GIFT OF. SOLID GEOMETRY BY C. A. HART INSTRUCTOR OF MATHEMATICS, WADLEI^H HIGH SCHOOL, NEW YORK CITY AND DANIEL D. FELDMAN HEAD OF DEPARTMENT OF MATHEMATICS, ERASMUS HALL HIGH SCHOOL, BROOKLYN WITH THE EDITORIAL COOPERATION OF J. H. TANNER AND VIRGIL SNYDER PROFESSORS OF MATHEMATICS IN CORNELL UNIVERSITY NEW YORK : CINCINNATI : CHICAGO AMERICAN BOOK COMPANY \[ COPYRIGHT, 1912, BY AMERICAN BOOK COMPANY. ENTERED AT STATIONERS' HALL, LONDON. H.-F. SOLID GEOMETRY. W. P. I PREFACE IN addition to the features of the Plane Geometry, which are emphasized in the Solid as well, the chief characteristic of this book is the establishment, at every point, of the vital relation between the Solid and the Plane Geometry. Many theorems in Solid Geometry have been proved, and many problems have been solved, by reducing them to a plane, and simply applying the corresponding principle of Plane Geometry. Again, many proofs of Plane Geometry have been made to serve as proofs of corresponding theorems in Solid Geometry by merely mak- ing the proper changes in terms used. (See 703, 786, 794, 813, 853, 924, 951, 955, 961, etc.) Other special features of the book may be summarized as follows : The student is given every possible aid in forming his early space concepts. In the early work in Solid Geometry, the average student experiences difficulty in fully comprehending space relations ; that is, in seeing geometric figures in space. The student is aided in o/e.Toroing this difficulty by the intro- duction of many easy and 1 practical questi6ris j and exercises, as well as by being encouraged to make his frgirrae. (See 605.) As a further aid in this direction, reproductions of models made by students themselves are shown in a group (p. 302), and at various points throughout Book VI. The student's fund of knowledge is constantly drawn upon. In the many questions, suggestions, and exercises, his knowledge of the things about him has been constantly appealed to. Especially is this true pf the work on the sphere, where the sa 249124 iv PREFACE student's knowledge of mathematical geography has been ap- pealed to in making clear the terms and the relations of figures connected with the sphere. The treatment of the Solid Geometry is logical. The same logical rigor that characterizes the demonstrations in the Plane Geometry is used consistently throughout the Solid. If a pos- tulate is needed to make a proof complete, it is clearly stated, as in 615. In the mensuration of the prism and the pyra- mid, the same general plan has been followed as that used in Book IV; in the mensuration of the cylinder, the cone, and the sphere, the method pursued is similar to that used in the mensuration of the circle. ' More proofs and parts of proofs are left to the student in the Solid, than in the Plane Geometry ; but in every case in which the proof is not complete, the incompleteness is specifically stated. TJie treatment of the polyhedral angle (p. 336), of the prism (p. 345), and of the pyramid (p. 350), is similar to that of the cylinder and the cone. This is in accordance with the recom- mendations of the leading Mathematical Associations through- out the country. The complete collection of formulas of Solid Geometry at the end of the book, it is hoped, will be found helpful to teacher and student alike. The grateful acknowledgment of the authors is due to many friends for helpful ' suggestions ; especially to Miss Grace A. Brace, of the Wadleigk 'High School, New York ; to Mr. Edward B. PArsJofus: of th-3 .Boys' High School, Brooklyn ; and to Professor'Mc'Mahbri, of Cornell University. CONTENTS SOLID GEOMETRY PAGE SYMBOLS AND ABBREVIATIONS . . . . . vi REFERENCES TO PLANE GEOMETRY vii BOOK VI. LINES, PLANES, AND ANGLES IN SPACE . 299 Lines and Planes ......... 301 Dihedral Angles 322 Polyhedral Angles 336 BOOK VII. POLYHEDRONS 343 Prisms 345 Pyramids 350 Mensuration of the Prism and Pyramid 354 Areas .354 Volumes . . t 358 Miscellaneous Exercises ........ 381 BOOK VIII. CYLINDERS AND CONES 383 Cylinders 383 Cones . 388 Mensuration of the Cylinder and Cone ..... 392 Areas 392 Volumes 406 Miscellaneous Exercises 414 BOOK IX. THE SPHERE 417 Lines and Planes Tangent to a Sphere 424 Spherical Polygons . . .429 Mensuration of the Sphere ........ 444 Areas ........... 444 Volumes 457 Miscellaneous Exercises on Solid Geometry .... 467 FORMULAS OF SOLID GEOMETRY 471 APPENDIX TO SOLID GEOMETRY 474 Spherical Segments ......... 474 The Prismatoid . . , ....... 475 Similar Polyhedrons . . . . . . . . 477 INDEX 481 v SYMBOLS AND ABBREVIATIONS = equals, equal to, is equal to. rt. right. = does not equal. str. straight. > greater than, is greater than. ext. exterior. < less than, is less than. int. interior. ~ equivalent, equivalent to, is equiva- alt. alternate, lent to. def. definition. ~ similar, similar to, is similar to. ax. axiom. S2 is measured by. post. postulate. _L perpendicular, perpendicular to, is hyp. hypothesis, perpendicular to. prop, proposition. Js perpendiculars. prob. problem. || parallel, parallel to, is parallel to. th. theorem. ||s parallels. cor. corollary. . . . and so on (sign of continuation). cons, construction. since. ex. exercise. .. therefore. fig. figure. ~ arc ; AB, arc AB. id en. identity. O, [U parallelogram, parallelograms. com P- complementary. O, circle, circles. sup. supplementary. Z, A angle, angles. ad J- adjacent. A, & triangle, triangles. homol. homologous. Q.E.D. Quod erat demonstrandum, which was to be proved. Q.E.F. Quod erat faciendum, which was to be done. The signs -fj , x , -f- have the same meanings as in algebra. vi REFERENCES TO THE PLANE GEOMETRY Note. The following definitions, theorems, etc., from the Plane Geometry which are referred to in the Solid Geometry are here collected for the convenience of the student. (The numbers below refer to articles in the Plane Geometry.) 18. Def. Two geometric figures are equal if they can be made to coincide. 26. Two intersecting straight lines can have only one point in common; i.e. two intersecting straight lines determine a point. 34. Def. A plane surface (or plane) is a surface of unlimited extent such that whatever two of its points are taken, a straight line joining them will lie wholly in the surface. ASSUMPTIONS 54. 1. Things equal to the same thing, or to equal things, are equal to each other. 2. If equals are added to equals, the sums are equal. 3. If equals are subtracted from equals, the remainders are equal. 4. If equals are added to unequal s, the sums are unequal in the same order. 5. If equals are subtracted from unequals, the remainders are unequal in the same order. 6. If unequals are subtracted from equals, the remainders are unequal in the reverse order. 7. (a) If equals are multiplied by equals, the products are equal ; (5) if unequals are multiplied by equals, the products are unequal in the same order. 8. (a) If equals are divided by equals, the quotients are equal ; (6) if unequals are divided by equals, the quotients are unequal in the same order. 9. If unequals are added to unequals, the less to the less and the greater to the greater, the sums are unequal in the same order. 10. If three magnitudes of the same kind are so related that the first is greater than the second, and the second greater than the third, then the first is greater than the third. vii viii REFERENCES TO THE PLANE GEOMETRY 11. The whole is equal to the sum of all its parts. 12. The whole is greater than any of its parts. 13. Like powers of equal numbers are equal, and like roots of equal numbers are equal. 14. Transference postulate. Any geometric figure may be moved from one position to another without change of size or shape. 15. Straight line postulate I. A straight line may be drawn from any one point to any other. 16. Straight line postulate II. A line segment may be pro- longed indefinitely at either end. 17. Revolution postulate. A straight line may revolve in a plane, about a point as a pivot, and when it does revolve continuously from one position to another, it passes once and only once through every intermediate position.. 62. At every point in a straight line there exists only one perpen- dicular to the line. 63. At every point in a straight line there exists one and only one perpendicular to the line. 65. If one straight line meets another straight line, the sum of the two adjacent angles is two right angles. 76. If two adjacent angles are supplementary, their exterior sides are collinear. 77. If two straight lines intersect, the vertical angles are equal. 92. Def. A polygon of three sides is called a triangle; one of four sides, a quadrilateral; one of five sides, a pentagon; one of six sides, a hexagon; and so on. 105. Two triangles are equal if a side and the two adjacent angles of one are equal respectively to a side and the two adjacent angles of the other. 107. Two triangles are equal if two sides and the included angle of one are equal respectively to two sides and the included angle of the other. 110. Homologous parts of equal figures are equal. 111. The base angles of an isosceles triangle are equal. 116. Two triangles are equal if the three sides of one are equal respectively to the three sides of the other, REFERENCES TO THE PLANE GEOMETRY ix 122. Circle postulate. A circle may be constructed having any point as center, and having a radius equal to any finite line. 124. To construct an equilateral triangle, with a given line as side. 134. Every point in the perpendicular bisector of a line is equi- distant from the ends of that line. 139. Every point equidistant from the ends of a line lies in the perpendicular bisector of that line. 142. Two points each equidistant from the ends of a line deter- mine the perpendicular bisector of the line. 148. To construct a perpendicular to a given straight line at a given point in the line. 149. From a point outside a line to construct a perpendicular to the line. 153. If one side of a triangle is prolonged, the exterior angle formed is greater than either of the remote interior angles. 154. From a point outside a line there exists only one perpendic- ular to the line. 156. If two sides of a triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less side. 161. () In the use of the indirect method the student should give, as argument 1, all the suppositions of which the case he is considering admits, including the conclusion. As reason 1 the number of such possible suppositions should be cited. (&) As a reason for the last step in the argument he should state which of these suppositions have been proved false. 167. The sum of any two sides of a triangle is greater than the third side. 168. Any side of a triangle is less than the sum and greater than the difference of the other two. 173. If two triangles have two sides of one equal respectively to two sides of the other, but the third side of the first greater than the third side of the second, then the angle opposite the third side of the first is greater than the angle opposite the third side of the second. 178. Parallel line postulate. Two intersecting straight lines cannot both be parallel to the same straight line. X REFERENCES TO THE PLANE GEOMETRY 179. The following form of this postulate is sometimes more con- venient to quote : Through a given point there exists only one line parallel to a given line. 180. If two straight lines are parallel to a third straight line, they are parallel to each other. 187. If two straight lines are perpendicular to a third straight line, they are parallel to each other. 190. If two parallel lines are cut by a transversal, the correspond- ing angles are equal. 192. If two parallel lines are cut by a transversal, the sum of the two interior angles on the same side of the transversal is two right angles. 193. A straight line perpendicular to one of two parallels is per- pendicular to the other also. 194. If two straight lines are cut by a transversal making the sum of the two interior angles on the same side of the transversal not equal to two right angles, the lines are not parallel. 198. Two angles whose sides are parallel, each to each, are either equal or supplementary. 206. In a triangle there can be but one right angle or one obtuse angle. 209. Two right triangles are equal if the hypotenuse and an acute angle of one are equal respectively to the hypotenuse and an acute angle of the other. 211. Two right triangles are equal if the hypotenuse and a side of one -are equal respectively to the hypotenuse and a side of the other. 215. An exterior angle of a triangle is equal to the sum of the two remote interior angles. ^216. The sum of all the angles of any polygon is twice as many right angles as the polygon has sides, less four right angles. 220. Def. A parallelogram is a quadrilateral whose opposite sides are parallel. 228. Def. Any side of a parallelogram may be regarded as its base, and the line drawn perpendicular to the base from any point in the opposite side is then the altitude. 232. The opposite sides of a parallelogram are equal. REFERENCES TO THE PLANE GEOMETRY xi 234. Parallel lines intercepted between the same parallel lines are equal. 240. If two opposite sides of a quadrilateral are equal and par- allel, the figure is a parallelogram. 252. The two perpendiculars to the sides of an angle from any point in its bisector are equal. 253. Every point in the bisector of an angle is equidistant from the sides of the angle. 258. The bisectors of the angles of a triangle are concurrent in a point which is equidistant from the three sides of the triangle. 276. Def. A circle is a plane closed figure whose boundary is a curve such that all straight lines to it from a fixed point within are equal. 279. (a) All radii of the same circle are equal. (&) All radii of equal circles are equal, (c) All circles having equal radii are equal. 297. Four right angles contain 360, angle degrees, and four right angles at the center of a circle intercept a complete circumference ; therefore, a circumference contains 360 arc degrees. Hence, a semi- circumference contains 180 arc degrees. 298. In equal circles, or in the same circle, if two chords are equal, they subtend equal arcs ; conversely, if two arcs are equal, the chords that subtend them are equal. 307. In equal circles, or in the same circle, if two chords are equal, they are equally distant from the center ; conversely, if two chords are equally distant from the center, they are' equal. 308. In equal circles, or in the same circle, if two chords are unequal, the greater chord is at the less distance from the center. 309. Note. The student should always give the full statement of the substitution made ; for example, " Substituting A E for its equal CD." 310. In equal circles, or in the same circle, if two chords are une- qually distant from the center, the chord at the less distance is the greater. 313. A tangent to a circle is perpendicular to the radius drawn to the point of tangency. REFERENCES TO THE PLANE GEOMETRY 314. A straight line perpendicular to a radius at its outer ex- tremity is tangent to the circle. 321. To inscribe a circle in a given triangle. 323. To circumscribe a circle about a given triangle. 324. Three points not in the same straight line determine a circle. 328. If two circumferences intersect, their line of centers bisects their common chord at right angles. 335. Def. To measure a quantity is to find how many times it contains another quantity of the same kind. The result of the measurement is a number and is called the numerical measure, or measure-number, of the quantity which is measured. The measure employed is called the unit of measure. 337. Def. Two quantities are commensurable if there exists a measure that is contained an integral number of times in each. Such a measure is called a common measure of the two quantities. 339. Def. Two quantities are incommensurable if there exists no measure that is contained an integral number of times in each. 341. Def. The ratio of two geometric magnitudes may be defined as the quotient of their measure-numbers, when the same measure is applied to each. 349. Def. If a variable approaches a constant in such a way that the difference between the variable and the constant may be made to become and remain smaller than any fixed number previously assigned, however small, the constant is called the limit of the variable. 355. If two variables are always equal, and if each approaches a limit, then their limits are equal. 358. An angle at the center of a circle is measured by its inter- cepted arc. 362. (a) In equal circles, or in the same circle, equal angles are measured by equal arcs ; conversely, equal arcs measure equal angles. (I) The measure of the -) , J- of two angles is equal to J difference v the } , > of the measures of the angles. J difference ( REFERENCES TO THE PLANE GEOMETRY xiii (c) The measure of any multiple of an angle is equal to that same multiple of the measure of the angle. 373. To construct a tangent to a circle from a point outside. 399. If four numbers are in proportion, they are in proportion by division ; that is, the difference of the first two terms is to the first (or second) term as the difference of the last two terms is to the third (or fourth) term. 401. In a series of equal ratios the sum of any number of antecedents is to the sum of the corresponding consequents as any antecedent is to its consequent. 409. A straight line parallel to one side of a triangle divides the other two sides proportionally. 410. A straight line parallel to one side of a triangle divides the other two sides into segments which are proportional. 419. Def. Two polygons are similar if they are mutually equiangular and if their sides are proportional. 420. Two triangles which are mutually equiangular are similar. 422. Two right triangles are similar if an acute angle of one is equal to an acute angle of the other. 424. (1) Homologous angles of similar triangles are equal. (2) Homologous sides of similar triangles are proportional. (3) Homologous sides of similar triangles are the sides opposite equal angles. 435. In two similar triangles any two homologous altitudes have the same ratio as any two homologous sides. 438. If two polygons are composed of the same number of triangles, similar each to each and similarly placed, the polygons are similar. 443. In a right triangle, if the altitude upon the hypotenuse is drawn : I. The square of the altitude is equal to the product of the seg- ments of the hypotenuse. II. The square of either side is equal to the product of the whole hypotenuse and the segment of the hypotenuse adjacent to that side. 444. If from any point in the circumference of a circle a per- pendicular to a diameter is drawn, and if chords are drawn from the point to the ends of the diameter : xiv REFERENCES TO THE PLANE GEOMETRY I. The perpendicular is a mean proportional between the seg- ments of the diameter. II. Either chord is a mean proportional between the whole diameter and the segment of the diameter adjacent to the chord. 478. The area of a square is equal to the square of its side. 479. Any two rectangles are to each other as the products of their bases and their altitudes. 480. (a) Two rectangles having equal bases are to each other as their altitudes, and (b) two rectangles having equal altitudes are to each other as their bases. 481. The area of a parallelogram equals the product of its base and its altitude. 482. Parallelograms having equal bases and equal altitudes are equivalent. 483. Any two parallelograms are to each other as the products of their bases and their altitudes. 484. (a) Two parallelograms having equal bases are to each other as their altitudes, and (b) two parallelograms having equal altitudes are to each other as their bases. 485. The area of a triangle equals one half the product of its base and its altitude. 491. The area of a triangle is equal to one half the product of its perimeter and the radius of the inscribed circle. 492. The area of any polygon circumscribed about a circle is equal to one half its perimeter multiplied by the radius of the in- scribed circle. 498. Two triangles which have an angle of one equal to an angle of the other are to each other as the products of the sides including the equal angles. 503. Two similar triangles are to each other as the squares of any two homologous sides. 517. If the circumference of a circle is divided into any number of equal arcs : (a) the chords joining the points of division form a regu- lar polygon inscribed in the circle ; (6) tangents drawn at the points of division form a regular polygon circumscribed about the circle. REFERENCES TO THE PLANE GEOMETRY xv 538. The perimeters of two regular polygons of the same number of sides are to each other as their radii or as their apothems. 541. I. The perimeter and area of a regular polygon inscribed in a circle are less, respectively, than the perimeter and area of the regu- lar inscribed polygon of twice as many sides. II. The perimeter and area of a regular polygon circumscribed about a circle are greater, respectively, than the perimeter and area of the regular circumscribed polygon of twice as many sides. 543. By repeatedly doubling the number of sides of a regular poly- gon inscribed in a circle, and making the polygons always regular : I. The apothem can be made to differ from the radius by less than any assigned value. II. The square of the apothem can be made to differ from the square of the radius by less than any assigned value. 546. By repeatedly doubling the munber of sides of regular cir- cumscribed and inscribed potygons of the same number of sides, and making the polygons always regular : I. Their perimeters approach a common limit. II. Their areas approach a common limit. 550. Def. The length of a circumference is the common limit which the successive perimeters of inscribed and circumscribed regu- lar polygons (of 3, 4, 5, etc., sides) approach as the number of sides is successively increased and each side approaches zero as a limit. 556. Any two circumferences are to each other as their radii. 558. Def. The area of a circle is the common limit which the successive areas of inscribed and circumscribed regular polygons ap- proach as the number of sides is successively increased and each side approaches zero as a limit. 559. The area of a circle is equal to one half the product of its circumference and its radius. 586. If a variable can be made less than any assigned value, the quotient of the variable by any constant, except zero, can be made less than any assigned value. 587. If a variable can be made less than any assigned value, the product of that variable and a decreasing value may be made less than any assigned value. xvi REFERENCES r l O THE PLANE GEOMETRY 590. The limit of the product of a variable and a constant, not zero, is the limit of the variable multiplied by the constant. 592. If two variables approach finite limits, not zero, then the limit of their product is equal to the product of their limits. 593. If each of any finite number of variables approaches a finite limit, not zero, then the limit of their product is equal to the product of their limits. 594. If two related variables are such that one is always greater than the other, and if the greater continually decreases while the less continually increases, so that the difference between the two may be made as small as we please, *then the two variables have a common limit which lies between them. 599. An angle can be bisected by only one line. SOLID GEOMETRY BOOK VI LINES, PLANES, AND ANGLES IN SPACE 602. Def . Solid geometry or the geometry of space treats of figures whose parts are not all in the same plane. (For defini- tion of plane or plane surface, see 34.) 603. From the definition of a plane it follows that : \ptylftwo points of a straight line lie in a plane, the whole line lies in that plane. (b) A straight line can intersect a plane in not more than one point. 604. Since a plane is unlimited in its two dimensions (length and breadth) __ only a portion of it can be shown in a figure. This is usually repre- sented by a quad- rilateral drawn as a j parallelogram. Thus MN represents a plane. Sometimes, however, conditions make it necessary to represent a plane by a figure other than a parallelogram, as in 617. Ex. 1152. Draw a rectangle freehand which is supposed to lie : (a) in a vertical plane ; (6) in a horizontal plane. May the four angles of the rectangle of (a) be drawn equal ? those of the rectangle of (6) ? 299 300 ' ' ' SOLID GEOMETRY 605. ' Note. In tlfe'figMres in solid geometry dashed lines will be used to represent all auxiliary lines and lines that are not supposed to be visible but which, for purposes of proof, are represented in the figure. All other lines will be continuous. In the earlier work in solid geometry the stu- dent may experience difficulty in imagining the figures. If so, he may find it a great help, for a time at least, to make the figures. By using pasteboard to represent planes, thin sticks of wood or stiff wires to repre- sent lines perpendicular to a plane, and strings to represent oblique lines, any figure may be actually made with a comparatively small expenditure of time and with practically no expense. For reproductions of models actually made by high school students, see group on p. 302 ; also 622, 633, 678, 756, 762, 770, 797. 606. Assumption 20. Revolution postulate. A plane may revolve about a line in it as an axis, and as it does so revolve, it can contain any particular point in space in one and only one position. 607. From the revolution postulate it follows that : Through a given straight line any number of planes may be For, as plane MN revolves about AB as an axis ( 606) it may occupy an unlimited num- ber of positions each of which will represent a different plane through AB. 608. Def . A plane is said to be determined by given condi- tions if that pfane and no other plane fulfills those conditions. 609. From 607 and 608 it is seen that: A straight line does not determine a plane. Ex. 1153. How many planes may be passed through any two points in space ? why ? Ex. 1154. At a point P in a given straight line AB in space, con- struct a line perpendicular to A B. How many such lines can be drawn ? BOOK VI 301 / LINES AND PLANES ** PROPOSITION I. THEOREM 610. A plane is determined by a straight line and a point not in the line. S A - M R N ^-B Given line AB and P, a point not in AB. To prove that AB and P determine a plane. ARGUMENT 1. Through AB pass any plane, as MN. 2. Kevolve plane MN about AB as an axis until it contains point P. Call the plane in this position RS. 3. Then plane RS contains line AB and REASONS 1. 607. 2. 606. 3. Arg. 2. point P. 4. Furthermore, in no other position can 4. 606. plane MN, in its rotation about AB, contain point P. 5. .. RS is the only plane that can contain 5. Arg. 4. AB and P. 6. .'. AB and P determine a plane. Q.E.D. 6. 608. 611. Cor. 1. 1* A plane is determined by three points in the same straight line. HINT. Let A, B, and C be the three given points. Join A and B by a straight line, and apply 610. 612. Cor. II. A plane is determined by two intersec- ting straight lines. 1 ' * oo oW not 2/" . ^ 302 SOLID GEOMETRY ( f 613. Cor. III. A plane is determined by two parallel straight lines. Ex. 1155. Given line AB in space, and P a point not in AB. Con- struct, through P, a line perpendicular to AB. Ex. 1156. Hold two pencils so that a plane can be passed through them. In how many ways can this be done, assuming that the pencils are lines? why? Ex. 1157. Can two pencils be held so that no plane can be passed through them ? If so, how ? Ex. 1158. In measuring wheat with a half bushel measure, the meas- ure is first heaped, then a straightedge is drawn across the top. Why is the measure then even full ? Ex. 1159. Why is a surveyor's transit or a photographer's camera always supported on three legs rather than on two or four ? Ex. 1160. How many planes are determined by four straight lines, no three of which lie in the same plane, if the four lines intersect : (1) at a common point ? (2) at four different points ? 614. Def. The intersection of two surfaces is the locus of all points common to the two surfaces. 615. Assumption 21. Postulate. Two planes having one point in common also have another point in common. Reproduced from Models made by High School Students BOOK VI PROPOSITION It. THEOREM 303 If two planes intersect, their intersection is a straight line. S N Given intersecting planes UN and RS. To prove the intersection of MN and RS a str. line. ARGUMENT 1. Let A and B be any two points common to the two planes MN and RS. 2. Draw str. line AB. 3. Since both A and B lie in plane MN, str. line AB lies in plane MN. 4. Likewise str. line AB lies in plane RS. 5. Furthermore no point outside of AB can lie in both planes. 6. .. AB is the intersection of planes MN and RS. 7. But AB is a str. line. 8. .-. the intersection of MN and RS is a str. line. Q.E.D. REASONS 1. 615. 2. 54, 15. 3. 603, -a. 4. 603, a. 5. 6J.O. 6. 614. 