THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA SANTA BARBARA COLLEGE PRESENTED BY Mrs. S. 0. Werner ELEMENTS OF SOLID GEOMETRY BY W. H. BRUCE, A.M., PH.D. PRESIDENT NORTH TEXAS STATE NORMAL COLLEGE ABTD C. C. CODY, A.M., PH.D. PROFESSOR OF MATHEMATICS, SOUTHWESTERN UNIVERSITY DALLAS, TEXAS THE SOUTHEKN PUBLISHING COMPANY 1914 COPYRIGHT, 1912, BY THE SOUTHERN PUBLISHING COMPANY. UNIVERSITY OP SANTA BARBARA COLLEGE LIBRARY CONTENTS BOOK VI PAGE Definitions 1 Lines and Planes in Space 10 Dihedral and Polyhedral Angles 16, 26 BOOK VII Definitions ........... 31 Polyhedrons 31 Prisms 32 Pyramids .46 Regular Polyhedrons 63 Cylinders 65 Cones 69 BOOK VIII THE SPHERE Definitions 75 Spherical Angles 82 Spherical Polygons 83 Measurement of Spherical Surfaces 95 Spherical Volumes 102 iii INDEX OF DEFINITIONS SECTION Angle of lune 785 Angle of spherical polygon . 763 Axis of cone 701 Axis of cylinder 682 Axis of pyramid 639 Bi-rectangular spherical tri- angle 775 Cone 698-702 Conical surface 696 Convex polyhedron .... 599 Convex spherical polygon . . 755 Cube 611 Cylinder ...... 678-682 Cylindrical surface .... 677 Diagonal of spherical polygon 754 Diameter of sphere .... 724 Dihedral angle 558 Directrix 677, 696 Distance 632, 738 Dodecahedron 600 Edge of dihedral angle . . . 559 Edge of polyhedral angle . . 686 Edge of polyhedron .... 597 Element 677, 696 Equal solids 613 Face 559, 586, 697 Face angle 586 Frustum of cone 711 Frustum of pyramid . . . 643 Generatrix 677, 696 Great circle of sphere . . . 733 Hexahedron 600 SECTION Icosahedron 600 Inclination 582 Lateral area . . . . 601, 638 Lateral edge .... 601, 638 Lateral face .... 601,638 Lateral surface . . . 678, 698 Lune 784 Nappe 697 Octahedron 600 Parallelepiped 608 Plane 522 Plane angle of dihedral angle 561, 664 Polar distance 740 Polar triangle 768 Poles of circle 735 Polyhedral angle 686 Polyhedron 597-599 Prism 601-607 Projecting plane 680 Projection 675 Pyramid 638-643 Radius of cylinder .... 682 Radius of sphere 723 Regular polyhedron .... 673 Right section of cylinder . . 687 Right section of prism . . . 606 Section 598 Side of spherical polygon . . 753 Similar cones 704 Similar cylinders 683 Similar polyhedrons .... 666 iv INDEX OF DEFINITIONS SECTION Slant height . . . 642, 644, 703 Small circle of sphere . . . 734 Sphere 722 Spherical angle 750 Spherical excess of polygon . 800 Spherical excess of triangle . 777 Spherical polygon . . . . 753 Spherical pyramid ' . . . . 802 Spherical sector 804 Spherical segment .... 806 Symmetrical polyhedral angles 591 SECTION Symmetrical spherical polygons 758 Tangent . . . 684, 705, 727, 728 Tetrahedron 600 Trihedral angle 589 Tri-rectangular spherical tri- angle 776 Vertex . . . 586, 597, 638, 763 Vertical polyhedral angles . 592 Vertical spherical polygons . 757 Volume 612 Zone . 786 SYMBOLS AND ABBREVIATIONS. ax. ... axiom. O . . . circle. cor. . . . corollary. + . . . plus. def. . . . definition. ... minus. iden. . . . identity. x . . . multiplied by. prop. . . . proposition. -*-,/,:. . . divided by. post. . . . postulate. = . . .is equal to or equivalent to. cons. . . . construction. ~ ... is similar to. hyp. . . . hypothesis. s . . .is congruent to. rect. . . . rectangle. > ... is greater than. rt. ... right. < ... is less than. st. ... straight. JL . . . is perpendicular to, or a ... angle. perpendicular. A ... triangle. II ... is parallel to, or a parallel. O ... parallelogram. Q. E. D. (quod erat demonstrandum), which was to be proved. Q. E. F. (quod erat faciendum), which was to be done. NOTE. The foregoing are used also in the plural, as = means " are equal to," as well as "is equal to." REFERENCES TO PLANE GEOMETRY 63. The sum of all the angles about a point is equal to two straight angles. 65. If two straight lines intersect, the vertical angles are equal. 85. In congruent figures homologous parts are equal. 86. Any side of a triangle is less than the sum of the other two, and greater than their difference. 91. Two triangles are congruent if they have two sides and the included angle of the one equal, respectively, to two sides and the included angle of the other. 92. Two right triangles are congruent if the legs of the one are equal, respectively, to the legs of the other. 94. Two triangles are congruent if they have two angles and the included side of the one equal, respectively, to two angles and the included side of the other. 95. Two right triangles are congruent if a leg and an adja- cent acute angle of the one are equal, respectively, to a leg and an adjacent acute angle of the other. 108. The perpendicular bisector of a line is the locus of points equidistant from the extremities of the line. 109. Two points each equidistant from the extremities of a line determine the perpendicular bisector of the line. 115. Only one perpendicular can be drawn from a given ex- ternal point to a given straight line. 116. The perpendicular is the shortest line that can be drawn from a given point to a given line. 117. Two oblique lines from the same point in the perpen- dicular to a given line, cutting off equal segments from the vii viii BOOK VI. SOLID GEOMETRY. foot of the perpendicular, are equal ; and of two lines cutting off unequal segments from the foot of the perpendicular, the one cutting off the greater segment is the greater line. 121. Two right triangles are congruent if they have the hy- potenuse and an acute angle of the one equal, respectively, to the hypotenuse and an acute angle of the other. 122. Two right triangles are congruent if the hypotenuse and a leg of the one are equal, respectively, to the hypotenuse and a leg of the other. 123. Two triangles are congruent if they have the three sides of the one equal, respectively, to the three sides of the other. 132. If two triangles have two sides of the one equal, re- spectively, to two sides of the other, but the third side of the first greater than the third side of the second, the angle oppo- site the third side of the first is greater than the angle opposite the third side of the second. 137. Two straight lines in the same plane perpendicular to the same straight line are parallel. 139. If a straight line is perpendicular to one of two par- allels, it is perpendicular to the other. 151. Two angles having their right sides respectively par- allel and also their left sides parallel are equal, whereas if the right side of each is parallel to the left side of the other they are supplementary. 155. The sum of three angles of a triangle is equal to a straight angle. 166. A diagonal divides a parallelogram into congruent tri- angles. 167. The opposite sides and the opposite angles of a paral- lelogram are equal. 169. Parallels included between parallels are equal. REFERENCES TO PLANE GEOMETRY. ix 173. If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram. 176. Two parallelograms are congruent if two sides and the included angle of one are equal, respectively, to two sides and the included angle of the other. 213. Radii of the same circle or of equal circles are equal. 233. In the same circle, or in equal circles, equal arcs are sub- tended by equal chords and intercepted by equal central angles. 234. In the same circle, or in equal circles, equal chords subtend equal central angles and equal arcs. 240. Ten propositions, in all, may be obtained by selecting any two of the following conditions for the hypothesis and any one of the remaining three for the conclusion; 1. passes through the center of the circle. 2. bisects the chord. A straight line that 3. is perpendicular to the chord. 4. bisects the minor arc. 5. bisects the major arc. 246. Through three points not in the same straight line one circle, and only one, can be drawn. 275. If two variables are constantly equal and each ap- proaches a limit, their limits are equal. 277. The limit of the product of a constant and a variable is the product of the constant by the limit of the variable. 283. In the same circle, or in equal circles, central angles have the same ratio as their intercepted arcs. 287. A central angle is measured by its intercepted arc. 330. In any proportion the terms are in proportion by alter- nation. 331. In any proportion the terms are in proportion by in- version. ' X BOOK VI. SOLID GEOMETRY. 338. In any proportion like powers of the terms are in pro- portion. 340. A line parallel to one side of a triangle divides the other sides proportionally. 343. If a straight line parallel to the side BC of. a triangle ABC cuts AB at D and AC at E, then DE : BC = AD : AB. 363. Similar polygons are those whose homologous angles are equal and whose homologous sides are proportional. 369. Two triangles are similar if their sides are respectively proportional. 370. Two triangles are similar if the sides of the one are, respectively, parallel or perpendicular to the sides of the other. 374. The homologous altitudes of two similar triangles have the same ratio as any two homologous sides. 376. Two similar polygons may be divided into the same number of similar triangles, similarly placed. 391. The square of the hypotenuse of a right triangle is equal to the sum of the squares of the two legs. 392. The square of either leg of a right triangle is equal to the difference of the square of the hypotenuse and the square of the other leg. 420. The area of a parallelogram is equal to the product of its base and altitude. 421. Parallelograms having equal bases and equal altitudes are equivalent. 425. The area of a triangle is equal to half the product of its base and altitude. 428. Triangles of equal altitudes are to each other as their bases. REFERENCES TO PLANE GEOMETRY. xi 430. Triangles of equal bases and altitudes are equivalent. 4.34. The areas of two triangles, having an angle in the one equal to an angle in the other, are to each other as the products of the sides including the equal angles. 436. The areas of two similar polygons are to each other as the squares of any two homologous sides. 486. If the number of sides of an inscribed polygon be in- definitely increased, the apothem of the polygon will approach the radius of the circle as a limit. 488. The circle is the limit of the perimeters of regular in- scribed and circumscribed polygons and the area of the circle is the limit of the areas of these polygons when the number of their sides is indefinitely increased. SOLID GEOMETRY. BOOK VI. Solid geometry, or geometry of three dimensions, treats of figures whose elements are not all in the same plane. ( 35.) 522. A plane is a surface such that the straight line joining any two of its points, lies wholly in the surface. ( 16.) While regarded as indefinite in extent, it is usually represented in diagrams, by parallelograms that lie in the plane. A plane is said to be determined by points or lines, when these points or lines fix its position in space. 523. A plane is determined by any three given points which are not in the same straight line. Two points determine a line ( 9), but do not deter- mine a plane, because a ^^ ' p plane may be rotated about / | any given line, assuming, in turn, an indefinite number of / f / ' y positions, but a third point, witho plane. without this line, fixes the / P ~7 / 4 - */ 2 BOOK VI. SOLID GEOMETRY. 524. Because two points determine a line and three points a plane, so a plane is determined by the equivalent of three points ; namely, a straight line and a point without this line; two intersecting straight lines; two parallel straight lines. 525. The point in which a straight line meets a plane is called the foot of the line. 526. A straight line is perpendicular to a plane when it is perpendicular to every line in the plane drawn through its foot. 527. A straight line is parallel to a plane, or a plane is parallel to a straight line, when the two will not meet if pro- duced indefinitely. 