GIFT OF ROBBINS'S NEW SOLID GEOMETRY BY EDWARD RUTLEDGE ROBBINS, A.B. FORMERLY OF LAWRENCEVILLE SCHOOL AMERICAN BOOK COMPANY NEW YORK CINCINNATI CHICAGO COPYEIGHT, 1916, BY EDWARD RUTLEDGE ROBBINS. ROBBINS'8 NEW SOLID GEOMETRY. W. P. I FOR THOSE WHOSE PRIVILEGE IT MAY BE TO ACQUIRE A KNOWLEDGE OP GEOMETRY THIS VOLUME HAS BEEN WRITTEN AND TO THE BOYS AND GIRLS WHO LEARN THE ANCIENT SCIENCE FROM THESE PAGES, AND WHO ESTEEM THE POWER OF CORRECT REASONING THE MORE BECAUSE OF THE LOGIC OF PURE GEOMETRY THIS VOLUME IS DEDICATED 459960 PREFACE THIS " New Solid Geometry " is not only the outgrowth of the author's long experience in teaching geometry, but has profited further by suggestions from teachers who have used Robbins's " Solid Geometry " and by the recommendations of the " National Committee of Fifteen." While many new and valuable features have been added in the reconstruction, yet all the characteristics that met with widespread favor in the old book have been retained. Among the features of the book that make it sound and teach- able may be mentioned the following : 1. The book has been written for the pupil. The objects sought in the study of Geometry are (1) to train the mind to accept only those statements as truth for which convincing reasons can be provided, and (2) to cultivate a foresight that will appreci- ate both the purpose in making a statement and the process of reasoning by which the ultimate truth is established. Thus, the study of this formal science should develop in the pupil the ability to pursue argument coherently, and to establish geometric truths in logical order. To meet the requirements of the various degrees of intellectual capacity and maturity in every class, the reason for every statement is riot printed in full but is indicated by a reference. The pupil who knows the reason need not con- sult the paragraph cited ; while the pupil who does not know it may learn it by the reference. It is obvious that the greater progress an individual makes in assimilating the subject and in entering into its spirit, the less need there will be for the printed* reference. 2. Every effort has been made to stimulate the mental activity of the pupil. To compel a young student, however, to supply his own demonstrations frequently proves unprofitable as well as vi PREFACE arduous, and engenders in the learner a distaste for a study in which he might otherwise take delight. This text does not aim to produce accomplished geometricians at the completion of the first book, but to aid the learner in his progress throughout the volume, wherever experience has shown that he is likely to require assistance. It is designed, under good instruction, to develop a clear conception of the geometric idea, and to produce at the end of the course a rational individual and a friend of this particular science. 3. The theorems and their demonstrations the real subject matter of Geometry are introduced as early in the study as possible. 4. The simple fundamental truths are explained instead of being formally demonstrated. 5. The original exercises are distinguished by their abundance, their practical bearing upon the affairs of life, their careful gradation and classification, and their independence. Every ex- ercise can be solved or demonstrated without the use of any other exercise. Only the truths in the numbered paragraphs are nec- essary in working originals. 6. The exercises are introduced as near as practicable to the theorems to which they apply. 7. Emphasis is given to the discussion of original constructions. 8. The historical notes give the pupil a knowledge of the devel- opment of the science of geometry and add interest to the study. 9. The attractive open page will appeal alike to pupils and to teachers. 10. The Solid Geometry formulas are grouped together at the end of the text, as a ready means of reference. The author desires to extend his sincere thanks to those friends and fellow teachers who, by suggestion and encourage- ment, have inspired him in the preparation of the text, as well as to Clement B. Davis for his original and skillful treatment of the illustrations. EDWARD R. ROBBINS. CONTENTS PAGE REFERENCES TO "NEW PLANE GEOMETRY" , ix BOOK VI. LINES, PLANES, AND ANGLES IN SPACE DEFINITIONS 261 PRELIMINARY THEOREMS 262 THEOREMS AND DEMONSTRATIONS 264 ORIGINAL EXERCISES . . 287 DIHEDRAL ANGLES 290 ORIGINAL EXERCISES . 303 POLYHEDRAL ANGLES 305 ORIGINAL EXERCISES 312 BOOK VII. POLYHEDRONS DEFINITIONS 313 PRISMS . . . . 314 PRELIMINARY THEOREMS . 316 THEOREMS AND DEMONSTRATIONS 317 ORIGINAL EXERCISES . 330 PYRAMIDS 332 PRELIMINARY THEOREMS 333 THEOREMS AND DEMONSTRATIONS 334 REGULAR AND SIMILAR POLYHEDRONS 353 ORIGINAL EXERCISES 360 BOOK VIII. CYLINDERS AND CONES DEFINITIONS 367 CYLINDERS 367 PRELIMINARY THEOREMS 368 vii viii CONTENTS PAGE THEOREMS AND DEMONSTRATIONS . 369 ORIGINAL EXERCISES 376 CONES 377 PRELIMINARY THEOREMS 379 THEOREMS AND DEMONSTRATIONS 381 ORIGINAL EXERCISES 388 BOOK IX. THE SPHERE DEFINITIONS 893 PRELIMINARY THEOREMS 395 THEOREMS AND DEMONSTRATIONS 397 CONSTRUCTION PROBLEMS . . 404 SPHERICAL TRIANGLES 408 PRELIMINARY THEOREMS 410 THEOREMS AND DEMONSTRATIONS 411 ORIGINAL EXERCISES 424 SPHERICAL AREAS AND VOLUMES 427 PRELIMINARY THEOREMS 429 THEOREMS AND DEMONSTRATIONS 431 ORIGINAL EXERCISES 443 SUMMARY OF FORMULAS 451 INDEX OF DEFINITIONS 453 REFERENCES TO ROBBINS'S "NEW PLANE GEOMETRY" NOTE. Many of the theorems in the " New Solid Geometry " make reference in their proof to theorems and definitions in the " New Plane Geometry." These references are here collected for the convenience of the pupil. 16. One line is perpendicular to another if they meet at right angles. Either line is perpendicular to the other. 21. Parallel lines are straight lines that lie in the same plane and that never meet, however far they are extended in either direction. 23. A triangle is a portion of a plane bounded by three straight lines. 26. Two geometric figures are said to be equal if they have the same size or magnitude. Two geometric figures are said to be congruent if, when one is super- posed upon the other, they coincide in all respects. 27. Homologous parts of congruent figures are equal. 29. Symbols. The usual symbols and abbreviations employed in geometry are the following : + plus. minus. = equals, is equal to. =f=. does not equal. ^ congruent, or is con- gruent to. > is greater than. < is less than. .'. hence, therefore, con- sequently. _L perpendicular. Ji perpendiculars. O circle. (D circles. Ax. axiom. /. angle. Hyp. hypothesis. A angles. comp. complementary. rt. /. right angle. supp. supplementary. rt. A right angles. Const. construction. A triangle. Cor. corollary. A triangles. St. straight. rt. & right triangles. rt. right. || parallel. Def. definition. Us parallels. alt. alternate. 7 parallelogram. int. interior. H7 parallelograms. ext. exterior. x REFERENCES TO "NEW PLANE GEOMETRY" 30. An axiom is a statement admitted without proof to be true. 31. AXIOMS. 1. Magnitudes that are equal to the same thing, or to equals, are equal to each other. 2. If equals are added to, or subtracted from, equals, the results are equal. 3. If equals are multiplied by, or divided by, equals, the results are equal. [Doubles of equals are equal ; halves of equals are equal.] 4. The whole is equal to the sum of all of its parts. 5. The whole is greater than any of its parts. 6. A magnitude may be replaced by its equal in any process. [Briefly called " substitution."] 7. If equals are added to, or subtracted from, unequals, the results are unequal in the same order. 8. If unequals are added to unequals in the same order, the results are unequal in that order. 9. If unequals are subtracted from equals, the results are unequal in the opposite order. 10. Doubles or halves of unequals are unequal in the same order. Also, unequals multiplied by equals are unequal in the same order. 11. If the first of three magnitudes is greater than the second, and the second is greater than the third, the first is greater than the third. 12. A straight line is the shortest line that can be drawn between two points. 13. Only one line can be drawn through a point parallel to a given line. 14. A geometrical figure may be moved from one position to another without any change in form or magnitude. REFERENCES TO -NEW PLAXE GEOMETRY ~ xi 39. Only one straight line can be drawn between two points. 42. All right angles are equal. 43. Only one perpendicular to a line can be drawn from a point in the line. 44. If two adjacent angles have their exterior sides in a straight line, they are supplementary. 45. If two adjacent angles are supplementary, their exterior sides are in the same straight line. 46. The sum of all the angles on one side of a straight line at a point equals two right angles. 47. The sum of all the angles about a point in a plane is equal to four right angles. 49. Angles that have the same supplement are equal. Or, supplements of the same angle, or of equal angles, are equal. 51. If two straight lines intersect, the vertical angles are equal. 52. Two triangles are congruent if two sides and the included angle of one are equal respectively to two sides and the included angle of the other. 53. Two right triangles are congruent if two legs of one are equal respectively -to two legs of the other. 54. Only one perpendicular can be drawn 'to a line from an external point. 55. The angles opposite the equal sides of an isosceles triangle are equal. 62. Two lines in the same plane and perpendicular to the same line are parallel. 64. If a line is perpendicular to one of two parallels, it is per- pendicular to the other also. 66. If a transversal intersects two parallels, the alternate interior angles are equal. 67. If a transversal intersects two parallels, the corresponding angles are equal. xii REFERENCES TO "NEW PLANE GEOMETRY" 76. Two triangles are congruent if a side and the two angles adjoining it in the one are equal respectively to a side and the two angles adjoining it in the other. 77. Two right triangles are congruent if a leg and the adjoin- ing acute angle of one are equal respectively to a leg and the ad- joining acute angle of the other. 78. Two triangles are congruent, if the three sides of one are equal respectively to the three sides of the other. 80. Any point in the perpendicular bisector of a line is equally distant from the extremities of the line. 81. Any point not in the perpendicular bisector of a line is not equally distant from the extremities of the line. 82. If a point is equally distant from the extremities of a line, it is in the perpendicular bisector of the line. 83. Two points each equally distant from the extremities of a line determine the perpendicular bisector of the line. 84. Two right triangles are congruent if the hypotenuse and a leg of one are equal respectively to the hypotenuse and a leg of the other. 