7. Ar'g.2. .8. Args. 6 and 7. Ex. 1161. Is it possible for more than two planes to intersect in a straight line ? Explain. Ex. 1162. By referring to 26 and 608, give the meaning of the expression, "Two planes determine a straight line." Ex. 1163. Is the statement in Ex. 1162 always true ? Give reasons for your answer. 304 SOLID GEOMETRY PROPOSITION III. THEOREM 617. If three planes, not passing through the same line, intersect each other, their three lines of intersection are concurrent, or else they are parallel, each to each. M N M FIG. 1. FIG. 2. Given planes MQ, PS, and EN intersecting each other in lines MN, PQ, and ES-, also : I. Given MN and ES intersecting at (Fig. 1). To prove MN, PQ, and ES concurrent. ARGUMENT 1. v is in line MN, it lies in plane MQ. 2. v is in line ES, it lies in plane PS. 3. .: O, lying in planes MQ and PS, must lie in their intersection, PQ. 4. .-. PQ passes through 0; i.e. MN, PQ, and ES are concurrent in 0. Q.E.D. II. Given MN II ES (Fig. 2). To prove PQ II MN and ES. ARGUMENT 1. PQ and MN are either II or not II. 2. Suppose that PQ intersects MN ; then MN also intersects ES. 3. But this is impossible, for MN II ES. 4. .-. PQ II MN. 5. Likewise PQ II ES. Q.E.D. REASONS 1. 603, a. 2. 603, a. 3. 614. 4. Arg. 3. REASONS 1. 161, a. 2. 617, I. 3. By hyp. 4. 161, b. 5. By steps sim- ilar to 1-4. BOOK VI 305 Cor. If two straight lines cure parallel to a third straight line, they are parallel to each oilier. Given lines AB and CD, each II EF. To prove AB II CD. ARGUMENT 1. AB and CD are either || or not II. 2. Through AB and EF pass plane AF, and through CD and #F pass plane CF. 3. Pass a third plane through AB and point (7, as plane BC. 4. Suppose that ^45 is not II CD ; then plane BC will intersect plane CF in some line other than (7Z>, as OT. 5. Then CH II ^. 6. But CD II J^F. 7. .*. Off and <7D, two straight lines plane CF, are both II EF. 8. This is impossible. 9. .'. AB II CD. Q.E.D n REASONS 1. 161, a. 2. 613. 3. 610. 4. 613. 5. 617,11. 6. By hyp. 7. Args. 5 and 6. 8. 178. 9. 161, 6. 619. Def. A straight line is perpendicular {o a plane if it is perpendicular to every straight line in the plane passing through the point of intersection of the given line and plane. 620. Def. A plane is perpendicular to a straight line if the line is perpendicular to the plane. ^ 621. Def. If a line is perpendicular to a plane, its point of intersection with the plane is called the foot of the perpendicular. 306 SOLID GEOMETRY PROPOSITION IV. THEOREM 622& 1 If a straight line Is perpendicular to each of two Intersecting straight lines at their point of Inter- section, It Is perpendicular to the plane of those lines. Given str. line FBj_AB and to BC at B, and plane MN con- taining AB and BC. To prove FB J_ plane MN. OUTLINE OF PROOF 1. In plane MN draw AC, through B draw any line, as BH, meeting AC at H. 2. Prolong FB to E so that BE = FB-, draw AF, HF, CF, AE, HE, CE. 3. AB and BC are then _L bisectors of FE ; i.e. FA AE, FC = CE. 4. Prove A AFC = A EAC ; then Z JL4.F = Z .EL4#. 5. Prove A HA F A EAH ; then HF = HE. 6. .-.BH^FE; i.e. FB.BH, any line in plane MN passing through B. 7. .'. FBA.MN. 623. Cor. All the perpendiculars that can be drawn to a straight line at a given point In the line lie in a plane perpendicular to the line at the given point. BOOK VI 307 PROPOSITION V. PROBLEM 624. Through a given point to construct a plane per- pendicular to a given line. FIG. 1. FIG. 2. Given point P and line AB. To construct, through P, a plane _L AB. I. Construction 1. Through line AB and point P pass a plane, as APD (in Fig. 1, any plane through AB}. 607, 610. 2. In plane APD construct PD, through P, _L AB. 148, 149. 3. Through AB pass a second plane, as ABC. 607. 4. In plane ABC, through the foot of PD, construct a J_ to AB (PC in Fig. 1, DC in Fig. 2). 148. 5. Plane MN, determined by C, D, and P, is the plane required. II. The proof is left as an exercise for the student. HINT. Apply 623". III. The discussion will be given in 625. Ex. 1164. Tell how to test whether or not a flag pole is erect. Ex. 1165. Lines AB and CD are each perpendicular to line EF. Are AB and CD necessarily parallel ? Explain. Do they necessarily lie in the same plane ? why or why not ? 308 SOLID GEOMETRY PROPOSITION VI. THEOREM 625. Through a given point there exists only one plane perpendicular to a given line. / ---"" ~7" / "-- / / ** y /* - 1 - -y jf/ / M L 1 f FIG. 1. FIG. 2. Given plane MN 9 through P, *1~AB. To prove 3/JV the only plane through P _L AS. ARGUMENT ONLY 1. Either MN is the only plane through P J_ AB or it is not. 2. In MN draw a line through P intersecting Jine A#, as PR. 3. Let plane determined by AB and Ptf be denoted by APR. 4. Suppose that there exists another plane through P J 4#; let this second plane intersect plane APR in line PS. 5. Then AB J_ P# and also PS; i.e. PR and PS are _L ^/?. 6. This is impossible. 7. .-. MN is the only plane through P J_ ^5. Q.E.D. 626. Question. In Fig. 2, explain why AB J_ P#. 627. 624 and 625 may be combined in one statement: Through a given point there exists one and only one plane per- pendicular to a given line. 628. Cor. I. The locus of all points in space equidis- tant from the extremities of a straight line segment is the plane perpendicular to the segment at its mid-point. 629. Def. A straight line is parallel to a plane if the straight line and the plane cannot meet. 630. Def. A straight line is oblique to a plane if it is neither perpendicular nor parallel to the plane. 631. Def. Two planes are parallel if they cannot meet. BOOK VI 309 / PROPOSITION VII. THEOREM V 632. Two planes perpendicular to the same straight line are parallel. A M' R Given planes MN and RS, each _L line AB. To prove MN II #S. HINT. Use indirect proof. Compare with 187. PROPOSITION VIII. THEOREM 633. If a plane intersects two parallel planes, the lines of intersection are parallel. Given II planes MN and RS, and any plane PQ intersecting MN and RS in AB and CD, respectively. To prove AB II CD. HINT. Show that AB and CD cannot meet. 634. Cor. I. Parallel lines intercepted between the same parallel planes are equal. (HINT. Compare with 234.) Ex. 1166. State the converse of Prop. VIII. Is it true ? SOLID GEOMETRY PROPOSITION IX. THEOREM 635. // two angles, not in the same plane, have their sides parallel respectively, and lying on the same side of the line joining their vertices, they are equal-* R Given Z ABC in plane MN and Z DEF in plane RS with BA and BC II respectively to ED and EF, and lying on the same side of line BE. To prove Z ABC= Z DEF. 1. 2. 3. 4. 5. 6. 7: 8. 9. 10. 11. 12. 13. ARGUMENT Measure off BA = ED and BC = EF. Draw AD, CF, AC, and DF. - BA II ED and BC II ^. Then ADEB and C^J^B are /17. /. AD = BE and CF = BE. .'.AD=CF. Also ^ II BE and OF II #E. .'. AD II C y ^. /. ACFD is a O. But BA = ED and C7= EF. .'.A ABC = A DEF. .'. Z ABC= Z Q.E.D. REASONS 122. 54, 15. 3. By hyp. 4. 240. 232. 54, 1. 220. 618. 240. 232. Arg. 1. 116. 110. 1. 2. 5. 6. 7. 8. 9. 10. 11. 12. 13. Ex. 1167. Prove Prop. IX if the angles lie on opposite sides of l!h\ * It will also be seen ( 645) that the planes of these angles are parallel. BOOK VI PROPOSITION X. THEOREM 311 J 636. If one of two parallel lines is perpendicular to a plane, the other also is perpendicular to the plane. M Given AB II CD and AB _L plane MN. To prove CD J_ plane MN. ARGUMENT 1. Through D draw any line in plane MN, as DF. 2. Through B draw BE in plane JOT II DF. 3. Then Z ABE = Z CZXF. 4. But Z .45.E is a rt. Z. .'. Z CD^ is a rt. Z; ie. CD _L DF, any line in plane MN through D. .'. CD _L plane JfJV. 5. 6. REASONS 1. 54, 15. 2. 179. 3. 635. 4. 619. 5. 54, 1. 6. 619. Ex. 1168. In the accompanying diagram AS and CD lie in the same plane. Angle CBA = 35, angle BCD = 35, angle ABE = 90, BE lying in plane MN. Is CD necessarily perpendicular to plane MN? Prove your answer. Ex. 1169. Can a line be perpendicular to each of two intersecting planes ? Prove. Ex. 1170. If one of two planes is per- pendicular to a given line, but the other is not, the planes are not parallel. Ex. 1171. If a straight line and a plane are each perpendicular to the same straight line, they are parallel to each other. M 812 SOLID GEOMETRY PROPOSITION XI. PROBLEM 637. Through a given point to construct a line per- pendicular to a given plane- FIG. 1. FIG. 2. Given point P and plane 1T#. To construpt, through P, a line _L plane JfJ\T. I. Construction 1. In plane MN draw any convenient line, as AB. 2. Through P construct plane PQ 1_ AB. 624. 3. Let plane PQ intersect plane MN in CD. 616. 4. In plane P Q construct a line through P J_ CD, as PR. \ 148, 149. 5. P# is the perpendicular required. II. Proof ARGUMENT 1. Through the foot of PR (P in Fig. 1, R in Fig. 2) in plane MN, draw EF II AB. 2. AB _L plane PQ. 3. .-. EF plane PQ. 4. .-. EFPR-, i.e. PR J_ tfJ 1 . 5. But PR J_ (7Z). 6. .-. PR JL plane MN. Q.E.D. III. The discussion will be given in 639. REASONS 1. 179. 2. By cons. 3. 636. 4. 619. 5.. By cons. 6. 622. BOOK VI 313 PROPOSITION XII. THEOREM \J 638. Through a given point there exists only one line perpendicular to a given plane. Given point P and line PJ, through P, _L plane MN. To prove PA the only line through P J_ MN. ARGUMENT 1. Either PA is the only line through P _L MN or it is not. 2. Suppose there exists another line through P _L MN, as PB then PA and PB determine a plane. 3. Let this plane intersect plane MN in line CD. 4. Then PA and PB, two lines through P and lying in the same plane, are _L CD. 5. This is impossible. 6. .-. PA is the only line through P _1_ MN. Q.E.D. REASONS 1. 161, a. 2. 612. 3. 616. 4. 619. 5. 62, 154. 6. 161, b. 639. 637 and 638 may be combined in one statement as follows : Through a given point there exists one and only one line per- pendicular to a given plane. Ex. 1172. Find the locus of all points in a plane that are equidistant from two given points not lying in the plane. 314 SOLID GEOMETRY PROPOSITION XIII. THEOREM \^p40. Two straight lines perpendicular to the same plane are parallel. * M 7 Given str. lines AB and CD J_ plane MN. To prove AB II CD. The proof is left as an exercise for the student. HINT. Suppose that AB is not || CZ>, but that some other line through B, as BE, is II CD. Use 638. PROPOSITION XIV. THEOREM 641. If a straight line is parallel to a plane, the in- tersection of the plane with any plane passing through the given line is parallel to the given line. Ar \B Given line AB II plane MN, and plane AD, through AB, inter- secting plane MN in line CD. To prove AB II CD. The proof is left as an exercise for the student. HINT. Suppose that AB is not II CD. Show that AB will then meet plane MN. BOOK VI 315 642. Cor. I. If a plane intersects one of two parallel lines, it must, if suffi- ciently extended, inter- sect the other also. HINT. Pass a plane through AB and CD and let it intersect plane MN in EF. Now if MN does not intersect CD, but is II to it, then EF II CD, 641. Apply 178. D 643. Cor. II. If two in- tersecting lines are each parallel to a given plane, the plane of these lines is parallel to the given plane. HINT. If plane MN, determined by AB and CD, is not I). to plane RS, it will intersect it in some line, as EF. What is the relation of EF to AB and CD ? 644. Cor. III. Prob- lem. Through a given point to construct : (a) A line parallel to a given plane. (V) A plane parallel to a given plane. HINT, (a) Let A be a point outside of plane MN. Through A con- struct any plane intersecting plane MN in line CD. Complete the con- struction. 645. Cor. IV. If two angles, not in the same plane, have their sides parallel respectively, their planes are parallel. Ex. 1173. Hold a pointer parallel to the blackboard. Is its shadow on the blackboard parallel to the pointer ? why ? \ Ex. 1174. Find the locus of all straight lines passing through a given point and parallel to a given plane. 316 SOLID GEOMETRY J PROPOSITION XV. THEOREM 646. If two straight lines are parallel, a plane con- taining one of the lines, and only one, is parallel to the other. - Given II lines AB and CD, and plane MN containing CD. To prove plane MN II AB. ARGUMENT 1. Either plane MN is II A B or it is not. 2. Suppose MN is not II AB ; then plane MN will intersect AB. 3. Then plane JOT must also intersect CD. 4. This is impossible, for MN contains CD. 5. .-. plane MN II AB. Q.E.D. REASONS 1. 161, a. 2. 629. 3. 642. 4. By hyp. 5. 161, 6. 647. Cor. I. Problem. Through a given line to construct a plane parallel to another given line- HINT. Through E, any point in <7Z>, construct a line HK II AB. 648. Cor. II. Problem. Through a given point to con- struct a plane parallel to any two given straight lines in space. BOOK VI 317 PROPOSITION XVI. THEOREM 649. If two straight lines are intersected by three par- allel planes, the corresponding segments of these lines are proportional- Given II planes MN, PQ, and RS intersecting line AB in A, E, B and line CD in (7, H, D, respectively. AE CH To prove = EB HD ARGUMENT 1. Draw AD intersecting plane PQ in F. 2. Let the plane determined by AB and AD intersect PQ in EF and RS in Bf). 3. Let the plane determined by AD and DC intersect PQ in FH and MN in AC. 4. .-. EF II BD and FH II AC. EB AE EB FD CH HD' HD FD Q.E.D. REASONS 54, 15. 612, 616. 3. 612, 616. 4. 633. 5. 410. 6. 54,1. 650. Cor. If two straight lines are intersected by three parallel planes, the lines are divided proportionally. Ex. 1175. If any number of lines passing through a common point are cut by two or more parallel planes, their corresponding segments are proportional. Ex. 1176. In the figure for Prop. XVI^ AE = 6, EB = 8, AD = 21, CD = 28. Find AF and HD. 318 SOLID GEOMETRY J PROPOSITION XVII. THEOREM 651. A straight line perpendicular to one of two par allel planes is perpendicular to the other also. A B Given plane MN II plane RS and line AB _L plane RS. To prove line AB J_ plane MN. ARGUMENT ONLY 1. In plane MN, through (7, draw any line (7Z), and let the plane determined by AC and CD intersect plane RS in EF. 2. Then CD II EF. 3. But AB J_ EF. 4. .. AB J_ CD, any line in plane MN passing through C. 5. /. line AB _L plane MN. Q.E.D. 652. Cor. I. Through a given point there exists only one plane parallel to a given plane. (HINT. Apply 638, 644&, 651, 625.) 653. 6446 and 652 may be combined in one statement : Through a given point there exists one and only one plane parallel to a given plane. 654. Cor. II. If two planes are each parallel to a third plane, they are parallel to each other. (HINT. See 180.) 655. Def. The projection of a point upon a plane is the foot of the perpendicular from the point to the plane. 656. Def. The projection of a line upon a plane is the locus of the projections of all points of the line upon the plane. BOOK VI 319 PROPOSITION XVIII. THEOREM 657. The projection upon a plane of a straight line not perpendicular to the plane is a straight line. Given str. line AB not _L plane MN. To prove the projection of AB upon MN a str. line. ARGUMENT 1. Through C, any point in AB, draw CD _L plane MN. 2. Let the plane determined by AB and CD intersect plane MN in the str. line EF. 3. From H, any point in AB, draw HK, in plane AF, II CD. 4. Then HK _L plane MN. 5. .-. K is the projection of H upon plane. 6. 7. MN. EF is the projection of AB upon plane MN. the projection of is a str. line. upon plane MN Q.E.D. REASONS 1. 639. 2. 612, 616. 3. 179. 4. 636. 5. 655. 6. 656. 7. Args. 2 and 6. Ex. 1177. Compare the length of the projection of a line upon a plane with the length of the line itself : (a) If the line is parallel to the plane. (6) If the line is neither parallel nor perpendicular to the plane, (c) If the line is perpendicular to the plane. 320 SOLID GEOMETRY PROPOSITION XIX. THEOREM 658. Of all oblique lines drawn from a point to a plane : I. Those having equal projections are equal- II. Those having unequal projections are unequal, and the one having the greater projection is the longer. P N Given line PO J_ plane MN and : I. Oblique lines PA and PB with projection OA = projec- tion OB. II. Oblique lines PA and PC with projection OC > projec- tion OA. To prove : I. PB = PA- II. PC > PA. The proof is left as an exercise for the student. 659. Cor. I. (Converse of Prop. XIX). Of all oblique lines drawn from a point to a plane : I. Equal oblique lines have equal projections. II. Unequal oblique lines have unequal projections, and the longer line has the greater projection. 660. Cor. II. The locus of a point in space equidistant from all points in the circumference of a circle is a straight line perpendicular to the plane of the circle and passing through its center. 661. Cor. III. The shortest line from a point to a given plane is tli^e perpendicular from that point to the plane. 662. Def. The distance from a point to a plane is the length of the perpendicular from the point to the plane. BOOK VI 321 663. Cor. IV. Two parallel planes are everywhere equally distant. (HINT. See 634.) 664. Cor. V. If a line is parallel to a plane, all points of the line are equally distant from the plane. Ex. 1178. In the figure of 658, if PO - 12 inches, PA = 15 inches, and PC = 20 inches, find OA and CA'. Ex. 1179. Find the locus of all points in a given plane which are at a given distance from a point outside of the plane. Ex. 1180. By applying 660, suggest a practical method of con- structing a line perpendicular to a plane : (a) Through a point in the plane ; (6) Through a point not in the plane. Ex. 1181. Find a point in a plane equidistant from all points in the circumference of a circle not lying in the plane. Ex. 1182. Find the locus of all points equidistant from two parallel planes. Ex. 1183. Find the locus of all points at a given distance d from a given plane MN. Ex. 1184. Find the locus of all points in space equidistant from two parallel planes and equidistant from two fixed points. Ex. 1185. A line and its projection upon a plane always lie in the same plane. Ex. 1186. (a) The acute angle that a straight line makes with its own projection upon a plane is the least angle j that it makes with any line passing K through its foot in the plane. |\\ (/>) With what line passing through / \~\\B~~E7 its foot and lying in the plane does it / C 1 \ ) / make the greatest angle ? /_ *# / HINT, (a) Measure off BD = BC. *f Which is greater, AD or AC ? By means of 173, prove Z ABC < ZABD. 665. Def. The acute angle that a straight line, not perpen- dicular to a given plane, makes with its own projection upon the plane, is called the inclination of the line to the plane. Ex. 1187. Find the projection of a line 12 inches long upon a plane, if the inclination of the line to the plane is 30 ; 45 ; 60. 322 SOLID GEOMETRY DIHEDRAL ANGLES 666. Defs. A dihedral angle is the figure formed by two planes that diverge from a line. The planes forming a di- hedral angle are called --its faces, and the intersection of these planes, its edge. 667. A dihedral angle may be designated by reading in order the two planes forming the angle; thus, an angle formed by planes AB and CD is angle AB-CD, and is usually written angle A-BC-D. If there is no other dihedral angle having the same edge, the line forming the edge is a sufficient designation, as dihedral angle BC. 668. Def. Points, lines, or planes lying in the same plane are said to be coplanar. 669. A clear notion of the magnitude of a dihedral angle may be obtained by imagining that its two faces, considered as finite portions of planes, were at first coplanar and that one of them has revolved about a line com- mon to the two. Thus in the figure we may imagine face CD first to have been in the position of face AB and then to have revolved about BC as an axis to the position of face CD. 670. Def. The plane angle of a dihedral angle is the angle formed by two straight lines, one in each face of the dihedral angle, perpendicular to its edge at the same point. Thus if EF, in face AB, is _L BC at F, and FH, in face CD, is _L BC at F, then Z EFHis the plane -Z of the dihedral Z A-BC-D. Ex. 1188. All plane angles of a dihedral angle are equal. Ex. 1189. Is the plane of angle EFH ( 667) perpendicular to the edge BC? Prove. State your result in the form of a theorem. Ex. 1190. Is Ex. 309 true if the quadrilateral is a quadrilateral in space, i.e. if the vertices of the quadrilateral are not all in the same plane ? Prove. LOOK VI 323 671. Def. Two dihedral angles are adjacent if they have a common edge and a common face which lies between them ; thus Z A-BC-D and Z D-CB-E are adj. dihe- dral Zs. 672. Def. If one plane meets another so as to make two adjacent dihe- dral angles equal, each of these angles is a right dihedral angle, and the planes are said to be perpendicular to each other. Thus if plane HP meets plane LM so that dihedral A H-KL-M and M-LK-N are equal, each Z is a rt. dihedral Z, and planes HP and LM are _L to each other. Ex. 1191. By comparison with the definitions of the corresponding terms in plane geometry, frame exact definitions of the following terms : acute dihedral angle ; obtuse dihedral angle ; reflex dihedral angle ; oblique dihedral angle ; vertical dihedral angles ; complementary di- hedral angles ; supplementary dihedral angles ; bisector of a dihedral angle ; alternate interior dihedral angles ; corresponding dihedral angles. Illustrate as many of these as you can with an open book. Ex. 1192. If one plane meets another plane, the sum of the two adjacent dihedral angles is two right dihedral angles. HINT. See proof of 65. Ex. 1193. If the sum of two adjacent dihedral angles is equal to two right dihedral angles, their exterior faces are coplanar. HINT. See proof of 76. Ex. 1194. If two planes intersect, the vertical dihedral angles are equal. HINT. See proof of 77. 324 SOLID GEOMETRY PROPOSITION XX. THEOREM 673. If two dihedral angles are equal, their plane angles are equal. C' Given two equal dihedral A BO and B'c 1 whose plane A are A MNO and M'N'O', respectively. To prove Z. MNO = Z. REASONS 1. 54, 14. 2. 670. ' 3. 62. 4. 670, 62. 5. 18. 674. Cor. I. T&e plane angle of a right dihedral angle is a right angle. 675. Cor. n. If two intersecting planes are each per- pendicular to a third plane, their intersections with, ////? third plane intersect each other. ARGUMENT 1. Superpose dihedral Z BC upon its equal, dihedral Z B'c', so that point N of edge BC shall fall upon point N 1 of edge B'C'. 2. Then MN and M'N 1 , two lines in plane AB, are _L BC at point N. 3. .'. JfA r and M'N' are collinear. 4. Likewise NO and JV'O' are collinear. 5. .'.Z MNO = Z M'N'O'. Q.E.D. BOOK VI 325 Given planes AB and CD J_ plane MN and intersecting each other in line DB ; also let AE and FC be the intersections of planes AB and CD with plane MN. To prove that AE and FC intersect each other. ARGUMENT 1. Either ^ II TO or AE and TO intersect each other. 2. Suppose ^ II FC. Then through 77, any point in DB, pass a plane HKL _L TO, intersecting FC in # and -4# in . 3. Then plane HKL is _L ^4# also. 4. /. Z .HJ5TL is the plane Z of dihedral Z TO, and Z KLH is the plane Z of di- hedral Z .4#. But dihedral A FC and AE are rt. dihe- dral A. .' . A HKL and -SXZT are rt. A. . ' . A FJPL contains two rt. A. 8. But this is impossible. 9. .'. AE and FC intersect each other. Q.E.D. REASONS 1. 161, a. 2. 627. 636. 670. 5. 672. 6. 674. 7. Arg. 6. 8. 206. 9. 161, 6. Ex. 1195. Find the locus of all points equidistant from two given points in space. Ex. 1196. Find the locus of all points equidistant from three given points in space. Ex. 1197. Are the supplements of equal dihedral angles equal ? complements ? Prove your answer. Ex. 1198. If two planes are each perpendicular to a third plane, can they be parallel to each other? Explain. If they are parallel to each other, prove their intersections with the third plane parallel. 326 SOLID GEOMETRY PROPOSITION XXI. THEOREM (Converse of Prop. XX) \j 676. If the plane angles of two dihedral angles are equal, the dihedral angles are equal. C" Given two dihedral A EG and B f C f whose plane A MNO and M'N'O' are equal. To prove dihedral Z BC = dihedral Z B'c'. ARGUMENT 1. Place dihedral Z BC upon dihedral ' so that plane Z.MNO shall be superposed upon its equal, plane Z M'N'O'. 2. Then BC and B'C' are both _L Jf# and NO at ^. 3. .-. BC and 'C Y ' are both J_ plane MNO at AT. 4. .*. .6(7 and -B'C' are collinear. 5. .-. planes AB and A'B', determined by MN and BC, are coplanar ; also planes CD and C'D', determined by BC and NO, are coplanar. 6. .-. dihedral Z. BC = dihedral Z '-,is a rt. dihedral Z, and plane PQ_L plane MJV. Q.E.D. REASONS 1. 619. 63. 670. 4. 619. 5. 677. Ex. 1211. If from the foot of a per- pendicular to a plane a line is drawn at right angles to any line in the plane, the line drawn from the point of intersec- tion so formed to any point in the per- pendicular is perpendicular to the line of the plane. HINT. Make KE = EH. Prove AK - AH, and apply 142. Ex. 1212. In the figure of Ex. 1211, if AB = 20, BE = 4 VTl, and EK= 10, find AK. BOOK VI 329 PROPOSITION XXIII. THEOREM \v679. //' two planes are perpendicular to each other, any line in one of them, perpendicular to their intersec- tion, is perpendicular to the other. * N B n / E; M Given plane PQ J_ plane MN, CD their line of intersection, and AB, in plane PQ, _L CD. To prove AB _L plane MN. ARGUMENT 1. Through. JS, in plane MN, draw BEJLCD. 2. Then Z ABE is the plane Z of the rt. dihedral Z Q-CD-M. 3. .-. Z J# is a rt. Z, and ^41? _L BE. 4. But .45 _L CD. 5. .-. ^ _L plane MN. Q.E.D. REASONS 1. 63. 2. 670. 3. 674. 4. By hyp. 5. 622. 680. Cor. If two planes are perpendicular to each other, a line perpendicular to one of them at any point in their line of intersection, lies in the other. HINT. Apply the indirect method, using 679 and 638. Ex. 1213. If a plane is perpendicular to the edge of a dihedral angle, is it perpendicular to each of the faces of the dihedral angle ? Prove your answer. Ex. 1214. The plane containing a straight line and its projection upon a plane is perpendicular to the given plane. Ex. 1215. If two planes are perpendicular to each other, a line per- pendicular to one of them from any point in the other lies in the other plane. 330 SOLID GEOMETRY PROPOSITION XXIV. THEOREM 681. If each of two intersecting planes is perpendicu- lar to a third plane; I. Their line of intersection intersects the third plane. II. Their line of intersection is perpendicular to the third plane. M Given planes PQ and RS _L plane MN and intersecting each other in line AB. To prove : I. That AB intersects plane MN. II. AB J_ plane MN. I. ARGUMENT 1. Let planes PQ and RS intersect plane MN in lines PD and RE, respectively. 2. Then PD and RE intersect in a point as C. 3. .. AB passes through G\ i.e. AB intersects plane MN. Q.E.D. II. ARGUMENT 1. Either AB _L plane MN or it is not. 2. Suppose AB is not J_ plane MN, but that some other line through <7, the point common to the three planes, is _L plane MN, as line OF. 3. Then CF lies in plane PQ, also in plane RS. 4. .-. CF is the intersection of planes PQ and RS. 5. .-. planes PQ and RS intersect in two str. lines, which is impossible. 'fe. .-. AB _L plane MN. Q.E.D. REASONS 1. 616. 2. 675. 3. 617, I. REASONS 1. 161, a. 2. 639. 3. 680. 4. 614. 5. 616. 6. 161, b. BOOK VI 331 PROPOSITION XXV. PROBLEM 682. Through any straight line, not perpendicular to a plane, to construct a plane perpendicular to the given plane. Given line AB not _L plane MN. To construct, through AB, a plane _L plane MN. The construction, proof, and discussion are left as an exer- cise for the student. HINT. Apply 678. For discussion, see 683. 683. Cor. Through a straight line, not perpendicular to a plane, there exists only one plane perpendicular to the given plane. HINT. Suppose there should exist another plane through AB _L plane MN. What would you know about AB ? 684. 682 and 683 may be combined in one statement as follows : Through a straight line, not perpendicular to a plane, there exists one and only one plane perpendicular to the given plane. Ex. 1216. Apply the truth of Prop. XXIV : (a) to the planes that intersect at the corner of a room ; (6) to the planes formed by an open book placed perpendicular to the top of the desk. Ex. 1217. If a plane is perpendicular to each of two intersecting planes, it is perpendicular to their intersection. 332 SOLID GEOMETRY PROPOSITION XXVI. PROBLEM 685. To construct a common perpendicular to any two 'straight lines in space. K X Given AB and CD, any two str. lines in space. To construct a line _L both to AB and to CD. I. Construction 1. Through CD construct plane MN II AB. 647. 2. Through AB construct plane AF J_ plane MN intersecting MN in EF, and CD in H. 682. 3. Through H construct HK, in plane AF, _L EF. 148. 4. HK is _L to both AB and C7) and is the line required. II. Proof ARGUMENT 1. ^.S II .plane MN. 2. .-. tfF II AS. 3. But #Jf _L EF. 4. .-. /r.g'J. AB. 5. Also JMT plane MN. 6. .-. HK _ CD. 7. .: HK _L to both J and Q.E.D. REASONS 1. By cons. 2. 641. 3. By cons. 4. 193. 5. 679. 6. 619. 7. Args. 4 and 6. III. The discussion will be given in 686. 686. Cor. Between two straight lines in space (not in the same plane} there exists only one common per- pendicular. BOOK VI 333 HINT. Suppose XF, in figure of 685, a second _L to AB and CD. Through T draw ZW \\ AB. What is the relation of XT to AB? to ZW? to CD? to plane MN? Through X draw XE _L EF. What is the relation of XB to plane MN ? Complete the proof. 