528. An oblique line is one which is neither perpendicular nor parallel to a plane. 529. Two planes are parallel if they will not meet if pro- duced indefinitely. 530. If two planes are not paral- lel, they must intersect in a line common to the two planes. For two planes cannot have, in common, a straight line and any point with- out that line. If they could, the two planes would coincide. ( 523.) That is, the common points, or intersection, cannot be other than a straight line. BOOK VI. SOLID GEOMETRY. PROPOSITION I. THEOREM. 531. TJie perpendicular is the shortest line that can be drawn from a point to a plane. B Given the point P, the line PA _L the plane MN, and any other line PB from P to MN. To prove PA < PB. Proof. Through A, the foot of PA, draw the line AB. Then PA is J_ AB. 526 .-. PA< PB. 116 (The _L is the shortest line that can be drawn from a given point to a given line.') Q. E. D. 532. The distance from a point to a plane is the length of the perpendicular from the point to the plane. BOOK VI. SOLID GEOMETRY. PROPOSITION II. THEOREM. 533. Oblique lines draivn from a point to a plane, meeting the plane at equal distances from the foot of the perpendicular, are equal ; and of two oblique lines meeting the plane at unequal distances from the foot of the perpendicular, the more remote is the greater. Given PO JL plane MN, and AO = BO, and OD > OB. To prove I. PA = PB. II. PD > PB. Proof. I. The rt. As POA and POB are congruent, having Hyp. 92 PO common and OA = OB. .:PA = PB. II. On OD take OC = OB and draw PC. Then PD > PC. But PC=PB. .-. PD > PB. 85 117 Proof I Q. E. D. 534. COR. Conversely : Equal oblique lines from a point to a plane meet the plane at equal distances from the foot of the perpendicular; and of two unequal lines the greater meets the plane at the greater distance from the foot of the perpendicular. BOOK VI. SOLID GEOMETEY. 5 PROPOSITION III. THEOREM. 535. A straight line perpendicular to each of two straw^ lines at their point of intersection is perpen- dicular to the plane of those lines. M Given AB BC and BD at B. To prove AB JL MN, the plane of these lines. Proof. Draw BE, any other line, in the plane MN. Draw DC meeting BE at E, and prolong AB to A' so that BA' = AB. Join A and A' with D, E, and C. BD and BC are each _L AA' at its mid-point. Hyp. and Cons. Hence in As ADC and A 1 DC, AD = DA', 108 also, AC = A'C, 108 and DC = DC. Jden. .-. A ADC ^ A A'DC. 123 .-. ADE = % A'DE. 85 Hence A ADE ^ A A'DE. 91 .-. AE = .EM', and # _L AA' at B. 109, 85 .. AB _L any line in MN passing through its foot. /. it is J_ the plane MN. 526 Q. E. D. BOOK VI. SOLID GEOMETRY. PROPOSITION IV. THEOREM. 536. All the perpendiculars to a straight line at the same point lie in a plane perpendicular to that line (Converse of 535). Given PB ; PC, PD, each _L line AP at P. To prove PB, PC, PD,lie in a common plane _L AP. Proof. Let MN be the plane of PB and PC. 524 Then AP plane MN. 535 Let the plane of AP and PD intersect the plane MN in the line PD'. 530 Then AP JL PD 1 . 526 But AP _L PD. Hyp. Since in the plane APD but one _L can be drawn to AP at P, 115 PD' must coincide with PD. .'. PB. PC, and PD lie in a common plane _L AP. Q.E.D. 537. COR. TJirough a given point but one plane can be passed perpendicular to a given line, and the plane whicJi is the perpendicular bisector of a straight line is the locus of points equi- distant from the extremities of the line. BOOK VI. SOLID GEOMETRY. PROPOSITION V. PROBLEM. 538. Through a given point, to draw a line perpen- dicular to a given plane. I. When the given point is within the plane. II. When the given point is without the plane. I. Given point P within the plane MN. Required through P to draw a _L MN. Construction. From P in plane MN draw any line PB. In same plane draw PD _L PB. 113 Through P pass a plane intersecting MN in the line PB, and in this plane draw PL _L PB. In US, the plane of PD and PL, draw AP _L PD. 113 Then PA _L MN. Proof. Because PB is _L PD and PL, it is _L their plane. 535 .-.PBA.AP. 526 And because AP is _L PB and PD, it is J_ MN, their plane. 535 Ex. 556. Find the locus of points in space equidistant from all points of a circle. 8 BOOK VI. SOLID GEOMETRY. II. Given point P without the plane MN. Eequired through P to draw a A. to MN. Construction. Pass through P a plane intersecting MN in some line BC, and in this plane draw BP _L EC. 114 In the plane MN draw BD _L BC. 113 And in the plane of BD, BP, draw PA _L BD. 114 Then is PA _L MN. Proof. The As PAB, ABC, and PBC are rt. As. Cons. .-. PA 2 = PB*- AB 2 . 392 C 2 . 391 C\ 391 Adding, PA 2 + AC 2 = PC\ .-. APACis art. A. 391 .-. PA AC. .-.PAMN. Q.E.F. 539. COR. Through a given point but one perpendicular to a plane can be drawn. Ex. 557. Find the locus of a point in space equidistant from three given points not in the same straight line. Ex. 558. Find a point equidistant from four given points not all iu the same plane. BOOK VI. SOLID GEOMETRY. 9 PROPOSITION VI. THEOREM. 540. If from the foot of a perpendicular to a plane a line be drawn at right angles to any line of the plane, and the point of intersection be joined with any point of the perpendicular, the last line will be perpendic- ular to the line of the plane. Given AF _!_ plane MN, FP _L any line CB in plane MN, and PA drawn from P to any point of AF. To prove Proof. Then Hence APA.BC. From P on CB take PC = PB. Join A and F with C and B. FC=FB. AP _L BC. 108 533 109 Q. E. D. 10 BOOK VI. SOLID GEOMETRY. V / PROPOSITION VII. THEOREM. 541. Two perpendiculars to the same plane are parallel. c 'F 113 540 Given AB and CD J_ the plane MN. To prove AB II CD. Proof. Join D to A and B. Draw EF JL BD. .'.ADA.EF. Because BD, AD, and CD are each _L EF, they lie in same plane. 536 AB also lies in this plane. Because the three points A, B, and D determine a plane. 523 .'. AB and CD lying in the same plane and JL the same line, BD, are I!. 137 Q. E. D. . A.\ 542. COR. 1. Tjf one of two parallels is perpendicular to a plane, the other is perpendicular to the same plane. If CD is not MN, draw C^ _L MN. ThenCE\\AB. 541. .'. CE and CD must coincide. /. CD _L JOT. BOOK VI. SOLID GEOMETRY. 11 543. COR. 2. If two lines are II a third line they are II each other. If MN is drawn _L AB, it must be _L CD and also _L EF. .'. the lines are II. PROPOSITION VIII. THEOREM. 544. A straight line parallel to a line in a plane is parallel to the plane. Given AB I! CD in plane MN. To prove AB II plane MN. Proof. AB and CD being II lie in the same plane AD. 524 .'. if AB meets MN, it must meet it in line CD. 530 But AB and CD are II and cannot meet. .'. AB II plane MN. Q.E.D. 545. COR. 1. If a line is parallel to a plane, the intersection of the plane with any plane through the line is parallel to the line. 546. COR. 2. If two intersecting lines are each parallel to a given plane, the plane of these lines is parallel to the given plane. 12 BOOK VI. SOLID GEOMETRY. TWO PLANES. PROPOSITION IX. THEOREM. 547. 7V?o planes perpendicular to same straight line are parallel. Given planes MN and PQ _L AB. To prove the plane MN II the plane PQ. Proof. If MN and PQ are not II, they will meet if sufficiently produced. Suppose them to meet. We would then have two planes through the same point _!_ to the same line, which is impos- sible. 537 Therefore, plane MN II plane PQ. Q. E. D. 548. COR. If a straight line and a plane are perpendicular to the same straight line, they are parallel. \ BOOK VI. SOLID GEOMETRY. PROPOSITION X. THEOREM. 13 549. The intersections of two parallel planes by a third plane are parallel lines. Given two planes, MN and PQ, intersected by the plane RS in AB and CD. To prove AB II CD. Proof. AB and CD lie in the same plane RS. Because they also lie in the planes MN and PQ they can- not meet. 529 .-. AB II CD. Q. E. D. 550. COR. 1. Parallel lines included between parallel planes are equal. 551. COR. 2. Parallel planes are everywhere equally distant. 14 BOOK VI. SOLID GEOMETRY. PROPOSITION XI. THEOREM. A 552. A straight line perpendicular to one of two parallel planes is perpendicular to the other, also. Q Given MN and PQ, II planes, and AB JL plane PQ. To prove AB JL plane MN. Proof. Draw BE and BF, any two lines in PQ intersecting at B. Suppose planes passed through AB-BE and AB-BF, inter- secting MN in AC and AD, respectively. Then AC II BE and AD II BF. 549 But AB J_ BE and to BF. 526 .-. AB _L AC and to AD. 139 .-. AB plane MN. 535 Q. E.D. 553. COR. 1. Reciprocally, a plane perpendicular to one of two parallel lines is perpendicular to the other, also. 554. COR. 2. Through a given point one plane, and only one, can be drawn parallel to a given plane. BOOK VI. SOLID GEOMETRY. PROPOSITION XII. THEOREM. 15 555. Two angles not in the same plane) having their sides respectively parallel, right side to right side and left side to left side, are equal and their planes are parallel. Given ^s o and o' lying in planes MN and PQ, respectively, and having sides AB II A'B', AC II A'C'. To prove %o = %o' and MN II PQ. Proof. Take A'B' = AB and A'O = AC. (7(7'. Because AB and AC are respectively II and AA'B'B and AA'C'C are Os. .-. A A 1 = and II BB' and A A' = and II CO. Draw AA' } BB', = A'B' and A'C', 173 Hence But Also BB' = and II CC'. CS'isaO, andJ5(7=5 .-. A ABC ^ A A'B'C'. ' o = o. o = o. .-. o = o". PQ is II AB and AC. .\MN\\PQ. 167 543 167 123 85 65 545 546 Q. . D. 16 BOOK VI. SOLID GEOMETRY. 556. COB. Two angles not in the same plane, having their sides respectively parallel, right side to left side and left side to right side, are supplementary and their planes are parallel. PROPOSITION XIII. THEOREM. 557. If two straight lines are intersected by parallel planes, the corresponding segments are proportional. . N Jf \ E [ S ^A \ "r r 'p \ \ \ * 1 -- -^ \ Jl Given AB and CD, intersected by planes MN, PQ, and RS, in A, E, B, and C, G, D, respectively. To prove = EB GD Proof. Draw AD intersecting PQ in F. Draw EF, BD, FG, AC. Then EF II BD and FG II AC. 549 AE AF __j CG AF ^ . AE = CG ' EB GD' Q.E.D. DIHEDRAL ANGLES. 558. When two planes intersect, their divergence from the line of intersection is called a dihedral angle. 559. The two planes are its faces and the line of intersection is its edge. BOOK VI. SOLID GEOMETRY. 17 D E 560. When a dihedral angle stands alone it may be desig- nated by the two letters on its edge, as , , the dihedral angle AB. When two or / . / - more dihedral angles have a common I/ l^r edge, each angle is designated by four A B letters, as angles C-AB-D, or D-AB-E. 561. The plane angle of a dihedral angle is the angle formed by two lines, one in each face, drawn perpendicular to the edge at the same point. Thus 2C abc and 2 def are plane angles of the dihedral C-AB-D. ^~ B A plane perpendicular to the edge of a dihedral angle inter- sects the faces in lines which form the ffl ^ plane angle of the dihedral. ~ ~ C 562. Any two plane angles of a dihe- dral angle or of equal dihedral angles ^ abc = "%. def, their sides being II. 555 Since the points b and e are taken anywhere on the edge, the plane angle of a dihedral angle is of the same magnitude at every point of the edge. Conversely, two dihedral angles are equal if their plane angles are equal. Given the plane %. BAG = DEF. The edge AH _L the plane BAC and EK J_ plane DEF. Then by superposition it may be shown that the dihedral B-RA-C = dihedral % D-KE-F. 524 563. A dihedral angle may be con- ceived as generated by the revolution of a plane about a line taken as an edge. As the magnitude of the plane angle depends upon the amount of the revolution of one of its sides, from an initial line, independent of the length of its 'JB K D 18 BOOK VI. SOLID GEOMETRY. sides ( 57), so the magnitude of a dihedral angle depends upon the amount of the revolution of a plane from an initial plane, and is independent of the ex- tent of the planes. 