87. The perpendicular is the shortest line that can be drawn from a point to a straight line. 88. If from any point in a perpendicular to a line two oblique lines are drawn, I. Oblique lines cutting off equal distances from the foot of the perpendicular are equal. II. Equal oblique lines cut off equal distances. III. Oblique lines cutting off unequal distances are unequal, and that one which cuts off the greater distance is the greater. 90. The method of exclusion consists in making all possible supposi- tions, leaving the probable one last, and then proving all these supposi- tions impossible, except the last, which must necessarily be true. REFERENCES TO "NEW PLANE GEOMETRY" xiii The method of proving the individual steps is called reductio ad absurdum (reduction to an absurd or impossible conclusion). This method consists in assuming as false the truth to be proved and then showing that this assumption leads to a conclusion altogether contrary to known truth or the given hypothesis. 92. If two triangles have two sides of one equal to two sides of the other, but the third side of the first greater than the third side of the second, the included angle of the first is greater than the included angle of the second. 94. Every point in the bisector of an angle is equally distant from the sides of the angle. 104. The sum of the angles of any triangle is two right angles ; that is, 180. 109. Each angle of an equiangular triangle is 60. 114. If two angles of a triangle are equal, the triangle is isosceles. 120. A parallelogram is a quadrilateral having its opposite sides parallel. 124. The opposite sides of a parallelogram are equal. 126. The diagonal of a parallelogram divides it into two con- gruent triangles. 128. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. 129. If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram. 133. Two parallelograms are congruent if two sides and the included angle of one are equal respectively to two sides and the included angle of the other. 134. Two rectangles are congruent if the base and altitude of one are equal respectively to the base and altitude of the other. 136. The line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. 138. The line bisecting one leg of a trapezoid and parallel to the base bisects the other leg, is the median, and is equal to half the sum of the bases. xiv REFERENCES TO "NEW PLANE GEOMETRY" 139. The median of a trapezoid is parallel to the bases and equal to half their sum. 143. The three medians of a triangle meet in a point which is two thirds the distance from any vertex to the midpoint of the opposite side. 145. The number of sides of a polygon is the same as the number of its vertices or the number of its angles. 160. Two polygons are congruent if they are mutually equiangular and their homologous sides are equal. 153. The sum of the interior angles of an n-gon is equal to (n-2) times 180. 155. Each angle of an equiangular n-gon = ( - n ~*> 18 . n 167. If three angles of a quadrilateral are right angles, the figure is a rectangle. 168. If the sides of a polygon are produced, in order, one at each vertex, the sum of the exterior angles of the polygon equals four right angles, that is, 360. 179. A circle is a plane curve all points of which are equally distant from a point in the plane, called the center. 180. The length of the circle is called the circumference. 183. Equal circles are circles having equal radii. 187. All radii of the same circle are equal. 188. All radii of equal circles are equal. 190. All diameters of the same or of equal circles are equal. 191. The diameter of a circle bisects the circle. 193. In the same circle (or in equal circles) equal central angles intercept equal arcs. 196. In the same circle (or in equal circles) equal chords sub- tend equal arcs. 197. In the same circle (or in equal circles) equal arcs are sub- tended by equal chords. REFERENCES TO "NEW PLANE GEOMETRY" xv 202. The line perpendicular to a radius at its extremity is tangent to the circle. 203. If a line is tangent to a circle, the radius drawn to the point of contact is perpendicular to the tangent. 208. In the same circle (or in equal circles) equal chords are equally distant from the center. 209. In the same circle (or in equal circles) chords which are equally distant from the center are equal. 210. In the same circle (or in equal circles) if two chords are un- equal, the greater chord is at the less distance from the center. 211. In the same circle (or in equal circles) if two chords are un- equally distant from the center, the chord at the less distance is the greater. 214. One circle and only one can be drawn through the vertices of a triangle. 219. If two circles intersect, the line joining their centers is the perpendicular bisector of their common chord. 224. To measure a quantity is to find the number of times it contains another quantity of the same kind, called the unit. This number is the ratio of the quantity to the unit. 225. Two quantities are called commensurable if there exists a com- mon unit of measure which is contained in each a whole (integral) number of times. Two quantities are called incommensurable if there does not exist a common unit of measure which is contained in each a whole number of times. 227. The limit of a variable is a constant, to which the variable can- not be equal, but from which the variable can be made to differ by less than any mentionable quantity. 229. If two variables are always equal and each approaches a limit, their limits are equal. 232. A central angle is measured by its intercepted arc. 233. A central right angle intercepts a quadrant of arc. 234. A right angle is measured by half a semicircle, that is, by a quadrant. xvi REFERENCES TO "NEW PLANE GEOMETRY" 240. All angles inscribed in a semicircle are right angles. 246. The locus of a point is the series of positions the point, must occupy in order that it may satisfy a given condition. It is the path of a point whose positions are limited or defined by a given condition, or given conditions. 280. In a proportion the product of the extremes is equal to the product of the means. 281. If the product of two quantities is equal to the product of two others, one pair may be made the extremes of a proportion and the other pair the means. 282. In any proportion the terms are also in proportion by alter- nation (that is, the first term is to the third as the second is to the fourth). 284. In any proportion the terms are also in proportion by compo- sition (that is, the sum of the first two terms is to the first, or the second, as the sum of the last two terms is to the third, or the fourth). 286. In any proportion the terms are also in proportion by division (that is, the difference between the first two terms is to the first, or the second, as the difference between the last two terms is to the third, or the fourth). 287. In any proportion, like powers of the terms are in propor- tion, and like roots of the terms are in proportion. 288. In two or more proportions the products of the correspond- ing terms are in proportion. 289. A mean proportional is equal to the square root of the product of the extremes. 291. In a series of equal ratios, the sum of all the antecedents is to the sum of all the consequents as any antecedent is to its conse- quent. 293. A line parallel to one side of a triangle divides the other sides into proportional segments. 294. If a line parallel to one side of a triangle intersects the other sides, it divides these sides proportionally. 301. Similar polygons are polygons that are mutually equiangular and whose homologous sides are proportional. REFERENCES TO "NEW PLANE GEOMETRY" xvii 304. Two right triangles are similar if an a'cute angle of one is equal to an acute angle of the other. 305. If a line parallel to one side of a triangle intersects the other sides, the triangle formed is similar to the original triangle. 308. If two triangles have their homologous sides proportional, they are similar. 310. If two triangles have their homologous sides perpendicular, they are similar. 313. In similar figures homologous sides are proportional. 318. If two polygons are similar, they may be decomposed into the same number of triangles similar each to each and similarly placed. 331. If in a right triangle a perpendicular is drawn from the vertex of the right angle upon the hypotenuse, I. The triangles formed are similar to the given triangle and similar to each other. II. The perpendicular is a mean proportional between the seg- ments of the hypotenuse. 333. The square of a leg of a right triangle is equal to the product of the hypotenuse and the projection of this leg upon the hypotenuse. 334. The sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. 357. The area of a rectangle is equal to the product of its base by its altitude. 359. The area of a parallelogram is equal to the product of its base by its altitude. 360. All parallelograms having equal bases and equal altitudes are equal in area. 364. The area of a triangle is equal to half the product of its base by its altitude. 366. All triangles having equal bases and equal altitudes are equal in area. ROBBINS'8 NEW SOLID GEOM. 2 xviii REFERENCES TO "NEW PLANE GEOMETRY" 368. Two triangles having equal altitudes are to each other as their bases. 372. The area of a trapezoid is equal to half the product of the altitude by the sum of the bases. 374. If two triangles have an angle of one equal to an angle of the other, they are to each other as the products of the sides includ- ing the equal angles. 376. Two similar polygons are to each other as the squares of any two homologous sides. 395. To construct a triangle equal to a given polygon. 422. If the number of sides of an inscribed regular polygon is indefinitely increased, the apothem approaches the radius as a limit. NOTE, page 232. It is evident that if the difference between two variables approaches zero, either ' (1) one is approaching the other as a limit, or (2) both are approaching some third quantity as their limit. 424. If the number of sides of an inscribed regular polygon and of a circumscribed regular polygon is indefinitely increased, I. The perimeter of each polygon approaches the circum- ference of the circle as a limit. II. The area of each polygon approaches the area of the circle as a limit. 