687. 685 and 686 may be combined in one statement as follows : Between two straight lines in space (not in the, same plane) there exists one and only one common perpendicular. Ex. 1218. A room is 20 feet long, 15 feet wide, and 10 feet high. Find the length of the shortest line that can be drawn on floor and walls from a lower corner to the diagonally opposite corner. Find" the length of the line that extends diagonally across the floor, then along the intersec- tion of two walls to the ceiling. Ex. 1219. If two equal lines are drawn from a given point to a given plane, the inclinations of these lines to the given plane are equal. If two unequal lines are thus drawn, which has the greater inclination ? Prove. Ex. 1220. The two planes determined by two parallel lines and a point not in their plane, intersect in a line which is parallel to each of the given parallels. Ex. 1221. If two lines are parallel, their projections on a plane are either the same line, or parallel lines. Ex. 1222. If each of three planes is perpendicular to the other two : (a) the intersection of any two of the planes is perpendicular to the third plane ; (6) each of the three lines of intersection is perpendicular to the other two. Find an illustration of this exercise in the classroom. Ex. 1223. If two planes are parallel, no line in the one can meet any line in the other. Ex. 1224. Find all points equidistant from two parallel planes and equidistant from three points : (a) if the points lie in neither plane ; (&) if the points lie. in one of the planes. Ex. 1225. Find all points equidistant from two given points, equi- distant from two parallel planes, and at a given distance d from a third plane. Ex. 1226. If each of two intersecting planes is parallel to a given line, the intersection of the planes is parallel to the line. Ex. 1227. Construct, through a point in space, a straight line that shall be parallel to two intersecting planes. 334 SOLID GEOMETRY PROPOSITION XXVII. THEOREM 688. Every point in the plane that bisects a dihedral angle is equidistant from the faces of the angle. M Given plane BE bisecting the dihedral Z formed by planes AG and CD ; also PH and PK Js from P, any point in plane BE, to faces AC and CD, respectively. To prove PH= PK. ARGUMENT 1. Through PH and PK pass plane MN intersecting plane AC in CH, plane CD in CK, plane BE in PC, and edge BC in C. 2. Then plane MN J_ planes AC and (77); t.e. planes AC and CZ) are _L plane JOT. 3. /. .BC _L plane MN. 4. .-. BC CH, CP, and C#. 5. .-. APCH and J5TCP are the plane A of the dihedral A E-BC-A and D-CB-E. 6. But dihedral Z E-BC-A = dihedral Z D-CB-E. 7. 8. 9. 10. .\ PH = PIT. Q.E.D. Also PC = PC. .'. rt. A PC/f= rt. A J5TCP. .\ PH = PIT. REASONS 1. 612, 616. 2. 678. 3. 681, II. 4. 619. 5. 670. 6. By hyp. 7. 673. 8. By iden. 9. 209. 10. 110. BOOK VI 335 689. Cor. I. Every point equidistant from the two faces of a dihedral angle lies in the plane bisecting the angle. 690. Cor. n. The plane bisecting a dihedral angle is the locus of all points in space equidistant from the. faces of the angle. 691. Cor. m. Problem. To construct the bisector of a given dihedral angle. Ex. 1228. Prove that a dihedral angle can be bisected by only one plane. HINT. See proof of 599. Ex. 1229. Find the locus of all points equidistant from two inter- secting planes. Of how many planes does this locus consist ? Ex. 1230. Find the locus of all points in space equidistant from two intersecting lines. Of how many planes does this locus consist ? Ex. 1231. Find the locus of all points in space equidistant from two parallel lines. Ex. 1232. Find the locus of all points in space equidistant from two intersecting planes and equidistant from all points in the circumference of a circle. Ex. 1233. Find the locus of all points in space equidistant from two intersecting planes and equidistant from two fixed points. Ex. 1234. Find the locus of all points in space equidistant from two intersecting planes, equidistant from two parallel planes, and equidistant from two fixed points. Ex. 1235. If from any point within a dihedral angle lines are drawn perpendicular to the faces of the angle, the angle formed by the perpen- diculars is supplementary to the plane angle of the dihedral angle. Ex. 1236. Given two points, P and Q, one in each of two intersecting planes, M and N. Find a point JTin the intersection of planes Mand N such that PX+XQ is a minimum. Ex. 1237. Given two points, P and Q, on one side of a given plane MN. Find a point X in plane MN such that PX -f XQ shall be a minimum. HINT. See Ex. 176. 336 SOLID GEOMETRY POLYHEDRAL ANGLES 692. Def. A polyhedral angle is the figure generated by a moving straight line segment that continually intersects the boundary of a fixed polygon and one extremity of which is a fixed point not in the plane of the given polygon. A poly- hedral angle is sometimes called a solid angle. ? 693- Defs. The moving line is called the generatrix, as V*A ; the fixed polygon is calLed the directrix, as polygon ABODE', the fixed point is called the vertex of the polyhedral- angle, as F. 694. Defs. The generatrix in any position is an element of the polyhedral angle ; the elements through the vertices of the poly- gon . are the edges, as VA, VB, etc. ; the portions of the planes determined by the edges of the polyhedral angle, and limited by them are the faces, as AVB, BVC, etc.; the angles formed by the edges are the face angles, as A AVB, BVC, etc.; the dihedral angles formed by the faces are called the dihedral angles of the poly- hedral angle, as dihedral A VA, VB, etc. 695. Def. The face angles and the dihedral angles taken together are sometimes called the parts of a polyhedral angle. 696. A polyhedral angle may be designated by a letter at the vertex and one on each edge, as V-ABCDE. If there is no other polyhedral -angle having the same vertex, the letter at the vertex is a sufficient designation, as V. 697. Def. A convex polyhedral angle is a polyhedral angle whose directrix is a convex polygon, i.e. a polygon no side of which, if prolonged, will enter the polygon; as V-ABCDE. In this text only convex polyhedral angles will be considered. 698. Defs. A trihedral angle is a polyhedral angle whose directrix is a triangle (ri-gon) ; a tetrahedral angle, a polyhe- dral angle whose directrix is a quadrilateral (tetra-gou) ; etc. BOOK VI 337 699. Defs. A trihedral angle is called a rectangular, birec- tangular, or trirectangular trihedral angle according as it contains one, two, or three right dihedral angles. 700. Def. An isosceles trihedral angle is a trihedral angle having two face angles equal. Ex. 1238. By holding an open book perpendicular to the desk, illus- trate birectangular and trirectangular trihedral angles. By placing one face of the open book on top of the desk and the other face along the side of the desk against the edge, illustrate a rectangular trihedral angle. Ex. 1239. Is every birectangular trihedral angle isosceles ? Is every isosceles trihedral angle birectangular ? 701. From the general definition of equal geometric figures ( 18) it follows that: Two polyhedral angles are equal if they can be made to coincide. PROPOSITION XXVIII. THEOREM 702. Two trihedral angles are equal : I. If a face angle and the two adjacent dihedral an- gles of one are equal respectively to a face angle and tlie two adjacent dihedral angles of tfo other; II. If two face angles and the included dihedral angle of one are equal respectively to two face angles and the included dihedral angle of the other: provided the equal parts are arranged in the same order. The proofs are left as exercises for the student. 703. Questions. Compare care- fully the wording of I above and the accompanying figures with the wording and figures of 105. What in I takes the place of A in 105? side? adj. A? What, in the accompanying figure, cor- responds to A ABC in the proof of 105? A DEF? AC? DF? point A? point C? If these and similar changes are made in the proof of 105, will it serve" as a proof of I above ? Compare II above with 107. 338 SOLID GEOMETRY PROPOSITION XXIX. THEOREM 704. Two trihedral angles are equal if the three face angles of one are equal respectively to the three face angles of the other, and the equal parts are arranged in the same order. Given trihedral A V-ABC and v'-A'B'c', Z A VB = Z A' Z BVC = Z B'V'C', Z CVA = Z c'v'A', and the equal face angles arranged in the same order. To prove trihedral Z V-ABC= trihedral Z v'-A'&C*. OUTLINE OF PROOF 1. Since, by hyp., any two face A of V-ABC, as A AVB and BVC, are equal, respectively, to the two corresponding face A of v'-A'B'C 1 , it remains only to prove the included dihe- dral A VB and V'B' equal. 702, II. (See also 705.) 2. Let face A AVB and BVC be oblique A] then from any point E in VB, draw ED and EF, in planes AVB and BVC, respec- tively, and _L VB. 3. Since A AVB and BVC are oblique A, ED and EF will meet VA and VC in D and F, respectively. Draw FD. 4. Similarly, lay off V'E' = VE and draw A D'E'F'. 5. Prove rt. &DVE=rt. AD 1 V'E'; thenVD=v'D',ED = E'D'. 6; Prove rt. A EVF= rt. A E'V'F'; then VF= V'F', EF=E'F'. 7. Prove A FVD = A F'V'D 1 ; then FD = F'D'. 8. . . A DEF = A D'E'F' ; then Z ZXE^ = Z BOOK VI 339 9. But A DEF and D'E'F' are the plane A of dihedral A VB and V'B', respectively. 10. .-. dihedral Z VB = dihedral Z V'B 1 . 11. .-. trihedral Z v -AB C = trihedral Z V'-A'B'C'. Q.E.D. 705. Note. If all the face A are rt. A, show that all the dihedral A are rt. dihedral A and hence that all are equal. If two face A of a trihe- dral Z are rt. A, show that the third face Z is the plane Z of the included dihedral Z, and hence that two homologous dihedral A, as VB and V'B 1 , are equal. It remains to prove that Prop. XXIX is true if only one face Z of the first trihedral Z and its homologous face Z of the other are rt. A, or if all face A are oblique. 706. Questions. State the proposition in Bk. I that corresponds to 704. What was the main step in the proof of that proposition? Did that correspond to proving dihedral Z VB of 704 = dihedral Z V'B' ? 707. Def. Two polyhedral angles are said to be sym- metrical if their corresponding parts are equal but arranged in reverse order. By making symmetrical polyhedral angles and comparing them, the student can easily satisfy himself that in general they cannot be made to coincide. 708. Def. Two polyhedral angles are said to be vertical if the edges of each are the prolongations of the edges of the other. It will be seen that two vertical, like two symmetrical, poly- hedral angles have their corresponding parts equal but arranged in reverse order. Two Equal Polyhedral Two Vertical Poly- Two Symmetrical Poly- Angles hedral Angles hedral Angles 340 SOLID GEOMETRY PROPOSITION XXX. THEOREM 709. Two trihedral angles are symmetrical: I. If a face angle and the two adjacent dihedral an- gles of one are equal respectively to a face angle and the two adjacent dihedral angles of the other; II. If two face angles and the included dihedral angle of one are equal respectively to two face angles and the included dihedral angle of the other; III. If the three face angles of one are equal respec- tively to the three face angles of the other : provided the equal parts are arranged in reverse order- \ \ The proofs are left as exercises for the student. HINT. Let V and V be the two trihedral A with parts equal but arranged in reverse order. Construct trihedral Z V" symmetrical to F. Then what will be the relation of V" to V ? of V to F? Ex. 1240. Can two polyhedral angles be symmetrical and equal ? vertical and equal ? symmetrical and vertical ? If two polyhedral angles are vertical, are they necessarily symmetrical ? if symmetrical, are they necessarily vertical ? Ex. 1241. Are two trirectangular trihedral angles necessarily equal? Are two birectangular trihedral angles equal ? Prove your answers. Ex. 1242. If two trihedral angles have three face angles of one equal respectively to three face angles of the other, the dihedral angles of the first are equal respectively to the dihedral angles of the second. BOOK VI 841 PROPOSITION XXXI. THEOREM 710. The sum of any two face angles of a trihedral angle is greater than the third face angle- V c Given trihedral Z V-ABC in which the greatest face Z is A vs. To prove Z BVC + Z CVA > Z ^IFB. OUTLINE OF PROOF 1. In face AVB draw FZ) making Z D Ffi = Z F(7, and through D draw any line intersecting VA in # and VB in .F. 2. On VC lay off F = VD and draw'^G and GE. 3. Prove A ^F = A DVF-, then ^ = FD. 4. But FG + GE > FD + DE ; .'. GE > #. 5. In A GVE and ^FD, pdft^*?Z > Z J?FD. 6. But Z ^F = ^LDVF. 1. .'. Z ^F + Z GVE >Z.EVD + Z.D%F; i.e. /. BVC + Z. CVA > Z ^FB. Q.E.D. 711. Question. State the theorem in Bk. I that^orrespouds to Prop. XXXI. Can that theorem be proved by a method similar to the one used here ? If so, give the proof. Ex. 1243. If, in trihedral angle V-ABC, angle SVC = 60, and angle CVA = 80, make a statement as to the number of degrees in angle AVB. Ex. 1244. Any face angle of a trihedral angle is greater than the difference of the other two. 342 SOLID GEOMETRY PROPOSITION XXXII. THEOREM 712. The sum of all the face angles of any convex poly- hedral angle is less than four right angles- V Given polyhedral Z V with n faces. To prove the sum of the face A at V Z 4 rt. A. HINT. Let a plane intersect the edges of the polyhedral Z. in A, B, (T, etc. From O, any point in polygon ABC . . ., draw OA, OB, OC, etc. How many A have their vertices at V? at O ? What is the sum of all the A of all the A with vertices at V ? at ? Which is the greater, Z ABV + Z VBG or Z ABO + Z OBC ? Then which is the greater, the sum of the base A of A with vertices at T 7 , or the sum of the base A of A with vertices at O ? Then which is greater, the sum of the face A about T, or the sum of the A about ? 713. Question. Is there a proposition in plane geometry corre- sponding to Prop. XXXII ? If so, state it. If not, state the one that most nearly corresponds to it. Ex. 1245. Can a polyhedral angle have for its faces three equi- lateral triangles ? four ? five ? six ? Ex. 1246. Can a polyhedral angle have for its faces three squares ? four ? Ex. 1247. Can a polyhedral angle have for its faces three regular pentagons ? four ? Ex. 1248. Show that the greatest number of polyhedral angles that can possibly be formed with regular polygons as faces is five. Ex. 1249. Can a trihedral angle have for its faces a regular decagon and two equilateral triangles? a regular decagon, an equilateral tri- angle, and a square ? two regular octagons and a square ? BOOK VII POLYHEDRONS 714. Def. A surface is said to be closed if it separates a finite portion of space from the remaining space. 715. Def. A solid closed figure is a figure in space composed of a closed surface and the finite portion of space bounded by it. 716. Def. A polyhedron is a solid closed figure whose bounding surface is composed of planes only. 717. Defs. The intersections of the bounding planes are called the edges ; the intersections of the edges, the vertices; and the portions of the bounding planes bounded by the edges, the faces, of the polyhedron. 718. Def. A diagonal of a polyhe- dron is a straight line joining any two vertices not in the same face, as AB. 719. Defs. A polyhedron of four faces is called a tetrahedron; one of six faces, a hexahedron; one of eight faces, an octahedron; one of twelve faces, a do- decahedron ; one of twenty faces, an icosahedron ; etc. Ex. 1250. How many diagonals has a tetrahedron ? a hexahedron ? Ex. 1251. What is the least number of faces that a polyhedron can have ? edges ? vertices ? Ex. 1252. How many edges has a tetrahedron ? a hexahedron ? an octahedron ? Ex. 1253. How many vertices has a tetrahedron ? a hexahedron ? an octahedron ? Ex. 1254. If E represents the number of edges, F the number of faces, and F the number of vertices in each of the polyhedrons mentioned in Exs. 1252 and 1253, show that in each case E + 2 = V + F. This result is known as Euler's theorem. 343 344 St)LID GEOMETRY Ex. 1255. Show that in a tetrahedron S= (F 2) 4 right angles, where S is the sum of the face angles and V is the number of vertices. Ex. 1256. Does the formula, S = ( V - 2) 4 right angles, hold for a hexahedron ? an octahedron ? a dodecahedron ? 720. Def. A regular polyhedron is a polyhedron all of whose faces are equal regular polygons, and all of whose polyhe- dral angles are equal. 721. Questions. How many equilateral triangles can meet to form a polyhedral angle ( 712) ? Then what is' the greatest number of regular polyhedrons possible having equilateral triangles as faces ? What is the greatest number of regular polyhedrons possible having squares as faces ? having regular pentagons as faces ? Can a regular polyhedron have as faces regular polygons of more than five sides ? why ? What, then, is the maximum number of kinds of regular polyhedrons possible ? 722. From the questions in 721, the student has doubtless drawn the conclusion that not more than five kinds of regular polyhedrons exist. He should convince himself that these five are possible by actually making them from cardboard as indicated below : Tetrahedron Hexahedron Octahedron Dodecahedron Icosahednm BOOK VII 345 723. Historical Note. The Pythagoreans knew that there were five regular polyhedrons, but it was Euclid who proved that there can be only jive. Ilippasus (arc. 470 B.C.), who discovered the dodecahedron, is said to have been drowned for announcing his discovery, as the Pythago- reans were pledged to refer the glory of any new discovery " back to the founder." PEISMS * 724. Def. A prismatic surface is a surf ace . generated by a moving straight line that continually intersects a fixed broken line and remains parallel to a fixed straight line not coplanar with the given broken line. B C Prism Prismatic Surface 725. Defs. By referring to 693, the student may give the definitions of generatrix and directrix of a prismatic surface. Point these out in the figure. 726. Def. A prism is a polyhedron whose boundary consists of a prismatic surface and two parallel planes cutting the generatrix in each of its positions. 727. Defs. The two parallel plane sections are the bases of the prism, as ABODE and FGHKL ; the faces forming the pris- matic surface are the lateral faces, as AG, BH, etc.; the inter- sections o*f the lateral faces are the lateral edges, as AF, BG, etc. In this text only prisms whose bases are convex polygons will be considered. * This treatment of prisms and pyramids is given because of its similarity to the treat- ment of cylinders and cones given in SID-S'2'2 and t>37-S40. 346 SOLID GEOMETRY 728. Def. A right section of a prism is a section formed by a plane which is perpendicular to a lateral edge of the prism and which cuts the lateral edges or the edges prolonged. Right Prism Regular Prism Oblique Prism 729. Def. A right prism is a prism whose lateral edges are perpendicular to the bases. 730. Def. A regular prism is a right prism whose bases are regular polygons. 731. Def. An oblique prism is a prism whose lateral edges are oblique to the bases. 732. Defs. A prism is triangular, quadrangular, etc., accord- ing as its bases are triangles, quadrilaterals, etc. 733. Def. The altitude of a prism is the perpendicular from any point in the plane of one base to the plane of the other base. 734. The following are some of the properties of a prism ; the student should prove the correctness of each : (a) Any two lateral edges of a prism are parallel. (6) The lateral edges of a prism are equal. (c) Any lateral edge of a right prism is equal to the altitude. (d) TJie lateral faces of a prism are parallelograms. (e) The lateral faces of a right prism are rectangles. (/) TJie bases of a prism are equal polygons. (g) TJie sections of a prism made by two parallel planes cutting all the lateral edges are equal polygons. (h) Every section of a prism made by a plane parallel to the base is equal to the base. BOOK VII PROPOSITION I. THEOREM 347 735. Two prisms are equal if three faces including a trihedral angle of one are equal respectively, and simi- larly placed, to three faces including a trihedral angle of the other. H A B A' B Given prisms AI and A 1 1 1 , face AJ= face A'j', face AG = face 'G'J face AD = face A'D'. To prove prism AI= prism A'l'. ARGUMENT 1. A BAF, FAE, and BAE are equal, respec- tively, to A B'A'F', F'A'E', and B'A'E'. 2. .-. trihedral Z A = trihedral Z A 1 . 3. Place prism AI upon prism A' I 1 so that trihedral Z A shall be superposed upon its equal, trihedral Z A'. 4. Faces AJ, AG, and AD are equal, re- spectively, to faces A '/', A'G', and A'D'. 5. . *. J, F, and G will fall upon /', F', and (?', respectively. 6. CH and ) Suppose that u is not a measure of AC, AF, and AG, re- spectively, but that some aliquot part of u is such a measure. The proof is left as an exercise for the student. II. If AC, AF, and AG are each incommensurable with u. ARGUMENT 1. Let m be a measure of u. Apply m as a measure to AC, AF, and AG, respec- tively, as many times as possible. There will be remainders, as MC, NF, and QG, each less than m. 2. Through M draw plane MK J_ AC, through N draw plane NK A. AF, and through Q draw plane QK _L AG. 3. Now ^3f, -4JV, and ^Q are each com- mensurable with the measure m, and hence with u, the linear unit. 4. .*. the volume of rectangular parallele- piped AK = AM AN - AQ. 5. Now take a smaller measure of u. No matter how small a measure of u is taken, when it is applied as a measure to AC, AF, and ^(?, the remainders, MC, NF, and QG, will be smaller than tho Tw.'isure taken. REASONS 1. 339. 2. 627. 3. 337. 4. 778,1. 5. 335. BOOK VII 363 ARGUMENT 6. .-. the difference between AM and AC, the difference between AN and AF, and the difference between AQ and AG, may each be made to become and remain less than any previously assigned segment, however small. 7. .-. AM approaches AC as a limit, AN approaches AF as a limit, and AQ approaches AG as a limit. 8. .-. AM- AN- AQ approaches AC' AF AG as a limit. 9. Again, the difference between rectan- gular parallelopiped AK and rec- tangular parallelepiped AD may be made to become and remain less than any previously assigned volume, however small. 10. .'. the volume of rectangular parallelo- piped AK approaches the volume of rectangular parallelopiped AD as a limit. 11. But the volume of AK is always equal to AM- AN - AQ. 12. .-. the volume of AD=AC- AF- AG. Q.E.D. 11. 12. REASONS Arg. 5. 7. 349. 593. Arg. 5. 10. 349. Arg. 4. 355. III. If AC is commensurable with u but AF and AG are in- commensurable with u. IV. If AC and AF are commensurable with u but AG is in- commensurable with u. The proofs of III and IV are left as exercises for the student. 779. Cor I. The volume of a cube is equal to the cube of its edge. HINT. Compare with 478. 364 SOLID GEOMETRY 780. Cor. II. Any two rectangular parallelepipeds are to each other as the products of their three di- mensions. (HINT. Compare with 479. ) 781. Note. By the product of a surface and a line is meant the product of the measure-numbers of the surface and the line. 782. Cor. III. The volume of a rectangular parallelo- piped is equal to the product of its base and its altitude. 783. Cor. IV. Any two rectangular parallelepipeds are to each other as the products of their bases and their altitudes. (HINT. Compare with 479.) 784. Cor. V. (a) Two rectangular parallelepipeds having equivalent bases are to each other as their altitudes; (&) two rectangular parallelepipeds having equal altitudes are to each other as their bases. (HINT. Compare with 480.) 785. Cor. VI. (a} Two rectangular parallelepipeds hav- ing two dimensions in common are to each other as their third dimensions, and (b) two rectangular parallelepi- peds having one dimension in common are to each other as the products of their other two dimensions. 786. Questions. What is it in Book IV that corresponds to volume in Book VII ? to rectangular parallelepiped ? State the theorem and corol- laries in Book IV that correspond to 778, 779, 782, 783, and 784. Will the proofs given there, with the corresponding changes in terms, apply here ? Compare the entire discussion of 466-480 with 769-785. Ex. 1304. Find the volume of a cube whose diagonal is 5\/3 ; d. Ex. 1305. The volume of a rectangular parallelepiped is V ; each side of the square base is one third the altitude of the parallelepiped. Find the side of the base. Find the side of the base if V= 192 cubic feet. Ex. 1306. The dimensions of two rectangular parallelepipeds are 6, 8, 10 and 5, 12, 16, respectively. Find the ratio of their volumes. Ex. 1307. The total area of a cube is 300 square inches ; find its volume. Ex. 1308. The volume of a certain cube is V\ find the volume of a cube whose edge is twice that of the given cube. Ex. 1309. The edge of a cube is a ; find the edge of a cube twice as large ; i.e. containing twice the volume of the given cube. BOOK VII 365 PLATO 787. Historical Note. Plato (429-348 B.C.) was one of the first to discover a solution to that famous problem of antiquity, the duplication of a cube, i.e. the finding of the edge of a cube whose volume is double that of a given cube. There are two legends as to the origin of the problem. The one is that an old tragic poet rep- resented King Minos as wishing to erect a tomb for his son Glau- cus. The king being dissatis- fied with the dimensions (100 feet each way) proposed by his architect, exclaimed : " The in- closure is too small for a royal tomb ; double it, but fail not in the cubical form." The other legend asserts that the Athenians, who were suf- fering from a plague of typhoid fever, consulted the oracle at Delos as to how to stop the plague. Apollo replied that the Delians would have to double the size of his altar, which was in the form of a cube. A new altar was constructed having its edge twice as long as that of the old one. The pestilence became worse than before, whereupon the Delians ap- pealed to Plato. It is therefore known as the Delian problem. Plato was born in Athens, and for eight years was a pupil of Socrates. Plato possessed considerable wealth, and after the death of Socrates in 399 B.C. he spent some years in traveling and in the study of mathe- matics.. It was during this time that he became acquainted with the members of the Pythagorean School, especially with Archytas, who was then its head. No doubt it was his association with these people that gave him his passion for mathematics. About 380 B.C. he returned to his native city, where he established a school. Over the entrance to his school was this inscription: "Let none ignorant of geometry enter my door." Later an applicant who knew no geometry was actually turned away with the statement : " Depart, for thou hast not the grip of philosophy." Plato is noted as a teacher, rather than an original discoverer, and his contributions to geometry are improvements in its method rather than additions to its matter. He valued geometry mainly as a "means of ed- ucation in right seeing and thinking and in the conception of imaginary processes." It is stated on good authority that "Plato was almost as important as Pythagoras to the advance of Greek geometry." 366 SOLID GEOMETRY PROPOSITION VI. THEOREM 788. An oblique prism is equivalent to a right prism, whose base is a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. 4, B C Given oblique prism AD' ; also rt. prism KN f with base KN a rt. section of AD', and with KK f , LL f , etc., lateral edges of KN', equal to AA', BB 1 , etc., lateral edges of AD\ To prove oblique prism AD 1 =c= rt. prism KN 1 . OUTLINE OF PROOF 1. In truncated prisms AN and A'N', prove the A of face BK equal, respectively, to the A of face B'K'. 2. Prove the sides of face BK equal, respectively, to the sides of face B'K'. 3. .-. face BK=f&ce B'K'. 4. Similarly face 53f=face B'M', and face BE = face B'E'. 5. .-. truncated prisrn AN = truncated prism A'N' ( 737). 6. But truncated prism A'N = truncated prism A'N. 7. .-. oblique prism AD' =c=rt. prism KN'. Q.E.D. 789. Question. Is there a theorem in Book IV that corresponds to Prop. VI ? If not, formulate one and see if you can prove it true. BOOK VII 367 PROPOSITION VII. THEOREM 790. The volume .of any parallelepiped is equal to the product of its base and its altitude. Given parallelepiped I with its volume denoted by F, its base by Ji, and its altitude by H. To prove V = B H. ARGUMENT 1. Prolong edge 4C and all edges of I II AC. 2. On the prolongation of AC take DF = AC, and through D and ^ pass planes J_ ^, forming rt. parallelepiped 17. 3. Then I =c= 77. 4. Prolong edge -FJT and all edges of // II FK. 5. On the prolongation of FK take MN = FK, and through J/ and N pass planes J_ FN, forming rectangular parallele- piped ///. 6. Then II ^ III. 7. .: 1=0=1/7. 8. Again, J? =0= B' = B". 9. Also Hj the altitude of 7, = the altitude of ///. 10. But the volume of 777= B" H. 11. .-. v B H. Q.E.D. REASONS 1. 54,16. 2. 627. 3. 788. 4. 54, 16. 5. 627. 6. 788. 7. 54,1. 8. 482. 9. 