564. By passing a plane perpendicular to the edge of the dihedral angle it is evident that the plane angle is the measure of the dihedral angle. A dihedral angle is acute, right, obtuse, straight, or reflex, according as its plane angle is acute, right, obtuse, straight, or reflex. 565. Dihedral angles are adja- cent, vertical, complementary, or supplementary just as their plane angles hold these relations. 566. Many of the theorems of lines and angles have analo- gous theorems with planes and dihedral angles. By forming right sections of the dihedral angles the proofs of the follow- ing theorems may be obtained from the corresponding theorems of plane geometry : 1. The supplements of equal dihedral angles are equal. 2. The complements of equal dihedral angles are equal. 3. Vertical dihedral angles are equal. 4. If two parallel planes are cut by a third plane, the alternate interior dihedral angles are equal ; the interior dihedral angles on the same side of the transverse plane are supplementary ; the corresponding dihedral angles are equal ; and conversely. 5. Two dihedral angles whose faces are respectively parallel are either equal or supplementary. 6. Two dihedral angles whose edges are parallel and whose faces are respectively perpendicular are either equal or supple- mentary. BOOK VI. SOLID GEOMETRY. 19 PROPOSITION XIY. THEOREM. 567. If a straight line is perpendicular to a plane, every plane containing that line is perpendicular to the plane. Given AB _L plane MN, and PQ any plane passing through AB intersecting MN in CQ. To prove plane PQ J_ plane MN. Proof. In Jtfy draw BD _L CQ. But AB _L CQ. .-. ABD is the plane of the dihedral And ABD is a rt. . /. plane PQ JLpZane MN. 113 526 P-QC-N. 561 526 564 Q.E.D. 568. COR. A plane perpendicular to the edge of a dihedral angle is perpendicular to its faces. PROPOSITION XV. THEOREM. 569. If two planes are perpendicular to each other, any straight line in one plane drawn perpendicular to their intersection is perpendicular to the other. 20 BOOK VI. SOLID GEOMETRY. Pr- D. A SB E Given plane PQ -L plane MN, and AB in PQ _L DE, their in. tersection. To prove AB _L plane MN. Proof. In plane MN draw EG JL DE. 113 Then ABC is the plane of the dihedral P-DE-M. 561 But dihedral P-DE-M is a rt. dihedral . Hyp. . ' . ^ ABC, its measure, is a rt. 2. 564 Also AB JL DE. Hyp. .'. AB, being _L JB(7 and DE at their intersection, is _1_ their plane. 535 .' AB J_ plane JOT. Q. E. D. 570. COK. 1. If tivo planes are perpendicular to each other, and a perpendicular to one of them is drawn from any point in their intersection, it will lie in the other plane. 571. COR, 2. If two planes are perpendicular to each other, and a perpendicular to one of them is drawn from any point in the other plane, it icill lie in that plane. ^ PROPOSITION XVI. THEOREM. 572. If two intersecting planes are each perpendicu- lar to a third plane, the line of intersection is perpen- dicular to that plane. BOOK VI. SOLID GEOMETRY. 21 M, Given planes PQ and RS, each _L plane MN, and intersecting in AB. To prove AB _L plane MN. Proof. At B in plane MN erect a J_ to MN. This _L lies in each of the planes BS and PQ. 570 .'.it coincides with AB, their intersection. 530 . ' . AB _L plane MN. Q. E D 573. COB. Conversely, if a plane is perpendicular to each of two intersecting planes, it is perpendicular to their line of inter- section. PROPOSITION XVII. THEOREM. 574. The plane bisecting a dihedral angle is the locus of points equidistant from its faces. 22 BOOK VI. SOLID GEOMETRY. I. Given the dihedral A-BG-C, the plane BD bisecting this %., any point P in the plane BD, PE, and PF, _Ls drawn from P to the faces AB and BC. To prove P is equidistant from the faces AB and BC, or PE=PF. Proof. 1. Through PE and PF pass a plane intersecting AB in EG, BC in OK, and BD in PG. Because PE J_ AB and PF _L BC, Cons. the plane PEF _L BG, their intersection. 573 .-. BG -L EG, PG, and FG. 535 .-. s EGP and PGF are the plane s of the dihedral s A-BG-D and D-BG-C. 561 But the A-BG-D = D-BG-C. Hyp. .-. EGP= % PGF. 562 .-. rt. A PEG ^ rt. A PFG. 121 = PF. 85 II. Given P' any point without the bisecting plane BD. To prove P 1 unequally distant from the faces AB and AC. Proof. Draw P"K. BC, P'E _L AB, and from P draw PF _L BC. Pass a plane through P'E and PK intersecting AB in EG, BD in P#, and BC in #/ Then PP' + PF>P'F. (86.) Also P'F>P'K 531 .-. PP' + PF> P'/f and PF= PE. Proof I .-. PP' + PE> P'K or PE > P7iT. .-. any point without the bisecting line is unequally distant from the faces of the dihedral angle. Q. E. D. 575. DBF. The projection of a point on a plane is the foot of the perpendicular from the point to the plane. BOOK VI. SOLID GEOMETRY. 23 The projection of a line on a plane is the locus of the projec- tion of its points on the plane. This may be illustrated by a shadow cast by a ruler on a wall, when the rays of light casting the shadow fall perpen- dicularly upon the wall. PROPOSITION XVIII. PROBLEM. 576. Through a straight line not perpendicular to a given plane to pass a plane perpendicular to that plane. A r i j c * \ K N F Given the line AB not J_ plane MN. Required to pass a plane through AB _L MN. Construction. From P, any point in AB, draw PK _L plane MN. Through PK and AB pass a plane AF. AF is the plane required. Proof. The plane AF through AB intersects MN. .-. plane AF J_ plane MN. 538 Cons. 567 Q. E. F. 577. COR. 1. TJirough a straight line not perpendicular to a given plane, only one plane can be passed perpendicular to that plane. 578. COR. 2. The projection of a straight line, not perpen- dicular to a plane, upon that plane, is a straight line. 24 BOOK VI. SOLID GEOMETRY. 579. COB. 3. The projection of a straight line, perpendicular to a plane, is a point. 580. DEF. The plane containing all the perpendiculars drawn from a straight line to a plane is called the projecting plane. PROPOSITION XIX. THEOREM. 581. The acute angle which a line makes with its pro- jection on a plane is the least angle which it makes ivith any line of that plane. Given BC the projection of the line AB on the plane MN, and BD any other line, drawn in plane MN, through B. To prooe ABC < % ABD. Proof. Take BD = BC, draw AD and AC. In the As ABC and ABD, AB = AB. BD = BC. But AC AVB or BVC. To prove % AVB + %BVC>%AVC. Proof. In % AVC draw VD, making A VD = J. VB. Through D, any point in VD, draw .4(7. Take VB = VD, and draw AB and .BO. Then A^FB^A^FD. 91 .'.AB=AD. 85 In A ABO, AB + BC> AC. 86 But AB = AD. By subtraction, BC > (AC -AD) or BC > DC. In As BVC and DFC, F<7 is common, VB = VD, and BC > DC. But Adding, %AVB=%AVD. AVB + %BVC> >XAVC. Cons. Just proved 132 Cons. DVC. Q. E. D. PROPOSITION XXII. THEOREM. 594. The sum of the face angles of any convex poly- hedral angle is less than two straight angles. v BOOK VI. SOLID GEOMETRY. 29 Given the polyhedral V. To prove the sum of the ^.s at V less than 2 st. 2s. Proof. Pass a plane cutting the polyhedral %. in the polygon ABODE. From 0, any point in this polygon, draw OA, OB, OC, OD, OE. The number of 2s having a common vertex is the same as the number of ^s having a common vertex V. .'. the sum of all the 2s in the As having a vertex at = the sum of all the 2s of the As having a vertex at V. 155 But YEA + VBC > % ABC, and 4 VCB + VCD > BCD, etc. 593 .'. the sum of the base 2s of the As having a common vertex at Vis > the sum of the base 2s of the As having a common vertex at 0. .'. the sum of the 2s at the vertex at V < the sum of the 2s at the vertex 0. But the sum of the angles at the vertex = 2 st. 2s. 63 .'. the sum of the ^s at V < 2 st. ^!s. Q. E. D. PROPOSITION XXIII. THEOREM. 595. If two trihedral angles have the three face angles of one equal to the three face angles of the other, they are either congruent or symmetrical. 30 BOOK VI. SOLID GEOMETRY. Given the trihedral ^s V and V, having the face s AVB, AVC, and BVC = the face s A'V'B', A'V'C', and B'V'C', respectively: To prove the corresponding dihedral ^.s of V and V are =, and that the trihedral %.s V and V are congruent or symmetrical. Proof. On the edges of V and V take VA, VB, VC, V'A', VB', V'O all equal, and draw AB, BC, AC, A'B', B'C', A'C 1 . Then As VAB, VBC, VAC ^ As V'A'B', V'B'C 1 , V'A'C'. 91 /. A ABC 3* A A'B'C'. 123 In the edges VC, V'C 1 , take VD= V'D' and in the faces VBC, VAC, and V'B'C', V'A'C', draw DE, DF, and D'E 1 , D'F' _L the edges VC and F'C" respectively, meeting BC, AC, B'C 1 , and A'C' in E, F, E', and F 1 . Draw EF, E'F'. Then A CED =* A C'E'D'. 95 .-.CE = C'E', and DE = D'E'. Likewise CF= C'F', and DF= D'F'. .'.A CFE ^ A C'F'E'. 91 .'.FE = F'E'. 85 .'. A FED ^ A F'E'D'. 123 . . -4 FED = % F'E'D'. 85 /. dihedral VC= dihedral V'O. In like manner it can be proved that dihedral %. VB = dihe- dral VB' and dihedral VA = dihedral V'A'. 562 .*. the trihedrals V and V' are congruent or symmetrical according as the equal face angles are arranged in the same or reverse order. Q.ED. 596. COB. If two trihedral angles have three face angles of the one equal to the three face angles of the other, then the dihedral angles of the one are respectively equal to the dihedral, angles of the other. BOOK VII. POLYHEDKONS. 597. A polyhedron is a solid tiounded by planes. These planes are the faces; the intersections of the faces, the edges, and the intersections of the edges, the vertices of the polyhe- dron. A line joining any two vertices, not in the same face, is a diagonal. A polyhedron cannot have less than four faces. 598. A section of a polyhedron is the figure formed by its intersection with a plane. 599. A polyhedron is convex when every section of it, formed by the intersection of a plane with its faces, is a convex polygon. All polyhedrons considered in this book are convex. 600. A polyhedron of four faces is called a tetrahedron ; of six faces, a hexahedron ; of eight faces, an octahedron ; of twelve faces, a dodecahedron ; of twenty faces, an icosahedron. Tetrahedron. Hexahedron. 31 Octahedron. 32 BOOK VII. SOLID GEOMETRY. Dodecahedron. Icosahedron. Prism. PRISMS AND PARALLELOPIPEDS. 601 . A prism is a polyhedron, of which two faces are congruent polygons in parallel planes, and the other faces are parallelo- grams. The congruent polygons are the bases and the parallelograms are the lateral faces of the prism. The intersections of the lateral faces are the lateral edges. The sum of the areas of the lateral faces is the lateral area. The per- pendicular distance between its bases is the altitude. The lateral edges of a prism are all equal, because parallel lines included between parallel planes are equal. 602. A prism whose lateral edges are perpendicular to its bases is a right prism. Hence, the lateral edge of a right prism is equal to its alti- tude. 603. A right prism whose bases are regular polygons is a regular prism. 604. A prism whose lateral edges are not perpendicular to its bases is an oblique prism. 605. A prism is triangular, quadran- gular, etc., according as the bases are triangles, quadrilaterals, etc. POLYHEDRONS. 33 606. The section of a prism made by a plane perpendicular to its lateral edges is a right section. 607. A truncated prism is that part of a prism included between a base and a section made by a plane oblique to the base. 608. A parallelepiped is a prism whose bases are parallelograms. Hence all the faces of a paral- lelepiped are parallelograms. 609. A right parallelepiped is a parallelepiped whose lateral edges are perpendicular to its bases. 610. A rectangular parallelepiped is a right parallelepiped whose bases are rectangles. Hence all its faces are rectangles. 611. A cube is a parallelepiped whose faces are squares. Hence all of its edges are equal. 