428. Let C = circumference and R = radius. Then, C = 2-rrR. 430. Let S = area of O, C = its circumference, and R = its radius. Then, S = SOLID GEOMETRY BOOK VI LINES, PLANES, AND ANGLES IN SPACE 465. A solid is any limited portion of space. The bound- aries of a solid are surfaces. A plane is a surface in which, if any two points are taken, the straight line connecting them lies wholly in that surface. Solid geometry is a science that treats of magnitudes, the parts of which are not all in the same plane. 466. The intersection of two surfaces is the line, or the lines, all of whose points lie in both surfaces. The inter- section of a line and a surface is the point, or the points, common to both the line and the surface. The foot of a line intersecting a plane is their point of intersection. 467. A straight line is perpendicular to a plane if the line is perpendicular to every straight line in the plane drawn through its foot. A normal is a straight line perpendicular to a plane. 468. A straight line is parallel to a plane if the line and the plane never meet, when indefinitely extended. A straight line is oblique to a plane if it is neither perpendicular nor parallel to the plane. Two planes are parallel if they never meet when indefinitely extended. 261 262 VI. SOLID GEOMETRY 469. The projection of a point upon a plane is the foot of the perpendicular from the point to the plane. The projection of a line upon a plane is the line formed by the projections of all the points of the given line. 470. A plane is determined if its position is fixed and if that position can be occupied by only one plane. PRELIMINARY THEOREMS 471. If two points of a straight line are in a plane, the whole line is in the plane. (Def. 465.) 472. A straight line can intersect a plane in only one point. (471.) 473. If a line is perpendicular to a plane, it is perpendicular to every line in the plane drawn through its foot. (467.) 474. Through one straight line any number of planes may be passed. Because, if we consider a plane containing a line AB to revolve about AB, it may occupy an indefinitely great number of positions. Each of these will be a different plane con- taining AB. 475. Through a fixed straight line and an external point a plane can be passed. Because, if we pass a plane containing this line AB, it may be revolved about AB until it contains the given point. 476. A straight line and an external point determine a plane. (See 475, 470.) LINES AND PLANES 263 477. Three points not in a straight line determine a plane. Because two of the points may be joined by a line ; then this line and the third point determine a plane. (See 476.) 478. Two parallel lines determine a plane. (21, 476.) 479. Two intersecting straight lines determine a plane. Because one of these lines and a point in the second line determine a plane ; and this plane contains the second line. (476, 471.) 480. If two planes are parallel, no line in the one can meet any line in the other. (Def. 468.) NOTE. A plane is represented to the eye by a quadrilateral. In some positions it appears to be a parallelogram, and in others, a trapezoid. The eye, however, must be aided by the imagination in really under- standing the diagrams of solid geometry. Thus, in the adjoining figure, the line CN is perpendicular to the plane FR, and to every line in FR drawn through N. Consider several lines drawn through a point on the floor, and a cane, CN, occupying a verti- cal position, so that it is perpendicular to all these lines. Then every angle CNX is a right angle, though to the unskilled eye they do not all appear to be right angles in the diagram. The object of all geometrical diagrams is that the eye may assist the rnind in grasping truths or in developing logical demonstrations, and the student should thoroughly examine every figure until he completely understands the relative positions of its parts. A photograph, like a geometrical diagram, represents three dimensions in a plane, and we should be as familiar with the significance of one as with the other. When, during the process of a demonstration or elsewhere, it becomes necessary to employ a plane not already indicated, it is customary to pass such a plane, or to conceive it constructed. Ex. 1. How many planes can be passed through two points? through three points in the same straight line ? Ex. 2. Hold two pencils (representing lines) so that a plane can be passed containing both; so that no plane can be passed containing both. 264 ' BOOK VI. SOLID GEOMETRY THEOREMS AND DEMONSTRATIONS POINTS, LINES, AND PLANES PROPOSITION I. THEOREM 481. If two planes intersect, their intersection is a straight line. Given : Intersecting planes MN and RS. To Prove : Their intersection is a straight line. Proof : Suppose A and B are two points common to both planes. Draw straight line AB. Now AB is in plane RS (471). And AB is in plane MN (?). That is, AB is common to both planes. Again, if there were a point outside of AB in both planes, these planes would coincide (476). Hence AB contains all points common to the two given P lanes * .-. AB is the intersection (466). That is, the intersection of the two planes is a straight line. Q.E.D. Ex. 1. Do any two planes intersect? Explain. Ex. 2. What is meant by the statement " Two planes determine a line "? Is this universally true? Ex. 3. What kind of lines are the folds in your letter paper? in a pamphlet? in the edges of a box or a brick? Explain. LINES AND PLANES 265 PROPOSITION II. THEOREM 482. If two straight lines are parallel, a plane containing one, and only one, is parallel to the other line. A Given : II lines AB and CD ; plane MN containing CD. To Prove : plane MN II to line AB. Proof: AB and CD are in the same plane AD Plane AD intersects plane MN in CD (481). Now as AB cannot meet MN in CD( hyp.) it can never meet MN. .-. AB is II to MN (468). Q.E.D. PROPOSITION III. THEOREM 483. If a straight line is parallel to a plane, and another plane containing this line intersects the given plane, the inter- section is parallel to the given line. C M B Given : AB II to MN ; plane AD containing AB and inter- secting plane MN in CD. To Prove : AB II to CD. Proof: AB and CD are in the same plane AD (Hyp.). Now a,sAB cannot meet Jfjr(hyp.)it can never meet CD in MN. .'. AB is II to CD (21). Q.E.D. 266 BOOK VI. SOLID GEOMETRY PROPOSITION IV. THEOREM 484. The intersections of two parallel planes by a third plane are parallel lines. Given : if planes AB and CD cut by plane ES in lines LM and PQ. To Prove : LM II to PQ. Proof : LM and PQ are in the same plane BS Also LM and PQ can never meet .-. LM is II to PQ (Hyp.)- (480). (21), Q.E.D. Ex. 1. Hold a pencil parallel to the blackboard, so that its shadow falls on the blackboard. Is this shadow parallel to the pencil ? Why ? Ex. 2. Can a plane intersect two planes that are not parallel so that the intersections are parallel? Illustrate your answer by passing a plane across the room so that it cuts the end and a side of the room. How must this plane be passed so that the intersections are parallel lines? Ex. 3. Draw a line on the blackboard that will never meet the plane of the ceiling. How was it drawn ? Ex. 4. Draw a line on the blackboard and one on the floor that will meet if extended. How must these lines be drawn ? Ex. 6. When will a line on the ceiling be parallel to a line on the floor? LINES AND PLANES PROPOSITION V. THEOREM 267 485. A straight line perpendicular to each of two straight lines at their intersection is perpendicular to the plane of the lines. M Given: -4.FJ.to BF and CF at F-, plane MN containing BF and CF. To Prove : AF JL to plane MN. Proof: In plane MN draw BC\ draw also DF from F to any point, D, in BC. Prolong AF to X, making FX = to AF, and draw AB, AD, AC, BX, DX, CX. BF and CF are _L bisectors of. AX (Hyp. and Const.). In A ABC and BXC, AB = BX and AC = CX BC=BC .-. A ABC ^ A BXC Also in A ^UD and JTD, ABC = /. CBX BD = BD And AB BX .'. AABD ^ .'. AD = DX DF is -L to AX Hence That is, (80). (?) (78). (27). (?) CO- (52). (?) (83). AF is J_ to all lines in MN through F. AF is to plane Jf^r (467). Q.E.D. 268 BOOK VI. SOLID GEOMETRY PROPOSITION VI. THEOREM 486. All straight lines perpendicular to a line at one point are in one plane, which is perpendicular to this line at this point. A N Given : AB _L to BC, .BD, BE, etc. ; plane MN containing BC and BD. To Prove : BE is in the plane MN and MN is -L to AB at B. Proof: Pass plane AE containing AB and BE, and inter- secting plane MN in line BX. Now AB is -L to plane MN (485). That is, plane MN is J_ to AB .-. AB is -L to BX (473). But AB is -L to BE (Hyp.)- That is, BX and BE are both in plane AX and J- to AB at B. . . BX and BE coincide (43). That is, BE is in plane MN. Q.E.D. Ex. 1. Can a line be perpendicular to two other lines if these two do not intersect ? Illustrate. Ex. 2. If two lines are perpendicular to a third line are they neces- sarily in the same plane? Are the two lines necessarily parallel? Illus- trate. Ex. 3- Give a reason why the " corners " of a building are perpen- dicular to the horizontal plane of a level street. LINES AND PLANES 269 487. COROLLARY. Through a point in a straight line one plane can be passed perpendicular to the line, and only one. (486.) PROPOSITION VII. THEOREM 488. Through an external point one plane can be passed perpendicular to a given straight line, and only one. IB Given : The line AB and point P outside of AB. To Prove : Through P, one plane can be passed J_ to AB, and only one. Proof: I. Draw from P, PC _L to AB, and at C draw CX, another line J_ to AB. PC and CX determine a plane MN (479). Plane MN contains P and is -L to AB (485). II. Only one line J_ to AB can be drawn from P (54). And only one plane _L to AB can be passed at C (487). That is, MN is the only plane _L to AB that can be passed through P. Q.E.D. Ex. 1. Considered as lines, why are the spokes of a cart wheel per- pendicular to the axle? Ex. 2. As the hand of a clock revolves, what may it be said to describe, if it is considered of indefinite length? Why? Ex. 3. Illustrate Proposition VI by revolving a carpenter's square, holding one straight edge against the wall or the floor. 2TO BOOK VI. SOLID GEOMETRY PROPOSITION VIII. THEOREM 489. Two planes perpendicular to the same straight line are parallel. B Given : Planes MN and OP J_ to AB. To Prove : Plane MN \\ to plane OP. Proof: If the planes MN and OP are not ||, they will meet when sufficiently extended (Def. 468). Then there would be two planes from the same point _L to AB (-L by hyp.). But this is impossible (488). . *. the planes never meet and are parallel (468). Q.E.D. PROPOSITION IX. THEOREM 490. At a given point in a plane one line can be drawn perpendicular to the given plane, and only one. N M Given : Plane MN and point P within it. LINES AND PLANES 271 To Prove : One line can be drawn _L to plane 'MN at P, and only one. Proof: I. In plane MN draw any line AB, through P. Suppose plane CD is passed _L to AB at P, meeting the plane MN in CE. In plane CD draw PE J_ to CE, from P. Now AB is J- to plane CD (Const.). .