663. 10. 11. 782. 309. 368 SOLID GEOMETRY 791. Cor. I. Parallelopipeds having equivalent bases and equal altitudes are equivalent. 792. Cor. II. Any two parallelepipeds are to each other as the products of their bases and their altitudes. 793. Cor. III. (a) Two parallelepipeds having equiva- lent bases are to each other as their altitudes, and (b} two parallelepipeds having equal altitudes are to each other as their bases. 794. Questions. What expression in Book IV corresponds to vol- ume of a parallelepiped 9 Quote the theorem and corollaries in Book IV that correspond to 790-793. Will the proofs given there, with the corresponding changes, apply here ? Ex. 1310. Prove Prop. VI by subtracting the equal truncated prisms of Arg. 5 from the entire figure. Ex. 1311. The base of a parallelepiped is a parallelogram two adja- cent sides of which are 8 and 15, respectively, and they include an angle of 30. If the altitude of the parallelepiped is 10, find its volume. Ex. 1312. Four parallelepipeds have equivalent bases and equal lat- eral edges. In the first the lateral edge makes with the base an angle of 30 ; in the second an angle of 45 ; in the third an angle of 60 ; and in the fourth an angle of 90. Find the ratio of the volumes of the four parallelepipeds. Ex. 1313. Find the edge of a cube equivalent to a rectangular par- allelepiped whose edges are 6, 10, and 15 ; whose edges are a, &, and c. Ex. 1314. Find the diagonal of a cube whose volume is 512 cubic inches ; a cubic inches. Ex. 1315. The edge of a cube is a. Find the area of a section made by a plane through two diagonally opposite edges. Ex. 1316. How many cubic feet of cement will be needed to make a box, including lid, if the inside dimensions of the box are 2 feet 6 inches, 3 feet, and 4 feet 6 inches, if the cement is 3 inches thick ? HINT. In a problem of this kind, always find the volume of the whole solid, and the volume of the inside solid, then subtract. Ex. 1317. The volume of a rectangular parallelepiped is 2430 cubic inches, and its edges are in the ratio of 3, 6, and 6. Find its edges. Ex. 1318. In a certain cube the area of the surface and the volume have the same numerical value. Find the volume of the cube. BOOK VII 369 PROPOSITION VIII. THEOREM 795. The plane passed through two diagonally opposite edges of a parallelopiped divides it into two equivalent triangular prisms Given plane AG passed through edges AE and CG of paral- lelopiped BH dividing it into the two triangular prisms ABC-F and CDA-H. To prove prism ABC-F o prism CDA-H. ARGUMENT ONLY 1. Let MNOP be a rt. section of parallelopiped BH, cutting the plane AG in line MO. 2. Face AF II face DG and face AH\\ face BG. 3. . . MN II PO and MP II NO. 4. . . MNOP is a O. 5. .'.A MNO = A 0PM. 6. Now triangular prism ABC-F =c= a rt. prism whose base is A MNO, a rt. section of prism ABC-F, and whose altitude is AE, a lateral edge of prism ABC-F. 7. Likewise triangular prism CDA-Hoa, rt. prism whose base is A 0PM and whose altitude is AE. 8. But two such prisms are equivalent. 9. .-. prism ABC-F =c= prism CDA-H. Q.E.D. 796. Questions. Is there a theorem in Book I that corresponds to Prop. VIII ? If so, state it. Could an oblique prism exist such that a right section, as MNOP, might intersect either base ? If so, draw a figure to illustrate. 370 SOLID GEOMETRY PROPOSITION IX. THEOREM 797. The volume of a triangular prism/ is equal to the product of its base and its altitude. A C Given triangular prism A CD-X with its volume denoted by F, its base by B, and its altitude by H. To prove F = B H. The proof is left as an exercise for the student. 798. Questions. What proposition in Book IV corresponds to Prop. IX above ? Can you apply the proof there given ? What is the name of the figure CZ in 797 ? What is its volume ? What part of CZ is AGD-X ( 795) ? Ex. 1319. The volume of a triangular prism is equal to one half the product of any lateral face and the perpendicular from any point in the opposite edge to that face. HINT. The triangular prism is one half of a certain parallelepiped ( 795). Ex. 1320. The base of a coal bin which is 8 feet deep is a triangle with sides 10 feet, 15 feet, and 20 feet, respectively. How many tons of coal will the bin hold considering 35 cubic feet of coal to a ton ? Ex. 1321. One face of a triangular prism contains 45 square inches ; the perpendicular to this face from a point in the opposite edge is 6 inches. Find the volume of the prism. Ex. 1322. During a, rainfall of \ inch, how many barrels of water will fall upon a ten-acre field, counting 7 gallons to a cubic foot and 31^ gallons to a barrel ? Ex. 1323. The inside dimensions of an open tank before lining are 6 feet, 2 feet inches, and 2 feet, respectively, the latter being the height. Find the number of pounds of zinc required to line the tank with a coat- ing \ inch thick, a cubic foot of zinc weighing 6860 ounces. BOOK vn THEOREM 371 PROPOSITION X. 799. The volume of any prism is equal to the product of its base and its altitude, c> Given prism AM with its volume denoted by F, its base by B, and its altitude by H. To prove F = B H. ARGUMENT 1. From any vertex of the lower base, as A y draw diagonals AD, AF, etc. 2. Through edge AI and these diagonals pass planes AK, AM, etc. 3. Prism AM is thus divided into triangu- lar prisms. 4. Denote the volume and base of trian- gular prism ACD-J by v l and bi', of ADF-K by v 2 and 6 2 ; etc. Then v = V = B - H. etc. Q.E.D. REASONS 1. 54,15. 2. 612. 3. 732. 4. 797. 5. 54, 2. 6. 309. 800. Cor. I. Prisms having equivalent bases and equnl altitudes are equivalent. 801. Cor. n. Any two prisms are to each other as the products of their bases and their altitudes. 802. Cor. m. (a) Two prisms having equivalent bases are to each other as their altitudes; (&) tivo prisms having equal altitudes are to each other as their bases. 372 SOLID GEOMETRY PROPOSITION XI. THEOREM 803. Two triangular pyramids having equivalent bases and equal altitudes are equivalent. o R Given triangular pyramids 0-ACD and O'-A'C'D' with base ACD =c= base A'C'D', with altitudes each equal to QR, and with volumes denoted by Fand F', respectively. To prove V V 1 . ARGUMENT 1. V F', F< F', or F> F'. 2. Suppose F < F', so that v 1 V = ~k, a constant. For convenience, place the two pyramids so that their bases are in the same plane, MN. 3. Divide the common altitude QR into 71 equal parts, as QX, XY, etc., and through the several points of division pass planes II plane MN. 4. Then section FGU =e= section F'G'U', section JKW =c= section J'K' w', etc. 5. On FGU, JKW, etc., as upper bases, con- struct prisms with edges II DO and with altitudes = QX. Denote these prisms by II, III, etc. 0. On A'c'D', F'G'u', etc., as lower bases, construct prisms with edges II D'o' and with altitudes = QX. Denote these prisms by I', II' ; etc. REASONS 1. 161, a. 2. 54, 14. 3. 653. 4. 759. 5. 726. 6. 726. BOOK VII 373 9. 10. 11. ARGUMENT 7. Then prism II =0= prism II', prism III =c= prism III', etc. 8. Now denote the sum of the volumes of prisms II, III, etc., by /S; the sum of the volumes of prisms I', II', III', etc., by S 1 ; and the volume of prism I' by v'. Then S' - S = v'. But V 1 < S' and S < F. .-. V* + S < F + S'. .'. V 1 V < S' S ; i.e. V* VH; .-. V= B- H. Q.K.D. Ex. 1328. Find the volume of a regular tetra- hedron whose edge is 0. HINT. 0, the foot of the _L fromFto the plane of base ABC, is the center of A ABC ( 752). Hence OA = f of the altitude of A ABC, and a _L from O to any edge of the base, as OD = \ of OA. Ex. 1329. Find tlie volume of a regular tetra- hedron with slant height 2 V3 ; with altitude a. BOOK VII 375 PROPOSITION XIII. THEOREM 805. The volume of any pyramid is equal to one third the product of its base and its altitude, O A C D Given pyramid 0-ACDFG with its volume denoted by F, its base by n, and its altitude, OQ, by H. To prove F= ^ B H. The proof is left as an exercise for the student. HINT. See proof of Prop. X. 806. Cor. I. Pyramids having equivalent bases and equal altitudes are equivalent. 807. Cor. II. Any two pyramids are to each other as the products of their bases and their altitudes. 808. Cor. III. (a) Two pyramids having equivalent bases are to each other as their altitudes, and (b} two pyramids having equal altitudes are to each other as their bases. Ex. 1330. In the figure of 805, if the base = 250 square inches, OO = 18 inches, and the inclination of OC to the base is 60, find the volume. Ex. 1331. A pyramid and a prism have equivalent bases and equal altitudes ; find the ratio of their volumes. 809. Historical Note. The proof of the proposition that "every pyramid is the third part of a prism on the same base and with the same altitude " is attributed to Eudoxus (408-355 B.C.), a great mathematician of the Athenian School. In a noted work written by Archimedes (287- 212 B.C.), called Sphere and Cylinder, there is also found an expression for the surface and volume of a pyramid. (For a further account of Archimedes, see 542, 896, and 973.) Later a solution of this problem was given by Brahmagupta, a noted Hindoo writer born about 598 A. DO 376 SOLID GEOMETRY PROPOSITION XIV. THEOREM 810. Two triangular pyramids, having a trihedral angle of one equal to a trihedral angle of the other, are to each other as the products of the edges including the equal trihedral angles. D Given triangular pyramids 0-ACD and Q-FGM with tri- hedral Z == trihedral Z Q, and with volumes denoted by F and F', respectively. F OA - OO- OD To prove , = . F' QF-QG-QM ARGUMENT REASONS 1. Place pyramid Q-FGM so that trihedral 1. 54, 14. Z Q shall coincide with trihedral Z 0. Eepresent pyramid Q-FGM in its new position by O-F'G'M'. 2. From J9 and J/' draw DJ and J/'.K' _L plane OA C. 2. 639. 3 Then F A OAC ' DJ A ^ a - * 7 Q S 807 J. F' ~~ A OF'G 1 3/'# A OF'G' M'K o 3 ou i . 4 But A ^ ^ ' C - 4 S 4.Q8 A O^'G?' OF' - OG' T:. S i tt/o. 5. Again let the plane determined by DJ and 5. 613,616. M'K intersect plane OA C in line OKJ. 6. Then rt. A DJO ~ rt. A M*KO. 6. 422. 7. DJ _ OD ' M'K~~ OM 1 ' 7. 424, 2. ft V OA-OC OD OA-OC-OD ' ~~~ ~~ Q E D Q S 30Q o. F' OF'-OG' OM' QF.QG-QM' O. J5 <./\/ 7 811. Def. Two polyhedrons are similar if they have the same number of faces similar each to each and similarly placed, and have their corresponding polyhedral angles equal. BOOK VII 377 PROPOSITION XV. THEOREM 812. The volumes of two similar tetrahedrons are to each other as t}ie cubes of any two homologous edges. M Given similar tetrahedrons 0ACD and Q-FGM with volumes denoted by F and F*, and with OA and QF two homologous edes. _ Y iy_i3. To prove = . V QF 3 ARGUMENT 1. Trihedral Z O = trihedral Z Q. 2 F _ O4 - OC - OP _ OA OC OP ' V' ~ QF - QG - QM ~~ QF QG QJ/ ' 3. But^ = ^ = ^. QF QG QM V OA OA OA ~OA 3 4. .'. = . . = . Q.E.D. V' QF QF QF Q F 3 813. Question. Compare 810 and 812 with 498 and 503. Are the same general methods used in the two sets of theorems ? 814. Note. The proposition, " two similar convex polyhedrons are to each other as the cubes of any two homologous edges," will be assumed at this point, and will be applied in some of the exercises that follow. For a complete discussion of this principle see Appendix, 1022-1029. REASONS 1. 811. 2. 810. 3. 424, 2. 4. 309. Ex. 1332. The edges of two regular tetrahedrons are 6 centimeters and 8 centimeters, respectively. Find the ratio of their volumes. Ex. 1333. The volumes of two similar polyhedrons are 343 cubic inches and 512 cubic inches, respectively: (a) an edge of the first figure is 14 inches, find the homologous edge of the second ; (&) the total area of the first figure is 280 square inches, find the total area of the second. 378 SOLID GEOMETRY PROPOSITION XVI. THEOREM 815. The volume of a frustum of any pyramid is equal to one third the product of its altitude and the sum of its lower base, its upper base, and the mean pro- portional between its two bases. C D Given frustum AM, of pyramid 0-AF, with its volume de- noted by F, its lower base by B, its upper base by b, and its altitude by H. To prove V= $H(B + V 'B b). ARGUMENT 1. Frustum AM = pyramid 0-AF minus pyramid 0-RM. 2. Let H' denote the altitude of 0-RM. Then F^= 1 B(H + H') 1 b H' = i HB 4- i H'(B b). It now remains to find the value of H'. b H' 2 3. 4. 5. Whence H 1 = G. .-. v = B (H+ H 1 )'' II' i.e. F= J //( - b ; lf -6). Q.K.I). REASONS 1. 54, 11. 2. 805. 3. 757. 4. 54, 13. 5. Solving for if'. G. 309. BOOK VII 379 PROPOSITION XVII. THEOREM 816. Jl truncated triangular prism is equivalent to three triangular pyramids whose bases are the base of the frustum and whose vertices are the three vertices of the inclined section. K Given truncated triangular prism ACD-FGK. To prove ACD-FGK =c= F-ACD + G-ACD + K-ACD. ARGUMENT 1. Through A, D, F and K, D, F pass planes dividing frustum ACD-FGK into three triangular pyramids F-ACD, F-ADK, and F-DGK. Since F-ACD is one of the required pyramids, it re- mains to prove F-ADK =0= K-ACD and F-DGK =0= G-ACD. 2. CF II plane AG. 3. .-. the altitude of pyramid F-ADK = the altitude of pyramid C-ADK. 4. .-. F-ADK =o C-ADK. 5. But in C-ADK, A CD may be taken as base and K as vertex. 6. .'. F-ADK =0= K-ACD. 7. Likewise F-DGK =c= C-DGK = K-CDGj and K-CDG =0= A-CDG = G-ACD.\ 8. .'. F-DGK =c= G-ACD. 9. .-. .4<7Z> TOS- =0= F-ACD + O-^CD + K-ACD. Q.E.D. REASONS 1. 611. 646. 664. 4. 806. 5. 749,750. 309. By steps sim- ilar to 3-7. 54, 1. 309. 380 SOLID G-EOMETRY 817. Cor. I. The volume of a truncated right triangu- lar prism is equal to one third the product of its base and the sum of its lateral edges. 818. Cor. H. The volume of any truncated triangular prism is equal to one third the product of a right sec- tion and the sum of its lateral edges. HINT. Rt. section ACD divides truncated triangular prism QE into two truncated right triangular prisms. Ex. 1334. The base of a truncated right triangular prism has for its sides 13, 14, and 15 inches ; its lateral edges are 8, 11, and 13 inches. Find its volume. Ex. 1335. In the formula of 815 : (1) put 6 = and compare result with formula of 805 ; (2) put b = B and compare result with formula of 799. Ex. 1336. A frustum of a square pyramid has an altitude of 13 inches ; the edges of the bases are 2| inches and 4 inches, respectively. Find the volume. Ex. 1337. The edges of the bases of a frustum of a square pyramid are 3 inches and 5 inches, respectively, and the volume of the frustum is 204 cubic inches. Find the altitude of the frustum. Ex. 1338. The base of a pyramid contains 144 square inches, and its altitude is 10 inches. A section of the pyramid parallel to the base divides the altitude into two equal parts. Find : (a) the area of the section; (6) the volume of the frustum formed. Ex. 1339. A section of a pyramid parallel to the base cuts off a pyra- mid similar to the given pyramid. Ex. 1340. The total areas of two similar tetrahedrons are to each other as the squares of any two homologous edges. Ex. 1341. The altitude of a pyramid is 6 inches. A plane parallel to the base cuts the pyramid into two equivalent parts. Find the altitude of the frustum thus formed. Ex. 1342. Two wheat bins are similar in shape ; the one holds 1000 bushels, and the other 800 bushels. If the first is 15 feet deep, how deep is the second ? Ex 1343. A plane is passed parallel to the base of a pyramid cut- ting the altitude into two equal parts. Find : (a) the ratio of the section to the base ; (6) the ratio of the pyramid cut off to the whole pyramid. BOOK VII 381 MISCELLANEOUS EXERCISES Ex. 1344. Find the locus of all points equidistant from the three edges of a trihedral angle. Ex. 1345. Find the locus of all points equidistant from the three faces of a trihedral angle. Ex. 1346. () Find the ratio of the volumes and the ratio of the total areas of two similar tetrahedrons whose homologous edges are in the ratio of 2 to 5. (6) Find the ratio of their homologous edges and the ratio of their total areas if their volumes are in the ratio of 1 to 27. Ex. 1347. (a) Construct three or more equivalent pyramids on the same base. (6) Find the locus of the vertices of all pyramids equivalent to a given pyramid and standing on the same base. HINT. Compare with Exs. 821 and 822. Ex. 1348. The altitude of a pyramid is 12 inches. Its base is a regular hexagon whose side is 5 inches. Find the area of a section paral- lel to the base and 4 inches from the base ; 4 inches from the vertex. Ex. 1349. A farmer has a corn crib 20 feet long, a cross section of which is represented in the figure, the numbers denoting feet. If the crib is entirely filled with corn in the ear, how many bushels of corn will it contain, counting 2 bushels of corn in the ear for 1 bushel of shelled corn. (Use the approxi- mation, 1 bushel = l\ cubic feet. For the exact volume of a bushel, see Ex. 1439.) Ex. 1350. A wheat elevator in the form of a frus- tum of a square pyramid is 30 feet high ; the edges of its bases are 12 feet and 6 feet, respectively. How many bushels of wheat will it hold? (Use the approximation given in Ex. 1349. ) Ex. 1351. A frustum of a regular square pyramid has an altitude of 12 inches, and the edges of its bases are 4 inches and 10 inches, respec- tively. Find the volume of the pyramid of which the frustum is a part. Ex. 1352. In a frustum of a regular quadrangular pyramid, the sides of the bases are 10 and 6, respectively, and the slant height is 14. Find the volume. Ex. 1353. Find the lateral area of a regular triangular pyramid whose altitude is 8 inches, and each side of whose base is 6 inches. Ex. 1354. The edge of a cube is a. Find the edge of a cube 3 times as large ; n times as large. 382 SOLID GEOMETRY 36 50 Ex. 1355. A berry box supposed to contain a quart of berries is in the form of a frustum of a pyramid 5 inches square at the top, 4| inches square at the bottom, and 2 inches deep. The United States dry quart contains 67.2 cubic inches. Does the box contain more or less than a quart ? Ex. 1356. The space left in a basement for a coal bin is a rectangle 8 x 10 feet. How deep must the bin be made to hold 10 tons of coal ? Ex. 1357. The figure represents a barn, the numbers denoting the dimensions in feet. Find the number of cubic feet in the barn. Ex. 1358. Let AB, BC, and BD, the dimensions of the barn in Ex. 1357, be denoted by a, ft, and c, respectively. Substitute the values of a, 6, and c in Ahmes' formula given in 777. Compare your result with the result obtained in Ex. 1357. Would Ahmes' formula have been correct if the Egyptian barns had been similar in shape to the barn in Ex. 1357? Ex. 1359. How much will it cost to paint the barn in Ex. 1357 at 1 cent per square foot for lateral surfaces and 2 cents per square foot for the roof ? ^ Ex. 1360. The barn in Ex. 1357 has a stone foundation 18 inches wide and 3 feet deep. Find the number of cubic feet of masonry if the outer surfaces of the walls are in the same planes as the sides of the barn. Ex. 1361. The volume of a regular tetrahedron is *f-V2. Find its edge, slant height, and altitude. Ex. 1362. The edge of a regular octahedron is a. Prove that the volume equals V2. 3 Ex. 1363. The planes determined by the diagonals of a parallelo- piped divide the parallelepiped into six equivalent pyramids. Ex. 1364. A dam across a stream is 40 feet long, 12 feet high, 7 feet wide at the bottom, and 4 feet wide at the top. How many cubic feet of material are there in the dam ? how many loads, counting 1 cubic yard to a load ? Give the name of the geometrical solid represented by the dam. Ex. 1365. Given 8 the lateral area, and H the altitude, of a regular square pyramid, find the volume. Ex. 1366. Find the volume F, of a regular square pyramid, if its total surface is 7 1 , and one edge of the base is a. BOOK VIII CYLINDERS AND CONES CYLINDERS 819. Def. A cylindrical surface is a surface generated by a moving straight line that continually intersects a fixed curve and remains parallel to a fixed straight line not coplanar with the given curve. N Cylindrical Surface FIG. 2. Cylinder 820. Defs. By referring to 693 and 694, the student may give the definitions of generatrix, directrix, and element of a cylindrical surface. Point these out in the figure. The student should note that by changing the directrix from a broken line to a curved line, a prismatic surface becomes a cylindrical surface. 821. Def. A cylinder is a solid closed figure whose boun- dary consists of a cylindrical surface and two parallel planes cutting the generatrix in each of its positions, as DC. 383 384 SOLID GEOMETRY 822. Defs. The two parallel plane sections are called the bases of the cylinder, as AC and DF (Fig. 4) ; the portion of the cylindrical surface between the bases is the lateral surface of the cylinder ; and the portion of an element of the cylin- drical surface included between the bases is an element of the cylinder, as MN. 823. Def . A right cylinder is a cylinder whose elements are perpendicular to the bases. FIG. 3. Right Cylinder FIG. 4. Oblique Cylinder 824. Def. An oblique cylinder is a cylinder whose elements are not perpendicular to the bases. 825. Def. The altitude of a cylinder is the perpendicular from any point in the plane of one base to the plane of the other base, as HK in Figs. 3 and 4. 826. The following are some of the properties of a cylinder ; the student should prove the correctness of each : (a) Any two elements of a cylinder are parallel and equal ( 618 and 634). (b) Any element of a right cylinder is equal to its altitude. (c) A line drawn through any point in the lateral surface of a cylinder parallel to an element, and limited by the bases, is itself an element ( 822 and 179). BOOK VIII 385 PROPOSITION I. THEOREM 827. The bases of a cylinder are equal. Given cylinder AD with bases AB and CD. To prove base AB = base CD. ARGUMENT ONLY 1. Through any three points in the perimeter of base AB, as E, F, and G, draw elements EH, FK, and GL. 2. Draw EF, FG, GE, HK, KL, and LH. 3. EH is II and = FK\ .-. EK is a O. 4. .: EF= HK; likewise FG = KL and GE =LH. 5. .'.A EFG = A JZKL. 6. .*. base AB may be placed upon base CD so that E, F, and (? will fall upon H, K, and L, respectively. 7. But E, F, and # are any three points in the perimeter of base AB ; i.e. every point in the perimeter of base AB will fall upon a corresponding point in the perimeter of base CD. 8. Likewise it can be shown that every point in the perim- eter of base CD will fall upon a corresponding point in the perimeter of base AB. 9. .*. base AB may be made to coincide with base CD. 10. .. base AB = base CD. Q.E.D. 828. Cor. I. The sections of a cylinder made by two parallel planes cutting all the elements are equal. 829. Cor. n. Every section of a cylinder made by a plane parallel to its base is equal to the base. 386 SOLID GEOMETRY Ex. 1367. Every section of a cylinder made by a plane parallel to its base is a circle, if the base is a circle. Ex. 1368. If a line joins the centers of the bases of a cylinder, this line passes through the center of every section of the cylinder parallel to the bases, if the bases are circles. 830. Def. A right section of a cylinder is a section formed by a plane perpendicular to an element, as section EF. FIG. 1. Cylinder with Circular Base AB Circular Cylinder FIG. 3. Eight Cir- cular Cylinder 831. Def. A circular cylinder is a cylinder in which a right section is a circle ; thus, in Fig. 2, if rt. section EF is a O, cylinder AD is a circular cylinder. 832. Def. A right circular cylinder is a right cylinder whose base is a circle (Fig. 3). 833. Questions. In Fig. 1, is rt. section EF a O ? In Fig. 2, is base AB a O ? In Fig. 3, would a rt. section be a O ? 834. Note. The theorems and exercises on the cylinder that follow will be limited to cases in which the bases of the cylinders are circles. When the term cylinder is used, therefore, it must be understood to mean a cylinder with circular bases. See also 846. Ex. 1369. Find the locus of all points at a distance of 6 inches from a straight line 2 feet long. Ex. 1370. Find the locus of all points : (a) 2 inches from the lateral surface of a right circular cylinder whose altitude is 12 inches and the radius of whose base is 5 inches ; (6)2 inches from the entire surface. Ex. 1371. A log is 20 feet long and 30 inches in diameter at the smaller end. Find the dimensions of the largest piece of square timber, the same size at each end, that can be cut from the log. BOOK VIII PROPOSITION II. THEOREM 387 835. Every section of a cylinder made by a plane pass- ing through an element is a parallelogram. (See 834.) Given cylinder AB with base AK, and CDEF a section ma'de by a plane through element CF and some point, as D, not in CF, but in the circumference of the base. To prove CDEF a O. ARGUMENT 1. Through D draw a line in plane DF II CF. 2. Then the line so drawn is an element; i.e. it lies in the cylindrical surface. 3. But this line lies also in plane DF. 4. /.it is the intersection of plane DF with the cylindrical surface, and coin- cides with DE. 5. .'. DE is a str. line and is II and = CF. 6. Also CD and EF are str. lines. 7. .'. CDEF is a O. Q.E.D. KEASONS 179. 826, c. Arg. 1. 614. 826, a. 616. 240. 836. Cor. Every section of a right circular cylinder made by a plane passing through an element is a rec- tangle. Ex. 1372. In the figure of Prop. II, the radius of the base is 4 inches, element CF is 12 inches, CD is 1 inch from the center of the base, and CF makes with CD an angle of 60. Find the area of section CDEF. Ex. 1373. Every section of a cylinder, parallel to an element, is a parallelogram. How is the base of this cylinder restricted ? (See 834.) 388 SOLID GEOMETRY Conical Surface CONES 837. Def. A conical surface is a surface generated by a moving straight line that continually intersects a fixed curve and that passes through a fixed point not in the plane of the curve. 838. Defs. By referring to 693, 694, and 746, the stu- dent may give the definitions of generatrix, directrix, vertex, element, and upper and lower nappes of a conical surface. Point these out in the figure. The student should observe that by changing the directrix from a broken line to a curved line, a pyramidal surface becomes a conical surface. 839. Def. A cone is a solid closed figure whose boundary consists of the .portion of a conical surface extending from its vertex to a plane cutting all its elements, and the section formed by this plane. 840. Defs. By referring to 748 and 822, the student may give 'the definitions of vertex, base, lateral surface, and element of a cone. Point these out in the figure. 841. Def. A circular cone is a cone containing a circular section such that a line joining the vertex of the cone to the center of the section is perpendicular to the section. Thus in Fig. 4, if section AB of cone V-CD is a O with center 0, such that VO is _L the section, cone V-CD is a circular cone. 842. Def. The altitude of a cone is the perpendicular from its vertex to the plane of its base, as VC in Fig. 3 and VO in Fig. 5, Fio. 2. Cone BOOK VIII 389 843. Defs. In a cone with a circular base, if the line join- ing its vertex to the center of its base is perpendicular to the FIG. 3. Cone with Circular Base FIG. 4. Circular Cone V FIG. 5. Right Circu- lar Cone plane of the base, the cone is a right cir- cular cone (Fig. 5). If such a line is not perpendicular to the plane of the base, the cone is called an oblique cone (Fig. 3). 844. Def. The axis of a right circular cone is the line joining its vertex to the center of its base, as FO, Fig. 5. 845. The following are some of the properties of a cone; the student should prove the correct- ness of each : (a) The elements of a right circular cone are equal. (b) TJie axis of a right circular cone is equal to its altitude. (c) A straight line drawn from the vertex of a cone to any point in the perimeter of its base is an element. 846. Note. The theorems and exercises on the cone that follow will be limited to cases in which the bases of the cones are circles, though not necessarily to circular cones. When the term cone is used, therefore, it must be understood to mean a cone with circular base. See also 834. Ex. 1374. What is the locus of all points 2 inches from the lateral surface, and 2 inches from the base, of a right circular cone whose alti- tude is 12 inches and the radius of whose base is 5 inches ? 390 SOLID GEOMETRY PROPOSITION III. THEOREM 847. Every section of a cone made by a plane passing through its vertex is a triangle. V B Given cone VAB with base AB and section VCD made by a plane through V. To prove VCD a A. ARGUMENT 1. From V draw str. lines to C and D. 2. Then the lines so drawn are elements ; i.e. they lie in the conical surface. 3. But these lines lie also in plane VCD. .-. they are the intersections of plane VCD with the conical surface, and coincide with VC and VD, respectively. Also CD is a str. line. .-. VC, VD, and CD are str. lines and VCD is a A. Q.E.D. 4. REASONS 1. 54, 15. 2. 845, c. 3. 603, a. 4. 614. 5. 616. 6. 