612. The volume of any solid is the ratio of the solid to another solid taken arbitrarily as the unit of volume. Because of its convenience a cube whose edge is a linear unit is adopted as the unit of volume. 613. Equal solids are solids whose volumes are equal. 34 BOOK VII. SOLID GEOMETRY. PROPOSITION I. THEOREM. 614. Sections of a prism made by parallel planes cutting all the lateral edges are congruent polygons. Given the prism PM cut by II planes making the sections AD and A'D'. To prove section AD ^ section A'D'. Proof. AB, BC, CD, etc. II A'B', B'C', CD', etc., respec- tively. 549 .-. AB, BO, CD, etc. = A'B', B'C 1 , CD', etc., respectively. 169 Also s ABC, BCD, etc. = s A'B'C 1 , B'CD', etc., respec- tively. 555 .-. polygon AD ^ polygon A'D'. (for the polygons have their sides and 2s equal each to each, and are therefore congruent}. Q. E. D. 615. COR. Any section of a prism made by a plane parallel to the base is equal to the base; also all right sections of a prism are equal. POLYHEDRONS. 35 PROPOSITION II. THEOREM. 616. The lateral area of a prism is equal to the product of the perimeter of a right section by a lateral edge. ar Given the prism PM. Denote its lateral area by s, a lateral edge by e, and the perimeter of a right section by p. To prove Proof. Also and s=p X e. PF= KG = LH, etc. = e. AB J_ KG, BC J_ LH, etc. . area O FK= AB x PF, = ABxe, area O GL = BC x e, area O #. = CD x e, etc. 601 606 420 But the sum of these Os equals the lateral area s. And the sum of AB + BC + CD + etc. =p. .-. s=p X e. Q. E D. 617. COR. The lateral area of a right prism equals the prod- uct of the perimeter of the base by the altitude. 36 BOOK VII. SOLID GEOMETRY. PROPOSITION III. THEOREM. 618. Tivo prisms are congruent when the three faces, including a trihedral angle of the one, are congruent, re- spectively, to the three faces, including a trihedral angle of the other, and are similarly placed. 'D 'D' Given the prisms, AK and A'K', having the faces AD, AL, AG = respectively to faces A'D', A'L', A'G', and similarly placed. To prove AK ^ A'K'. Proof. The face 2s BAE, EAF, BAF are equal respectively to the face s B'A'E', E'A'F', B'A'F'. Hyp. .-. trihedral A = trihedral A'. 595 Apply the prism F*D' to the prism FD so that trihedral 2 A' will fall on its congruent trihedral %. A, and the faces of trihedral 2 A' shall coincide with the congruent faces of tri- hedral A, and the points C, D shall fall on the points C", D'. Since the points L', F 1 , G' coincide with L, F, G, the planes F'K ! and FK coincide ( 523), because the lateral edges of the prism are II. .-.the edges C'H 1 , D'K' coincide with CH, DK, and the points H', K' coincide with H, K. .'.AK^zA'K'. Q.E.D. POLYHEDRONS. 37 619. COR. 1. Two truncated prisms are congruent when the three faces including a trihedral angle of the one are congruent to the three faces including a trihedral angle of the other. 620. COB. 2. Two prisms with equivalent bases and equal altitudes are equal. PROPOSITION IV. THEOREM. 621. An oblique prism is equivalent to a right prism whose base is a right section of the oblique prism, and, whose altitude is a lateral edge of the oblique prism. B c Given the oblique prism PM with the right section A'M' and lateral edge PA ; also the right prism P'M' with base A'M' and lateral edges each equal to PA. To prove PM = PM 1 . Proof. PA' = PA. Subtracting PA' from each of these equals, P'P = A' A. R'R = B'B. PR = AB and P'R = A'B'. s of PR' = s of AB'. .: face PR' ^ face AB'. face RS' ^ face BC 1 . Likewise Also And Hyp- 167 151 176 Similarly, 38 BOOK VII. SOLID GEOMETRY. And base PN '= AM. 601 .-. truncated prism PN' = truncated prism AM' , 619 Hence P M PN ' = PM AM', or PM=P'M'. Q.E.D. PROPOSITION V. THEOREM. 622. The opposite lateral faces of a parallelepiped are parallel and congruent. H E 0- Given parallelepiped AG. To prove O AF II and ^ O DG, and O ^fl" II and s O #. Proof. Because .4(7 is a O, 608 .-. AB II and = DC. 167 But AH" is a O, 608 .-. AE\\ and = DH. 167 . V T? A 7? V ttT\ri . ^. MI^MJ =^ A. xnj\jj and plane -4F II plane DG 1 . 555 .-. n AF ?* n DG. 17C Similarly, O AH" II and ^ O jB#. Q. E. D. 623. COR. Any two opposite faces of a parallelopiped may be taken as bases. POLYHEDRONS. 39 PROPOSITION VI. THEOREM. 624. The plane passed through two diagonally oppo- site edges of a parallelopiped divides it into two equivalent triangular prisms. Given the plane QSBC passing through the diagonally opposite edges QC and SB of the parallelopiped PD. To prove PD is divided by the plane into two equivalent tri- angular prisms, ACB-S and BCD-R. Proof. Let EFGH be a right section of PD intersected by the plane QSB C in the diagonal HF. Then EF II HG and EH II FG. .: EFGH is a O. .-. AEFH^AHFG. 549 164 166 The triangular prism ACB-S = a right prism whose base is EHF and altitude AP, and the prism BCD-R = a right prism whose base is FHG and altitude AP. But these right prisms are equal, having equal bases and equal altitudes. 621 .-. ACB-P= BCD-R. Q. E. D. 40 BOOK VII. SOLID GEOMETRY. PROPOSITION VII. THEOREM. 625. Tivo rectangular parallelopipeds having con gruent bases are to each other as their altitudes. \ \ \ \ N P \^ D C X \ X Q X, Given two rectangular parallelopipeds P and Q having congruent bases and the altitudes AB and CD. To prove CD CASE I. When AB and CD are commensurable. Proof. Let AKbe the common measure of AB and CD. Suppose it is contained m times in AB and n times in CD. Then CD n At the points of division of AB and CD pass planes II to the bases. These planes divide P into m and Q into n equal parallelo- pipeds. 620 . P = m P^AB ' n g CD' POLYHEDRONS. CASE II. When AB and CD are incommensurable. 41 N \ \ S N P V E C N Q \ Q' Proof. Divide AB into any number of equal parts and apply one of these parts, as AK, to CD as a unit of measure. Since AB and CD are incommensurable, there will be a re- mainder ED less than one of the parts. Through E pass a plane II the bases of Q and let Q' be the parallelepiped between this plane and the lower base of Q. Then P AB Case I If the unit of measure for AB be continually decreased, the remainder ED, which is always less than the unit of measure, may be made smaller than any assignable quantity, but not equal to zero, since AB and CD are incommensurable. .-. CE will approach CD as a limit. .. will approach as a limit. Qt Q .. -SL w iH approach - as a limit. P AB P_ = AB Q CD' 275 331 Q. ED 42 BOOK VII. SOLID GEOMETRY. 626. DBF. The three edges of a rectangular parallelepiped meeting at a common vertex are called its dimensions. 627. COR. Two rectangular parallelepipeds which have two dimensions in common are to each other as their third dimension. PROPOSITION VIII. THEOREM. 628. Two rectangular parallelopipeds having equal altitudes are to each other as their bases. C V Given two rectangular parallelopipeds P and Q, having a com- mon altitude a and the dimensions of their bases b, c, and b', c', respectively. To prove = l2<_2.. Q, b 1 X c' Proof. Construct a third rectangular parallelopiped R, hav- ing the same altitude a and the dimensions of the base c and b'. Then Q and R have two dimensions, a and 6' in common. R_ ' Q = Likewise P and R have two dimensions a and c in common. 6 627 . P 'R Multiplying the equations, 627 Q b'xcf' Q.B.D. POLYHEDRONS* 43 629. COK. Two rectangular parallelepipeds having one di- mension in common are to each other as the products of the other two dimensions. PROPOSITION IX. THEOREM. 630. Two rectangular parallelepipeds are to each other as the products of their three dimensions. Given the two rectangular parallelepipeds P and Q having the dimensions a, b, c and a', b', c', respectively. To prove P _ axbxc Q~a' x b' xc 1 ' Proof. Construct a third rectangular parallelepiped E having the dimensions a, b, and c'. Then E axb Q a'xV and E c 1 Multiplying the equations, axbxc a'xb'xc 1 ' 629 627 Q.E.D. 44 BOOK VII. SOLID GEOMETRY. PROPOSITION X. THEOREM. 631. The volume of a rectangular parallelopiped is equal to the product of its three dimensions. u Given any rectangular parallelopiped P, with dimensions a, b, c, and the cube U, the unit of volume, whose edge is the linear unit. To prove Proof. P = a x b x c. P_aX& X c U Ixlxl 630 P . Because U is the unit of volume, is the numerical meas- ases are the upper base, the lower base of the frustum, and a mean proportional between them. Given the frustum of any triangular pyramid, F-ADC, with lower base B, upper base B'. and altitude H. POLYHEDRONS 55 To prove, FADO= three pyramids whose altitude is H and whose bases are B, B', and a mean proportional between B and B'. Proof. Divide the frustum, by passing planes through E and FC, E and AC, into three triangular pyramids: E-ADC, C-EFG, and E-AFC. Denoting these pyramids by P, Q, and R, respectively. (P and Q have a common altitude, H.) Then P = %Hx B, and Q = ^Hx B'. 654 It remains to prove that E-AFC or R = a pyramid whose altitude is H and whose base is a mean proportional between B and B' or V-B x B'. Because P and R have a common vertex C, and their bases in the same plane AFED, . P A AED R - But As AED and AEF have a common altitude. . A AED AD . . - = - A AEF EF P R EF Likewise pyramids Q, and R may be considered as having a common vertex E, with their bases in the same plane AFGC. R A AFC and ' Q AFGC' AAFC = AC AFGC FG' R = AC *' Q FG' 56 BOOK VII. SOLID GEOMETRY. Because A ADC ~ A FOE, AD = AC EF OF !-!-**-*. 363 B'. .-. R = | H^/B x B', or E-AFC= pyramid with altitude H and baseV.B x B'. .-. frustum F-ABC= H (B + B' + X B'). Q. E. D. 660. Formula for volume of frustum of a triangular pyra- mid is V= | H (B + B' + V x -B')- PROPOSITION XVIII. THEOREM. 661. A frustum of any pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the upper l}ase, the lower base of the frustum, and a mean propor- tional between them. s T POLYHEDRONS. 57 Given the frustum of any pyramid Ad, with, its upper base B', its lower base B, its altitude H, and its volume V. To prove V = 1 H ( B + B' + V-B x B'}. Proof. Produce the lateral edges of the frustum .Ad to meet mS. Construct a triangular pyramid T-FKG with altitude = H and base = ABCDE and lying in the same plane. Produce the plane of abcde to cut the pyramid T-FKG in f/ig. .'. f kg = abcde. 651 But S-ACDE = T-FKG (1), 653 and S-abcde = T-fhg (2). Subtracting (2) from (1), frustum aD = frustum fG. But volume fG = \ H(B + B' + V-B x B'). 660 Q. E. D. Ex. 570. The diagonal of a cube is to its edge as V3 is to 1. Ex. 571. Find the diagonal of the rectangular parallelepiped whose edges are 8 ft., 9 ft., and 12 ft., respectively. Ex. 572. Find the area of the entire surface of a triangular pyramid each of whose edges is 10 ft. Ex. 573. Find the volume of a regular triangular pyramid each of whose edges is 8 ft. Ex. 574. The volume of a truncated parallelepiped is equal to the area of a right section multiplied by one-fourth the sum of the lateral edges. Ex. 575. Find the volume of a right prism, if its altitude is 15 ft., and the sides of its base are, respectively, 10, 17, and 21 ft. Ex. 576. A regular pyramid, whose base is an equilateral triangle, each side of which is 12 ft., has an altitude of 20 ft. Find its volume. 58 BOOK VII. SOLID GEOMETRY. PROPOSITION XIX. THEOREM. 662. A truncated triangular prism is equivalent to the sum of three pyramids ivhose common base is the base of the prism, and whose vertices are the vertices of the inclined section. Given the truncated triangular prism P, with base ABC and in- clined section DEF. To prove P = pyramid F-ABG + pyramid D-ABC -f- pyra- mid E-ABC. Proof. Let the planes determined by DE and C, and by AC and E divide the truncated prism into three pyramids, E-FDC, E-DAC, and E-ABC. 1. E-FDC = B-ACF, for their bases DFC and AFC are equal ( 430) and their altitudes are equal because their ver- tices E and B lie in the edge of the prism which is II to the face in which the bases lie. But B-ACF is identical with F-ABC. .