-. AB is _L to PR (473). PE is _L to CE (Const.). .-. PE is _L to plane MN (485). Q.E.D. II. Suppose another line PX to be _L to plane MN at P. Then PX and PE determine a plane CD (479). And plane CD intersects plane MN in line CE (481). Then PX and PE would both be -L to CE at P (473). But this is impossible (43). That is, PX and PE coincide and PE is the only J_ to plane MN at P. Q.E.D. Ex. 1. How many positions can a flagpole occupy without being per- fectly erect? How many positions may it assume and be perfectly erect? Ex. 2. Name all the right angles at P, in figure of 490, and tell why each is a right angle. Ex. 3. What information can the mason or the surveyor obtain from a plumb bob? Does he obtain this information when the bob is swing- ing or when it is stationary ? Ex. 4. In transplanting a tree to a horizontal lawn a gardener may use a carpenter's square to make certain that the tree is perpendicular to the lawn. In how many different positions must he place the square against the tree to ascertain its erectness? Why? Ex. 5. In the diagram of 490, if PX is in plane CD, is it perpen- dicular to CE ? ' Why ? Ex. 6. In the same diagram, if PX is perpendicular to CE, is it in plane GDI Why? Ex. 7. How many planes are determined by four random fixed points (that is, not all in one plane) ? 272 BOOK VI. SOLID GEOMETRY PROPOSITION X. THEOREM 491. Through a given external point one line can be drawn perpendicular to a given plane, and only one. G Given : Plane MN and point P outside of it. To Prove : One line can be drawn through P _L to plane MN, and only one. Proof : I. In plane MN draw any line AB. Suppose a plane GH is passed through P _L to AB, meeting plane MN in KC, and AB at C. In plane GH draw PR J_ to KG and prolong PR to X, mak- ing RX = PR. Draw RD to any point in AB, except C. Draw PC, PD, CX, DX. Now RC is .1 to PX at its midpoint (Const.). Also AB is J_ to plane GH (Const.). Hence A DCP and DCX are rt. A (473). In rt. A DCP and DCX, DC = DC (?). PC=CX (80). .*. A DCP^A DCX (53). .-. JH> is JL to PX (83). That is, PR is -L to RC and RZ>, in plane MN. .-. PR is J_ to plane JfJV from P (485). Q.E.D. II. Suppose there is another line PL, _L.to plane MN from P. Then PR and PL determine a plane GH (479). LINES AND PLANES 273 This plane intersects plane MN in KC (481). PR and PL would then both be J_ to KG (473). But this is impossible (54). That is, PR and PL coincide, and therefore PR is the only line J- to plane MN from P. Q.E.D. Ex. Name all the right angles at C in the figure of 491. 492. COROLLARY. If a plane is perpendicular to a line in another plane, any line in the first plane perpendicular to the intersection of the planes is perpendicular to the second plane. Proof : Identical with the proof of 491, I. PROPOSITION XI. THEOREM 493. If a plane is perpendicular to one of two parallel lines, it is perpendicular to the other also. N Given: Plane MN J_ to line AB, and AB II to CP. To Prove : CP -L to plane MN. Proof: AB and CP determine a plane. (478.) Pass this plane BC, intersecting plane MN in line BP. Draw BX J_ to BP, in plane MN. AB is J- to BX (473). . . BX is -L to plane BC (485). But AB is J. to BP (473). .-. BP is J. to CP (64). That is, plane BC is J_ to BX, and CP, in plane BC, is _L to the intersection BP. .-. CP is J_ to plane MN (492). Q.E.D. 274 BOOK VI. SOLID GEOMETRY PROPOSITION XII. THEOREM 494. Two lines perpendicular to the same plane are parallel. M, B 'N Given : Lines AB and CD _L to plane MN. To Prove : AB\\io CD. Proof : Through D, the foot of CD, draw DX II to AB. Then DX is J_ to plane MN (493). But CD is -L to plane MN at D. (Hyp.) . . DX and DC coincide (490). That is, AB is II to CD. Q.E.D. PROPOSITION XIII. THEOREM 495. Two straight lines that are parallel to a third straight line are parallel to each other. M, A E 1 \ j j s \ i ! B D Given: Lines CD and .Breach II to AB. To Prove : CD II to EF. Proof : Suppose plane MN is .passed _L to AB. .-. MN is _L to CD and to EF .-. CD is II to EF (^94), (493). Q.E.I). LINES AND PLANES 275 PROPOSITION XIV. THEOREM 496. A line perpendicular to one of two parallel planes is perpendicular to the other also. Given : Plane MN II to plane ES ; AP -L to plane RS. To Prove : AP JL to plane MN. Proof: Through AP pass any two planes, AB and AC, intersecting MN in AD and AE, and intersecting RS in PB and PC, respectively. AD is II to PB, and AE is II to PC (484). AP is J. to PB and PC (473). . . AP is _L to AD and AE (64). .-. AP is _L to plane MN (485). Q.E.D. 497. COROLLARY. If two planes are each parallel to a third plane, they are parallel to each other. Proof : Draw a line _L to the third plane. This line is J_ to each of the other planes (496). .-. The two planes are II (489). Ex. 1. Can a line be perpendicular to both of two planes if they are not parallel ? Prove. Ex. 2. Are three lines that are perpendicular to the same plane necessarily parallel ? BOBBINS 1 S NEW SOLID GEOM. 3 276 BOOK VI. SOLID GEOMETRY PROPOSITION XV. THEOREM 498. If two intersecting lines are each parallel to a plane, the plane of these lines is parallel to the given plane. Given: Intersecting lines AB and AC in plane MN; each line II to plane PQ. To Prove : Plane MN II to plane PQ. Proof: Draw AE -L to MN at A, meeting PQ at R. Through AE and AB pass plane AS, and through AB and AC pass plane AT, intersecting plane PQ in RS and ET> respectively. Now AB is II to ES, and AC is II to ET (483). AE is J_ to AB and AC (473). .-. AE is _L to ES and ET (64). Hence AE is _L to plane PQ (485). .-. plane MN is II to plane PQ (489). Q.E.D. Ex. 1. Can one line be perpendicular to two other lines that do not intersect? How? Ex. 2. Can one line be perpendicular to two other lines that do in- tersect? How? Ex. 3. Could 484 be quoted correctly as the reason that AB is par- allel to E S in Proposition XV ? Ex. 4. Could 483 be quoted correctly as the reason that AD is par- allel to BP in Proposition XIV? Ex. 6. Prove Proposition XV by drawing AR perpendicular to PQ from A. LINES AND PLANES 277 PROPOSITION XVI. THEOREM 499. If two angles, not in the same plane, have their sides parallel each to each, and extending in the same directions from their vertices, the angles are equal and the planes are parallel. N Given: /.BAG in plane MN and Z EDF in plane PQ; AB II to DE-, AC II to DF, and extending in the same directions. To Prove : I. Z BAG = Z EDF. II. Plane MN II to plane PQ. Proof: I. Take DE and AB equal, and DF and AC equal. Draw AD, BE, CF, BC, EF. The figure ABED is a O (129). .-. AD = BE (124). Also ACFD is a O (?). AD = CF (?). .-. BE=CF (?). Again, AD is II to BE and AD is II to CF (120). .-. BE is II to CF (495). .-. BCFEis a O (129). Now in &ABC and DEF, AB = DE', AC = DF (Const.). Also BC = EF (124). (78). (27). Q.E.D. II. AB is II to plane PQ and AC is II to plane PQ (482). .-. plane MN is II to plane PQ (498). Q.E.D. 278 BOOK VI. SOLID GEOMETRY PROPOSITION XVII. THEOREM 500. If three parallel planes intersect two straight lines, the corresponding intercepts are proportional. 5 Given: Parallel planes, L3/, NP, QR, intersecting line AB at A, E, J5, and CD at C, F, D, respectively. To Prove : AE : EB = CF : FD. Proof : Draw BC, meeting plane NP at G. Through AB and BC pass a plane cutting LM in AC and NP in EG. Through BC and CD pass a plane cutting NP in GF and QR in BD. Now EG is II to ^1(7 and GF is II to D (484). .EB GJ5 Consequently AE : En = CF : FD (Ax. 1). Q.E.D. Also AE + EB : AE = CF + FD Or AE + J : EB = C.P + FD .'. AB : AE= CD : CF } Or AB:EK=CD:FD CD CF FD ... : CF 1 : FD J Ex. 1. If any number of lines which meet at a point are cut by two parallel planes, the corresponding intercepts are proportional. Ex. 2. In the above diagram, why are not AC and BD parallel? Under what condition would EOF be a straight line ? LINES AND PLANES 279 PROPOSITION XVIII. THEOREM 501. The projection of a straight line upon a plane is a straight line.* T R J N Given : Line ^17? and plane MN. To Prove : The projection of AB on MN is a straight line. Proof : Draw PJ _L to plane MN from any point P, in AB. AB and PJ determine a plane. (479). Plane AD cuts plane MN in a straight line CD. (481). Now in plane AD, draw XE II to PJ from X, any other point in AB. XE is -L to plane MN. (493). Now E is the projection of X. (469). .*. CD is the projection of AB. (469). That is, the projection of AB upon the plane MN is a straight line. Q.E.D. Ex. 1. What is the length of the projection of a 5 ft. rod, inclined at an angle of 45 ? Ex. 2. A ladder 26 ft. long leans against the wall of a house, at a point 10 ft. from the ground. What is the length of the projection of the ladder on the ground? 502. COROLLARY. A straight line and its projection upon a plane are in the same plane. * Except only if the given line is a normal to the given plane. 280 BOOK VI. SOLID GEOMETRY PROPOSITION XIX. THEOREM 503. A line not parallel to a plane is longer than its projection upon the plane. N Given: A plane and line LN not II to the plane, and DE the projection of LN upon the plane. To Prove : LN > DE. Proof : Draw LD and NE. Draw LX -L to NE from z, in the plane LE. LD and NE are JL to the plane. (Def. of projection, 469). LXED is a rectangle. (157). Now LN > LX (87). But LX= DE (124). .-. LN > DE (Ax. 6). Q.E.D. Ex. If a line is parallel to a plane, all points of the line are equally distant from the plane. PROPOSITION XX. THEOREM 504. Of all lines that can be drawn to a plane from a point : I. The perpendicular is the shortest. II. Oblique lines having equal projections are equal. HI. Equal oblique lines have equal projections. IV. Oblique lines having unequal projections are unequal, and the line having the greater projection is the longer. V. Unequal oblique lines have unequal projections, and the longer line has the greater projection. LINES AND PLANES P 281 'N I. Given : Plane MN ; point P ; PR _L to MN ; any other line from P to plane MN, as PA. To Prove : PR < PA. Proof: Draw AE. Now PB is J. to AE (473). And PA is not _L to ^412 (54). .-. P < P^i (87). Q.E.D. II. Given: Oblique lines PA and PB whose projections, AE and BE, are equal. To Prove : PA = PB. Proof: The right A PAR and PEB are ^ (?). III. Given : Equal oblique lines PA and PB. To Prove : Their projections, AE and BR, are equal. Proof : The right A PAR and PEB are ^ (?). IV. Given : Oblique lines PC and PA ; proj. EC > proj. l?