92. Ex. 1375. What kind of triangle in general is the section of a cone through the vertex, if the cone is oblique ? if the cone is a right circular cone ? Can any section of an oblique cone be perpendicular to the base of the cone? of a right circular cone ? Explain. Ex. 1376. Find the locus of all straight lines making a given angle with a given straight line, at a given point in the line. What will this locus be if the given angle is 90 ? Ex. 1377. Find the locus of all straight lines making a given angle with a given plane at a given point. What will the locus be if the given angle is 90 ? BOOK VIII 391 PROPOSITION IV. THEOREM 848. Every section of a cone made by a plane parallel to its base is a circle. (See 846.) V Given CD a section of cone V-AB made by a plane II base AB. To prove section CD a O. OUTLINE OF PROOF 1. Let R and S be any two points on the boundary of section CD; pass planes through OF and points R and S. 2. Prove A VOM ~ A VPR and A VON ~ A VPS. 3. Then =!2 and ON _ VO . OM _ ON PR VP PS ~ VP ' PR ~ PS 4. But OM= ON-, .-. PS = PR; i.e. P is equidistant from any two points on the boundary of section CD. 5. .-. section CD is a O. Q.E.D. 849. Cor. Any section of a cone parallel to its base is to the base as the square of its distance from the vertex is to the square of the altitude of the cone. OUTLINE OF PROOF section CD PR By 563, Prove Then base AB PR OM section CD base AB VP vo VE VE VF For applications of 848 and 849, see EXS T 1380-1384. 392 SOLID GEOMETRY MENSURATION OF THE CYLINDER AND CONE AREAS 850. Def. A plane is tangent to a cylinder if it contains an element, but no other point, of the cylinder. 851. Def. A prism is inscribed in a cylinder if its lateral edges are elements of the cylinder, and the bases of the two figures lie in the same plane. 852. Def. A prism is circumscribed about a cylinder if its lateral faces are all tangent to the cylinder, and the bases of the two figures lie in the same plane. Ex. 1378. How many planes can be tangent to a cylinder ? If two of these planes intersect, the line of intersection is parallel to an element. How are the bases of the cylinders in 850-852 restricted ? (See 834.) 853. Before proceeding further it might be well for the student to review the more important steps in the develop- ment of the area of a circle. In that development it was shown that : (1) The area of a regular polygon circumscribed about a circle is greater, and the area of a regular polygon inscribed in a circle less, than the area of the regular circumscribed or in- scribed polygon of twice as many sides ( 541). (2) By repeatedly doubling the number of sides of regular circumscribed and inscribed polygons of the same number of sides, and making the polygons always regular, their areas approach a common limit ( 546). (3) This common limit is defined as the area of the circle ( 558). (4) Finally follows the theorem for the area of the circle ( 559). It will be observed that precisely the same method is used throughout the mensuration of the cylinder and the cone. Compare carefully the four articles just cited with 854, 855, 857, and 858. BOOK VIII 393 PROPOSITION V. THEOREM 854. I. The lateral area of a regular prism circum- scribed about a right circular cylinder is greater than the lateral area of the regular circumscribed prism whose base has twice as many sides. II. The lateral area of a regular prism inscribed in a right circular cylinder is less than the lateral area of the regular inscribed prism whose base has twice as many sides. The proof is left as an exercise for the student. HINT. See 541. Let the given figures represent the bases of the actual figures. Ex. 1379. A regular quadrangular and a regular octangular prism are inscribed in a right circular cylinder with altitude 25 inches and radius of base 10 inches. Find the difference between their lateral areas. Ex. 1380. The line joining the vertex of a cone to the center of the base, passes through the center of every section parallel to the base. Ex. 1381. Sections of a cone made by planes parallel to the base are to each other as the squares of their distances from the vertex. Ex. 1382. The base of a cone contains 144 square inches and the altitude is 10 inches. Find the area of a section of the cone 3 inches from the vertex ; 5 inches from the vertex. Ex. 1383. The altitude of a cone is 12 inches. How far from the vertex must a plane be passed parallel to the base so that the section shall be one half as large as the base ? one third ? one nth ? Ex. 1384. The altitude of a cone is 20 inches ; the area of the section parallel to the base and 12 inches from the vertex is 90 square inches. Find the area of the base. 394 SOLID GEOMETRY PROPOSITION VI. THEOREM 855. By repeatedly doubling the number of sides of the bases of regular prisms circumscribed about, and in- scribed in, a right circular cylinder, and malting the bases always regular polygons, their lateral areas approach a common limit. Given H the common attitude, R and r the apothems of the bases, P and p the perimeters of the bases, and S and s the lateral areas, respectively, of regular circumscribed and in- scribed prisms whose bases have the same number of sides. Let the given figure represent the base of the actual figure. To prove that by repeatedly doubling the number of sides of the bases of the prisms, and making them always regular polygons, S and s approach a common limit. 1. 2. 3. 4. 5. ARGUMENT S = P H and S = p - H. S_P s p' p s s S-s r R r R r REASONS 1. 763. 2. 54, 8 a. 3. 538. 4. 54, 1. 5. 399. BOOK VIII 395 6. 8. 9. 11. s= S ARGUMENT R r B But by repeatedly doubling the num- ber of sides of the bases of the prisms, and making them always regular polygons, R r can be made less than any previously assigned value, however small. .\ - - can be made less than any previously assigned value, however small. R S R can be made less than any previously assigned value, however small, S being a decreasing variable. 10. .-. S s, being always equal to S , can be made less than any previously assigned value, however small. S and s approach a common limit. Q.E.D. REASONS 6. 54, 7 a. 7. ' 543, I. 8. 586. 9. 587. 10. 309. 11. 594. 856. Note. The above proof is limited to regular prisms, but it can be shown that the limit of the lateral area of any inscribed (or circum- scribed) prism is the same by whatever method the number of the sides of its base is successively increased, provided that each side approaches zero as a limit. (See also 549.) Compare the proof of 855 with that of 546, I. 857. Def. The lateral area of a right circular cylinder is the common limit which the successive lateral areas of circum- scribed and inscribed regular prisms (having bases containing 3, 4, 5, etc., sides) approach as the number of sides of the bases is successively increased and each side approaches zero as a limit. 396 SOLID GEOMETRY PROPOSITION VII. THEOREM 858. The lateral area of a right circular cylinder is equal to the product of the circumference of its base and its altitude. Given a rt. circular cylinder with its lateral area denoted by , the circumference of its base by (7, and its altitude by //. To prove S = C H. ARGUMENT 1. Circumscribe about the rt. circular cylinder a regular prism. Denote its lateral area by S', the perimeter of its base by P, and its altitude by H. 2. Then S' = P - H. 3. As the number of sides of the base of the regular circumscribed prism is repeatedly doubled, P approaches O as a limit. 4. .-. P - H approaches C H as a limit. 5. Also S' approaches S as a limit. G. But S 1 is always equal to P H. 7. .-. 8= C - H. Q.E.D. REASONS 1. 852. 2. 763. 3. 550. 4. 590. 5. 857. 6. Arg. 2. 7. 355. 859. Cor. // S denotes the lateral area, T the total area, H the altitude, and E the radius of the base, of a right circular cylinder, BOOK VIII 397 T= = 2 860. Note. Since the lateral area of an oblique prism is equal to the product of the perimeter of a right section and a lateral edge ( 762), the student would naturally infer that the lateral area of an oblique cylinder with circular bases is equal to the product of the perimeter of a right sec- tion and an element. This statement is true. But the right section of an oblique cylinder with circular base is not a circle. And since the only curve dealt with in elementary geometry is. the circle, this theorem and its applications have been omitted here. Ex. 1385. Find the lateral area and total area of a right circular cylinder whose altitude is 20 centimeters and the diameter of whose base is 10 centimeters. Ex. 1386. How many square inches of tin will be required to make an open cylindrical pail 8 inches in diameter and 10 inches deep, making no allowance for waste ? Ex. 1387. In a right circular cylinder, find the ratio of the lateral area to the sum of the two bases. What is this ratio if the altitude and the radius of base are equal ? Ex. 1388. Find the altitude of a right circular cylinder if its lateral area is 8 and the radius of its base B. Ex. 1389. Find the radius of the base of a right circular cylinder if its total area is T and its altitude is H. 861. Def. Because it may be generated by a rectangle re- volving about one of its sides as an axis, a right circular cylinder is sometimes called a cylinder of rev- olution. 862. Questions. What part of the cylinder will side CD, opposite the axis AB, generate? What will AD and BC generate ? What will the plane AC gen- erate ? What might CD in any one of its positions be called ? 863. Def. Similar cylinders of revolution are cylinders generated by similar rectangles revolving about homologous sides as axes. 398 SOLID GEOMETRY PROPOSITION VIII. THEOREM 864. The lateral areas, and the total areas, of two similar cylinders of revolution are to each, other as the squares of their altitudes, and as the squares of the radii of their bases. Given two similar cylinders of revolution with their lateral areas denoted by S and s', their total areas by T and T 1 , their altitudes by IT and H', and the radii of their bases by .Rand R r , respectively. To prove: (.) -- H' R' 2 ARGUMENT = 27rRH and S f = 2 -rrR'H'. 2 # R' 2 Q.E.D. REASONS 8. 54, 8 a. 9. 401. 10. 309. 11. 309. Ex. 1390. The altitudes of two similar cylinders of revolution are 5 inches and 7 inches, respectively, and the total area of the first is 676 square inches. Find the total area of the second. Ex. 1391. The lateral areas of two similar cylinders of revolution are 320 square inches and 500 square inches, and the radius of the base of the larger is 10 inches. Find the radius of the base of the smaller. Ex. 1392. Two adjacent sides of a rectangle are a and b; find the lateral area of the cylinder generated by revolving the rectangle : (1) about a as an axis ; (2) about b as an axis. Put the results in the form of a general statement. Have you proved this general statement ? 865. Def. The slant height of a right circular cone is a straight line joining its vertex to any point in the circumference of its base. Thus any element of such a cone is its slant height. 866. Def. A plane is tangent to a cone if it contains an element, but no other point, of the cone. 867. Def. A pyramid is inscribed in a cone if its base is inscribed in the base of the cone and its vertex coincides with the vertex of the cone. 868. Def. A pyramid is circumscribed about a cone if its base is circumscribed about the base of the cone and its vertex coincides with the vertex of the cone. 869. The student may state and prove the theorems on the right circular cone corresponding to those mentioned in 854. Ex. 1393. How many planes can be tangent to a cone ? Through what point must each of these planes pass ? Prove. How are the bases of the cones in 860-868 restricted ? (See 846.) 400- SOLID GEOMETRY PROPOSITION IX. THEOREM 870. By repeatedly doubling the number of sides of the bases of regular pyramids circumscribed about, and in- scribed in, a right circular cone, and making the bases always regular polygons, their lateral areas approach a common limit. Given H the common altitude, L and I the slant heights, R and r the apothems of the bases. P and p the perimeters of the bases, and S and s the lateral areas, respectively, of regular circumscribed and inscribed pyramids whose bases have the same number of sides. Let the given figure represent the base of the actual figure. To prove that by repeatedly doubling the number of sides of the bases of the pyramids, and making them always regular polygons, S and s approach a common limit. ARGUMENT 1. S = 1 PL and s = 1 pi. p - ~ - s pi p I 2 3. i O / 1 J J 1\ LJ ' s r I rl' s RL rl 5. .-. RL REASONS 1. 766. 2. 54, 8 a. 3. 538. 4. 309. 5. 399. BOOK VIII 401 ARGUMENT 9. 10. 11. 12. 13. 14. 15. 16. 17. ,_. = RL Now L l'2 R' 2 R' R' 3 6. But, from Arg. 4, f- = . 7. ...JU* F f ^ _ ' 3 ' Q. E. D. REASONS 1. 890. 2. 54, 8 a. 3. 863. 4. 419. 5. 309. 6. 54, 13. 7. 309. Ex. 1410. The volumes of two similar cylinders of revolution are 135 cubic inches and 1715 cubic inches, respectively, and the altitude of the first is 3 inches. Find the altitude of the second. 410 SOLID GEOMETRY Ex. 1411. A cylinder of revolution has an altitude of 12 inches and a base with a radius of 5 inches. Find the total area of a similar cylinder whose volume is 8 times that of the given cylinder. Ex. 1412. The dimensions of a rectangle are 6 inches and 8 inches, respectively. Find the volume of the solid generated by revolving the rectangle : (a) about its longer side as an axis ; (6) about its shorter side. Compare the ratio of these volumes with the ratio of the sides of the rectangle. Ex. 1413. Cylinders having equal bases and equal altitudes are equivalent. Ex. 1414. Any two cylinders are to each other as the products of their bases and their altitudes. Ex. 1415. (a) Two cylinders having equal bases are to each other as their altitudes, and (&) having equal altitudes are to each other as their bases. Ex. 1416. The volume of a right circular cylinder is equal to the product of its lateral area and one half the radius of its base. Ex. 1417. Cut out a rectangular piece of paper 0x9 inches. Roll this into a right circular cylinder and find its volume (two answers). Ex. 1418. A cistern in the form of a right circular cylinder is to be 20 feet deep and 8 feet in diameter. How much will it cost to dig it at 5 cents a cubic foot ? Ex. 1419. Find the altitude of a right circular cylinder if its volume is Fand the radius of its base R. Ex. 1420. In a certain right circular cylinder the lateral area and the volume have the same numerical value, (a) Find the radius of the base. (&) Find the volume if the altitude is equal to the diameter of the base. Ex. 1421. A cylinder is inscribed in a cube whose edge is 10 inches. Find : (a) the volume of each ; (ft) the ratio of the cylinder to the cube. Ex. 1422. A cylindrical tin tomato can is 4^ inches high, and the diameter of its base is 4 inches. Does it hold more or less than a liquid quart, i.e. ^f- 1 cubic inches ? 892. The student may : (a) State and prove the theorems on the cone corresponding to those given in 885 and 886. (b) State, by aid of 888, the definition of the volume of a cone. BOOK VIII 411 PROPOSITION XVII. THEOREM 893. The volume of a cone is equal to one third the product of its base and its altitude. Given a cone with its volume denoted by F, its base by B, and its altitude by H. To prove V = J B - H. The proof is left as an exercise for the student. 894. Question. What changes must be made in the proof of Prop. XV to make it the proof of Prop. XVII ? 895. Cor. If V denotes the volume, H the altitude, and R the radius of tlie base of a cone, Ex. 1423. Any two cones are to each other as the products of their bases and altitudes. Ex. 1424. The slant height of a right circular cone is 18 inches and makes with the base an angle of 60 ; the radius of the base is 8 inches. Find the volume of the cone. Ex. 1425. The base of a cone has a radius of 12 inches ; an element of the cone is 24 inches long and makes with the base an angle of 30. Find the volume of the cone. Ex. 1426. The hypotenuse of a right triangle is 17 inches and one side is 15 inches. Find the volume of the solid generated by revolving the triangle about its shortest side as an axis. Ex. 1427. A cone and a cylinder have equal bases and equal alti- tudes. Find the ratio of their volumes. 896. Historical Note. To Eudoxus is credited the proof of the proposition that "every cone is the third part of a cylinder on the same base and with the same altitude." Proofs of this proposition were also given later by Archimedes and Brahmagupta. (Compare with 809. ) 412 SOLID GEOMETRY PROPOSITION XVIII. THEOREM 897. The volumes of two similar cones of revolution are to each other as the cubes of their altitudes, as the cubes of their slant heights, and as the cubes of the ra- dii of their bases. Given two similar cones of revolution with their volumes denoted by V and V', their altitudes by H and H\ their slant heights by L and L', and the radii of their bases by R and ', respectively. To prove Z* = 4 * V 1 H 13 L' 3 R' 3 The proof is left as an exercise for the student. HINT. Apply the method of proof used in Prop. XVI. Ex. 1428. If the altitude of a cone of revolution is three fourths that of a similar cone, what other fact follows by definition ? Compare the circumferences of the two bases ; their areas. Compare the total areas of the two cones ; their volumes. Ex. 1429. If the lateral area of a right circular cone is 1 T 9 times that of a similar cone, what is the ratio of their volumes ? of their altitudes ? Ex. 1430. Through a given cone X two planes are passed parallel to the base ; let Y denote the cone cut off by the upper plane, and Z the entire cone cut off by the lower plane. Prove that Y and Z are to each other as the cubes of the distances of the planes from the vertex of the given cone X. HINT. See Ex. 1381. Ex. 1431. Show that 897 is a special case^f Ex. 1430. Ex. 1432. The lateral area of a cone of revolution is 144 square inches and the total area 240 square inches. Find the volume. BOOK VIII 413 PROPOSITION XIX. THEOREM 898. The volume of a frustum of a cone is equal to one third the product of its altitude and the sum of its lower base, its upper base, and the mean proportional between its two bases. Given frustum AM, of cone 0-AF, with its volume denoted by F, its lower base by B y its upper base by 6, and its altitude by H. To prove F= -J- H(B + b + V^6). The proof is left as an exercise for the student. HINT. In the proof of 815, change "pyramid" to "cone." Cor. If v denotes the volume, H the altitude, and and r the radii of the bases of a frustum of a cone, Ex. 1433. Make a frustum of a right circular cone as indicated in Ex. 1401, and of the same dimensions. Find its volume. Ex. 1434. A tin pail is in the form of a frustum of a cone ; the diameter of its upper base is 12 inches, of its lower base 10 inches. How high must the pail be to hold 2j gallons of water ? (See Ex. 1422. ) Ex. 1435. A cone 6 feet high is cut by a plane parallel to the base and 4 feet from the vertex ; the volume of the frustum formed is 456 cubic inches. Find the volume of the entire cone. Ex. 1436. Find the ratio of the base to the lateral area of a right circular cone whose altitude is equal to the diameter of its base. 414 SOLID GEOMETRY MISCELLANEOUS EXERCISES Ex. 1437. The base of a cylinder is inscribed in a face of a cube whose edge is 10 inches. Find the altitude of the cylinder if its volume is equal to the volume of the cube. Ex. 1438. A block of marble in the form of a regular prism is 10 feet long and 2 feet 6 inches square at the base. Find the volume of the largest cylindrical pillar that can be cut from it. Ex. 1439. The Winchester bushel, formerly used in England, was the volume of a right circular cylinder 18| inches in internal diameter and 8 inches in depth. Is this the same volume as the bushel used in the United States (2150.42 cubic inches) ? Ex. 1440. To determine the volume of an irregular body, it was immersed in a vessel containing water. The vessel was in the form of a right circular cylinder the radius of whose base was 8 inches. On placing the body in the cylinder, the surface of the water was raised 10| inches. Find the volume of the irregular solid. Ex. 1441. In draining a certain pond a 4-inch tiling (i.e. a tiling whose inside diameter was 4 inches) was used. In draining another pond, supposed to contain half as much water, a 2-inch tiling was laid. It could not drain the pond. What was the error made ? Ex. 1442. A grain elevator in the form of a frustum of a right cir- cular cone is 25 feet high ; the radii of its bases are 10 feet and 5 feet, respectively ; how many bushels of wheat will it hold, counting 1|- cubic feet to a bushel ? Ex. 1443. The altitude of a cone with circular base is 16 inches. At what distance from the vertex must a plane be passed parallel to the base to cut the cone into two equivalent parts ? Ex. 1444. Two sides of a triangle including an angle of 120 are 10 and 20, respectively. Find the volume of the solid generated by revolv- ing the triangle about side 10 as an axis. Ex. 1445. Find the volume of the solid generated by revolving the triangle of Ex. 1444 about side 20 as an axis. Ex. 1446. Find the volume of the solid generated by revolving the triangle of Ex. 1444 about its longest side as an axis. Ex. 1447. The slant height of a right circular cone is 20 inches, and the circumference of its base 4 IT inches. A plane parallel to the base cuts off a cone whose slant height is 8 inches. Find the lateral area and the volume of the frustum remaining. BOOK VIII 415 Ex. 1448. A cone has an altitude of 12.5 feet and a base whose radius is 8.16 feet; the base of a cylinder having the same volume as the cone has a radius of 6.25 feet. Find the altitude of the cylinder. Ex. 1449. A log 20 feet long is 3 feet in diameter at the top end and 4 feet in diameter at the butt end. (a) How many cubic feet of wood does the log contain ? (6) How many cubic feet are there in the largest piece of square timber that can be cut from the log? (c) How many cubic feet in the largest piece of square timber the same size throughout its whole length ? (d) How many board feet does the piece of timber in (c) contain, a board foot being equivalent to a board 1 foot square and 1 inch thick ? HIXT. In (6) the larger end is square ABCD. What is the smaller end ? In (c) one end is square EFGH. What is the other end ? Ex. 1450. The base of a cone has a radius of 16 inches. A section of the cone through the vertex, through the center of the base, and per- pendicular to the base, is a triangle two of whose sides are 20 inches and 24 inches, respectively. Find the volume of the cone. Ex. 1451. The hypotenuse of a right triangle is 10 inches and one side 8 inches ; find the area of the surface generated by revolving the tri- angle about its hypotenuse as an axis. Ex. 1452. A tin pail in the form of a frustum of a righi circular cone is 8 inches deep ; the diameters of its bases are 8 inches and 10* inches, respectively. How many gallons of water will it hold ? (One liquid gallon contains 231 cubic inches.) Ex. 1453. The altitude of a cone is 12 inches. At what distances from the vertex must planes be passed parallel to the base to divide the cone into four equivalent parts ? HINT. See Ex. 1430. Ex. 1454. Find the volume of the solid generated by an equilateral triangle, whose side is a, revolving about one of its sides as an axis. Ex. 1455. Regular hexagonal prisms are inscribed in and circum- scribed about a right circular cylinder. Find (a) the ratio of the lateral areas of the three solids ; (6) the ratio of their total areas ; (c) the ratio of their volumes. 416 SOLID GEOMETRY Ex. 1456. How many miles of platinum wire ^ of an inch in diam- eter can be made from 1 cubic foot of platinum ? Ex. 1457. A tank in the form of a right circular cylinder is 5 feet long and the radius of its base is 8 inches. If placed so that its axis is horizontal and filled with gasoline to a depth of 12 inches, how many gallons of gasoline will it contain ? HINT. See Ex. 1024. Ex. 1458. Find the weight in pounds of an iron pipe 10 feet long, if the iron is inch thick and the outer diameter of the pipe is 4 inches. (1 cubic foot of bar iron weighs 7780 ounces.) Ex. 1459. In a certain right circular cone whose altitude and radius of base are equal, the total surface and the volume have the same numeri- cal value. Find the volume of the cone. Ex. 1460. Two cones of revolution lie on opposite sides of a common base. Their slant heights are 12 and 5, respectively, and the sum of their altitudes is 13. Find the radius of the common base. Ex. 1461. The radii of the lower and upper bases of a frustum of a right circular cone are R and R', respectively. Show that the area of a TT C K -4- 7?'^2 section midway between them is -^ "*" ; . Ex. 1462. A plane parallel to the base of a right circular cone leaves three fourths of the cone's volume. How far from the vertex is this plane ? How far from the vertex is the plane if it cuts off half the volume ? Answer the same questions for a cylinder. Ex. 1463. Is every cone cut from a right circular cone by a plane par- allel to its base necessarily similar to the original cone ? why ? How is it with a cylinder ? why ? Ex. 1464. Water is carried from a spring to a house, a distance of f mile, in a cylindrical pipe whose inside diameter is 2 inches. How many gallons of water are contained in the pipe ? Ex. 1465. A square whose side is 6 inches is revolved about one of its diagonals as an axis. Find the surface and the volume of the solid generated. Can you tind the volume of the solid generated by revolving a cube about one of its diagonals as an axis ? HINT. Make a cube of convenient size from pasteboard, pass a hat- pin through two diagonally opposite vertices, and revolve the cube rapidly. Ex. 1466. Given V the volume, and R the radius of the base, of a right circular cylinder. Find the lateral area and total area. Ex. 1467. Given the total area 2\ and the altitude //, of a right circular cylinder. Find the volume. BOOK IX THE SPHERE 900. Def. A sphere is a solid closed figure whose boundary is a curved surface such that all straight lines to it from a fixed poiot within are equal. 901. Defs. The fixed point within the sphere is called its center; a straight line joining the center to any point on the surface is a radius ; a straight line passing through the center and having its extremities on the surface is a diameter. 902. From the above definitions and from the definition of equal figures, 18, it follows that : (a) All radii of the same sphere, or of equal spheres, are equal. (b) All diameters of the same sphere, or of equal spheres, are equal. (c) Spheres having equal radii, or equal diameters, are equal. (d) A sphere may be generated by the revolution of a circle about a diameter as an axis. Ex. 1468. Find the locus of all points that are 3 inches from the sur- face of a sphere whose radius is 7 inches. Ex. 1469. The three edges of a trihedral angle pierce the surface of a sphere. Find the locus of all points of the sphere that are : (a) Equidistant from the three edges of the trihedral angle, (6) Equidistant from the three faces of the trihedral angle. Ex. 1470. Find a point in a plane equidistant from three given points in space. Ex. 1471. Find the locus of all points in space equidistant from the three sides of a given triangle. 417 418 SOLID GEOMETRY PROPOSITION I. THEOREM 903. Every section of a sphere made by a plane is a circle. Given AMBN a section of sphere made by a plane. To prove section AMBN a O. ARGUMENT 1. From draw OQ _L section AMBN. 2. Join Q to C and D, any two points on the perimeter of section AMBN. Draw OC and OD. 3. Ill rt. A OQC and OQD, OQ = OQ. 4. OC = OD. 5. /.A OQC = A OQD. 6. /. QC = QD-, i.e. any two points on the perimeter of section AMBN are equi- distant from Q. 1. .'. section AMBN is a O. Q.E.D. 904. Def. A great circle of a sphere is a section made by a; plane which passes through the center of the sphere, as O CRDS. 905. Def. A small circle of a sphere is a section made by a plane which does not pass through the center of the sphere, as O AMBN. REASONS 1. 639. 2. 54, 15. 3. By iden. 4. 902, a. 5. 211. 6. 110. 7. 276. BOOK IX 419 906. Def. The axis of a circle of a sphere is the diameter of the sphere which is perpendicular to the plane of the circle. 907. Def. The poles of a circle of a sphere are the extremi- ties of the axis of the circle. Ex. 1472. Considering the earth as a sphere, what kind of circles are the parallels of latitude ? the equator ? the meridian circles ? What is the axis of the equator ? of the parallels of latitude ? What are the poles of the equator ? of the parallels of latitude ? Ex, 1473. The radius of a sphere is 17 inches. Find the area of a section made by a plane 8 inches from the center. Ex. 1474. The area of a section of a sphere 45 centimeters from the center is 784 TT square centimeters. Find the radius of the sphere. Ex. 1475. The area of a section of a sphere 7 inches from the center is 576 TT. Find the area of a section 8 inches from the center. 