-. E-FDC =F- ABC. 2. E-DAC = B-DAC, having same base and equal altitudes because their vertices E and B lie in the edge of the prism II the face opposite. But B-DA C = D-ABC, Iden. . -. E-DA C = D-ABC. 3. E-ABC has the required base and vertex. . . P = F-ABC + D-ABC + E-ABC. Q. E. D. POLYHEDRONS. 59 663. COR. 1. The volume of a right truncated triangular prism is equal to the product of its base by one third the sum of its lateral edges. 664. COR. 2. The volume of any truncated triangular prism is equal to the product of its right section by one third the sum of its lateral edges. PROPOSITION XX. THEOREM 665. Tetrahedrons having a trihedral angle of one equal to a trihedral angle of the other are to each other as the products of the edges about the equal trihedral angles. A D Given two tetrahedrons T-ABC and T'-DEF, with equal trihedral %s at T and T' and volumes V and V, respectively. Toprove V = TAxTBxTG V T'DxTCxTD' Proof. Apply- the tetrahedron T-ABC to T-DEF so that the equal trihedral ^s T and T' coincide. From C" and F drop J_s upon the plane TED. The three points T, K, and G lie in a straight line, 578 (the projection of a straight line upon a plane is a straight line"). Then with TDE and TA'B 1 as the bases and FG and OK the altitudes of the tetrahedrons, V_ = TA'B' x OK V TA'B' OK FDExFG " TDE FG' 656 60 BOOK VII. SOLID GEOMETRY. TA'B' _ TA' x TB' ~TDE~ T'DxTE' In the right similar As T'C'/i 7 and TFG, VK = TV FG T'F ' V _ TA' x T'B' TC 1 = TAxTBxTC '' y> T'DxTE TF T'D x TE x TF' 374 Q.E.D. 666. Polyhedrons are similar if they have the same number of faces, similar each to each, arid similarly placed, and have their corresponding polyhedral 2s equal. Ex. 577. The altitude of the frustrum of a given pyramid is 18 ft. The lower base is a triangle whose sides are, respectively, 8 ft., 26 ft., and 30 ft. The shortest side of the upper base is 4 ft. Find the volume of the frustrum. Ex. 578. The diagonal of one of the faces of a cube is d. Find the volume of the cube. Ex. 579. The diagonal of a cube is D. Find the volume of the cube. Ex. 580. Find the lateral area of a regular pyramid if its base is a square 16 ft. to the side, and the slant height 28 ft. Ex. 581. The specific gravity of mercury being 14, and water weigh- ing 62 Ib. per cubic foot, what is the edge of a cubical box that would hold 40 Ib. of mercury ? Ex. 582. Find the area of the surface of a regular icosahedron, whose edge is 2 in. Ex. 583. In a regular pyramid with square base the lateral edge is 41 ft. and the slant height is 40 ft. Find the volume of the pyramid. POLYHEDRONS. 61 PROPOSITION XXL THEOREM. 667. Tlie homologous edges of similar polyhedrons are proportional. E\ E F' A' C' B' Given the similar polyhedrons P and P', with edges AB and CH homologous to A'B' and C'H'. AB CH To prove A'B' C'H' Proof. Because the face ABGF ~ A'B'G'F 1 ; AB = BG ' A'B 1 B'G 1 ' and because face BCHG ~ B'C'ffG', . BG = CH ' B'G' C'H'' AB CH A'B' C'H 666 363 666 363 Q. E. D. Ex. 584. The base of a regular pyramid is an equilateral triangle whose side is 6 ft. The altitude of the pyramid is 20 ft. Find the volume. 62 BOOK VII. SOLID GEOMETRY. 668. COR. 1. Any two homologous lines of similar polyhe- drons are proportional to any other two homologous lines of the polyhedrons. 669. COB. 2. Any two homologous faces of similar polyhe- drons are proportional to the squares of any two homologous lines of the polyhedrons, and the total surface of two similar polyhedrons are proportional to the squares of any two homologous lines of the polyhedrons. PROPOSITION XXII. THEOREM. 670. Similar tetrahedrons are to each other as the cubes of their homologous edges. Given two similar tetrahedrons T and T', whose volumes ere V and V, and whose altitudes are H and H', respectively. V A& V A 1 ^' 3 ' V base ABC X H To prove Proof. But And V' ba.se A B'C' base_ABC_ = A&_ base ^''(7 ~~ J^j' 2 ' If ^15 H 1 A'B V ^AB* V AB' 2 xA'B' AB 13 656 669 668 Q E.D POLYHEDRONS. 63 PROPOSITION XXIII. THEOREM. 671. Similar polyhedrons may be divided into the same number of tetrahedrons similar each to each and similarly placed. Given the two similar polyhedrons, P and P'. To prove P and P' may be divided into the same number of tetrahedrons similar each to each, and similarly placed. Proof. Take any homologous trihedral 2s in P and P 1 , as B and B', and through points G, A, C pass a plane ; also through points G 1 , A', C' pass a plane. Then in the two tetrahedrons thus cut off, G-ABC and G'-A'B'C', AG BG CG AC A'G' B'G' C'G 1 A'C' 363 .'.the faces BAG, BAC, and BGC are similar to B'A'G', B'A'C 1 , and B'G'C' respectively. 376 .-. the homologous faces of these tetrahedrons are similar. 369 But the homologous trihedral 2s of these tetrahedrons are equal. 595 .-. tetrahedron G-ABC- tetrahedron G'-A'B'C 1 . 666 After removing tetrahedron G-ABC from P and G'-A'B'C' from P' the polyhedrons which remain will be similar, for their faces are similar and the polyhedral 2s are equal. By 64 BOOK VII. SOLID GEOMETRY. this process P and P' may be divided into the same number of tetrahedrons, similar each to each and similarly placed. Q.E D. 672. COR. The volumes of any two similar polyhedrons are to each other as the cubes of any trvo homologous lines of the polyhedrons. 673. A regular polyhedron has all its faces congruent regular polygons and all its polyhedral angles congruent. PROPOSITION XXIV. THEOREM. 674. Only five regular polyhedrons are possible. Given, congruent regular polygons of any number of sides. To prove, only five regular polyhedrons, with congruent regular polygons for faces, can be constructed. Proof. A polyhedron 2 must have at least three faces and the sum of the face 2s must be < 360. 1. With equilateral As where each 2 is 60; 60 x 3 = 180; 60 X 4=240; 60 x 5=300; but 60 X 6 =360. .-. only three regular polyhedrons can be formed with equi- lateral As for faces. 2. With squares where each 2 is 90, 90 x 3 = 270 ; but 90 x 4 = 360. .. only one regular polyhedron can be formed with squares for faces. 3. With regular pentagons where each %. is 108. 108 x 3 = 324 ; but 108 x 4 = 432 .-. only one regular polyhedron can be formed with regular pentagons for faces. 4. But with regular hexagons where each 2 is 120, because 120 x 3 = 360, no regular polyhedron can be formed. Hence POLYHEDRONS. 65 no regular polyhedron can be formed with polygons having more than five sides. .. only five regular polyhedrons are possible. Q. E. D. 675. NOTE. The five regular polyhedrons are the tetrahedron, the octahedron, and the icosahedron from equilateral triangles; the hexahe- dron or cube from squares^" and the dodecahedron from pentagons. 676. These regular polyhedrons may be formed by cutting out cardboard as indicated in the following diagrams. Cut entirely through on the full lines, half through on the broken lines, and bring the edges together. The edges may be held in place by pasting narrow strips of paper over them. 66 BOOK VII. SOLID GEOMETRY. CYLINDERS. 677. A cylindrical surface is a curved surface generated by a moving straight line that constantly touches a given curve and is always parallel to a fixed straight line. The moving line is called the gen- eratrix and the given curve the direc- trix. The generatrix in any position is called an element of the cylindrical surface. 678. A cylinder is a solid bounded by a cylindrical surface, called the lateral surface and two parallel planes which are the bases of the cylinder. 679. The altitude of a cylinder is the perpendicular distance between the bases. 680. Because parallel lines included between parallel planes are all equal, the elements of a cylin- drical surface are all equal. 681. A circular cylinder is a cylinder whose bases are circles. The term "cylinder," as hereafter used, will mean circular cylinder, as the circle is the only curve discussed in elementary plane geometry. 682. A right cylinder is a cylinder whose elements are perpendicular to its bases. An oblique cylinder is one whose elements are oblique to its bases. A right cylinder is called a cylinder of revolution because it may be generated by the revolution of a CYLINDERS. 67 rectangle about one of its sides as an axis. The radius of the base is the radius of the cylinder. 683. Similar cylinders of revolution are cylinders gen- erated by similar rectangles revolving about correspond- ing sides as axes. 684. A plane is tangent to a cylinder when it passes through one element of the cylinder but does not cut it. 685. A prism is inscribed in a cylinder when its base is a polygon inscribed in the base of the cylinder and its lateral edges are elements of the cylinder. 686. A prisin is circumscribed about a cylinder when its bases are polygons circumscribed about the bases of the cylinder and its lateral faces are tangent to the cylinder. 687. A right section of a cylinder is the figure formed by the intersection of a plane with the cylinder perpendicular to its elements. The right section of a cylinder is a circle. 688. Because the circle is the limit of the perimetej-sjof-segu- Jar inscribed and circumscribed polygons and the area of a circle is__he limit of the areas of these polygons when the number of their sides is indefinitely increased, 488, hence: 1. The circle of a right section of a cylinder is the limit of the perimeter of a right section of an inscribed or circum- scribed prism. 2. The lateral area of a cylinder is the limit of the lateral area of the inscribed or circumscribed prism. 3. The volume of a cylinder is the limit of the volume of an inscribed or circumscribed prism. 68 BOOK VII. SOLID GEOMETRY. PROPOSITION XXV. THEOREM. 689. The lateral area of a cylinder is equal to the product of the circle of a right section by an element. Given the cylinder AB, with lateral area S, circle of a right section C, and element H. To prove S=CxH. Proof. Inscribe in the cylinder a regular prism, with lateral area /S", and perimeter of a right section P. Its lateral edge is E. Then S' = P x H. 616 Let the number of lateral faces of the prism be indefinitely increased. Then S' approaches S as a limit. 688, 2 P approaches (7 as a limit 688, 1 and P x H approaches C X H as a limit. 277 But S' = PxH always. 616 .-. S=CxH. 275 Q. E. D. 690. COR. 1. The lateral area of a cylinder of revolution is equal to the circle of the base multiplied by the altitude. 691. COR. 2. If S is the lateral area, T the total area, and H an element, then T=2*Rx H+ 2 irR 2 =2 TrR(H+ R). CYLINDERS. 69 692. COR. 3. Lateral areas or total areas of two similar cylinders are to each other as the squares of their like dimensions. PROPOSITION XXVI. THEOREM. 693. The volume of a cylinder is equal to the product of its base and altitude. Given the cylinder AB, with volume V, base B, and altitude H. To prove V= B x H. Proof. Inscribe in the cylinder a regular prism with volume V and base B'. Its altitude is equal to H. Then V'=B'xH. 636 Let the number of the lateral faces of the prism be indefi- nitely increased. Then and . But V approaches V as a limit, B' approaches B as a limit, B r x H approaches B X H as a limit. F^-B'x-H" always. .% V=BxH. 688, 3 688, 1 277 636 275 Q.E D. If the radius of the cylinder is R, then V= irR 2 H. 695. COR. 2. Volumes of two similar cylinders are to each other as the cubes of their like dimensions. 694. COR. 1. 70 BOOK VII. SOLID GEOMETRY. CONES. 696. A conical surface is a curved surface generated by a Inoving straight line that constantly touches a given curve and passes through a fixed point. The moving line is called the generatrix, the fixed point, the vertex, and the given curve, the directrix. The generatrix in any position is called an element of the conical surface. 