^. To Prove : PC > PA. Proof : In A PRC, take on RC, EX = EA, and draw PX. Now PC > PX (88, III). But PA = PX (504, II). .-. PC > PA (Ax. 6). Q.E.D. V. Given : Unequal oblique lines, PC > PA. To Prove : Projection EC > projection EA. Proof: By method of exclusion (See 90). 2*2 BOOK VI. SOLID GEOMETRY PROPOSITION XX F. THEOREM 505. The acute angle that a line makes with its own projec- tion upon a plane is the least angle that the line makes with any line of the plane. A Given : AB, any line meeting plane MN at B ; BP, its projec- tion upon MN ; BD, any other line in JfJV, through B. To Prove : Z ABP < Z. ABD. Proof : On BD take BX = BP and draw AX. In A APB and ABX, AB = AB (?). BPBX (Const.). But AP < AX (504, I). .-. Z-ABP < Z.ABD (92). Q.E.D. Ex. 1. If PR is 12 in., and AP 13 in., find the length of AR, in 504. Ex. 2. Does the longer of two lines always have the longer projection on the same plane? Could they have equal projections? Illustrate. Ex. 3. With what line in a plane does a line oblique to that plane make the greatest angle ? Ex. 4. With what line in a plane does a line oblique to that plane make right angles ? Ex. 5. How do you construct the projection of a curved line upon a plane? When is this a straight line? LINES AND PLANES 283 PROPOSITION XXII. THEOREM 606. Through a given point one plane can be passed parallel to any two given non-parallel lines in space, and only one. B A M Given: Point P; two lines, AB and CD. To Prove : Through P one plane can be passed II to AB and CD, and only one. Proof: I. Through P draw a line II to AB and another II to CD. Pass a plane MN, containing these lines. MN is II to both AB and CD (482). II. Only one line can be drawn through P II to AB, and only one II to CD (Ax. 13). .-. there is only one plane (479). Q.E.D. 507. COROLLARY. If two lines are not in the same plane, one plane and only one can be passed through one of these lines parallel to the other. [Through a point in one line draw a line || to the other line, etc.] Ex. 1. Is Proposition XXII true if the given lines intersect ? Is it true if they are parallel ? Ex. 2. Explain, so that a blind boy could understand, how to pass a plane through a given point and parallel to two pencils he may hold in his outstretched hands. Ex. 3. Can two lines be parallel to a plane and not be parallel to each other? Illustrate, by means of pencils and the ceiling. 284 BOOK VI. SOLID GEOMETRY PROPOSITION XXIII. THEOREM 508. Through a given point one plane can be passed parallel to a given plane, and only one. / Given: (?). To Prove: (?). Proof : I. Suppose PR is drawn J_ to plane AB ; and plane XY is passed J_ to PR at P. Then XY is II to AB (489). II. Only one line -L to AB can be drawn from P (491). Only one plane _L to PR can be passed at P (^87). .. only one plane can contain P and be II to AB. Q.E.D. PROPOSITION XXIV. THEOREM 609. Parallel lines included between parallel planes are equal. Given: (?). To Prove: (?). Proof: The plane determined by AB and CD intersects RS and PQ in lines AC and BD, which are II (484). .-. ABDC is a O (Def.). Hence AB = CD (124). Q.E.D. LINES AND PLANES 285 PROPOSITION XXV. THEOREM 610. The plane perpendicular to a line at its midpoint is the locus of points in space equally distant from the extremities of the line. A^ M ,'P' B Given : Plane RS _L to AB at its midpoint, M. To Prove : Plane RS is the locus of points in space equally distant from A and B. Proof: (1) Take P, any point in RS. Draw PM, PA, PB. Now PM is J. to AB (473). .-. PA = PB (80). That is, any point in RS is equally distant from A and B. (2) Take P', any point outside of RS. Draw P f M. Now P'M is not -L to AB (486). .. P', any point outside of plane RS, is not equally distant from A and B (81). .-. Plane RS is the locus of points in space, equally dis- tant from A and B (246). Q.E.D. Ex. 1. How can you find a line in a plane such that each of its points is equally distant from two given points? Ex. 2. How can you find that point in a given line which is equally distant from two given points ? Ex. 3. There are two definite lines in space. It is desired to find all points that are equally distant from the ends of one line and at the same time equally distant from the ends of the other. How can this be done? 286 BOOK VI. SOLID GEOMETRY PROPOSITION XXVI. THEOREM 511. The locus of points in space equally distant from all the points in the circumference of a circle is the line perpen- dicular to the plane of the circle at its center. A N M Given: (?). To Prove: (?). Proof: I. Any point in AC is equally distant from all the points in the circumference of the circle (504, II). II. Any point equally distant from all points of the circumference of the circle is in AC (504, III). .-. AC is the required locus (246). Q.E.D. Ex. 1. What is the locus of points equally distant from two given points ? Ex. 2. What is the locus of points equally distant from three given points ? Ex. 3. Draw a triangle on the blackboard and a definite straight line on the floor. Tell how to find the one point which is both equally dis- tant from the vertices of the triangle and from the ends of the line. Is there always one point? 512. The distance from a point to a plane is the length of the perpendicular from the point to the plane. Thus, the word "distance," referring to the shortest line from a point to a plane, implies the perpendicular. ^ The inclination of a line to a plane is the angle between the line and its projection upon the plane. ORIGINAL EXERCISES 287 ORIGINAL EXERCISES 1. Through one straight line a plane can be passed parallel to any other straight line in space, and only one. Through a point of the first line draw a line II to the second. 2. Two parallel planes are everywhere equally distant. 3. If a line and a plane are parallel, another line parallel to the given line and through any point in the given plane lies wholly in the given plane. Through the given line and the point P pass a plane cutting the given plane in PX. [Use 483.] 4. A straight line parallel to the intersection of two planes, but in neither, is parallel to both planes. 5. If two straight lines are parallel and two intersecting planes are passed, each containing one of the lines, the intersection of these planes is parallel to each of the given lines. 6. If three straight lines through a point meet the same straight line, these four lines all lie in the same plane. 7. If a straight line meets two parallel planes, its inclinations to the planes are equal. 8. Two parallel planes can be passed, each containing one of two given lines in space. Is this ever impossible? 9. If each of three straight lines intersects the other two, the three lines all lie in a plane. 10. The projections of two parallel lines on a plane are parallel. Proof: AB is II to CD (?). AE is II to CG (?). . .-. planes AF and CH are II (?) ; etc. N 11. If two lines in space are equal and parallel, their projections on a plane are equal and parallel. 12. If a plane is parallel to one of two parallel lines, it is parallel to the other. 13. If a straight line and a plane are perpendicular to the same straight line, they are parallel. 14i Equal oblique lines drawn to a plane from one point have equal inclinations with the plane. 15. If a line and a plane are both parallel to the same line, they are parallel to each other. M i j JQ /sUp 288 BOOK VI. SOLID GEOMETRY 16. Four points in space, A, B, C, D, are joined, and these four lines are bisected. Prove that the four lines joining (in order) the four midpoints of the first lines form a parallelogram. Proof: Pass plane DP through points A, D, B, and plane DX through points B, C, D, these planes intersecting in BD. ST is || to BD and = ?): etc. 17. If a plane is passed containing a diagonal of a parallelogram and perpendiculars are drawn to the plane from the other vertices of the parallelogram, they are equal. To Prove: AE = CF. Proof: Draw diagonal AC. Draw EO and OF in plane MN. EO, OF, and EOF are projections; etc. 18. If from the foot of a perpendicular to a plane, a line is drawn at right angles to any line in the plane, the line connecting this point of inter- section with any point in the perpendicular is per- pendicular to the line in the plane. Given: AE _L to plane RS ; EC L to DE in the plane; PC drawn from C to P, in AE. To Prove : PC is J_ to DE. Proof : Take CD = CE, draw PD, PE, BD, BE. EC is _L to DE at its midpoint (?). .-. ED = BE (?). PD = PE (?) (504, II). .-. PC is to DE (?) (83). 19. A line PE is perpendicular to a plane at B, and a line is drawn from B meeting any line DE, of the plane, at C. If PC is perpendicular to DE, BCis perpendicular to DE. 20. Are two planes that are parallel to the same straight line necessarily parallel? 21. -If each of two parallel lines is parallel to a plane, is the plane of these lines also parallel to the given plane? 22. Is a three-legged stool always stable on the floor? Why? Is a four-legged chair always stable? Why? 23. What is the locus in space of points equally distant from two parallel planes? from two parallel lines? 24. What is the locus of points in space at a given distance from a given plane? ORIGINAL EXERCISES 289 26. What is the locus of points in a plane at a given distance from an external point? 26. What is the locus of points in space equally distant from two points and equally distant from two parallel planes? 27. What is the locus of points in space equally distant from the ver- tices of a triangle ? 28. What is the locus of all straight lines perpendicular to a given straight line at a given point ? 29. What is the locus of all lines parallel to a given plane and drawn through a given point? 30. If the points in a line satisfy one condition and the points in a plane satisfy another condition, what will be true of their intersection? What will be true if they do not intersect? 31. If the points in one plane satisfy one condition and the points in another plane satisfy another condition, what is true of their intersection V What is true if the planes are parallel? 32. Construct a plane perpendicular to a given line at a given point in the line. 33. Construct a plane perpendicular to a given line through a given external point. 34. Construct a line perpendicular to a given plane through a given point in the plane ; through a given external point. 36. Construct a plane parallel to a given plane through a given point. 36. Construct a number of equal oblique lines to a plane from a given external point. 