 361 bases of 361 circumscribed about cyl- inder 405 inscribed in cylinder . . 405 lateral area of . . . .361 lateral edges of . . .361 .lateral faces of ... 361 PAGE Prism, oblique 362 quadrangular .... 362 regular 362 right 362 right section of . . .361 triangular 362 truncated 362 Prismatoid 389 bases of 389 formula for volume of . 390 Prismoid 389 Projection of line on plane 338 of point on plane . . . 338 Pyramid 377 altitude of . . . . . . 377 axis of 378 base of 377 circumscribed about cone 414 frustum of .... 378 inscribed in cone . . .414 lateral area of . . . . 377 lateral edges of . . .377 lateral faces of ... 377 quadrangular .... 377 regular 377 regular, slant height of . 378 spherical 461 triangular . . . . . 377 truncated 378 vertex of 377 Pythagoras . 491, 492, 495, 497 Radius of sphere . . . 425 Reciprocity, principle of . 486 Rectilinear figures . . . 494 Rhetorical methods . . .491 Right section of cylinder . 405 of prism 361 Romans .... 494,496 Round Bodies, The Three 498 INDEX OF DEFINITIONS AND FORMULAS 523 PAGE Section of Polyhedron . . 360 of sphere 461 Segment of sphere . . . 462 Similar cones of revolution 413 cylinders of revolution . 404 Sectors of circles . . .275 Segments of circles . . . 275 Slant height of cone . .413 of frustum of cone . .414 of frustum of pyramid . 378 of regular pyramid . .377 Solid geometry . . ' . .319 Solids, equivalent . . . 363 Spaces, curved .... 488 Sphere 425 axis of 427 center of 425 circumscribed about poly- hedron 432 diameter of .... 425 formula for area of sur- face of 455 formula for volume of, . 463 great circle of .... 427 inscribed in polyhedron . 432 poles of 427 radius of 426 small circle of ... 427 Spherical angle .... 433 degrees 452 excess .... 444, 460 Spherical polygon . . . 438 angles of 438 sides of 438 vertices of 438 Spherical pyramid . . . 461 base of 461 vertex of 461 Spherical sector . . . .461 base of 461 PAGE Spherical sector, formula for volume of .... 464 Spherical segment . . . 462 altitude of 462 bases of 462 formulas for volumes of . 465 of one base .... 462 Spherical triangle . . . 438 bi-rectangular .... 443 formula for area of, in square units of area . 460 tri-rectangular .... 443 Spherical triangles, supple- mental 442 symmetrical .... 444 Spherical wedge . . . .461 Spherid 452 Surface, conical . . . .413 cylindrical 403 Tangent, line to sphere plane to cone . plane to cylinder . plane to sphere spheres .... Tetrahedral angle Tetrahedron . Thales . . . Triangle, polar spherical Trihedral angle bi-rectangular isosceles rectangular tri-rectangular 491, . 425 . 413 . 404 . 426 . 426 . 349 . 360 495, 496 . 441 , . 438 . 349 . 350 . 350 . 350 . 350 Ungula 461 Unit of volume .... 363 Units of spherical surface . 452 524 INDEX OF DEFINITIONS AND FORMULAS. Vertex of cone of polyhedral angle . of pyramid PAGE . 412 . 349 . 377 Wedge, spherical Xenodorus PAGE . . . 461 . . . 496 of spherical pyramid . Vertices of polyhedron . 461 . 360 . . . 452 of spherical polygon . 438 altitude of . 452 bases of 452 Volume of solid . . 363 of one base . . . . 452 RETURN TO: CIRCULATION DEPARTMENT 198 Main Stacks LOAN PERIOD 1 Home Use 2 3 4 5 6 ALL BOOKS MAY BE RECALLED AFTER 7 DAYS. Renewals and Recharges may be made 4 days prior to the due date. Books may be renewed by calling 642-3405. DUE AS STAMPED BELOW. JAN l ft ?UQ i FORM NO. 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