697. The conical surface may consist of two parts, one above and the other below the vertex, called the upper and lower nappes, respectively. 698. A cone is a solid bounded by a coni- cal surface, called the lateral surface and a plane which is the base of the cone. 699. The altitude of a cone is the per- pendicular from the vertex to the plane of the base. 700. A circular cone is a cone whose base is circular. The term " cone " as hereafter used, will mean a circular cone of one nappe. 701. The axis of a cone is the straight line from the vertex to the center of the base. 702. A right cone is a cone in which the axis is perpendicular to the base. A right cone is also called a cone of revolution because it may be generated by revolving a right triangle about one of its legs as an axis. CONES. 71 703. The slant height is an element of the conical surface. 704. Similar cones of revolution are cones generated by simi- lar right triangles revolving about corresponding legs. 705. A plane is tangent to a cone when it touches it in one element of the cone but does not cut it. 706. A pyramid is inscribed in a cone when its base is a polygon inscribed in the base of the cone and its lateral edges are elements of the cone. 707. A pyramid is circumscribed about a cone when its base is a polygon circum- scribed about the base of the cone, and its lateral faces are tangent to the cone. 708. When the number of sides of regu- lar inscribed or circumscribed polygons is indefinitely increased, 488, 688, 1. The circle of a right section of a cone is the limit of the perimeter of a right section of an inscribed or circumscribed pyramid. 2. The lateral area of a cone is the limit of the lateral area of an inscribed or circumscribed pyramid. 3. The volume of a cone is the limit of the volume of an inscribed or circumscribed pyramid. 709. The axis of a right cone is its altitude. 710. The elements of a right cone are all equal. 711. The frustum of a cone is the part of a cone contained between its base and a plane parallel to its base. The base of the cone is the lower base of the frustum and the section made by the plane parallel to the base is the upper base of the frustum. 72 BOOK VII. SOLID GEOMETRY. PROPOSITION XXVII. THEOREM. 712. The lateral area of a cone of revolution is equal to half the product of the slant height by the circle of the base. Given V-EFG a cone of revolution, L its slant height, C the circle of its base, and S its lateral area. To prove S = | L x C. Proof. Circumscribe about the cone a regular pyramid, de- noting its lateral area by S' and the perimeter of its base by P. Then S' = LxP. 646 Let the number of lateral faces of the pyramid be indefi- nitely increased. Then S' will approach S as a limit, 708, 2 P f will approach P as a limit, 708, 1 and \ L x P will approach \ L x C as a limit. 277 But S' = % L x P always. .-. S = L X C. Q.E.D. CONES. 73 713. COR. 1. If R is the radius of a cone and T is its total surface, because the circle of the base is 2 irli, and the area is =ir L = TrRL. T= 714. COR. 2. The lateral areas or the total areas of two similar cones are to each other as the squares of their like dimen- sions. PROPOSITION XXVIIL 715. The lateral area of the frustum of a right cone is equal to half the product of the slant height of the frustum by the sum of the circles of the bases. Given L' the slant height, c the upper base of the frustum, C the lower base, and S' the lateral area. To prove S' = $(c+C)L'. Proof. The lateral area of the frustum of the inscribed pyramid is 647 When the number of lateral faces of the inscribed frustum of the pyramid is indefinitely increased, P approaches C, p ap- proaches c, L approaches L', and OE. 531 .'. the point B is without the sphere. 722 But B is any point in the plane MN other than A. .'. the plane MNis tangent to the sphere. Q. E. D. 744. COR. 1. A plane tangent to a sphere is perpendicular to the radius drawn to the point of tangency. 745. COR. 2. A line in a tangent plane drawn through the point of tangency is tangent to the sphere at that point. 746. COR. 3. A line tangent to a circle of a sphere lies in the plane that is tangent to the sphere at the point of contact. Ex. 603. If the volumes of two similar prisms are to each other as 8 to 27, what is the ratio of their altitudes ? THE SPHERE. 83 PROPOSITION V. THEOREM. 747. Through any four points, not all in the same plane, one, and only one, sphere may be passed. c Given A, B, C, and D, four points, not all in the same plane. To prove one sphere, and only one, may be passed through A, B, C, and D. Proof. Let F and G be the centers of Os circumscribing As BCD and ACD, respectively. Let FK be J_ the plane BCD and GH _L the plane ACD. Then every point in FK is equidistant from B, C, and D, and every point in GH is equidistant from A, C, and D. 533 Join F and G to E, the mid-point of CD. Then FE and GE are each _L CD. 240 /. the plane GEF _L CD. 535 /.the plane GEF planes BCD and ACD. 567 Then, since GH is _L plane ACD by constniction, GH lies in the plane GEF. 569 Similarly, FK lies in the plane GEF. Therefore _Ls GH and and FK lie in the same plane and being to non-parallel planes, they meet in some point, as 0. .'. lies in _L GH and FK equidistant from B, C, and D, and from A, C, and D. 533 .'. is equidistant from A, B, (7, and D. 84 BOOK VIII. SOLID GEOMETRY. Hence the sphere whose center is and radius OA will pass through A, B, C, and D. Again, since the center of any sphere through A, B, C, and D must lie in GH and FK ( 534), their intersection is the center of the only sphere that will pass through the four points A, B, (7, and D. Q.E.D. 748. COB. Four points not all in the same plane determine a sphere. SPHERICAL ANGLES. 749. DBF. The angle formed by two intersecting curves is the angle formed by the tangents to the curves at the point of intersection. 750. DBF. The angle formed by two intersecting great circles of a sphere is called a spherical angle. PROPOSITION VI. THEOREM. 751. A spherical angle is measured by the arc of a great circle described from the vertex as a pole and in- cluded betiveen its sides, or its sides produced. E Given the great Os BCA and BDA, intersecting at A, and CD, the arc of a great O described with A as a pole. To prove the spherical angle CAD is measured by the arc CD. THE SPHERE. 85 Proof. Draw radii OC and OD and tangents AE and AF. Arcs AC and AD are quadrants. 216 OC and OD each JL OA, and COD = iMF. 555 .'. the spherical ^ CAD is measured by the arc CD. Q.E. D. COR. A spherical angle has the same measure as the dihedral angle formed by the planes of the two circles. SPHERICAL POLYGONS. 753. DEF. A spherical polygon is a portion of the surface of a sphere bounded by three or more great circles. The bounding arcs are called the sides, their points of inter- section, the vertices, and the spherical angles formed by the -sides, the angles of the spherical polygon. 754. DEF. The diagonal of a spherical polygon is the arc of a great circle drawn between two non-consecutive vertices. 755. The planes of the sides of a spherical polygon form a polyhedral angle at the center of the sphere. A spherical polygon is .convex if the corresponding polyhedral angle is con- vex. Unless stated otherwise a spherical polygon is assumed to be convex. From any property of polyhedral angles may be inferred an analogous property of spherical polygons, and con- versely. 756. The measures of the sides of a spherical polygon are usually expressed in degrees. 757. DEF. Two spherical polygons are vertical when their corresponding polyhedral angles are vertical. 758. Two spherical polygons are symmetrical when the parts of one are, respectively, equal to the parts of the other and arranged in reverse order. 86 BOOK VIII. SOLID GEOMETRY. PROPOSITION VII. THEOREM. 759. Two vertical spherical triangles are symmetrical. Given two vertical spherical triangles, ABC and A'B'C . To prove spherical As ABC and A'B'C 1 are symmetrical Proof. Let be the center of the sphere. Plane As AOC and A'OO have two sides and the included ^s respectively equal, and are, therefore, congruent. 65, 91 .-. chord AC = chord A'C. 85 .-. arc AC = arc A'C 1 . 234 Similarly, arc BC = arc B' ?, arc AB = arc A'B', etc. .*. spherical As ABC and A'B'C' are symmetrical. Q. E. D. Ex. 604. Find the volume of a right circular cylinder the diameter of whose base is 12 ft. and whose altitude is 20 ft. Ex. 605. If the diameter of a right circular cylinder is 8 ft., and its total surface is 128 sq. ft., what is its altitude ? Ex. 606. Reckoning 7 gal. to the cubic foot, how many gallons will a cylindrical standpipe hold if its diameter is 20 ft. and the altitude 60ft.? Ex. 607. Given the lateral surface of a right circular cylinder, S, and altitude H, to find the volume. THE SPHERE 87 760. Two symmetrical spherical polygons may be placed in position such that each is the vertical of the other. PROPOSITION VIII. THEOREM. 761. The sum of two sides of a spherical triangle is greater than the third side. Given the spherical A ABC. To prove AB + BC > AC. Proof. Draw radii OAj OB, and 0(7. Then in the trihedral 0-ABC, %AOB + BOG > % AOC. 593 But the central angle is measured by the intercepted arc. .-. arc AB + arc BC > arc AC. Q. E. D. Ex. 608. The slant height of a regular pyramid is divided by a plane parallel to the base in the ratio 1 : 4, the longer segment being next the base. What is the ratio of the section to the base ? Ex. 609. Given F, the volume, and the altitude equal to the radius, of a right circular cylinder, to find the entire surface. Ex. 610. A pyramid is divided into two parts by a plane parallel to the base and bisecting the altitude. What is the ratio of the two parts ? Ex. 611. Two similar right circular cones have their volumes in the ratio 8 : 27. What is the ratio of their lateral surfaces ? 88 BOOK VIII. SOLID GEOMETRY. PROPOSITION IX. THEOREM. 762. The sum of the sides of a spherical polygon is less than 360. Given the spherical polygon ABCD. To prove AB + BC+CD + DA< 360. Proof. Arcs AB, BC, CD, DA are the measures of the central angles AOB, BOC, COD, and DO A. But the sum of these central angles < 360. 594 .-. AB + BC + CD + DA < 360. Q.E.D. 763. COR. TJie sum of the sides of a spherical, polygon is less than a great circle. PROPOSITION X. THEOREM. 764. The shortest line that can be drawn on the sur- face of a sphere between two points on the surface is the arc of a great circle not greater than a semicircle. Given two points, A and B, on the surface of a sphere, and AB, the arc of a great circle not greater than a semicircle. THE SPHERE. 89 To prove AB is the shortest line on the surface of the sphere between A and B. Proof. Take any point C on the arc AB, and with A and B as poles with radii equal, respectively, to AC and BC, describe Os. These Os cannot meet in any other point, for should they meet in any point, as K, the spherical A ABK would have the sum of two sides, AK and BK=AC, which is impossible. 761 Therefore, any other line, as ADEB, between A and B must intersect the Os in two points, as D and E. But ADEB cannot be shorter than AB, for by revolving AD about A and BE about B until D and E coincide with C there would be a line between A and B shorter than ADEB by the part EF. Hence the shortest line between A and B must pass through a But by hypothesis C is any point on the great O arc AB. .-. AB is the shortest line that can be drawn on the surface between A and B. Q.E.D. Ex. 612. Given the lateral surface, S, and altitude, equal to radius of the base, of a right circular cone, to find the volume. Ex. 613. Two similar right circular cones have their altitudes in the ratio 6 : 7. What is the ratio of their volumes ? Ex. 614. The altitude of the frustum of a pyramid is 36 ft. The lower base is a triangle whose sides are, respectively, 8, 26, and 30 ft. The longest side of the upper base is 15 ft. Find the volume of the frustum. Ex. 615. The altitude of the frustum of a cone is 24 ft. The diam- eters of the bases are, respectively, 32 ft. and 18 ft. How far from the lower base must a plane parallel to the base be passed to divide the frus- tum into two equivalent frustums ? 