37. Construct a line through a given point parallel to a given plane. 38. Construct through a given point a line parallel to each of two given intersecting planes. 39. Construct a plane containing one given line and parallel to another. 40. Construct a plane through a given point parallel to any two given lines in space. 41. Construct a line through a given point in space which intersects two given lines not in the same plane. When is there no such line ? Is there ever more than one ? 42. Find a point in a plane such that the sum of the two lines join- ing it to two fixed points on one side of the plane is the least possible. 290 BOOK VI. SOLID GEOMETRY Construction: Draw A(.' -L to plane ^fN and yB prolong it to A", making- CX = AC. Draw BX, A / meeting plane MN at P. Draw A P. Take any . \ / .N other point R in plane MX. I \ \/ R\ Statement : .4P + PJ3 < .4 A' + 725. Etc. / P \ 43. Find a point in a given plane equally distant yj/ from three given points. Is this ever impossible? 44. Find the one point equally distant from four given points not in the same plane. Construction: Pass plane CM, containing points A, B, C, and plane CN, containing A, Z>, C. Find 0, the center of the O containing A, B, C. Find P, simi- larly. Draw the locus of points equally distant from A, By C. (Consult 511.) Draw the locus of points equally distant from A, D, C. The plane _L to AC at its midpoint contains both these loci. (Explain.) Hence OX and PA' intersect (?). .. X is the required point. DIHEDRAL ANGLES 513. A dihedral angle is the E^^^B^MK - 1 A amount of divergence of two inter- secting planes. The edge of the ft dihedral angle is the line of inter- QJ^ _Jp section of the planes. The faces of the dihedral angle are the planes. The intersecting planes AG and ED form the dihedral angle whose edge is EG, which is named A-GE-D; or, when there is only one dihedral angle at the edge, "the angle EG." 514. Adjacent dihe- dral angles are two dihedral angles that have the same edge and a common face between them. DIHEDKAL ANGLES 291 Vertical dihedral angles are two dihedral angles that have the same edge, the faces of one being the extensions of the faces of the other. 515. The plane angle of a dihedral angle is the angle formed by two straight lines, one in each face, and perpen- dicular to the edge at the same point. If PM is in plane AG and perpen- dicular to EG, and PN is in plane ED and perpendicular to EG at P, the angle MPN is the plane angle of the dihedral angle EG. 516. If one plane meets another, making the adjacent dihedral angles equal, these angles are right dihedral angles. One plane is perpendicular to another plane if the dihedral angle formed by the two planes is a right dihedral angle. 517. Two dihedral angles are equal if they can be made to coincide. A dihedral angle is acute, right, or obtuse according as its plane angle is acute, right, or obtuse. Dihedral angles are complementary or supplementary, cor- responding, alternate-interior, etc., according as their plane angles are complementary or supplementary, corresponding, alternate-interior, etc. NOTE. An open book often assists a student to a clear apprehension of the magnitude of dihedral angles. Thus he can see the angle increase during the act of opening the book, and observe the acute, right, and obtuse dihedrals. With the aid of two books he can understand better, perhaps, the meaning of complementary dihedrals, supplementary di- hedrals, corresponding dihedrals, alternate-interior dihedrals, etc. ROBBINS'S NEW SOLID GEOM. 4 292 BOOK VI. SOLID GEOMETRY PROPOSITION XXVII. THEOREM 518. The plane angles of a dihedral angle are all equal. A E R Given : Z EFG, the plane Z of dihedral Z .BC7, at .F, and Z RST, the plane Z at ' ~< T = T ^AX. o). dih. /.A'-B'C'-D' 4 . . dih. Z A-BC-D : dih. Z A'-B'c'-D 1 = Z ACE : Z A'C'E' (Ax. 1). II. If the plane angles are incommensurable. C B There does not exist a common unit. Suppose ACE to be divided into equal parts (any number of them). Apply one of these as a unit of measure to Z. A'C'E'. There is a remainder, XC'E', left over (because the A are incommensurable). Pass a plane C'Y, determined by B'C' and C'X. DIHEDRAL ANGLES 295 Now dih. /.A-BC-D /.ACE di\}./.A'-B'C'-Y /.A'C'X (Commensurable Indefinitely increase the number of subdivisions of /. A CE. Then each part, that is, our unit or divisor, is indefinitely decreased. Hence XC'E', the remainder, is indefinitely decreased. That is, /. XC'E' approaches zero as a limit. And dih. /.X-B'C'-D 1 approaches zero as a limit. .-. /.A'C'X approaches A'C'E' as a limit; and dih. /.A'-B'C'-Y approaches dih. /.A'-B'C'-D' as a limit. dih. /. A'-B'C'-Y ~ dih. Z A'-B'C'-D' approaches- r^r4r.as a limit, dih. /.A-BC-D /.ACE dih.^A'-B'C'-D' /.A'C'E' (229). Q.E.D. PROPOSITION XXXI. THEOREM 525. If a straight line is perpendicular to a plane, any plane containing this line is perpendicular to the given plane. Given: Line AB _L to plunc MN ; plane PQ containing AB and intersecting plane MN in RS. To Prove : Plane PQ is _L to plane MN. Proof : In plane M N draw BC -L to RS. Now AB is -L to RS. (473). .-. Z ABC is the plane Z of dih. Z p-SR-N. (515). But Z ^BC is a rt. Z. (473). .-. PQ is J. to MN (516). Q.E.D. 296 BOOK VI. SOLID GEOMETRY 526. COROLLARY. If a plane is perpendicular to the edge of a dihedral angle, it is perpendicular to each face. (See 525.) PROPOSITION XXXII. THEOREM 527. If one plane is perpendicular to another, any line in either plane, perpendicular to their intersection, is perpendicular to the other plane. [Converse of 525.] Given: Plane PQ to plane MN; AB in plane PQ _L to tl.. intersection, RS. To Prove : AB _L to plane MN. Proof : In plane MN draw BC J_ to RS. Now Z ABC is the plane angle of the dih. Z P-SR-N. (515). .-. Z ABC is art. Z (523). .-. AB is J_ to BC (16). But AB is _L to RS (Hyp.)- .-. AB is J_ to plane MN (485). Q.E.D. Ex. 1. In the figure of 527 prove BC perpendicular to the plane PQ. Also prove RS perpendicular to the plane ABC. Ex. 2. In the figure of 527 prove plane ABC perpendicular to the planes MN and PQ. Ex. 3. Under what condition will a line in one face of a dihedral angle meet a line in the other face ? DIHEDRAL ANGLES 297 PROPOSITION XXXIII. THEOREM 528. If one plane is perpendicular to another, a line drawn from any point in their intersection and perpendicular to one plane, lies in the other. Given: Plane PQ _L to plane MN, intersecting in R8-, AS (left-hand) J_ to plane MN from A, in RS. To Prove : AB is in plane PQ. Proof : At A erect in plane PQ, AX JL to RS. Then AX is -L to plane MN But AB is -L to plane MN at A .. AB and AX coincide That is, AB lies in plane PQ. PROPOSITION XXXIV. THEOREM (527). (Hyp.)- (490). Q.E.D. 529. If one plane is perpendicular to another, a line drawn from any point in one plane, and perpendicular to the other, lies hi the first plane. Given: Plane PQ _L to plane MN ; AB (right-hand) _L to plane MN from A, any point in plane PQ. To Prove : AB lies in plane PQ. Proof: From A draw in plane PQ, AX A. to RS. Then AX is -L to plane MN (527). .-. AB and AX coincide (491). That is, AB lies in plane PQ. Q.E.D. 298 BOOK VI. SOLID GEOMETRY PROPOSITION XXXV. THEOREM 530. If two planes are perpendicular to a third plane, their intersection also is perpendicular to that plane. Given : Planes LM and 2VP, each _L to plane ES. To Prove : The intersection AB is _L to plane BS. Proof : If, at A, a line is erected _L to plane RS, it will lie in plane LM (528). This _L will lie also in plane NP (?). .-. this -L is the intersection AB (466). That is, AB is _L to plane RS. Q.E.D. 531. COROLLARY. If a plane is perpendicular to each of two intersecting planes, it is perpendicular to their intersection. (The same truth as 530.) Ex. 1. If each of three planes is perpendicular to the other two, each of the three intersections is perpendicular to the remaining plane, and perpendicular to the other two intersections. [Use figure of 530.] Ex. 2. If two parallel planes are each perpendicular to a third plane, their intersections with that plane are parallel. Ex. 3. Is Proposition XXXV true in the case of the intersecting walls of a building? Ex. 4. What is the " exterior point " that, together with the plumb- bob, determines the vertical plane for the mason ? Ex. 5. If the builder keeps the edge of a building perpendicular to the plane of a level street, the two intersecting walls will also be perpen- dicular to the street. Why ? DIHEDRAL ANGLES 299 PROPOSITION XXXVI. THEOREM 532. Through a given line not perpendicular to a plane, one plane can be passed perpendicular to that plane, and only one. Given : AB not J_ to plane MN. To Prove : Through AB one plane can be passed J_ to MN, and only one. Proof : I. From P, any point in AB, draw PX _L to MN. Through AB and PX pass plane AC. Plane AC is J_ to plane MN (525). II. Suppose another plane containing AB is -L to plane MN. Then the intersection AB, of these two planes, which are J_ to plane MN, will be -L to plane MN (530). But AB is not -L to plane MN (Hyp.). .*. there is only one plane containing AB that is _L to plane MN. Q.E.D. 533. COROLLARY. The plane containing a straight line and its projection upon a plane is perpendicular to the given plane. 534. COROLLARY. If a line meets its projection on a plane, any line of the plane perpendicular to one of these lines at their intersection is perpendicular to the other also. Proof: Use fig. of 505. Plane ABP is J_ to plane MN (533). A line -L to plane ABP at B will lie in plane MN (528). .-. this line is J. to both AB and PB (473). 300 BOOK VI. SOLID GEOMETRY PROPOSITION XXXVII. THEOREM 535. Between any two straight lines not in the same plane, one and only one common perpendicular can be drawn, and this common perpendicular is the shortest line that can be drawn between the two lines. B Given : Lines AB and CD not in the same plane. To Prove : I. One line can be drawn -L to AB and CD. I 1 . Only one -L can be drawn. III. This -L is the shortest line that can be drawn be- tween AB and CD. Proof : I. At P, any point in CD, draw EF II to AB. Pass plane JfJV, containing CD and EF. Pass plane AH through AB and _L to plane MN, intersect- ing plane MN in GH, and CD at L. In plane AH draw EL _L to GH. Plane MN is li to AB (482). GH is II to AB (483). EL is -L to GH (Const.). .-. EL is -L to plane MN (527). .-. EL is _L to CD (473). EL is J- to AB (64). Also That is, EL is JL to both the given lines. Q.E.D. DIHEDRAL ANGLES 301 II. If another line can be drawn J. to AB and CD, sup- pose SP is this J_. In plane AH draw SX _L to GU. Then SX is J. to plane MN (527). But if SP is -L to AB, it is -L to EF (64). .