908. The following are some of the properties of a sphere ; the student should prove the correctness of each : (a) In equal spheres, or in the same sphere, if two sections are equal, they are equally distant from the center, and conversely. HINT. Compare with 307. (6) In equal spheres, or in the same sphere, if two sections are unequal, the greater section is at the less distance from the center, and conversely. (HINT. See 308, 310.) (c) In equal spheres, or in the same, sphere, all great circles are equal. (HINT. See 279, c.) (d) 'TJie axis of a small circle of a sphere passes through the center of the circle, and conversely. (e) Any two great circles of a sphere bisect each other. (/) Every great circle of a sphere bisects the surface and the sphere. () a plane through R tangent to sphere Q. HINT. Compare with 373. Ex. 1494. Two planes tangent to a sphere at the extremities of a diameter are parallel. Ex. 1495. If the straight line joining the centers of two spheres is equal to the sum of their radii, the spheres are tangent to each other. HINT. Show that the radius of the O of intersection of the two spheres ( 915) is zero. 926. Def. A polyhedron is circumscribed about a sphere if each face of the polyhedron is tangent to the sphere. 927. Def. If a polyhedron is circumscribed about a sphere, the sphere is said to be inscribed in the polyhedron. 928. Def. A polyhedron is inscribed in a sphere if all its vertices are on the surface of the sphere. 929. Def. If a polyhedron is inscribed in a sphere, the sphere is said to be circumscribed about the polyhedron. Ex. 1496. Find the edge of a cube inscribed in a sphere whose radius is 10 inches. Ex. 1497. Find the volume of a cube : (a) circumscribed about a sphere whose radius is 8 inches ; (6) inscribed in a sphere whose radius is 8 inches. Ex. 1498. A right circular cylinder whose altitude is 8 inches is in- icribed in a sphere whose radius is 6 inches. Find the volume of the cylinder. Ex. 1499. A right circular cone, the radius of whose base is 8 inches, is inscribed in a sphere with radius 12 inches. Find the volume of the cone. Ex. 1500. Find the volume of a right circular cone circumscribed about a regular tetrahedron whose edge is . 426 SOLID GEOMETRY PROPOSITION VII. PROBLEM 930. To inscribe a sphere in a given tetrahedron, V D Given tetrahedron V-ABC. To inscribe a sphere in tetrahedron V-ABC. I. Construction 1. Construct planes RABS, SBCT, and TCAR bisecting dihedral A whose edges are AB, BC, and CA, respectively. 691. 2. From 0, the point of intersection of the three planes, construct OF A. plane ABC. 637. 3. The sphere constructed with as center and OF as radius will be inscribed in tetrahedron V-ABC. II. Proof ARGUMENT 1. Plane RABS, the bisector of dihedral Z. AB, lies between planes ABV and ABC; i.e. it intersects edge VC in some point as D. 2. .-. plane RABS intersects plane BCV in line BD and plane ACV in line AD. 3. Plane SBCT lies between planes BCV and ABC; i.e. it intersects plane RASB in a line through B between BA and BD, as & 4. Similarly plane TCAR intersects plane RABS iii a line through A between REASONS 1. By cons. 2. 616. 3. By cons. 4. By steps sim- ilar to 1 .",. BOOK IX 427 ARGUMENT AB and AD, as AR ; and plane SECT intersects plane TCAR in a line through C as CT. 5. But AR and BS pass through the in- terior of A ABD. 6. .-. AR and BS intersect in some point as 0, within A ABD. 1. .-. AR, BS, and CT are concurrent in point 0. 8. From draw OH, OK, and _L planes VAB, VBC, and FC4, respectively. 9. v is in plane OAB, OF = OH. 10. v is in plane OBC, OF = OK. 11. v is in plane OCA, OF = OL. 12. .-. OF= OH = OK= OL. 13. .-. each of the four faces of the tetra- hedron is tangent to sphere 0. 14. .-. sphere O is inscribed in tetrahedron V-ABC. Q.E.D. REASONS 5. Args. 3 & 4. 6. 194. 7. 617, I. 8. 639. 9. 688. 10. 688. 11. 688. 12. 54, 1. 13. 925. 14. 926, 927. III. Discussion The discussion is left as an exercise for the student. Ex. 1501. The six planes bisecting the dihedral angles of a tetrahe- dron meet in a point which is equidistant from the four faces of the tetrahedron. HINT. Compare with 258. Ex. 1502. Inscribe a sphere in a given cube. Ex. 1503. The volume of any tetrahedron is equal to the product of its surface and one third the radius of the inscribed sphere. HINT. See 491 and 492. Ex. 1504. Find a point within a triangular pyramid such that the planes determined by the lines joining this point to the vertices shall di- vide the pyramid into four equivalent parts. HINT. Compare with Ex. 1094. 428 SOLID GEOMETRY PROPOSITION VIII. PROBLEM 931. To circumscribe a sphere about a given tetrahe- dron. Given tetrahedron V-ABC. To circumscribe a sphere about tetrahedron V-ABC. I. Construction 1. Through D, the mid-point of AB, construct plane DO_L AB ; through E 9 the mid-point of BC, construct plane EOA.BC; and through F, the mid-point of VB, construct plane FO _L VB. 2. Join 0, the point of intersection of the three planes, to A. 3. The sphere constructed with as center and OA as radius will be circumscribed about tetrahedron V-ABC. II. Outline of Proof 1. Prove that the three planes OD, OE, and OF intersect each other in three lines. 2. Prove that these three lines of intersection meet in a point, as 0. 3. Prove that OA = OB = OC = OF. 4. . . sphere is circumscribed about tetrahedron V-ABC. Q.E.D. 932. Cor. Four points not in the same plane determine a sphere. 933. Questions. Are the methods used in 930 and 931 similar to those used in 321 and 323 ? In 930 could the lines forming the edges of the dihedral angles bisected be three lines meeting in one vertex ? In 931 could the three edges bisected be three lines lying in the same face ? BOOK IX 429 Ex. 1505. The six planes perpendicular to the edges of a tetrahedron at their mid-points meet in a point which is equidistant from the four ver- tices of the tetrahedron. Ex. 1506. The four lines perpendicular to the faces of a tetrahedron, and erected at their centers, meet in a point which is equidistant from the four vertices of the tetrahedron. Ex. 1507. Circumscribe a sphere about a given cube. Ex. 1508. Circumscribe a sphere about a given rectangular paral- lelepiped. Can a sphere be inscribed in any rectangular parallelepiped ? Explain. Ex. 1509. Find a point equidistant from four points in space not all in the same plane. SPHERICAL POLYGONS 934. Def. A line on the surface of a sphere is said to be closed if it separates a portion of the surface from the remain- ing portion. 935. Def. A closed figure on the surface of a sphere is a figure composed of a portion of the surface of the sphere and its bounding line or lines. 936. Defs. A spherical polygon is a closed figure on the surface of a sphere whose boundary is composed of three or more arcs of great circles, as ABCD. The bounding arcs are called the sides of the polygon, the points of intersection of the arcs are the vertices of the poly- gon, and the spherical angles formed by the sides are the angles of the polygon. 430 SOLID GEOMETRY 937. Def. A diagonal of a spherical polygon is an arc of a great circle joining any two non-adjacent vertices. 938. Def. A spherical triangle is a spherical polygon having three sides. Ex. 1510. By comparison with the definitions of the corresponding terms in plane geometry, frame exact definitions of the following classes of spherical triangles : scalene, isosceles, equilateral, acute, right, obtuse, and equiangular. Ex, 1511. With a given arc as one side, construct an equilateral spherical triangle. HINT. Compare the cons., step by step, with 124. Ex. 1512. With three given arcs as sides, construct a scalene spheri- cal triangle. 939. Since each side of a spherical polygon is an arc of a great circle ( 936), the planes of these arcs meet at the center of the sphere and form at that point a polyhedral angle, as polyhedral Z 0-ABCD. This polyhedral angle and the spheri- cal polygon are very closely related. The following are some of the more important relations; the student should prove the correctness of each : 940. (a) The sides of a spherical polygon have the same measures as the corresponding face angles of the poly- hedral angle. (b) The angles of a spherical polygon have the same measures as the corresponding dihedral angles of the polyhedral angle. Thus, sides AB, BC, etc., of spherical polygon ABGD have the same measures as face AAOB, BOG, etc., of polyhedral Z 0-ABCD ; and spherical A ABC, BCD, etc., have the same measures as the dihedral A whose edges are OB, OC, etc. These relations make it possible to establish certain prop- erties of spherical polygons from the corresponding known properties of the polyhedral angle, as in 941 and 942. BOOK IX PROPOSITION IX. THEOREM 431 941. The sum of any two sides of a spherical triangle is greater than the third side. Given spherical A ABC. To prove AB + BC > CA. ARGUMENT 1. Z AOB 4- Z 50C > Z COA. 2. Z ^1O.B QC AB, Z BOC&BC, Z COJ QC 3. .'. AB -f > Q.E.D. REASONS 1. 710. 2. 940, a, 3. 362, 6. PROPOSITION X. THEOREM 942. The sum of the sides of any spherical polygon is less than 360. Given spherical polygon ABC with n sides, To prove AB + BC + < 360. HINT. See 712 and 940, a. 432 SOLID GEOMETRY Ex. 1513. In spherical triangle ABC, arc AB = 40 and arc BC = 80. Between what limits must arc CA lie ? Ex. 1514. Any side of a spherical polygon is less than the sum of the remaining sides. Ex. 1515. If arc AB is the perpendicular bisector of arc CD, every point on the surface of the sphere and not in arc AB is unequally distant from C and D. Ex. 1516. In a spherical quadrilateral, between what limits must the fourth side lie if three sides are 60, 70, and 80? if three sides are 40, 60, and 70 ? Ex. 1517. Any side of a spherical polygon is less than 180. HINT. See Ex. 1514. 943. If, with A, B, and (7 the vertices of any spherical tri- angle as poles, three great circles are constructed, as B'c'ED, C'A'DJ and EA'B', the surface of the sphere will be divided into eight spherical triangles, four of which are seen on the hemi- sphere represented in the figure. Of these eight spherical tri- angles, A'B'C' is the one and only one that is so situated that A and A' lie on the same side of BC, B and B 1 on the same side of AC, and C and C' on the same side of AB. This particular tri- angle A'B'C' is called the polar triangle of triangle ABC. 944. Questions. In the figure above, A A'B'C', the polar of &ABC, is entirely outside of A ABC. Can the two & be so constructed that : (a) A'B'C 1 is entirely within ABC ? (6) A'B'C' is partly outside of and partly within ABC? Ex. 1518. What is the polar triangle of a spherical triangle all of whose sides are quadrants ? , BOOK IX PROPOSITION XI. THEOREM 433 945. If one spherical triangle is the polar of another, then the second is the polar of the first. Given A A'B'C' the polar of A ABC. To prove A ABC the polar of A A'B 1. A is the pole of B'c' ; i.e. AC' is a quad- rant. 2. B is the pole of A'C' ; i.e. BC 1 is a quad- rant. 3. /. C' is the pole of AB. 4. Likewise B' is the pole of AC, and A' is the pole of BC. 5. /.A ABC is the polar of A A'B'C'. Q.E.D. REASONS 1. 912. 2. 912. 3. 914. 4. By steps simi- lar to 1- 3. 5. 943. 946. Historical Note. The properties of polar triangles were dis- covered about 1626 A.D. by Albert Girard, a Dutch mathematician, born in Lorraine about 1595. They were also discovered independently and about the same time by Snell, an "infant prodigy," who at the age of twelve was familiar with the standard mathematical works of that time and who is remembered as the discoverer of the well-known law of refraction of light. Ex. 1519. Determine the polar triangle of a spherical triangle having two of its sides quadrants and the third side equal to 70 ; 110 ; (90 - a) ; (90 + a). 434 SOLID GEOMETRY PROPOSITION XII. THEOREM 947. In two polar triangles each angle of one and that side of the other of which its vertex is the pole are together equal, numerically, to 180 n - A Given- polar A ABC and A'B'c', with sides denoted by a, 6, c, and a', &', c', respectively. To prove: (a) Z ^+a' = 180, Z +6' = 180, Z <7+c' = lSO; (6) Z^'+a=180, Zj5'+&=180 , Z <7'+c=180. (a) ARGUMENT ONLY 1. Let arcs AB and AC (prolonged if necessary) intersect arc B'C' at D and E, respectively ; then c'D = 90 and EB' = 90. 2. .-. C'D + EB' = 180. 3. . ' . 180. 6. Again, a' -f 6' + c' > 0. 7. .-. Z'4 + Z + Z C<540. Q.E.D. 950. Cor. In a spherical triangle there can be one, two, or three right angles ; there can be one, two, or three obtuse angles. 951. Note. Throughout the Solid Geometry the student's attention has constantly been called to the relations between definitions and theorems of solid geometry and the corresponding definitions and theorems of plane geometry. In the remaining portion of the geometry of the sphere there will likewise be many of these comparisons, but here the student must be particularly on his guard for contrasts as well as comparisons. For ex- 436 SOLID GEOMETRY ample, while the sum of the angles of a plane triangle is equal to exactly 180, he has learned ( 949) that the sum of the angles of a spherical tri- angle may be any number from 180 to 540 ; while a plane triangle can have but one right or one obtuse angle, a spherical triangle may have one, two, or three right angles or one, two, or three obtuse angles ( 950). If the student will recall that, considering the earth as a sphere, the north and south poles of the earth are the poles of the equator, and that all meridian circles are. great circles perpendicular to the equator, it will make his thinking about spherical triangles more definite. 952. Def. A spherical triangle containing two right angles is called a birectangular spherical triangle. 953. Def. A spherical triangle having all of its angles right angles is called a trirectangular spherical triangle. Thus two meridians, as NA and NB, making at the north pole an acute or an obtuse Z, form with the equator a birectangular spherical A. If the Z between NA and NB is made a rt. Z, A ANB becomes a trirectangular spherical A. Ex. 1522. What kind of arcs are NA and NB ? Then what arc measures spherical angle ANB ? Are two sides of any birectangular spherical triangle quadrants? What is each side of a trirectangular spherical triangle ? Ex. 1523. If two sides of a spherical triangle are quadrants, the triangle is birectangular. (HINT. Apply 947.) Ex. 1524. What is the polar triangle of a trirectangular spherical triangle ? Ex. 1525. An exterior angle of a spherical triangle is less than the sum of the two remote interior angles. Compare this exercise with 2ir>, Make this new fact clear by applying it to a birortangular spherical triangle whose third angle is : (a) acute; (6) right; (c) obtuse. BOOK IX 487 PROPOSITION XIV. THEOREM 954. In equal spheres, or in the same sphere, two spher- ical triangles are equal : I. If a side and the two adjacent angles of one are equal respectively to a side and the two adjacent angles of the other; II. If two sides and the included angle of one are equal respectively to two sides and the included angle of the other ; III. // the three sides of one are equal respectively to the three sides of the other : provided the equal parts are arranged in the same order. The proofs are left as exercises for the student. HINT. In each of the above cases prove the corresponding trihedral A equal ( 702, 704) ; and thus show that the spherical & are equal. 955. Questions. Compare Prop. XIV, I and II, with 702, I and II, and with 105 and 107. Could the methods used there be employed in 954 ? Is the method here suggested preferable ? why ? 956. Def. Two spherical polygons are symmetrical if the corresponding polyhedral angles are symmetrical. 957. The following are some of the properties of symmet- rical spherical triangles ; the student should prove the correct- ness of each : (a) Symmetrical spherical triangles have their parts respectively equal, but arranged in reverse order. (6) Two isosceles symmetrical spherical triangles are equal. HINT. Prove (6) by superposition or by showing that the correspond- ing trihedral A are equal. 438 SOLID GEOMETRY PROPOSITION XV. THEOREM 958. In equal spheres, or in the same sphere, two sym- metrical spherical triangles are equivalent. A A' Given symmetrical spherical A ABC and A'B'C' in equal spheres and O 1 . To prove spherical AABC^ spherical A A'B'C'. ARGUMENT 1. Let P and P 1 be poles of small CD through A, B, C, and A 1 , B', C 1 , respectively. 2. Arcs AB, BC, CA are equal, respectively, to arcs A'B', B'c', C'A'. 3. .-. chords AB, BC, CA, are equal, respec- tively, to chords A'B', B'C', C'A'. 4. .-. plane A ABC = plane A A'B'C*. 5. .-. O ABC = O A' B'C'. 6. Draw arcs of great PA, PB, PC, P'A', P'B', and P'C 1 . 1. Then PA = PB = PC = P'A'= P'B'= P'C'. 8. .-. isosceles spherical A APB and A' P'B' are symmetrical. 9. .-. A APB = AA'P'B 1 . 10. Likewise A BPC = A tf'P'C' and A CPA = A C7'P'J'. 11. .-. spherical A 4C=D= spherical AA'B'C*. Q.E.D. REASONS 1. 908, h. 2. 957, a. 3. 298, IL 4. 116. 5. 324. 6. 908, g. 7. 913. 8. 956. 9. 957,6. 10. By steps simi- lar to 8-9. 11. 54, 2. BOOK IX 439 PROPOSITION XVI. THEOREM 959. In equal spheres, or in the same sphere, two spheri- cal triangles are symmetrical, and therefore equivalent : I. If a side and the two adjacent angles of one are equal respectively to a side and the two adjacent angles of the other ; II. If two sides and the included angle of one are equal respectively to two sides and the included angle of the other; III. If the three sides of one are equal respectively to the three sides of the other : provided the equal parts are arranged in reverse order. The proofs are left as exercises for the student. HINT. In each of the above cases prove the corresponding trihedral A symmetrical ( 709) ; and thus show that the spherical & are symmetrical. Ex. 1526. The bisector of the angle at the vertex of an isosceles spherical triangle is perpendicular to the base and bisects it. Ex. 1527. The arc drawn from the vertex of an isosceles spherical triangle to the mid-point of the base bisects the vertex angle and is per- pendicular to the base. Ex. 1528. State and prove the theorems on the sphere correspond- ing to the following theorems on the plane : (1) Every point in the perpendicular bisector of a line is equidistant from the ends of that line (134). (2) Every point equidistant from the ends of a line lies in the perpen- dicular bisector of that line ( 139). (3) Every point in the bisector of an angle is equidistant from the sides of the angle (253). HINT. In the figure for (3), corresponding to the figure of 252, draw PD JL AS and lay off BE = BD. Ex. 1529. The diagonals of an equilateral spherical quadrilateral are perpendicular to each other. Prove. State the theorem in plane geome- try that corresponds to this exercise. Ex. 1530. In equal spheres, or in the same sphere, if two spherical triangles are mutually equilateral, their polar triangles are mutually equi- angular ; and conversely. 440 SOLID GEOMETRY PROPOSITION XVII. THEOREM 960. In equal spheres, or in the same sphere, if two spherical triangles are mutually equiangular, they are mutually equilateral, and are either equal or symmet- rical. \P Given spherical A T and T 1 in equal spheres, or in the same sphere, and mutually equiangular. To prove T and T' mutually equilateral and either equal or symmetrical. ARGUMENT REASONS 1. Let A P and P' be the polars of A T and T 1 , respectively. 2. T and T' are mutually equiangular. 3. Then P and P' are mutually equi- lateral. 4. .*. P and P' are either equal or sym- metrical, and hence are mutually equiangular. 5. .*. T and T 1 are mutually equilateral. 6. .-. T and T 1 are either equal or sym- metrical. Q.E.D. 1. 943. 2. By hyp. 3. 947. 4. 954, III, and 959, III. 5. 947. 6. Same reason as 4. Ex. 1531. In plane geometry, if two triangles are mutually eqv angular, what can be said of them ? Are they equal ? equivalent ? Ex. 1532. Find the locus of all points of a sphere that are equidis- tant from two given points on the surface of the sphere ; from two givei points in space, not on the surface of the sphere. BOOK IX 441 PROPOSITION XVIII. THEOREM 961. The base angles of an isosceles spherical triangle are equal. Given isosceles spherical A ABC, with side AS = side BC. To prove Z. A = Z C. HINT. Compare with 111. PROPOSITION XIX. THEOREM 962. If two angles of a spherical triangle are equal, the sides opposite are equal, T \ Given spherical A RST with Z R = Z T. To prove r = t. ARGUMENT 1. Let A R's'T' be the polar of A SST. 2. Z R = Z r. 3. .-. r' = t'. 4. .-.R^T 1 . 5. .-. r = t. Q.E.D. REASONS 1. 943. 2. By hyp. 3. 947. 4. 961. 5. 947. 442 SOLID GEOMETRY PROPOSITION XX. THEOREM 963. // two angles of a spherical triangle are unequal, the side opposite the greater angle is greater than the side opposite the less angle. Given spherical A ABC with Z A > Z C. To prove BC> AB. ARGUMENT 1. Draw an arc of a great O AD making Z 1 = Z C. 2. Then ^Z> = <7. 3. But BD -f 2) > ifl. 4. . . S + Z>C' > 2e ; i.e. BC > AB. Q.E.D. REASONS 1. 908, g. 2. 962. 3. 941. 4. 309. Ex. 1533. In a birectangular spherical triangle the side included b] the two right angles is less than, equal to, or greater than, either of other two sides, according as the angle opposite is less than, equal to, 01 greater than 90. Ex. 1534. An equilateral spherical triangle is also equiangular. Ex. 1535. If two face angles of a trihedral angle are equal, the dihedral angles opposite are equal. Ex. 1536. State and prove the converse of Ex. 1534. Ex. 1537. State and prove the converse of Ex. 1535. Ex. 1538. The arcs bisecting the base angles of an isosceles spht cal triangle form an isosceles spherical triangle. Ex. 1539. The bases of an isosceles trapezoid are 14 inches and inches and the altitude 3 inches ; find the total area and volume of tl solid generated by revolving the trapezoid about its longer base as an axis. Ex. 1540. Find the total area and volume of the solid generated by revolving the trapezoid of Ex. 1539 about its shorter base as an axis. BOOK IX 443 PROPOSITION XXI. THEOREM 964. If two sides of a spherical triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less side. B Given spherical A ABC with BC > AB. To prove Z A > Z (7. The proof is left as an exercise for the student. HINT. Prove by the indirect method. 965. Questions. Could Prop. XXI have been proved by the method used in 156 ? Does reason 4 of that proof hold in spherical A ? See Ex. 1525. Ex. 1541. If two adjacent sides of a spherical quadrilateral are greater, respectively, than the other two sides, the spherical angle included between the two shorter sides is greater than the spherical angle between the two greater sides. HINT. Compare with Ex. 153. Ex. 1542. Find the total area and volume of the solid generated by revolving the trapezoid of Ex. 1539 about the perpendicular bisector of its bases as an axis. Ex. 1543. Find the radius of the sphere in- scribed in a regular tetrahedron whose edge is a. HINT. Let be the center of the sphere, A the center of face VED, and B the center of face EDF. Then OA = radius of inscribed sphere. Show that rt. A VAO and VBC are similar. Then VO : VC = VA : VS. VC, VA, and VB can be found (Ex. 1328). Find FO, then OA. Ex. 1544. Find the radius of the sphere circumscribed about a regu- lar tetrahedron whose edge is a. HINT. In the figure of Ex. 1543, draw AD and OD. 444 SOLID GEOMETRY MENSURATION OF THE SPHERE AREAS PROPOSITION XXII. THEOREM 966. If an isosceles triangle is revolved about a straight line lying in its plane and passing through its vertex but not intersecting its surface, the area of the surf ace gener- ated by the base of the triangle is equal to the product of its projection on the axis and the circumference of a circle whose radius is the altitude of the triangle- X X Given isosceles A AOB with base AB and altitude OE, a str. line XY lying in the plane of A AOB passing through and not intersecting the surface of A AOB, and CD the projection of AB on XY; let the area of the surface generated by AB be denoted by area AB. To prove area AB = CD 2 irOE. I. If AB is not II XY and does not meet XY (Fig. 1). ARGUMENT ONLY 1. From E draw EH _L XY. 2. Since the surface generated by AB is the surface of a frustum of a rt. circular cone, area AB = AB 2 TrEH. 3. From A draw AKA. BD. 4. Then in rt. A BAK and OEH, Z BAK /. OEH. BOOK IX 445 5. .-. ABAK~ AOEH. 6. .-. AB : AK = OE : EH', i.e. AB - EH = AK - OE. 7. But AK = CD ; .'. AB - EH= CD OE. 8. .*. area AB = CD 2 irOE. Q.E.D. II. If AB is not II XY and point A lies in XY (Fig. 2). III. If 4511 JKT (Fig. 3). The proofs of II and III are left as exercises for the student. HINT. See if the proof given for I will apply to Figs. 2 and 3. 967. Cor. I. If half of a regular polygon with an even number of sides is circumscribed about a semicircle, the area of the surface generated by its semiperimeter as it revolves about the diameter of the semicircle as an axis, is equal to the product of the diameter of the regular polygon and the circumference of a circle whose radius is E, the radius of the given semicircle. OUTLINE OF PROOF 1. Area 4 '5' = A'F' 2 -n-R; area B'c' = F'o' - 2 irR; etc. 'B'C'-" = (A'F' + F'O' + 968. Cor. H. If half of a regular polygon with an even number of sides is inscribed in a semicircle, the area of the surface generated by its semi- A perimeter as it revolves about the diameter of the semicircle as an axis, is equal to the product of the diameter of the semicircle and the circumference of a circle whose radius is the apothem of the regular polygon. j) HINT. Prove area AB (7 =AE-2ira. 446 SOLID GEOMETRY Cor. m. If halves of regular polygons with the same even number of sides are circumscribed about, and inscribed in, a semicircle, then by repeatedly doubling the number of sides of these polygons, and making the poly- gons always regular, the surfaces generated by the semi- perimeters of the polygons as they revolve about the diameter of the semicircle as an axis approach a com- mon limit. OUTLINE OF PROOF 1. If S and s denote the areas of the surfaces generated by the semi- perimeters A'B'c'-" and ABC--- as they revolve about A'E' as an axis, then S = A'E' - 2 -n-R ( 967) ; and s = AE 2 na ( 968). S A'E' 2 -rrR A'E' R AE a ~ polygon ABC 2. .-. 5 = S AE - 2 -rra 3. But polygon A'B'C'> ( 438). 4. 5. 6. A'E' A'B' R ... ^L = _ = - ( 419, 435). AE AB a^ S-S R R_ a a R 2 -a 2 R 2 S R 2 7. .-. by steps similar to Args. 6-11 ( 855), S and s ap- proach a common limit. Q.E.D. 970. Def. The surface of a sphere is the common limit which the successive surfaces generated by halves of regular polygons with the same even number of sides approach, if these eemipolygons fulfill the following conditions : (1) They must be circumscribed about, and inscribed in, a great semicircle of the sphere ; (2) The number of sides must be successively increased, each side approaching zero as a limit. BOOK IX PROPOSITION XXIII. THEOREM 447 971. The area of the surface of a sphere is equal to four times the area of a great circle of the sphere. Given sphere with its radius denoted by R, and the area of its surface denoted by S. To prove S = 4 TrJf? 2 . ARGUMENT 1. 2. In the semicircle ACE inscribe ABODE, half of a regular polygon with an even number of sides. Denote its apothem by a, and the area of the surface generated by the semiperime- ter as it revolves about AE as an axis by s'. Then s' = AE 2 wa ; i.e. S' = 2 R 2 no, = 4 irRa. 3. As the number of sides of the regular polygon, of which ABCDE is half, is repeatedly doubled, S 1 approaches S as a limit. 4. Also a approaches R as a limit. .-. 4 TTR - a approaches 4 irR R, i.e. 4 irR*, as a limit. But S' is always equal to 4 nR a. .: S = 4 TrR 2 . Q.E.D. 5. 6. 7. REASONS 1. 517, a. 2. 968. 3. 970. 4. 543, I. 5. 590. 6. Arg. 2. 7. 355. 448 SOLID GEOMETRY 972. Cor. I. The areas of the surfaces of two spheres are to each other as the squares of their radii and as the squares of their diameters. OUTLINE OF PROOF S' 4 TTtf' 2 R'* 2 But ^ = 4 * 2 = ( 2 *) 2 = D * L = **- R' 2 4*' 2 (2R') 2 D' 2 ' ' S' D'*' 973. Historical Note. Prop. XXIII is given as Prop. XXXV in the treatise entitled Sphere and Cylinder by Archimedes, already spoken of in 809. Ex. 1545. Find the surface of a sphere whose diameter is 16 inches. Ex. 1546. What will it cost to gild the surface of a globe whose radius is 1^ decimeters, at an average cost of f of a cent per square centimeter ? Ex. 1547. The area of a section of a sphere made by a plane 11 inches from the center is 3600 V square inches. Find the surface of the sphere. Ex. 1548. Find the surface of a sphere circumscribed about a cube whose edge is 12 inches. Ex. 1549. The radius of a sphere is It. Find the radius of a sphere whose surface is twice the surface of the given sphere ; one half ; one nth. Ex. 1550. Find the surface of a sphere whose diameter is 2 _R, and the total surface of a right circular cylinder whose altitude and diameter are each equal to 2 It. Ex. 1551. From the results of Ex. 1550 state, in the form of a theorem, the relation of the surface of a sphere to the total surface of the circumscribed cylinder. Ex. 1552. Show that, in Ex. 1550, the surface of the sphere is exactly equal to the lateral surface of the cylinder. 