90 BOOK VIII. SOLID GEOMETRY. PROPOSITION XI. THEOREM. 765. Two mutually equilateral triangles on the same sphere or equal spheres are mutually equiangular \ and are congruent or symmetrical. Given the spherical As ABC and A'B'C' on equal spheres, AB = A'B', BC = B'C', AC=A'C'. To prove As ABC and A'B'C' are either congruent or sym- metrical. Proof. The face angles of the corresponding polyhedral angles at the center of the spheres are, respectively, equal. 233 .'. the corresponding dihedral angles are equal. 596 .*. the angles of the spherical As are respectively equal. 752 Therefore the As ABC and A'B'C 1 are congruent or sym- metrical according as the equal sides are arranged in the same or reverse order. Q E.D. 766. COR. Two symmetrical isosceles triangles are congruent. Ex. 616. The altitude of the frustum of a cone is 24 ft. The diam- eters of the bases are, respectively, 31 ft. and 18 ft. How far from the lower base must a plane be passed in order to divide the frustum into two similar frustums ? Ex. 617. The edges of a rectangular parallelepiped are, respectively, 12 ft., 10 ft., and 21 feet. What is the area of the surface of a similar parallelepiped whose diagonal is 87 ft. ? THE SPHERE. 91 PROPOSITION XII. THEOREM. 767. Two symmetrical spherical triangles are equal in area. Given two symmetrical As, ABC and A'B'C'. To prove As ABO and A'B'C' are equal in area. Proof. Let Os be circumscribed about the plane As ABO and A'B'C'. 306 Let and 0', respectively, be the poles of these Os Then the chord AB = the chord A'B', . the chord BO = the chord B'C 1 , and the chord AC = the chord A'C*. 233 .-. O ABC = O A'B'C'. .'. radius of O ABC = radius of O A'B'C'. 213 .-. arcs AO, BO, CO, A'O', B'O', and (70' are all equal. 234 .'. the spherical As AOB and A'O'B 1 are congruent, there- fore equal in area. 766 Similarly, spherical As BOO and AOC are respectively equivalent in area to spherical AsB'O'C' and A'O'C. Whence, by addition, spherical As ABC and A'B'C' are equal in area. Q.E.D. Ex. 618. A regular cone 18 in. in height and 24 in. in diameter at the base is cut by a plane parallel to the base and 10 in. from it. Find the volume of the frustum so formed, 92 BOOK VIII. SOLID GEOMETRY. 768. DBF. If from the vertices of any spherical triangle as poles arcs of great circles are drawn, another triangle is formed which is called the polar triangle of the first triangle. Thus, if A is the pole of the great circle arc B'C', B the pole of the great circle arc A'C', and C the pole of the great circle arc A'B', the tri- angle A'B'C' is the polar triangle of the triangle ABC. 769. The great circles of which A'B', A'C', and B'C' are arcs form by their in- tersections eight spherical triangles. Of these eight triangles that one is the polar of ABC in which the vertex A', homologous to A, lies on the same side of the arc BC as the vertex A, etc. PROPOSITION XIII. THEOREM. 770. Two spherical triangles on the same or equal spheres are equal in area if they have two sides and the included angle, or two angles and the included side, of the one equal, respectively, to the corresponding 2^fts of the other. Proof. If the equal parts of spherical As ABC and A'B'C are arranged in the same order, they are superposable, as hi cases of plane triangles. 91, 94 THE SPHERE. 93 i If the equal parts are arranged in reverse order, the A ABC and the symmetrical A of A'B'C' will be superposable. But A A'B'C' and its symmetrical A are equal in area. 767 .*. A ABC and A A'B'C 1 are equal in area. Q.E.D. PROPOSITION XIV. THEOREM. 771. If one spherical triangle is the polar of another, then the second spherical triangle is the polar of the first. Given A A'B'C', the polar of ABC. To prove A ABC is the polar of A A'B' C*. Proof. Since A is the pole of B'C 1 , C is the pole of A'B'. 768 .*. B r is a quadrant's distance from A and C. 741 .'. B' is the pole of AC. Similarly, A' and C 1 are the poles of BC and AB, respec- tively. . . A ABC is the polar A of A A'B'C'. Q. B. D. Ex. 619. The radius of a sphere is given, J?, and the radius of a small circle of the sphere, r. Find the distance of the plane of the circle from the center of the sphere. Ex. 620. The base of a pyramid is a triangle whose sides are, respec- tively, 17 ft., 25ft., and 28 ft., and the altitude is 18 ft. Find the volume. Ex. 621. The base of a regular pyramid is a square whose side is 10 ft. and the lateral edge of the pyramid is 24 ft. Find the volume. 94 BOOK VIII. SOLID GEOMETRY. PROPOSITION XV. THEOREM. 772. In two polar triangles each angle of the one is the supplement of the side of the other of which it is the pole. Given the polar spherical triangles ABC and ATJ'C', A being the pole of B'C', etc. To prove %A + arc B'O = 180. Proof. Produce the sides of ^ A to meet B'C 1 at D and E. % A is measured by arc DE. 751 Since B' is the pole of arc AE and O of arc AD, arcs B'E and C'D are both quadrants. 741 But DE + B'C'=B'E+C'D = 180 , i.e. DE + B'C 1 = 180. c.B'C' = 180 . Q.E.D. PROPOSITION XVI. THEOREM. 773. The sum of the angles of a spherical triangle is less than 540 and greater than 180. Given a spherical A ABC, THE SPHERE. 95 To prove %A+%B+%C< 540 and > 180. Proof. Construct spherical A A'B'C 1 , the polar A of A ABO, and denote the sides B'C', A'C', A'B' } expressed in degrees by a, b, and c, respectively. Then A = 180 - a, 5 = 180 -&, and C = 180-c. 772 J5+ <7< 540. But a + 5 + c < 360. 762 Q.E.D. 774. COB. ^1 spherical triangle may have one, two, or three right angles ; also one, two, or three obtuse angles. 775. DEF. A birectangular spherical triangle is one that contains two right angles. 776. DEF. A trirectangular spherical triangle is one that contains three right angles. 777. DEF. The spherical excess of a spherical triangle is the difference between the sum of its angles and 180. 778. COB. 1. In a birectangular spherical triangle the sides opposite the right angles are quadrants, and the side opposite the third angle measures that angle. 779. COR. 2. Each side of a trirectangular triangle is a quadrant. 780. COR. 3. If three planes lie passed through the center of a sphere each perpendicular to the other two, they divide the sur- face of the sphere into eight congruent trirectangular spherical triangles. 96 BOOK VIII. SOLID GEOMETRY. PROPOSITION XVII. THEOREM. 781. Two mutually equiangular spherical triangles on the same sphere or equal spheres are mutually equi- lateral, and are either congruent or symmetrical. Given the spherical As ABC and DEF mutually equiangular, on equal spheres. To prove As ABC and DEF are mutually equilateral, and are either congruent or symmetrical. Proof. Let A'B'C' and D'E'F 1 be the polar As of ABC and DEF, respectively. Then because A ABC and A DEF are mutually equiangular, A A'B'C' and A D'E'F' are mutually equilateral. 772 /. As^4'.B'C" and D'E'F' are mutually equiangular. 765 But ABC is the polar triangle of A'B'C' and DEF is the polar triangle of D'E'F'. 771 /.As ABC and DEF are mutually equilateral. 772 /.As ABC and DEF are congruent or symmetrical. 765 Q.E. D. 782. COR. If two spherical triangles on the same or equal spheres are mutually equiangular, they are congruent if their equal parts are arranged in the same order, or symmetrical if their equal parts are arranged in reverse order. THE SPHERE. 97 PROPOSITION XVIII. THEOREM. 783. In an isosceles spherical triangle the angles opposite the equal sides are equal. Given the spherical triangle ABC, in which AB = AC. To prove y.B=y.C. Proof. Let AD, the arc of a great O, be drawn from C to D, the mid-point of the arc BC. Then As ABD and ACD are mutually equilateral. .*. As ABD and ACD are mutually equiangular 765 = a MEASUREMENT OF SPHERICAL SURFACES. 784. DBF. A lune is a portion of the surface of a sphere bounded by two great semicircles. 785. The angle of a lune is the spherical angle between the semicircles that bound it. It is evident that lunes on the same sphere are congruent if their angles are equal. Q.E.D. 98 BOOK VIII. SOLID GEOMETRY. 786. DEF. A zone is a portion of the surface of a sphere included between two parallel planes. 787. DEF. The common sections of the sphere and the planes are the bases of the zone, and the perpendicular distance be- tween the planes is the altitude of the zone. PROPOSITION XIX. THEOREM. 788. The area generated ~by the revolution of a straight line about an axis in its plane is equal to the product of the projection of the line on the axis by the circle whose radius is the perpendicular from the mid-point of the line terminated by the axis. Given AB, a line revolving about the line PQ in its plane, M, the mid-point of AB, DF, the projection of AB on PQ, and MO _L AB. To prove the area of surface generated by AB = DF X 2 vMO. Proof. Draw ME _L and AK II PQ. The area generated by AB is the lateral area of the frustum of a cone of revolution whose slant height is AB and altitude DF. .'. area AB = AB x 2 *ME. 716 THE SPHERE. 99 A ABK ~ A EOM. 370 .\AB :MO = AK:ME. Hence AB x ME = MO X AK= MO x DF. .-. area AB = DF x 2 irMO. Hence, if AB meets PQ or is II PQ, the result is the same. 712, 689 Q. E. D. PROPOSITION XX. THEOREM. 789. The area of the surface of a sphere is equal to the product of its diameter by a great circle. Let the sphere be generated by the revolution of a semicircle about the diameter PQ. Let be the center of the sphere, and let its radius OM be denoted by R and the surface by S. To prove S = PQx2-n-R. Proof. Let PA, AB, BC, CQ, be equal chords of the O. Draw BO, then BO _L PQ. 240 Draw AD, CF J_ PQ. Draw OM _L PA. OM bisects PA. 240 Then the area generated by PA = PD x 2 -n-OM. Similarly, the area generated by AB = DO x 2 irOM, etc. 788 100 v BOOK VIII. SOLID GEOMETRY. But the sum of the projections of PA, AB, etc., on PQ = PQ, the diameter. .'. the surface generated by the polygon PABCQ = PQ x Now, let the number of sides of the polygon be indefinitely increased, then the perimeter will approach the semicircle on PQ as a limit and OM will approach R as a limit. 488, 486 .'. the surface generated by the revolution of the polygo - will approach the surface of the sphere as a limit. .-.S = PQx2TrR. Q.K.D. 790. COR. 1. Since PQ = 2R,S = irR 2 . 791. COR. 2. The area of the surface of a sphere is equal to the area of four great circles. 792. COR. 3. The areas of the surfaces of two spheres are to each other as the squares of their radii or the squares of their diameters. 793. COR. 4. TJie area of a zone is equal to the product of its altitude by a great circle; Z=2 irRH: 794. COR. 5. TJie areas of zones on the same sphere or equal spheres are to each other as their altitudes. Ex. 622. Prove that the surface of a Sphere is equal to the lateral surface of the circumscribed cylinder. Ex. 623. The lateral edge of a regular pyramid is 73 ft. and its alti- tude 65 ft., the base of the pyramid being a square. Find its volume. Ex. 624. Find the volume of a truncated right triangular prism, the sides of the base being, respectively, 33 in., 34 in., and 65 in., and the lateral edges, respectively, 18 in., 21 in., and 27 in. Ex. 625. Find the capacity in bushels of a bin 12 ft. long, 10 ft. wide, 8 ft. high, a bushel being 2150.42 cu. in. Ex. 626. A zone whose altitude is 16 in. is one-third the surface of the sphere. What is the radius of the sphere ? THE SPHERE. 101 PROPOSITION XXL THEOREM. 