-. SP is -L to plane MN (485). Thus there are two J from s to plane M N (SX and SP). But this is impossible (491). .-. there can be no second JL to these two given lines. Q.E.D. III. Suppose SP is any other line between AB and CD. Now EL is II to sx (62). .-. EX is a O (120). .-. RL = SX (124). But sx < SP (504, I). .-. B < SP (Ax. 6). That is, EL is shorter than any other line between AB and CD. Q.E.D. Ex. 1. In the figure of 535 prove that a plane perpendicular to RL at its midpoint will be parallel to AB and CD. Ex. 2. Prove, also, that this plane will bisect SP. Ex. 3. Prove that if CD is not perpendicular to EF, no plane can be passed through AB, perpendicular to CD. Ex. 4. Tell how we can construct one plane perpendicular to another. Ex. 6. Tell how we can construct one plane through a given point, perpendicular to any two given planes. Ex. 6. Tell how we can construct a plane containing a given line and perpendicular to a given plane. Ex. 7. Tell, so that a blind boy could understand, how to draw a line perpendicular to any two lines in space (not in the same plane)., Ex. 8. If two planes are parallel, what is the form of the projection on one plane, of a circle in the other ? Ex. 9. If two planes are perpendicular, what is the form of the pro- jection on one plane, of a circle in the other? 302 BOOK VI. SOLID GEOMETRY PROPOSITION XXXVIII. THEOREM 536. Every point in a plane bisecting a dihedral angle is equally distant from the faces of the angle. Given: Plane AB, bisecting the dih. Z C-BD-E-, any point P in plane AB ; PF _L to face CB ; PH _L to face DE. To Prove: PF = PH. Proof : Pass plane MN, containing PF and Pfl, intersecting CB in FG, AB in PG, DE ill HG, BD at G. Now plane MN is _L to planes CB and DE (525). . . plane MN is _L to BD (531). .-. BG is -L to PG, PG, and HG (473). Hence Z PGF is the plane Z of dih. Z A-BD-C and Z PGH is the plane Z of dih. Z A-BD-E (515). These dih. z are = (Hyp.). .'. Z PGF= Z PGff (521). ^i PM? and PHG are rt. ^ (473). In the right A PFG and PGfl, P6? = PG (?). .'. PF=PH (?). Q.E.D. 537. COROLLARY. Any point in a dihedral angle and equally distant from its faces is in the plane bisecting the angle. To Prove : The plane AB, determined by the point P and the edge BD, bisects the dih. Z C-BD-E. 538. COROLLARY. The locus of points within a dihedral angle and equally distant from its faces is the plane bisecting that angle. (Proof : 536, 537.) ORIGINAL EXERCISES 303 ORIGINAL EXERCISES 1. Are two planes perpendicular to the same plane necessarily parallel? 2. A straight line and a plane perpendicular to the same plane are parallel. 3. A plane perpendicular to a line in another plane is perpendicular to that plane. 4. If three planes, all perpendicular to a fourth, intersect in three lines, these lines are parallel, in pairs. 6. If the projection of any line (straight or curved) upon a plane is a straight line, the line is entirely in one plane. 6. The angle between the normals drawn to the faces of a dihedral angle from a point within the angle is the supplement of the plane angle of the dihedral angle. 7. If a line is parallel to a plane, any plane perpendicular to the line is perpendicular also to the plane. [Construct the projection of the given line upon the given plane.] 8. What is the locus of points in space equally distant from two intersecting planes? 9. If from any point in a face of a dihedral angle, a normal is drawn to each face, the plane of these normals is perpendicular to the edge of the dihedral angle. 10. If from any point in a face of a dihedral angle, a normal is drawn to each face, the angle they form is equal to the plane angle of the dihedral angle. 11. If a line is perpendicular to a plane, any plane parallel to the line is also perpendicular to the plane. 12. If PA is a normal to plane MN, PB a M normal to plane ST, and BC a normal to plane MN, I AC is perpendicular to RS, the intersection of /_ planes MAT and ST. 13. The plane perpendicular to the line that is perpendicular to two lines in space, at its middle point, bisects every straight line having its extremities in these lines. 304 BOOK VI. SOLID GEOMETRY 14. The plane perpendicular to the plane of an angle and containing the bisector of the angle is the locus of points equally distant from the sides of the angle. Proof: The J from any point in plane NR to AB and EC will have equal projections. (Explain by use of 94.) .. these perpendiculars are equal (?). 16. What is the locus of points in space equally distant from two intersecting lines? 16. If A P is a normal to plane MN and if angle PBC, in plane MN, is a right angle, angle ABC also is a right angle. [Prove EC is JL to plane APB.] 17. If AP is a normal to plane MN, and Z.PBD, in plane MN, is obtuse, Z. ABD also is obtuse. Proof : Take EC = BD ; prove PD>PC. Then prove AD > A C, etc. 18. PA is perpendicular to plane RS; AB and PC are perpendicular to plane MR. Prove EC per- pendicular to RT. 19. If two parallel planes are cut by a third plane, the alternate-interior dihedral angles are equal; the corresponding dihedral angles are equal ; the alternate- exterior dihedral angles are equal; the adjoining in- terior dihedral angles are supplementary. 20. State and prove the converse theorems of those in No. 19. N 21. Construct a plane perpendicular to a given plane and containing a given line in that plane. 22. Construct a plane perpendicular to a given plane and containing a given line without that plane. 23. Construct through a given point a line which will intersect any two given lines in space. Construction : Pass a plane through the point and one of the lines. This plane intersects the other line at a point, etc. 24. To bisect a given dihedral angle. Construction : Pass a plane _L to the edge. Bisect the plane ^ of the given dihedral, etc. POLYHEDRAL ANGLES 305 25. Construct a line each of whose points shall be equally distant from the ends of a given line and also equally distant from the faces of a dihe- dral angle. 26. Find the locus of points equally distant from two points and equally distant from two intersecting planes. Discuss. 27. Find a point equally distant from three given points and equally distant from two intersecting planes. Is this problem ever impossible? Will there ever be two points? When will there be only one point? 28. Find a point equally distant from three given points and equally distant from two intersecting lines. Discuss fully. POLYHEDRAL ANGLES 539. If three or more planes meet at a point, they form a polyhedral angle. The opening partially surrounded by the planes is the polyhedral angle. The point common to all the planes is the vertex. The planes are the faces. The intersections of adjacent faces are the edges. The angles formed at the vertex, by adjacent edges, are the face angles. Thus, V - ABODE is a polyhedral angle; V is the vertex; A V, BV, etc., are edges ; planes A VB, BVC, etc., are faces ; A A VB, B VC, etc., are face angles. Vy 540. A plane section of a polyhedral angle is the plane figure bounded by the intersections of all the faces by a plane. Polygon LMNOP is a plane section of polyhedral angle V ABODE. A convex polyhedral angle is one whose plane sections are all convex. 306 BOOK VI. SOLID GEOMETRY B EQUAL POLYHEDRAL ANGLES VERTICAL POLYHEDRAL ANGLES VERTICAL DIHEDRAL ANGLES SYMMETRICAL POLYHEDRAL ANGLES 641. Two polyhedral angles are equal if they can be made to coincide in all particulars. That is, if two polyhe- dral angles are equal, their homologous dihedral angles are equal; their homologous face angles are equal, and they are arranged in the same order. The length of the edges or the extent of the faces does not affect the size of the angle. Two polyhedral angles are vertical if the edges of one are the prolongations of the edges of the other. Two polyhedral angles are symmetrical if all the parts of one are equal to the corresponding parts of the other, but arranged in opposite order. NOTE. It is apparent from the definitions that equal polyhedral angles are mutually equiangular as to the face angles and as to the dihe- dral angles. Vertical polyhedral angles are mutually equiangular as to their face angles and as to their dihedral angles, but the order is reversed. Symmetrical polyhedral angles are also mutually equiangular as to their face angles and as to their dihedral angles, but the order is reversed. Thus, if one follows around the polygon A'D' in alphabetical order, he is moving as the hands of a clock, if the eye is at the vertex 0' ; but if he follows around AD alphabetically, he is moving in a direction opposite to the motion of the hands of a clock, if the eye is at the vertex O. Hence, it is apparent that, in general, symmetrical polyhedral angles cannot be made to coincide. POLYHEDRAL ANGLES 307 642. A trihedral angle is a polyhedral angle having three and only three faces. A trihedral angle is rectangular if it contains a right di- hedral angle ; birectangular if it contains two right dihedral angles; trirectangular if it contains three right dihedral angles. A trihedral angle is isosceles if two of its face angles are equal. PRELIMINARY THEOREMS 543. Two vertical polyhedral angles are symmetrical. Proof : Their homologous face angles are equal and ar- ranged in reverse order, and their homologous dihedral angles are equal and arranged in reverse order. .-. they are symmetrical (Def. 541). 644. If two polyhedral angles are symmetrical, the vertical polyhedral angle of the one is equal to the other. Because the corresponding parts are equal and they are arranged in the same order. 645. Provided two trihedral angles have their parts arranged in the same order, they are equal : * I. 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. II. If a face angle and the two dihedral angles adjoining it of the one are equal respectively to a face angle and the two dihedral angles adjoining it, of the other. Proof : By method of superposition, as in plane A. Ex. Illustrate a birectanmilar trihedral angle by means of an open book standing on a table. Similarly illustrate a trirectangular trihedral angle. Similarly use two closed books to illustrate a rectangular trihe- dral angle. ROBBINS'8 NEW SOLID GEOM. 6 308 BOOK VI. SOLID GEOMETRY PROPOSITION XXXIX. THEOREM 546. Provided two trihedral angles have their parts arranged in the same order, they are equal, if the three face angles of one are equal respectively to the three face angles of the other. Given : Trih. A o and o' ; Z BOG = Z COA = Z C'0'^l'. To Prove: Trih. Z o = trih. Z o', that is, dih. Z o^i = dih. Z C/.4', etc. Proof: Take OA = O7? = OC = o'^i' = O'B' = o'c'. Draw 4, BC, AC, A'B', B'C', A'C'. Take, on edges AO and A'o', AP = A'P' and in face A OB draw PD J_ to .40. Z O.4.B is acute (A AOB is isosceles). .'