974. Historical Note. The discovery of the remarkable property that the surface of a sphere is two thirds of the surface of the circum- scribed cylinder (Exs. 1550 and 1551) is due again to Archimedes. The disco very of this proposition, and the discovery of the corresponding proposition for volumes ( 1001), were the philosopher's chief pride, and he therefore asked that a figure of this proposition be inscribed on his tomb. His wishes were carried out by his friend Marcellus. (For a further account of Archimedes, read also 542.) BOOK IX 449 975. Defs. A zone is a closed figure on the surface of a sphere whose boundary is composed of the circumferences of two circles whose planes are parallel. The circumferences forming the boundary of a zone are its bases. Thus, if semicircle NES is revolved about NS as an axis, arc AB will gen- erate a zone, while points A and B will generate the bases of the zone. 976. Def. The altitude of a zone is the perpendicular from any point in the plane of one base to the plane of the other base. 977. Def. If the plane of one of the bases of a zone is tan- gent to the sphere, the zone is called a zone of one base. Thus, arc NA or arc RS will generate a zone of one base. 978. Questions. Is the term " zone " used in exactly the same sense here as it is in the geography ? Name the geographical zones of one base ; of two bases. Name the five circles whose circumferences form the bases of the six geographical zones. Which of these are great circles ? * 979. Cor. n. The area of a zone is equal to the product of its altitude and the circumference of a great circle. OUTLINE *OF PROOF Let S denote the area generated by broken line A'B'C', s by broken line ABC, and Z by arc ABC; let DE, the altitude of the zone, be denoted by H. Then S = D'E' - 2 -n-R; S = DE 2 TTO,. .-. - = ^r- (See Args. 2-5, s a 2 969.) Then by steps similar to 969-971, Z = H-2TrR. * The student will observe that the projection used in this figure is different from that used in the other figures. 450 SOLID GEOMETRY 980. Cor. III. In equal spheres, or in the same sphere, the areas of two zones are to each other as their altitudes. 981. Question. In general, surfaces are to each other as the prod- ucts of two lines. Is 980 an exception to this rule ? Explain. Ex. 1553. The area of a zone of one base is equal to the area of a circle whose radius is the chord of the arc generating the zone. HINT. Use 979 and 444, II. Ex. 1554. Show that the formula of 971 is a special case of 979. Ex. 1555. Find the area of the surface of a zone if the distance between its bases is 8 inches and the radius of the sphere is 6 inches. Ex. 1556. The diameter of a sphere is 16 inches. Three parallel planes divide this diameter into four equal parts. Find the area of each of the four zones thus formed. Ex. 1557. Prove that one half of the earth's surface lies within 30 of the equator. Ex. 1558. Considering the earth as a sphere with radius .R, find the 7? 27? area of the zone adjoining the north pole, whose altitude is ; - Is the one area twice the other ? Ex. 1559. Considering the earth as a sphere with radius .R, find the area of the zone extending 30 from the north pole ; 60 from the north pole. Is the one area twice the other ? Ex. 1560. Considering the earth as a sphere with radius J?, find the area of the zone whose bases are parallels of latitude: (a) 30 and 45 C from the north pole ; (6) 30 and 45 from the equator. Are the two areas equal ? Explain your answer. Ex. 1561. How far from the center of a sphere whose radius is E must the eye of an observer be so that one sixth of the surface of the sphere is visible ? HINT. Let E be the eye of the observer. Then AB must =^. Find OB, then use 443, II. 3 Ex. 1562. What portion of the surface of a sphere can be seen if the distance of the eye of the observer from the center of the sphere is 2E ? 37?? n 2t? Ex. 1563. The radii of two concentric spheres are 6 inches and 10 inches. A plane is passed tangent to the inner sphere. Find : (a) the area of the section of the outer sphere made by the plane ; (6) the area of the surface cut off of the outer sphere by the plane. BOOK IX 451 982. Def. A lune is a closed figure on the surface of a sphere whose boundary is composed of two semicircumferences of great circles, as NA8B. 983. Defs. The two semicircumfer- ences are called the sides of the lune, as NAS and NBS-, the points of inter- section of the sides are called the vertices of the lune, as -ZV and S; the spherical angles formed at the vertices by the sides of the lune are called the angles of the lune, as A ANB and BSA. '984. Prove, by superposition, the following property of lunes : In equal spheres, or in the same sphere, two lunes are equal if their angles are equal. 985. So far, the surfaces considered in connection with the sphere have been measured in terms of square units, i.e. square inches, square feet, etc. For example, if the radius of a sphere is 6 inches, the surface of the sphere is 47r6 2 , i.e. 144 IT square inches. But, as the sides and angles of a lune and a spherical polygon are given in degrees and not in linear units, it will be necessary to introduce some new unit for determining the areas of these figures. For this purpose the entire surface of a sphere is thought of as being divided into 720 equal parts, and each one of these parts is called a spherical degree. Hence : 986. Def. A spherical degree is -^ of the surface of a sphere. Now if the area of a lune or of a spherical triangle can be ob- tained in spherical degrees, the area can easily be changed to square units. For example, if it is found that the area of a spherical triangle is 80 spherical degrees, its area is -f^, i.e. ^ of the entire surface of the sphere. On the sphere whose radius is 6 inches, the area of the given triangle will be J- of 144 TT square inches, i.e. 16 IT square inches. The following theorems are for the purpose of determining the areas of figures on the surface of a sphere in terms of spherical degrees. 452 SOLID GEOMETRY PROPOSITION XXIV. THEOREM 987. The area of a lune is to the area of the surface of the sphere as the number of degrees in the angle of the lune is to 360. N Q E FIG. 1. Given lune NASB with the number of degrees in its Z denoted by N, its area denoted by L, and the area of the surface of the sphere denoted by S ; let O EQ be the great O whose pole is N. L N To prove - = - s 360 I. If arc AB and circumference EQ are commensurable (Fig- 1). ARGUMENT REASONS 1. Let m be a common measure of arc AB 1. 335. and circumference EQ, and suppose that m is contained in arc AB r times and in circumference EQ t times, arc AB _ r t 2. Then circumference EQ 3. Through the several points of division on circumference EQ pass semicir- cumferences of great circles from N to S. 4. Then lime NASB is divided into r lunes and the surface of the sphere into t lunes, each equal to lune NCSS, 341. 908, h. 4. 984. BOOK IX 453 5. 6. L_ r s ~ t L S ARGUMENT arc AB 9. circumference EQ But arc AB is the measure of it contains N degrees. And circumference EQ contains 360. L N i.e. Q.E.D. REASONS 5. 341. 6. 54,1. 7. 918. 8. 297. 9. 309. II. If arc AB and circumference EQ are incommensurable (Fig. 2). The proof is left as an exercise for the student. HINT. The proof is similar to that of 409, II. 988. Cor. I. The area of a lune, expressed in spherical degrees, is equal to twice the number of degrees in its angle. OUTLINE OF PROOF 720 360 Ex. 1564. Find the area of a lune in spherical degrees if its angle is 35. What part is the lune of the entire surface of the sphere ? Ex. 1565. Find the area of a lune in square inches if its angle is 42 and the radius of the sphere is 8 inches. (Use ir = ^.) Ex. 1566. In equal spheres, or in the same sphere, two lunes are to each other as their angles. Ex. 1567. Two lunes in unequal spheres, but with equal angles, are to each other as the squares of the radii of their spheres. Ex. 1568. In a sphere whose radius is J?, find the altitude of a zone equivalent to a lune whose angle is 45. Ex. 1569. Considering the earth as a sphere with radius E, find the area of the zone visible from a point at a height h above the surface of the earth. 989. Def. The spherical excess of a spherical triangle is the excess of the sum of its angles over 180. 454 SOLID GEOMETRY PROPOSITION XXV. THEOREM 990. The area of a spherical triangle, expressed in spherical degrees, is equal to its spherical excess. Given spherical A ABC with, its spherical excess denoted by E. To prove area of A ABC = E spherical degrees. ARGUMENT 1. Complete the circumferences of which AB, BC, and CA are arcs. 2. A AB'c' and A'BC are symmetrical. 3. .'. A AB'C' =0= A A'BC. 4. .-. A ABC + A AB'C' o A ABC + A A'BC. 5. .-., expressed in spherical degrees, A ABC -f A AB'C' =0 lune A = 2 A ; A ABC + A AB'C = lune B = 2 B ; A ABC + A ABC' = lune C = 2 C. f>. .-. 2AABC+(AABC+AAB'C' + AAB'C + A ABC')= 2(A + B+C). 7. EutAABC+AAB'C'+AAB'C+AABC' = surface of a hemisphere = 360. 8. .-. 2 A ABC-\- 360 = 2(^4-^ + C). 9. .-. A^J5(7+180= 4 + #+ (7. 10. .-.A ABC = (A + B + C) 180, i.e. E spherical degrees. Q.E.D. REASONS 1. 908, g. 2. 95G. 3. 958. 4. 54, 2. 5. 988. 6. 54, 2. 7. 985. 8. 309. 9. 54, 8 a. 10. 54, 3. BOOK IX 455 991. In 949 it was proved that the sum of the angles of a spherical triangle is greater than 180 and less than 540. Hence the spherical excess of a spherical triangle may vary from to 360, from which it follows ( 990) that the area of a spherical triangle may vary from y|^ to ff-g- of the entire surface ; i.e. the area of a spherical triangle may vary from nothing to ^ the surface of the sphere. Thus in a spherical triangle whose angles are 70, 80, and 100, respectively, the spherical excess is (70 + 80 + 100) - 180 = 70 ; i.e. the area of the given, triangle is -f^ of the surface of the sphere. 992. Historical Note. Menelaus of Alexandria (circ. 98 A.D.) wrote a treatise in which he describes the properties of spherical triangles, although there is no attempt at their solution. The expression for the area of a spherical triangle, as stated in 990, was first given about 1626 A.D. by Girard. (See also 946.) This theorem was also discovered in- dependently by Cavalieri, a prominent Italian mathematician. Ex. 1570. If three great circles are drawn, each perpendicular to the other two, into how many trirectangular spherical triangles is the surface divided ? Then what is the area of a trirectangular spherical triangle in spherical degrees ? Test your answer by applying Prop. XXV. Ex. 1571. Find the area in spherical degrees of a birectangular spherical triangle one of whose angles is 70 ; of an equilateral spherical triangle one of whose angles is 80. What part of the surface of the sphere is each triangle ? Ex. 1572. The angles of a spherical triangle in a sphere whose sur- face has an area of 216 square feet are 95, 105, and 130. Find the number of square feet in the area of the triangle. Ex. 1573. In a sphere whose diameter is 16 inches, find the area of a triangle whose angles are 70, 86, and 120. Ex. 1574. The angles of a spherical triangle are 60, 120, and 160, and its area is lOOf square inches. Find the radius of the sphere. (Use =) Ex. 1575. The area of a spherical triangle is 90 spherical degrees, and the angles are in the ratio of 2, 3, and 5. Find the angles. Ex. 1576. Find the angle (1) of an equilateral spherical triangle, (2) of a lune, each equivalent to one third the surface of a sphere. Ex. 1577. Find the angle of a lune equivalent to an equilateral spherical triangle one of whose angles is 84. 456 SOLID GEOMETRY PROPOSITION XXVI. THEOREM 993. The area of a spherical polygon, expressed in spherical degrees, is equal to the sum of its angles dimin- ished by 180 taken as many times less two as the poly- gon has sides. Given spherical polygon ABCD with n sides; denote the sum of its angles by T. To prove area of polygon ABCD , expressed in spherical degrees, = T(n 2)180. ARGUMENT 1. From any vertex such as A, draw all possible diagonals of the polygon, forming n 2 spherical A, I, II, etc. 2. Then, expressed in spherical degrees, A I = (Z 1 + Z 2 + Z 3) - 180 ; AII=(Z 4-f-Z 5 + Z 6) -180; etc. 3. .-. AI + AII+ ... = r-(n-2)180. 4. .-. area of polygon ABCD = r-(w 2) 180. Q.E.D. 1. REASONS 937. 2. 990. 54, 2. 309. Ex. 1578. Prove Prop. XXVI by using a figure similar to that used in 216. Ex. 1579. Find the area of a spherical polygon whose angles are 80, 92, 120, and 140, in a sphere whose radius is 8 inches. Ex. 1580. Find the angle of an equilateral spherical triangle equiva- lent to a spherical pentagon whose angles are 00, 100, 110, 130, and 140. Ex. 1581. Find one angle of an equiangular spherical hexagon equivalent to six equilateral spherical triangles each with angles of 70. BOOK IX 457 1 K---.-X 1 Ex. 1582. The area of a section of a sphere 63 inches from the cen- ter is 256 TT square inches. Find the surface of the sphere. Ex. 1583. The figure represents a sphere inscribed in a cylinder, and two cones with the bases of the cylinder as their bases and the center of the sphere as their vertices. Any plane, as BS, is passed through the figure parallel to MN, the plane of the base. Prove that the ring between section AD of the cylinder, and section CD of the cone, is always equivalent to the section of the sphere. Ex. 1584. Find the volume of a barrel 30 inches high, 54 inches in circumference at the top and bottom, and 64 inches in circumference at the middle. HIXT. Consider the barrel as the sum of two frustums of cones. Ex. 1585. Given T the total area, and E the radius of the base, of a right circular cylinder. Find the altitude. Ex. 1586. Given $the lateral area, and E the radius of the base, of a right circular cone. Find the volume. Ex. 1587. Given 8 the lateral area, and T the total area, of a right circular cone. Find the radius and the altitude. VOLUMES 994. Note. The student should not fail to observe the striking similarity in the figures and theorems, as well as in the definitions, relat- ing to the areas and volumes connected with the measurement of the sphere. A careful comparison of the following articles will emphasize this similarity : AREAS 966 > 967, 968 969 970 971 972 975 977 VOLUMES 995 996, a 996, 6 996, c 997 999 1002 8 1003 AREAS 979 982 984 987 988 936 990 VOLUMES 5 1004, 1005 1006 1007 1008 1009 1010 1012 1013 458 SOLID GEOMETRY PROPOSITION XXVII. THEOREM 995. If an isosceles triangle is revolved about a straight line lying in its plane and passing through its vertex but not intersecting its surface, the volume of the solid gener- ated is equal to the product of the surface generated by the base of the triangle and one third of its altitude. \-D E B Y FIG. 2. FIG. 3. Given isosceles A A OB with altitude OE, and a str. line XY lying in the plane of A AOB, passing through and not inter- secting the surface of A AOB; let the volume of the solid gen- erated by A AOB revolving about XY as an axis be denoted by volume AOB. To prove volume AOB = area AB -J OE. I. If AB is not II XY and does not meet XY (Fig. 1). ARGUMENT ONLY 1. Draw AC and BD _L XY. 2. Prolong BA to meet XY at F. 3. Then volume .40.B = volume FOB volume FOA. 4. Volume FOB = volume FDB + volume DOB. . 5. .-. volume FOB = ^irlsi? - FD + I TT Z> 2 -DO = 1 TT BD\FD + 0) = ^irBD - BD But BD - FO twice area of A ^OJ5 = .ay 0#. .-. volume FOB = % TT BD BF OE = TT BD - BF J OE. But 7r## BF = area ^^. .-. volume FOB = area FB i O.E. FO. 6. 7. 8. 9. BOOK IX 459 10. Likewise volume FOA = area FA 1 OE. o 11. .-. volume AOB = area FB OE area FA - ^ OE = (area FB area FA) OE = area AB i OE. Q.E.D. II. If AB is not II XY and point A lies in XY (Fig. 2). The proof is left as an exercise for the student.* HINT. See Arg. 9 or Arg. 10 of 995, I. III. If AB II XY (Fig. 3). The proof is left as an exercise for the student. HINT. Volume AOB = volume ACDB twice volume CO A. 996. The student may : (a) State and prove the corollaries on the volume of a sphere corresponding to 967 and 968. (&) State and prove the theorem on the volume of a sphere corresponding to 969. 1. 3. OUTLINE OF PROOF V (area A'B'C' ^R = A'E' >27rR-R 2 J?2 A f T?rf ' ~o" TTXt ./I Ji ( 996, a and 967). C '< v = (area 4.B (7 )|- a 996, a and 968). * A'E' 2 AE B' D v -- Tra - AE a AE a 3 4. Proceed as in 855, observing that since the limit of =/?( 543, I), the limit of a s = R 3 ( 593); i.e. R 3 -a 3 may be made less than any previously assigned value, however small. (c) State, by aid of 970, the definition of the volume of a sphere. 460 SOLID GEOMETRY PROPOSITION XXVIII. THEOREM 997. The volume of a sphere is equal to the product of the area of its surface and one third its radius. B Given sphere with its radius denoted by R, the area of its surface by fif, and its volume by F. To prove V=S-R. ARGUMENT 1. In the semicircle ACE inscribe ABODE, half of a regular polygon with an even number of sides. Denote its apothem by a, the area of the surface generated by the semiperimeter as it revolves about AE as an axis by s', and the vol- ume of the solid generated by semi- polygon A B ODE by F'. 2. Then V 1 = S' J a. 3. As the number of sides of the regular polygon, of which ABODE is half, is repeatedly doubled, V 1 approaches F as a limit. 4. Also s' approaches S as a limit. 5. And a approaches R as a limit. 6. .'. S' a approaches S R as a limit. 7. .'. S 1 ' ^ a approaches S ^ R as a limit. 8. But V is always equal to s' J. 9. /. V=S I R. Q.K.I). REASONS 517, a. 996, a. 996, c. 970. 543, I. 592. 590. Arg. 2. BOOK IX 461 998. Cor. I. If v denotes the volume, R the radius, and D the diameter of a sphere, V=TTR*=lirD*. 999. Cor. n. The volumes of two spheres are to each other as the cubes of their radii and as the cubes of their diameters. (HINT. See 972.) 1000. Historical Note. It is believed that the theorem of 999 was proved as early as the middle of the fourth century B.C. by Eudoxus, a great Athenian mathematician already spoken of in 809 and 896. Ex. 1588. Find the volume of a sphere inscribed in a cube whose edge is 8 inches. Ex. 1589. The volume of a sphere is 1774| TT cubic centimeters. Find its surface. Ex. 1590. Find the radius of a sphere equivalent to a cone with alti- tude a and radius of base 6. Ex. 1591. Find the radius of a sphere equivalent to a cylinder with the same dimensions as those of the cone in Ex. 1590. Ex 1592. The metal cone and cylinder in the figure have their alti- tude and diameter each equal to 2 B, the diameter of the sphere. Place the sphere in the cylinder, then fill the cone with water and empty it into the cylinder. How nearly is the cylinder filled ? Next fill the cone with water and empty it into the cylinder three times. Is the cylinder filled ? Ex. 1593. From the results of Ex. 1592 state, in the form of a theorem, the relation of the volume of a sphere : (a) to the volume of a circumscribed cylinder; (6) to the volume of the corresponding cone. Prove these statements. 1001. Historical Note. The problem "To find a sphere equivalent to a given cone or a given cylinder" (Exs. 1590 and 1591), as well as the properties that the volume of a sphere is two thirds of the volume of the circumscribed cylinder and twice the volume of the corresponding cone (Exs. 1592 and 1593), are due to Archimedes. The importance attached to this by the author himself is spoken of more fully in 542 and 974. 462 SOLID GEOMETRY Ex. 1594. A bowl whose inner surface is an exact hemisphere is made to hold | gallon of water. Find the diameter of the bowl. Ex. 1595. A sphere 12 inches in diameter weighs, 93 pounds. Find the weight of a sphere of the same material 16 inches in diameter. Ex. 1596. In a certain sphere the area of the surface and the volume have the same numerical value. Find the volume of the sphere. Ex. 1597. Find the volume of a spherical shell 5 inches thick if the radius of its inner surface is 10 inches. Ex. 1598. A pine sphere 24 inches in diameter weighs 175 pounds. Find the diameter of a sphere of the same material weighing 50 pounds. Ex. 1599. The radius of a sphere is H. Find the radius of a sphere whose volume is one half the volume of the given sphere ; twice the vol- ume ; n times the volume. 1002. Defs. A spherical sector is a solid closed figure gen- erated by a sector of a circle revolving about a diameter of the circle as an axis. C FIG. 2. FIG. 3. The zone generated by the arc of the circular sector is called the base of the spherical sector. 1003. Def. If one radius of the circular sector generating a spherical sector is a part of the axis, i.e. if the base of the spherical sector is a zone of one base, the spherical sector is sometimes called a spherical cone. Thus if circular sector AOB (Fig. 1) revolves about diameter CD as an axis, arc AB will generate a zone which will be the base of the spherical sector generated by circular sector AOB (Fig. 2). If circular sector HOC revolves about diameter CD y the spherical sector generated, whose base is the zone generated by arc BC, will be a spherical cone (Fig. 3). BOOK IX 463 1004. Cor. HE. The volume of a spherical sector is equal to the product of its base and one third the radius of the sphere. OUTLINE OF PROOF Let V denote the volume generated by polygon OA'B'c 1 , v the volume generated by polygon OABC, S the area of the surface generated by broken line A'B'C', s the area of the surface generated by broken line ABC, and Z the base of the spherical sector generated by cir- cular sector AOC. Then V = S - 1 R, and v = s 4 a. But - = y (Args. 2-5, 969). S CL R? R a 2 a Then by steps similar to 996, b and c, and 997, the volume of the spherical sector generated by circular sector AOC = Z J R. 1005. Cor. IV. If v denotes the volume of a spherical sector, z the area of the zone forming its base, H the alti- tude of the zone, and R the radius of the sphere, V=Z-\R (1004) (979) Ex. 1600. Considering the earth as a sphere with radius B, find the volume of the spherical sector whose base is a zone adjoining the north 7? 97? pole and whose altitude is ; - 3 3 Is the one volume twice the other? Compare your results with those of Ex. 1558. Ex. 1601. Considering the earth as a sphere with radius 22, find the volume of the spherical sector whose base is a zone extending : (a) 30 from the north pole ; (ft) 60 from the north pole. Is the one volume twice the other ? Compare your results with those of Ex. 1559. 464 SOLID GEOMETRY Ex. 1602. Considering the earth as a sphere with radius J?, find the volume of the spherical sector whose base is a zone lying between the parallels of latitude : (a) 30 and 45 from the north pole ; (6) 30 and 45 from the equator. Are the two volumes equal ? Compare your results with those of Ex. 1560. Ex. 1603. Considering the earth as a sphere with radius R, find the area of the zone whose bases are the circumferences of small circles, one 30 north of the equator, the other 30 south of the equator. What part of the entire surface is this zone ? Ex. 1604. ' What part of the entire volume of the earth is that por- tion included between the planes of the bases of the zone in Ex.- 1603 ? HINT. This volume consists of two pyramids and a spherical sector. Ex. 1605. A spherical shell 2 inches in thickness contains the same amount of material as a sphere whose radius is 6 inches. Find the radius of the outer surface of the shell. Ex. 1606. A spherical shell 3 inches thick has an outer diameter of 16 inches. Find the volume of the shell. Ex. 1607. Find the volume of a sphere circumscribed about a rec- tangular parallelepiped whose edges are 3, 4, and 12. Ex. 1608. Find the volume of a sphere inscribed in a cube whose volume is 686 cubic centimeters. Ex. 1609. The surface of a sphere and the surface of a cube are each equal to 8. Find the ratio of their volumes. Which is the greater ? Ex. 1610. In a certain sphere the volume and the circumference of a great circle have the same numerical value. Find the surface and the vol- ume of the sphere. Ex. 1611. How many bullets \ of an inch in diameter can be made from a sphere of lead 10 inches in diameter ? from a cube of lead whose edge is 10 inches ? 1006. Defs. A spherical -wedge is a solid closed figure whose bounding surface consists of a June and the planes of the sides of the lune. The lune is called the base of the spherical wedge, and the angle of the lime the angle of the spherical wedge. 1007. Prove, by superposition, the following property of wedges : In equal spheres, or in the same sphere, two spherical wedges are equal if their angles are equal. BOOK IX 465 PROPOSITION XXIX. THEOREM 1008. The volume of a spherical wedge is to the volume of the sphere as the number of degrees in the angle of the spherical wedge is to 360. N N Q E Given spherical wedge NASB with the number of degrees in its Z denoted by N, its volume denoted by TT, and the volume of the sphere denoted by F; let O EQ be the great O whose pole is N. W N To prove = . v 360 The proof is left as an exercise for the student. HINT. The proof is similar to that of 987. 1009. Cor. I. The volume of a spherical wedge is equal to the product of its base and one third the radius of the sphere. OUTLINE OF PROOF S i R = L i R, where S represents w = -*-. r = -*- 360 360 the area of the surface of the sphere, and L the area of the lune, i.e. the area of the base of the spherical wedge. Ex. 1612. Tn a sphere whose radius is 16 inches, find the volume of a spherical wedge whose angle is 40. 466 SOLID GEOMETRY 1010. Defs. A spherical pyramid is a solid closed figure whose bounding surface consists of a spherical polygon and the planes of the sides of the spherical polygon. The spherical polygon is /^ the base, and the center of the sphere the vertex, of the spherical pyramid. 1011. By comparison with 957, b and 958, prove the following prop- erty of spherical pyramids : In equal spheres, or in the same sphere, two triangular spherical pyramids whose bases are symmetrical spherical triangles are equivalent. 1012. Cor. II. The volume of a spherical triangular pyramid is equal to the product of its base and one third the radius of the sphere. OUTLINE OF PROOF 1. Pyramid 0-AB'c' o pyramid O-A'BC ( 1011). 2. .-. pyramid 0-ABC -f pyramid < 0-AB'c' =o wedge A = 2 A 1 R ( 1009) ; pyramid 0-ABC + pyra- mid O-AB'C = wedge B = 2 B - ifl; pyramid 0ABC -f- pyramid 0ABC' = wedge C = 2 c - ^ R. 3. .-. twice pyramid 0-ABC + hemispnere = 2 (A -f- B + C) 1 R. 4. .-. twice pyramid 0-ABC -+- 360 1 R = 2(A -f B -f c) R. 5. .-. pyramid 0-ABC = (A + B -f C 180) | R = A ABC ^ R = K R. Q.E.D. 1013. Cor. in. The volume of any spherical pyramid is eqiuyl to the product of its base and one third the ra- dius of the sphere. (HINT. Compare with 805.) Ex. 1613. Show that the formula of 007 is a special case of 1004, 1000 and 1013. Ex. 1614. In a sphere whose radius is 12 inches, find the volume of a spherical pyramid whose base is a triangle with angles 70, 80, and 00. BOOK IX 467 MISCELLANEOUS EXERCISES ON SOLID GEOMETRY Ex. 1615. A spherical pyramid whose base is an equiangular penta- gon is equivalent to a wedge whose angle is 30. Find an angle of the base of the pyramid. Ex. 1616. The volume of a spherical pyramid whose base is an equi- angular spherical triangle with angles of 105 is 128 IT cubic inches. Find the radius of the sphere. Ex. 1617. In a sphere whose radius is 10 inches, find the angle of a spherical wedge equivalent to a spherical sector whose base has an alti- tude of 12 inches. Ex. 1618. Find the depth of a cubical tank that will hold 100 gallons of water. Ex. 1619. The altitude of a pyramid is H. At what distance from the vertex must a plane be passed parallel to the base so that the part cut off is one half of the whole pyramid ? one third ? one nth ? Ex. 1620. Allowing 550 pounds of copper to a cubic foot, find the weight of a copper wire \ of an inch in diameter and 2 miles long. Ex. 1621. Disregarding quality, and considering oranges as spheres, i.e. as similar solids, determine which is the better bargain, oranges averaging 2f inches in diameter at 15 cents per dozen, or oranges averaging 3^ inches in diameter at 30 cents per dozen. Ex. 1622. In the figure, J5, O, and D are the mid-points of the edges of the cube meet- ing at A. What part of the whole cube is the pyramid cut off by plane BCD ? HINT. Consider ABC as the base and D as the vertex of the pyramid. Ex. 1623. Is the result of Ex. 1622 the same if the figure is a rectangular parallelepiped ? any parallelepiped ? Ex. 1624. It is proved in calculus that in order that a cylindrical tin can closed at the top and having a given capacity may require the small- est possible amount of tin for its construction, the diameter of the base must equal the height of the can. Find the dimensions of such a can holding 1 quart ; 2 gallons. Ex. 1625. A cylindrical tin can holding 2 gallons has its height equal to the diameter of its base. Another cylindrical tin can with the same capacity has its height equal to twice the diameter of its base. Find the ratio of the amount of tin required for making the two cans. Is your answer consistent with the fact contained in Ex. 1624 ? 