795. Tlie area of a lune is to the area of the surface of a sphere as the number of degrees in its angle is to 360. B Given ACBE, a lune on the surface of the sphere whose center is 0. Let A denote the angle of the lune, L the area of the lune, and S the surface of the sphere. To prove L:S::A: 360. Proof. Let CEDF be the great O whose pole is A. CE measures A. 751 /. CE : O CEDF = A : 360. 287 If CE and O CEDF are commensurable, let their common measure be contained in CE m times and in O CEDF n times. Then arc CE : O CEDF= m : n. ..%A:360 = m:n. 287 By passing great Os through the points of division of the arc CE and the O CEDF, the arcs will divide the surface of the sphere into n equal lunes of which the lune ACBE will contain m. -re . . Jj : to = m : n. ..L:S = A:3W. If CE and O CDEF are incommensurable, by the method of limits as used in 283 the same conclusion is reached. Q E. D, 102 BOOK VIII. SOLID GEOMETRY. 796. COR. 1. The areas of two limes on the same or equal spheres are to each other as their angles. 797. COB. 2. If the right angle is the unit of angle and the trirectangular triangle the unit of surface, the area of the surface of the sphere being eight trirectangular triangles ( 780), then L : 8 = A : 4, or L = 2A. That is, the measure of the area of a lune is twice its angle. PROPOSITION XXII. THEOREM. 798. If the unit angle is the right angle and the unit surface the area of the trirectangular triangle, the area of a spherical triangle is equal to its spherical excess. Given the spherical A ABC. To prove area A ABC = % A + % B+%. C -2, the right A being the unit "%., and the trirectangular A being the unit surface. Proof. Complete the O of which BC is an arc, and let AB and AC intersect it again at B' and A'. Then, since As ABC and AB'C together form the lune whose angle is B, area A ABC + area A AB'C = 2%B. 796 Similarly, area A ABC + area A ABC 1 = THE SPHERE. 103 Also As ABC and AB'C' together form the lune whose angle is A. .-, area A ABC + area A AB'C' = 2%. A. .-. 2 A ABC + (A ABC + A AB'C+ A ABC' + A AB'C 1 ) But A ABC + A AB'C + A ABC' + A AB'C' make up the surface of a hemisphere. .-. A ABC + A AB'C + A ABC' + A AB'C' = 4 trirectangu- lar As. 779 .-. 2 A ABC + 4 = 2( ^+ .B + (7). Hence area A ABC = % A + % B+ % C 2. Q.E.D. PROPOSITION XXIII. THEOREM. 799. If the right angle is the angular unit and the trirectangular triangle the unit of surface, the area of a spherical polygon is equal to the sum of its angles diminished by the number of its sides less two. Given ABCDE. a spherical polygon of n sides. To prove area ABCDE = % A + B + -2(n-2). Proof. Draw all possible diagonals from the vertex A. These will divide the spherical polygon into (n 2) spheri- cal As. The area of each spherical A = the sum of its 2s less two. 798 104 BOOK VIII. SOLID GEOMETRY. /. area ABODE = -2(n-2). .Q.E.D. 800. DBF. The spherical excess of a spherical polygon- is the spherical excess of the triangles into which its diagonals divide it. 801. COR. The area of a spherical polygon is to the area of the sphere as the spherical excess of the polygon expressed in degrees is to 720. SPHERICAL VOLUMES. 802. DBF. A spherical pyramid is a portion of a sphere bounded by a spherical polygon and the faces of the corre- sponding polyhedral angle. 803. DBF. The spherical polygon is called the base of the pyramid. 804. DEF. A spherical sector is the vol- ume generated by the revolution of a circular sector about the diameter of the circle of which the sector is a part. 805. DEF. The base of a spherical sector is the zone gener- ated by the arc of the circular sector. 806. DEF. A spherical segment is a por- tion of a sphere bounded by two parallel planes. 807. DEF. The bases of a spherical seg- ment are the sections of the sphere made by the parallel planes. 808. DEF. The altitude of a spherical segment is the per- pendicular distance between its bases. If one of the planes is tangent to the sphere, the segment is called a segment of one base. THE SPHERE. 105 PROPOSITION XXIV. THEOREM. 809. The volume of a sphere is equal to the product of the area of its surface by its radius. Given V, the volume, S, the area of the surface, and R, the radius of a sphere. To prove V=\RS. Proof. Suppose the surface of the sphere to be divided into any number of congruent spherical polygons. Then let pyramids be formed by joining the vertices of the polygons successively and drawing radii of the sphere to the vertices. It is obvious that these pyramids will be congruent and will have equal altitudes. The volume of each pyramid is equal to its base by one third its altitude. 655 Therefore, the sum of the volumes of the pyramids is equal to the sum of the bases multiplied by one third the common altitude. Then let the number of spherical polygons be indefinitely increased. Then the sum of the bases of the pyramids will approach the surface of the sphere as a limit. .'. V=%RS. Q.B.D. Formula: V= 106 BOOK VIII. SOLID GEOMETRY. 810. COR. 1. The volume of a spherical pyramid is equal to the product of its base by one third its altitude. 811. COR. 2. TJie volumes of two spheres are to each other as the cubes of their radii or as the cubes of their diameters. 812. The volume of a spherical sector is equal to one third the area of the zone which forms its base multiplied by the radius of the sphere. For, if Z denote the area of the zone, H its altitude, R the radius of the sphere, and Fthe volume of the sector, then Z= 2 TrRH( 793), and F= 2 TrRH x R = f irR*H. PROPOSITION XXV. PROBLEM. 813. To find the volume of a spherical segment. Given a spherical segment generated by the revolution of arc ABCD about PQ as an axis, being the center of the sphere. Draw OB and OD. Denote the radius OD by R, AB by r, CD by r', and the volume of the segment by V. The volume generated by ABCD is equal to the spherical sector DBO + the cone generated by ODC the cone gen- erated by BAO. THE SPHERE. 107 Hence F= | TrR*H + 1 TrCL? x CO - TT Jj? x .40 812, 717 == | TT [2 72 2 # + (72 2 - CO 2 ) CO - ( 2 - A 0^0] = ^ ,-[2 R 2 H+R\CO -AO) - (CO 3 - ZO 3 )]. Factoring CO 3 ^1O 3 , and substituting H for its equal, CO - AO, V= % TT [2 ^.ff + .B 2 ^- JT(CO 2 + CO x AO - ^O 2 )] = ^7rlf[3^ 2 -(C0 2 + COx^O + Z0 2 )]. (1) But H* and CO + CO X .40 + -4( Z0 2 \ J Substituting this value in (1), 814. COR. Jn a spherical segment of one base, r' = 0. 108 BOOK VIII. SOLID GEOMETRY. Ex. 627. The section of a tunnel being a semicircle whose diameter is 50 ft., how many cubic feet of earth is removed in excavating the tun- nel 600 ft. ? Ex. 628. Find the volume of a sphere whose radius is 14 ft. Ex. 629. Find the surface of a sphere whose radius is 22 ft. Ex. 630. Find the radius of the sphere whose surface contains the same number of units of surface that its volume contains of units of volume. Ex. 631. What part of the surface of the sphere is the lune whose angle is 80 ? Ex. 632. How many spheres 2 ft. in diameter are required to equal in volume one sphere 6 ft. in diameter ? Ex. 633. If a sphere of iron 1 ft. in diameter weigh 230 lb., what would be the weight of a spherical iron shell whose inner diameter is 3 ft. and outer diameter 4 ft. ? Ex. 634. If the area of a lune whose angle is 30 is 200 sq. ft., what is the volume of the sphere ? Ex. 635. If the area of a lune whose angle is 60 is 360 sq. ft., what is the surface of the sphere ? Ex. 636. On a sphere whose surface is 600 sq. ft. the area of a lune is 200 sq. ft. What is the angle of the lune ? Ex. 637. What is the area of a spherical triangle whose angles are, respectively, 80, 100, and 130, if the surface of the sphere is 100 sq. ft.? Ex. 638. What is the area of a spherical triangle on a sphere whose radius is 30 yd., if the angles of the triangle are, respectively, 120, 130, and 140 ? Ex. 639. If the volume of a sphere is given 972 cu. ft., what is the area of a spherical triangle whose angles are, respectively, 140, 150, and 160? Ex. 640. What is the area of the zone whose altitude is 12 ft. on a sphere whose radius is 26 ft. ? Ex. 641. What part of the surface of the sphere is a zone whose altitude is one-third the diameter ? Ex. 642. If the radius of the earth be 4000 mi. and the altitude of the torrid zone 3200 mi., what part of the surface of the earth is the torrid zone ? THE SPHERE. 109 Ex. 643. The sides of a spherical triangle are 80, 110, and 120. What is the area of its polar triangle, the surface of the sphere being 144 sq. ft. ? Ex. 644. Find the area of the zone of a sphere of radius E, illumi- nated by a light at a distance d from the surface of the sphere. Ex. 645. At what distance from the surface of a sphere whose radius is 20 ft. must a light be placed so as to illuminate one-eighth its surface ? Ex. 646. What is the area of a spherical pentagon whose angles are, respectively, 80, 100, 130, 150, and 160, on a sphere whose radius is 18 ft. ? Ex. 647. If a light 20 ft. from the surface of a sphere illuminates one- eighth its surface, what is the volume of the sphere ? Ex. 648. A ball 3 in. in diameter is dropped into a conical glass 8 in. high and 5 in. in diameter at the top. What part of the volume of the glass does tLe ball occupy ? Ex. 649. What is the area of the surface of a spherical polygon of four sides, the angles being, respectively, 125, 135, 145, and 155, the diameter of the sphere being 80 ft. ? Ex. 650. What is the area of the section 8 in. from the center of a sphere whose radius is 17 in. ? Ex. 651. Find the volume of a spherical segment of one base which is 21 in. from the center of the sphere whose radius is 29 in. Ex. 652. The altitude of a zone is 6 ft. and its area 30 sq. ft. Find the area of a lune whose angle is GO 3 on the same sphere. Ex. 653. If the angle of a lune is 72 and equal to a zone with alti- tude 4 ft. on the same sphere, what is the diameter of the sphere ? Ex. 654. If a sphere is divided into two segments, the altitude of one being 2 ft. and the other 4 ft., what is the ratio of their volumes ? Ex. 655. The angles of a spherical triangle are, respectively, 130", 135, 140 ; its area is equivalent to that of a lune on the same sphere whose angle is how many degrees ? Ex. 656. Considering the moon as a sphere of diameter of 2160 mi., and whose surface as 240,000 mi. from the earth, what part of the surface of the moon can be seen ? Ex. 657. In a sphere whose radius is fi, what is the altitude of the zone whose area is equal to the area of a great circle of the sphere ? 110 BOOK VIII. SOLID GEOMETRY. Ex. 658. If a zone of one base is the mean proportional between the remainder of the surface of the sphere and the entire surface of the sphere, what is the distance of the base of the zone from the center of the sphere ? Ex. 659. A spherical triangle is one-tenth the area of the surface of the sphere. Two of its angles are right angles. How many degrees in the third angle ? Ex. 660. What fraction of the surface of the sphere is the spherical triangle whose angles are 100, 105, and 115, respectively? Ex. 661. What is the ratio of the area of a lune whose angle is 100 to that of an equiangular triangle on the same sphere, each of whose angles is 100? Ex. 662. The angles of a spherical quadrilateral are, respectively, 120, 130, 140, 150. Find the angle of an equivalent lune on the same sphere. Ex. 663. The volume of a sphere is to the volume of the circumscribed cube as TT is to 6. Ex. 664. The diameters of two spheres are to each other as 7 to 8 What is the ratio of their volumes ? Ex. 665. Prove that the volume of a sphere is two-thirds the volume of the circumscribing cylinder. Ex. 666. Prove that the surface of a sphere is two-thirds the entire surface of the circumscribing cylinder. Qpt 45 Efl University of California SOUTHERN REGIONAL LIBRARY FACILITY 405 Hilgard Avenue, Los Angeles, CA 90024-1388 Return this material to the library from which it was borrowed. t-0n JAN 2 3 95 3 1205 00434 5045