468 SOLID GEOMETRY Ex. 1626. A cannon ball 12 inches in diameter is melted, and the lead is cast in the form of a cube. Find the edge of the cube. Ex. 1627. The cube of Ex. 1626 is melted, and the lead is cast in the form of a cone, the diameter of whose base is 12 inches. Find the altitude of the cone. Ex. 1628. Find the weight of the cannon ball in Ex. 1626 if a cubic foot of iron weighs 450 pounds. Ex. 1629. The planes determined by the diagonals of a cube divide the cube into six equal pyramids. Ex. 1630. Let D, E, F, and G be the mid-points of VA, AB, EC, and OF, respectively, of triangular pyramid V-ABC. Prove DEFG a parallelogram. Ex. 1631. In the figure, is plane DEFG par- allel to edge AC? to edge VB? Prove that any section of a triangular pyramid made by a plane parallel to two opposite edges is a parallelogram. Ex. 1632. The three lines joining the mid- points of the opposite edges of a tetrahedron bisect each other and hence meet in a point. HINT. Draw DF and EG. Are these two of the required lines ? Ex. 1633. In a White Mountain two-quart ice cream freezer, the can is 4f- inches in diameter and Q\ inches high ; the tub is 6| inches in diameter at the bottom, 8 inches at the top, and 9f inches high, inside measurements, (a) Does the can actually hold 2 quarts ? (6) How many cubic inches of ice can be packed about the can ? Ex. 1634. Find the total area of a regular tetrahedron whose alti- tude is a centimeters. Ex. 1635. The lateral faces of a triangular pyramid are equilateral triangles, and the altitude of the pyramid is 6 inches. Find the total area. Ex. 1636. In the foundation work of the Woolworth Building, a 55- story building on Broadway, New York City, it was necessary, in order to penetrate the sand and quicksand to bed rock, to sink the caissons that contain the huge shafts of concrete to a depth, in some instances, of 131 feet. If the largest circular caisson, 19 feet in diameter, is 130 feet deep and was filled with concrete to within 30 feet of the surface, how many loads of concrete were required, considering 1 cubic yard to a load ? Ex. 1637. From A draw a line meeting line XY in B ; let G be the mid-point of AB. Find the locus of C as B moves in line XY. Ex. 1638. In Ex. 1637, let XY be a plane. Find the locus of C as B moves arbitrarily in plane XY. BOOK IX 469 Ex. 1639. A granite shaft in the form of a frustum of a square pyramid contains 161| cubic feet of granite ; the edges of the bases are 4 feet and 1 feet, respectively. Find the height of the shaft. Ex. 1640. The volume of a regular square pyramid is 42f cubic feet ; its altitude is twice one side of the base, (a) Find the total surface of the pyramid ; (6) find the area of a section made by a plane parallel to the base and one foot from the base. Ex. 1641. Allowing 1 cubic yard to a load, find the number of loads of earth in a railway cut \ mile in length, the average dimensions of a cross section being as represented in the figure, the numbers denoting feet. Give the name of the geometrical solid represented by the cut. Why is it not a frustum of a pyramid ? Ex. 1642. For protection against fire, a tank in the form of a frustum of a right circular cone was. placed in the tower room of a certain public building. The tank is 16 feet in diameter at the bottom, 12 feet in di- ameter at the top, and 16 feet deep. If the water in the tank is never allowed to get less than 14 feet deep, how many cubic feet of water would be available in case of an emergency ? how many barrels, counting 4^ cubic feet to a barrel ? Ex. 1643. A sphere with radius E is inscribed in a cylinder, and the cylinder is inscribed in a cube. Find : (a) the ratio of the volume of the sphere to that of the cylinder ; (&) the ratio of the cylinder to the cube ; (c) the ratio of the sphere to the cube. Ex. 1644. A cone has the same base and altitude as the cylinder in Ex. 1643. Find the ratio of the cone : (a) to the sphere ; (6) to the cylinder; (c) to the cube. Ex. 1645. In a steam-heated house the heat for a room was supplied by a series of 10 radiators each 3 feet high. gl The average cross section of a radiator is s\ ~ "TN shown in the figure, the numbers denoting yj 1 ^ *\J inches. It consists of a rectangle with a semicircle at each end. Find the total radiating surface in the room. Ex. 1646. A coffee pot is 5 inches deep, 4 inches in diameter at the top, and 5| inches in diameter at the bottom. How many cups of coffee will it hold , allowing 6 cups to a quart ? (Answer to nearest whole number. ) Ex. 1647. Any plane passing through the center of a parallelepiped divides it into two equivalent solids. Are these solids equal ? Ex. 1648. From two points, P and _Z?, on the same side of plane AB, two lines are drawn to point in plane AB, making equal angles with the plane. Find the locus of point O. (HINT. See Ex. 1237.) 470 SOLID GEOMETRY Ex. 1649. A factory chimney is in the form of a frustum of a regu- lar square pyramid. The chimney is 120 feet high, and the edges of its bases are 12 feet and 8 feet, respectively. The flue is 6 feet square throughout. How many cubic feet of material does the chimney contain? Ex. 1650. Find the edge of the largest cube that can be cut from a regular square pyramid whose altitude is 10 inches and one side of whose base is 8 inches, if one face of the cube lies in the base of the pyramid. Ex. 1651. Fig. 1 represents a granite monument, the numbers denoting inches. The main part of the stone is 5 feet high, the total height of the stone being 5 feet 6 inches. Fig. 2 represents a view of 54 FIG. 1. the main part of the stone looking directly from above. Fig. 3 repre- sents a view of the top of the stone looking directly from above. Calcu- late the volume of the stone. HINT. From Fig. 2 it is seen that the main part of the stone con- sists of a rectangular parallelepiped A, four right triangular prisms J5, and a rectangular pyramid at each corner. Fig. 3 shows that the top con- sists of a right triangular prism and two rectangular pyramids. Ex. 1652. The monument in Ex. 1651 was cut from a solid rock in the form of a rectangular parallelepiped. How many cubic feet of granite were wasted in the cutting ? Ex. 1653. In the monument of Ex. 1651 the two ends of the main part, and the top, have a rock finish, the front and rear surfaces of the main part being polished. Find the number of square feet of rock finish and of polished surface. Ex. 1654. The base of a regular pyramid is a triangle inscribed in a circle whose radius is J?, and the altitude of the pyramid is 211. Find the lateral area of the pyramid. Ex. 1655. Find the weight in pounds of the water required to fill the tank in Ex. 1323, if a cubic foot of water weighs 1000 ounces. BOOK IX 471 Ex. 1656. By using the formula obtained in Ex. 1543, find the vol- ume of the sphere inscribed in a regular tetrahedron whose edge is 12. Ex. 1657. By using the formula obtained in Ex. 1544, find the volume of the sphere circumscribed about a regular tetrahedron whose edge is 12. Ex. 1658. A hole 6 inches in diameter was bored through a sphere 10 inches in diameter. Find the vol- ume of the part cut out. HINT. The part cut out consists of two spherical cones and the solid generated by revolving isosceles A BOG about XY as an axis. Ex. 1659. Check your result for Ex. 1658 by finding the volume of the part left. Ex. 1660. Find the area of the spherical surface left in Ex. 1658. Ex. 1661. Four spheres, each with a radius of 6 inches, are placed on a plane surface in a triangular pile, each one being tangent to each of the others. Find the total height of the triangular pile. Ex. 1662. Find the total height of a triangular pile of spheres, each with radius of 6 inches, if there are three layers; four layers; n layers. FORMULAS OF SOLID GEOMETRY 1014. In addition to the notation given in 761, the follow- ing will be used : A, -B, (7, = number of degrees in the angles of a spherical polygon. a, 6, c, = sides of a spherical polygon. 7? = base of spherical sec- tor, wedge, and pyramid. C = circumference of base in general or of lower base of frus- tum of cone. c = circumference of up- per base of frustum of cone. Z> = diameter of a sphere. E = spherical excess of a spherical triangle. H = altitude of zone or spherical sector. K = area of a spherical triangle or spherical polygon. L = area of lune. N = number of degrees in the angle of a lune or wedge. E = radius of base in general, of lower base of frustum of cone, or of sphere. r = radius of upper base of frustum of cone. S = area of surface of a sphere. T = sum of the angles of a spheri- cal polygon. W volume of a wedge. Z = area of a zone. 472 SOLID GEOMETRY FIGURE FORMULA REFERENCE Prism. S = P E. 762. Right prism. S=P- H. 763. Regular pyramid. 8 = \ P- L. 766. Frustum of regular pyramid. S = I (P .+ p}L. 767. Rectangular parallelepiped. V= a b - c'. 778. Cube. V-E^. ' 779. Rectangular parallelepipeds. Rectangular parallelepiped. Rectangular parallelepipeds. Any parallelepiped. Parallelepipeds. Triangular prism. Any prism. Prisms. Triangular pyramid. Any pyramid. Pyramids. Similar tetrahedrons. Frustum of any pyramid. Truncated right triangular prism. Right circular cylinder. V a-b-c 780. 782. 783. 790. 792. 797. 799. 801. 804. 805. 807. 812. V a'-b' c' V=B- H; V B H V B' - H' V=B-H. v _ B.H . V B< . H 1 V=B-H. V=B-H. V B-H V B' H> V= i B- H. V B-H V B> H' V _ E* V E' s ' V=\B(E+E> + E" 8=C-H. b). 815. ). 817. 858. 859. 859. Similar cylinders of revolution. = L- = A. . 864. /S" H 1 ' 2 R' z T^W^W* Right circular cone. 8 = \ C L. 873. S = * R - L. 875. #) 875. Similar cones of revolution. , = -^ = ^ = -^ . 878. BOOK IX 473 FIGURE FORMULA REFERENCE Similar cones of revolution. T _ H* _ U- _ R* 878. T 1 H 1 ' 2 L 1 ' 2 R 1 ' 2 Frustum of right circular cone. 8 = l(C+c)L. 882. S = irL (R + r}. 883. - T ?rZ,(.R+r)-t-7r( J R 2 + r 2 ) . 883. , Cylinder with circular bases. V=B-H. 889. V = irR 2 H. 890. Similar cylinders of revolution. V l = H rs = W^ 891. Cone with circular base. V = I B H. 893. V = \ irR* H. 895. Similar cones of revolution. V H s L 8 R s 897. V ~ H' s ~~ L' 3 ~ R' s Frustum of cone with circular base. V = 1 H(B + b + V1T&). 898. V = \ irH (R 2 + r 2 + R r) . 899. Spherical triangle. a -f b > c. 941. Spherical polygon. a + 6 + c + < 360. 942. Polar triangles. A -f a' = 180, ^+&' = 180, . 947. Spherical triangle. A + B+ C > 180 and < 540. 949. Sphere. 8 = 4 TrR 2 . 971. Spheres. S /_ R 2 _ D 2 S' R 1 ' 2 D' 2 ' 972. Zone. Z = H-2irR. 979. Zones. Z _H 980. Z' H'' Lune. L _ N 987. 13 360 L = 2N. 988. Spherical triangle. K = (A+B+C)18QP=E. 990. Spherical polygon. K= T (n -2)180. 993. Sphere. V= S-%R.~ 997. R QQQ F7?3 713 yo. Spheres. -ft U ~Vl~ JJ/3 ~~ J)ls' 999. Spherical sector. V=Z-$R. 1004. V = | irR 2 H. 1005. Spherical wedge. W _ N 1008. V 360 W=L-$R. 1009. Spherical triangular pyramid. V= K- \R. 1012. Any spherical pyramid. V K- \R. 1013. 474 SOLID GEOMETRY APPENDIX TO SOLID GEOMETRY SPHERICAL SEGMENTS 1015. Defs. A spherical segment is a solid closed figure whose bounding surface consists of a zone and two parallel planes. Spherical Segment of Two Bases Spherical Segment of One Base The sections of the sphere formed by the two parallel planes are called the bases of the spherical segment. 1016. Defs. State, by aid of 976 and 977, definitions of: (a) Altitude of a spherical segment, (b) Segment of one base. PROPOSITION I. PROBLEM 1017. To derive a formula for the volume of a spher- ical segment in terms of the radii of its bases and its altitude. E Given spherical segment generated by ABCD revolving about EF as an axis, with its volume denoted by r, its altitude by h, and the radii of its bases by TI and r z , respectively. To derive a formula for V in terms of r 1} r 2 , and h. APPENDIX 475 Draw radii OC and OD. Then F= volume of spherical sector generated by COD + volume of cone generated by BOC vol- ume of cone generated by AOD. Denote OA by k, and the radius of the sphere by R. .-. v = I irR-li + ^ TTI\- (h -|- k) I TT r.:k But R 2 = ?y + & 2 ; and R 2 = rf + (h + fc) 2 . Solving these two equations for .R 2 and k, 2 rffi - 2 ^ V r 1018. Cor. I. Problem. To derive a formula for the volume of a spherical segment of one base : (a) In terms of its altitude and the radius of its base ; (6) In terms of its altitude and the radius of the sphere. (a) In 1017, put ^ = 0; then F=uT JJ 2 /i + -J-ir/i 3 . (b) If h represents the altitude of a segment of one base, and ?* 2 the radius of the base, then r 2 2 = h (2 # //,). 443, I. Q.E.F. Ex. 1663. A dumb-bell consists of the major portion of a sphere with diameter 6 inches attached to each end of a right circular cylinder 12 inches long and 2 inches in diameter. Find the volume of the segment cut from each sphere in fitting it to the cylinder. Ex. 1664. By means of the formulas given in 1017 and 1018, solve Exs. 1004 and 1658. THE PRISMATOID 1019. Def. A prismatoid is a polyhedron having for bases two polygons in parallel planes, and for lateral faces triangles or trapezoids with one side lying in one base, and the opposite vertex or side lying in the other base, of the polyhedron. 1020. Def. The altitude of a prismatoid is the length of the perpendicular from any point in the plane of one base to the plane of the other base. 476 SOLID GEOMETRY PROPOSITION II. PROBLEM 1021. To derive a formula for the volume of a prismatoid. FIG. 1. Given prismatoid CF with its volume denoted by v, its lower base by , its upper base by b, its altitude by H, and a section midway between the bases by M. To derive a formula for V in terms of B, b, H, and M. If any lateral face as AD is a trapezoid, divide it into two A by diagonal AD, intersecting NK at L. Let P be any point in M and join it to all vertices of the prismatoid. This will divide the prismatoid into pyramids having their vertices at P and having for their bases B, b, and the triangles forming the lateral faces of the prismatoid. The volume of pyramid P-B ^B-^H=^H-B; and the volume of pyramid P-b = %b>\ H = H-b. 805. Consider pyramid P-ADC. Draw PK, PL, and LC (Fig. 2). This divides pyramid P-ADC into three pyramids, D-KLP, C-KLP, and P-ALC. Denote A KLP by Wj. Then volume of pyramid D-KLP = ^ H m x ; and the vol- ume of pyramid C-KLP = J5T w^. 805. Pyramid P-ALC A ALC but A ALC Pyramid P-CLK (i.e. C-KLP) A CLK* A CLK .'. pyramid P-ALCo twice pyramid C-KLP. .*. volume of pyramid P-ALC = % H m 2 . AC LK APPENDIX 477 . . pyramid PADC = -J H m^-J- H m^ + 177 m l = ^ H 4 Wj. .*. the volume of all lateral pyramids = -J- # 4 3f. lf. Q.E.F. Ex. 1665. By substituting in the prismatoid formula, derive the formula for : (a) the volume of a prism ( 799) ; (6) the volume of a pyramid ( 805) ; (c) the volume of a frustum of a pyramid ( 815). Ex. 1666. Solve Ex. 1651 by applying the prismatoid formula to each part of the monument. SIMILAR POLYHEDRONS* 1022. The student should prove the following : (a) Any two homologous edges of two similar polyhedrons have the same ratio as any other two homologous edges. (b) Any two homologous faces of two similar polyhedrons have the same ratio as the squares of any two homologous edges. (c) The total surfaces of two similar polyhedrons have the same ratio as the squares of any two homologous edges. 1023. Def. The ratio of similitude of two similar polyhe- drons is the ratio of any two homologous edges. 1024. Def. If two polyhedrons ABCD and A'B'C'D' are so situated that lines from a point O to A', B', C', D' } etc., are divided by points A, B, C, D, etc., in such a mariner that 4_' _ OB' OA ~~ OB OC OD E' the two polyhedrons are said to be radially placed. Ex. 1667. Construct two polyhedrons radially placed and so that point lies between the two polyhedrons ; withm the two polyhedrons. * See 811. In this discussion only convex polyhedrons will be considered. 478 SOLID GEOMETRY PROPOSITION III. THEOREM 1025. Any two radially placed polyhedrons are similar. (See Fig. 2 below.) Given polyhedrons EC and E'c' radially placed with respect to point 0. To prove polyhedron EC ~ polyhedron E'c'. AB, BC, CD, and DA are II respectively to A'ti', B'c', C'D', and D'A'. 415. .-. ABCD II A'B'C'D', and is similar to it. 756, II. Likewise each face of polyhedron EC is ~ to the correspond- ing face of polyhedron E'c', and the faces are similarly placed. Again, face All II face A'n', and face AF II face A'F 1 . .: dihedral Z. AE= dihedral Z A'E 1 . Likewise each dihedral Z of polyhedron EC is equal to its corresponding dihedral Z. of polyhedron L'C'. .-. each polyhedral Z. of polyhedron EC is equal to its corre- sponding polyhedral Z of polyhedron E'c 1 . 18. .. polyhedron EC ~ polyhedron E'C'. 811. Q.E.D. PROPOSITION IY. THEOREM 1026. Any two similar polyhedrons may be radially placed. N Fia. 1. FIG. 2. Given two similar polyhedrons XM and E'C'. To prove that XM and E'c' may be radially placed. APPENDIX 479 OUTLINE OF PROOF 1. Take any point within polyhedron E'C 1 and construct polyhedron EC so that it is radially placed with respect to E'c' and so that OA' : OA = OB 1 : OS = = A'B' : KL. 2. Then polyhedron EC ~ polyhedron E'C'. 1025. 3. Prove that the dihedral A of polyhedron EC are equal, respectively, to the dihedral A of polyhedron XM, each being equal, respectively, to the dihedral A of polyhedron E'c 1 . 4. Prove that the faces of polyhedron EC are equal, respec- tively, to the faces of polyhedron XM. 5. Prove, by superposition, that polyhedron EC = XM. 6. .-. polyhedron JfJ/may be placed in the position of EC. 7. But EC and E'C' are radially placed. 8. .-. X M and E'c' may be radially placed. Q.E.D. PROPOSITION V. THEOREM 1027. If a pyramid is cut by a plane parallel to its base : I. The pyramid cut off is similar to the given pyra- mid. II. The two pyramids are to each other as the cubes of any two homologous edges. O A The proofs are left as exercises for the student. HINT. For the proof of II, pass planes through OS' and diagonals B'D', B'E', etc., dividing each of the pyramids into triangular pyramids. Then pyramid 0-BCD ~ pyramid O-B'C'D' ; pyramid 0-EBD ~ pyra- mid 0-E'B'D', etc. Use 812 and a method similar to that used in 505. 480 SOLID GEOMETRY PROPOSITION VI. THEOREM. 1028. Two similar polyhedrons are to each other as the cubes of any two homologous edges, ' Given two similar poly- hedrons XM and E'c', with their volumes denoted by V and F f , respectively, and with KL and A'B' two homol. edges. To prove = V' Place XM in position EC, so that XM and E'C' are radially placed with respect to point within both polyhedrons. 1026. Denote the volumes of pyramids 0-ABCD, 0-AEFB, etc., by VH v s , etc., and the volumes of pyramids 0-A'B'c'D 1 , 0-A'E'F'B 1 , etc., by Vi, v 2 ', etc. Then ; ^=^L ; etc. 1027,11. tt I . I |O AB A'B' 3 IV A' AE AE* But ^ = -^-=... (1022, a); .-. - - = ^- = .. tint A '~ci* ^ ' ' ~~ io r 3 ^1 /j ^1 JL A B A E V-i -f- Va ~\~ '" AB n t(\-t . l . V = -^. 401. AB polyhedron g(7 _ ~Alf m . V__ polyhedron E'c' ~~ A J B' 3 ' F ' KL Q.E.D. 1029. Note. Since 1028 was assumed early in the text (see 814), the teacher will find plenty of exercises throughout Books VII, VIII, and JX illustrating this principle. INDEX (The numbers refer to articles.) ART. Adjacent dihedral angles . 671 Altitude, of cone . . . . 842 of cylinder .... 825 of prism 733 of prismatoid .... 1020 of pyramid .... 751 of spherical segment . 1016 of zone 976 Angle, dihedral .... 666 magnitude of .... 669 of lune 983 ' of spherical wedge . . 1006 of two intersecting arcs 916 polyhedral 692 solid 692 spherical 917 tetrahedral .... 698 trihedral 698 Angles, designation of 667, 696 Axis, of circle of sphere . 906 of right circular cone . 844 Base, of cone 840 of pyramid .... 748 of spherical pyramid . 1010 of spherical sector . . 1002 Bases, of cylinder .... 822 of prism 727 of spherical segment . 1015 of zone 975 Birectangular spherical tri- angle 952 AKT Center of sphere .... 901 Circular cone 841 Circular cylinder .... 831 Circumscribed polyhedron . 926 Circumscribed prism . . 852 Circumscribed pyramid . 868 Circumscribed sphere . . 929 Closed figure 714, 715, 934, 935 Cone 839 altitude of 842 base of 840 circular 841. element of 840 frustum of 879 lateral surface of . . 840 oblique 843 of revolution .... 876 plane tangent to ... 866 right circular .... 843 spherical 1003 vertex of 840 volume of .... 832, 6. Cones, similar 877 Conical surface .... 837 directrix of .... 838 element of 838 generatrix of .... 838 lower nappe of ... 838 upper nappe of ... 838 vertex of 838 Convex polyhedral angle . 697 Coplanar points, lines, planes 668 481 482 INDEX AET. Cube ........ 742 Cylinder 821 altitude of 825 bases of 822 circular 831 element of ..... 822 lateral surface of . . 822 oblique 824 of revolution .... 861 plane tangent to ... 850 right 823 right circular .... 832 right section of ... 830 volume of 888 Cylinders, similar . . . 863 Cylindrical surface . . . 819 directrix of .... 820 element of 820 generatrix of .... 820 Degree, spherical . . . 986 Determined plane . . . 608 Diagonal of polyhedron . 718 Diameter of sphere . . . 901 Dihedral angle .... 666 edge of 666 faces of 666 plane angle of ... 670 right 672 Dihedral angles, adjacent . 671 Directrix, of conical surface 838 of cylindrical surface . 820 of polyhedral angle . . 693 of prismatic surface . 725 of pyramidal surface . 745 Distance, from point to plane 662 on surface of sphere . 909 polar 911 Dodecahedron .... 719 Edge of dihedral angle . . 666 Edges of polyhedral angle 694 ART. Edges of polyhedron . . 717 Element, of cone .... 840 of conical surface . . 838 of cylinder 822 of cylindrical surface . 820 of polyhedral angle . . 694 of pyramidal surface . 745 Equivalent solids . . . 776 Excess, spherical .... 989 Face angles of polyhedral angle 694 Faces, of dihedral angle . 666 of polyhedral angle . . 694 of polyhedron . . . 717 Foot of perpendicular . . 621 Formulas of Solid Geome- try 1014 Frustum of cone .... 879 slant height of ... 881 Frustum of pyramid . . 754 lateral area of ... 760 slant height of ... 765 Generatrix, of conical sur- face 838 of cylindrical surface . 820 of polyhedral angle . . 693 of prismatic surface . 725 of pyramidal surface . 745 Geometry, of space . . . 602 solid 602 Great circle 904 Hexahedron 719 Historical Notes Ahmes 777 Archimedes .... 809, 896, 973, 974, 1001 Archytas 787 Athenians . . 787 INDEX 483 ART. Historical Notes Brahmagupta . . 899, 896 Cavalieri 992 Egyptians 777 Euclid 723 Eudoxus . 809, 896, 1000 Girard, Albert . 946, 992 Hippasus 723 Menelaus of Alexandria 992 Plato 787 Pythagoras . . 723, 787 Snell 946 Socrates 787 Icosahedron 719 Inclination of line to plane 665 Inscribed polyhedron . . 928 Inscribed prism .... 851 Inscribed pyramid . . . 887 Inscribed sphere .... 927 Intersection of two surfaces 614 Lateral area, of frustum of pyramid .... 760 of prism 760 of pyramid .... 760 of right circular cone . 872 of right circular cylinder 857 Lateral edges, of prism . . 727 of pyramid .... 748 Lateral faces, of prism . . 727 of pyramid .... 748 Lateral surface, of cone . . 840 of cylinder 822 of frustum of pyramid . 760 Line, inclination of ... 665 oblique to plane . . . 630 parallel to plane . . . 629 perpendicular to plane 619 projection of .... 656 tangent to sphere . . 921 ART. Lune 982 angle of ..... 983 sides of 983 vertices of 983 Measure-number .... 770 Nappes, upper and lower 746, 838 Numerical measure . 770 Oblique cone . Oblique cylinder Octahedron 843 824 719 Parallel planes .... 631 Parallelepiped .... 739 rectangular . . . . 741 right 740 Perpendicular . . 619, 620, 672 foot of 621 Perpendicular planes . . 672 Plane, determined . . . 608 perpendicular to straight line 620 tangent to cone . . . 866 tangent to cylinder . . 850 tangent to sphere . . 921 Plane angle of dihedral angle 670 Planes, parallel .... 631 perpendicular .... 672 postulate of . . . . 615 Polar distance of circle . . 911 Polar triangle 943 Poles of circle 907 Polygon, spherical . . . 936 angles of 936 diagonal of .... 937 sides of 936 vertices of . . 936 484 INDEX AKT. Polyhedral angle .... 692 convex 697 dihedral angles of . . 694 edges of 694 element of . . . . . 694 face angles of .... 694 faces of 694 parts of 695 vertex of ..... 693 Polyhedral angles, sym- metrical .... 707 vertical 708 Polyhedron 716 circumscribed about sphere 926 diagonal of . . . . 718 edges of 717 faces of . . . . g . . 717 inscribed in sphere . . 928 regular 720 vertices of 717 Polyhedrons, radially placed 1024 similar . . . .811, 1022 Portrait of Plato .... 787 Postulate, of planes . . . 615 revolution 606 Prism 726 altitude of 733 bases of 727 circumscribed about cyl- inder 852 inscribed in cylinder . 851 lateral area of ... 760 lateral edges of ... 727 lateral faces of ... 727 oblique 731 quadrangular .... 732 regular 730 right 729 right section of ... 728 triangular 732 ABT. Prism, truncated .... 736 Prismatic surface . . . 724 Prismatoid 1019 altitude of 1020 Projection, of line .... 656 of point 655 Pyramid 747 altitude of 751 base of 748 circumscribed about cone 868 frustum of .... 754 inscribed in cone . . 867 lateral area of ... 760 lateral edges of ... 748 lateral faces of . . . 748 quadrangular .... 749 regujar 752 slant height of . . . . 764 spherical 1010 triangular .... 749, 750 truncated 753 vertex of ..... 748 Pyramidal surface . . . 744 Quadrangular prism . . 732 Quadrangular pyramid . . 749 Radially placed polyhedrons 1024 Radius of sphere .... 901 Ratio, of similitude . . . 1023 of two solids .... 775 Rectangular parallelepiped 741 Regular polyhedron . . . 720 Regular prism .... 730 Regular pyramid .... 752 Right circular cone . . . 843 axis of 844 lateral area of ... 872 slant height of ... 865 Right circular cylinder . . 832 lateral area of 857 INDEX 485 ART. Right cylinder .... 823 Right parallelepiped . . 740 Right prism 729 Right section, of cylinder . 830 of prism 728 Sector, spherical .... 1002 Segment of sphere .... 1015 Similar cones of revolution 877 Similar cylinders of revolu- tion 863 Similar polyhedrons . 811, 1022 Similitude, ratio of ... 1023 Slant height, of frustum of cone 881 of frustum of pyramid 765 of pyramid .... 764 of right circular cone . 865 Small circle of sphere . . 905 Solid angle 692 Solid geometry .... 602 Sphere 900 center of 901 circumscribed about polyhedron . . . 929 diameter of .... 901 great circle of ... 904 inscribed in polyhedron 927 line tangent to ... 921 plane tangent to . . . 921 radius of ..... 901 small circle of ... 905 surface of 970 volume of . . . . . 996, c. Spheres, tangent externally . 922 tangent internally . . 922 tangent to each other . 922 Spherical angle . . . . 917 Spherical cone .... 1003 Spherical degree .... 986 Spherical excess .... 989 AKT. Spherical polygon . . . 936 angles of 936 diagonal of .... 937 sides of 936 vertices of . . . . . 936 Spherical polygons, sym- metrical .... 956 Spherical pyramid . . . 1010 base of 10K vertex of 1010 Spherical sector .... 1002 base of 1002 Spherical segment . . .1015 altitude of 1016 bases of 1015 of one base .... 1016 Spherical surface .... 970 Spherical triangle . . . 938 birectangular .... 952 trirectangular . . . 953 Spherical wedge .... 1006 angle of ...... 1006 base of 1006 Straight line, oblique to plane 630 parallel to plane . . . 629 perpendicular to plane . 619 Supplemental triangles . . 948 Surface, closed .... 714 conical 837 cylindrical 819 of sphere 970 prismatic 724 pyramidal 744 Symmetrical polyhedral angles . . . . . 707 Symmetrical spherical poly- gons 956 Tetrahedral angle Tetrahedron . 719, 750 486 INDEX ART. Triangle, polar .... 943 spherical 938 Triangles, supplemental . 948 Triangular prism 62 Triangular pyramid . . 749, 750 Trihedral angle .... 698 birectangular .... 699 isosceles 700 rectangular .... 699 trirectangular . . . 699 Trirectangular spherical triangle .... 953 Truncated prism .... 736 Truncated pyramid . . . 753 Unit of volume .... 769 Vertex, of cone .... 840 of conical surface . . 838 of polyhedral angle . . 693 of pyramid .... 748 ART. Vertex, of pyramidal surface 745 of spherical pyramid . 1010 Vertical polyhedral angles . 708 Vertices, of lune .... 983 of polyhedron . . . 717 of spherical polygon . 936 Volume, of cone . . . 892, b. of cylinder 888 of rectangular parallele- piped 774 of solid ..... 769, 770 of sphere .... 996, c. unit of . 769 Wedge, spherical . 1006 Zone 975 altitude of . . . . . 976 bases ol 975 of one base 977 ROBBINS'S PLANE TRIGONOMETRY By EDWARD R. ROBBINS, S-iior Mathematical Mas- ter, William Penn Charter School, Philadelphia, Pa. $0.6o THIS book is intended for beginners. It aims to give a thorough familiarity with the essential truths, and a satisfactory skill in operating with those processes. It is illustrated in the usual manner, but the diagrams are more than usually clear-cut and elucidating. *f[ The work is sound and teachable, and is written in clear and concise language, in a style that makes it easily under- stood. Immediately after each principle has been proved, it is applied first in illustrative examples, and then further im- pressed by numerous exercises. Accuracy and rigor of treat- ment are shown in every detail, and all irrelevant and ex- traneous matter is excluded, thus giving greater prominence to universal rules and formulas. ^[ The .references to Plane Geometry preceding the first chapter are invaluable. 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