. PD will meet AB. In face AOC draw PEA. to .40, meeting .4(7 at E. Draw DJ. Similarly draw P'D', P'E', D'E'. Now AEPD and E 1 P'D 1 are the plane z of the dihedral A AO and .4'o' (515). To prove that these A are equal requires the proof that eight pairs of A are equal. .-.AB = A'B' (1) AO^B = A O'A'B' (Explain). (2) A OBC = A O'B'C' (Explain). (3) AOAC = A o'A'cf (Explain). BC = B'C' AC = A'C' Z OAB = z O'A'B', POLYHEDRAL ANGLES 309 (4) AAPD = A A 'P f D' (Explain). ST\ i N A Ei A Ei (Explain). pj^p/^etc. (6) AABC = AA'B'C' (Explain). .-. z^CAB = Z C f A r B r (?). (7) AAED = AA'E'D' (Explain). .'.ED=E'D' (?). (8) A PED = A P'E'D' (Explain). .-. Z.EPD = ^E f P r D r (?). Hence dih. Z^O = dih. ZA'O' (520). Similarly, one may prove the other pairs of homologous dihedral angles equal. .-. trihedral Z o = trihedral Z o' (541). Q.E.D. PROPOSITION XL. THEOREM 547. Provided two trihedral angles have then* parts arranged hi reverse order, they are symmetrical : I. 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. II. If a face angle and the two dihedral angles adjoining it of the one are equal respectively to a face angle and the two dihedral angles adjoining it of the other. III. If the three face angles of one are equal respectively to the three face angles of the other. Proof : In each case construct a third trihedral Z symmet- rical to the first. This third figure will have its parts = to the parts of the second, and arranged in the same order (Def . 541). .-. the third = the second (545 and 546). .*. the first is symmetrical to the second. (Ax. 6.) Q.E.D. Ex. The three planes bisecting the three dihedral angles of a tri- hedral angle intersect in a straight line. 310 BOOK VI. SOLID GEOMETRY PROPOSITION XLI. THEOREM 548. The sum of any two face angles of a trihedral angle is greater than the third face angle. Given: Trih. Z.O-RST .in which face angle EOT is the greatest. To Prove : Z EOS + Z SOT > Z ROT. Proof: Construct, in face EOT, Z ROD = Z.EOS. Take OD OB; draw ADC, meeting OT at C. Draw AB and BC. A AOD ^ A AOB (Explain). .'.AB = AD (?). Now AB + BC> AD + DC (Ax. 12). But AB =AD (?). Subtracting, BO DC (Ax. 7). Now OB = OD (?), OC OC (?) and BC > DC (Just proved). . . Z BOC > Z DOC (92). But /.AOB = Z^OD (?). Adding, Z ^iOB + Z BOC > Z ^loc (Ax. 7). That is, Z EOS + Z sor > Z HOT (Ax. 6). Ex. 1. Prove theorem of 548 for the case of an isosceles trihedral angle. Ex. 2. The three planes containing the three bisectors of the three face angles of a trihedral angle and perpendicular to those faces intersect in a straight line. POLYHEDRAL ANGLES 311 PROPOSITION XLII. THEOREM 549. The sum of the face angles of any polyhedral angle is less than four right angles or 360. O. Given : Polyhedral Z O, having n faces. To Prove : The sum of the face A at O < 4 rt. A or 360. Proof: Pass a plane AD, intersecting all the faces, and the edges at A, B, C, etc. In this section take any point X and join X to all the vertices of the polygon. (1) There are n face A having their vertices at O (Hyp.). (2) There are n base A having their vertices at X (Const.). (3) The sum of the A of the face A = 2n rt. A (104). (4) The sum of the A of the base A = 2w rt. A (104). (5) .*. the sum of the A of the face A = the sum of the A of the base A (Ax. 1). Now Z OAE + Z OAB > Z EAB (548). And Z OB A -f Z OBC > Z ABC (?), etc., etc. Adding, the sum of the base A of the face A > the sum of the base A of the base A (Ax. 8). Subtracting this inequality from equation (5) above, the sum of the face A at O < the sum of the A at X (Ax. 9). But the suni of all the A at X = 4 rt. A (47). .-. the sum of the face A at O < 4 rt. A, or 360 (Ax. 6). Q.E.D. 312 BOOK VI. SOLID GEOMETRY ORIGINAL EXERCISES 1. In the figure of 549, as the vertex O approaches the base, does the sum of the face angles at O increase or decrease ? What limit does this sum approach ? Does the sum ever become equal to this limit? 2. Can a polyhedral angle have for its faces three equilateral triangles? four? five? six? seven? 3. Can a polyhedral angle have for its faces four squares? five? three? 4. What other regular polygons can be used for the faces of a polyhedral angle ? 6. If two face angles of a trihedral angle are equal, the dihedral angles opposite them are equal. Given: /.RVS = SVT. To Prove : Dih. Z R V = dih. Z TV. Proof: Pass plane S VX bisecting dih. ZSV. Prove trih. A V-RSX and V-TSX are sym. by (547, 1). 6. An isosceles trihedral angle and its symmetri- cal trihedral angle are equal. 7. Find the locus of points equally distant from the three faces of a trihedral angle. 8. Find the locus of points equally distant from the three edges of a trihedral angle. 9. If the three face angles of a trihedral angle are equal, the three dihedral angles also are equal. 10. If the three face angles of a trihedral angle are right angles, the three dihedral angles also are right angles. 11. In any trihedral angle the greatest dihedral angle has the greatest face opposite it. 12. If the edges of one trihedral angle are perpendicular to the faces of a second trihedral angle, then the edges of the second are perpendicu- lar to the faces of the first. 13. Construct, through a given point, a plane which shall make, with the faces of a polyhedral angle having four faces, a section that is a parallelogram. Construction: Extend one pair of opp. faces to obtain their line of intersection. Similarly extend the other pair. Any plane section II to these lines will be a O. [Explain.] BOOK VII POLYHEDRONS 650. A polyhedron is a solid bounded by planes. The edges of a polyhedron are the intersections of the bounding planes. The faces are the portions of the bounding planes included by the edges. The vertices are the intersections of the edges. The diagonal of a polyhedron is a straight line joining two vertices not in the same face. POLY- TETRA- HEXAHEDRON OCTA- DODECA- ICOSA- HEDRON HEDRON CUBE HEDRON HEDRON HEDRON 651. A tetrahedron is a polyhedron having four faces. A hexahedron is a polyhedron having six faces. An octahedron is a polyhedron having eight faces. A dodecahedron is a polyhedron having twelve faces. An icosahedron is a polyhedron having twenty faces. 552. A polyhedron is convex if the section made by every plane is a convex polygon. Only convex polyhedrons are considered in this book. 313 314 BOOK VII. SOLID GEOMETRY PRISMS 553. A prism is a polyhedron two of whose opposite faces are congruent polygons in parallel planes, and whose other faces are all parallelograms. The bases of a prism are the congruent parallel polygons. The lateral faces of a prism are the parallelograms. The lateral edges of a prism are the intersections of the lateral faces. The lateral area of a prism is the sum of the areas of the lateral faces. The total area of a prism is the sum of the lateral area and the areas of the bases. The altitude of a prism is the perpendicular distance be- tween the planes of the bases. A triangular prism is a prism whose bases are triangles. PRISM TRIANGULAR REGULAR OBLIQUE PRISMS RIGHT SECTION PRISM PRISM TRUNCATED PRISMS 554. A right prism is a prism whose lateral edges are per- pendicular to the planes of the bases. A regular prism is a right prism whose bases are regular polygons. An oblique prism is a prism whose lateral edges are not per- pendicular to the planes of the bases. A truncated prism is the portion of a prism included between the base and a plane not parallel to the base. PRISMS 315 A right section of a prism is the section made by a plane perpendicular to the lateral edges of the prism. PARALLELEPIPED RIGHT CUBE RECTANGULAR PARALLELEPIPED PARALLELEPIPED 555. A parallelepiped is a prism whose bases are parallelo- grams. A right parallelepiped is a parallelepiped whose lateral edges are perpendicular to the planes of the bases. A rectangular parallelepiped is a right parallelepiped whose bases are rectangles. An oblique parallelepiped is a parallelepiped whose lateral edges are not perpendicular to the planes of the bases. A cube is a rectangular parallelepiped whose six faces are squares. 556. The unit of volume is a cube whose edges are each a unit of length. The volume of a solid is the number of units of volume it contains. The volume of a solid is the ratio of that solid to the unit 'of volume. The three edges of a rectangular parallelepiped meeting at any vertex are the dimensions of the parallelepiped. Equal solids are solids that have equal volumes. Congruent solids are solids that can be made to coincide. 316 BOOK VII. SOLID GEOMETRY Ex. What is the base of a rectangular parallelepiped V of a right parallelepiped ? of an oblique parallelepiped ? Illustrate, by removing the cover and bottom of an ordinary cardboard box and distorting the shape of the frame that remains, the three kinds of parallelepipeds. NOTE. The space that is bounded by the surfaces of a solid, independ- ent of the solid, is called a geometrical solid. That is, if a material or physical body should occupy a certain position and then be removed elsewhere, there is a definite portion of space that is the same shape and size as the solid, and can be conceived as bounded by exactly the same surfaces as bounded the solid when in that original position. In order that we may pass planes and draw lines through solids, and superpose one solid upon another, it is convenient in studying the properties of solids to consider them usually as geometric solids, the material body being removed for the time. PRELIMINARY THEOREMS 657. THEOREM. The lateral edges of a prism are equal. (?.) 658. THEOREM. Any two lateral edges of a prism are parallel. (495.) 559. THEOREM. Any lateral edge of a right prism equals the altitude. (509.) 560. THEOREM. The lateral faces of a right prism are per- pendicular to the bases. (525.) 561. THEOREM. The lateral faces of a right prism are rectangles. (Def. 554.) 562. THEOREM. The faces and bases of a rectangular parallelepiped are rectangles. (Def. 555.) 663. THEOREM. All the faces of any parallelepiped are parallelograms. (? ) 564. AXIOM. A polyhedron cannot have fewer than four faces. 565. AXIOM. A polyhedron cannot have fewer than three faces at each vertex. PRISMS 317 THEOREMS AND DEMONSTRATIONS PROPOSITION I. THEOREM 566. The sections of a prism made by parallel planes cutting all the lateral edges are congruent polygons. Given: Prism AB\ II sections CF and C'F 1 . To Prove: Polygon CF^ polygon C'F'. Proof: CD is II to C'D', DE is II to D'E', etc. .'. CD', DE', EF', etc. are 17 .*. CD = C'D', DE= D'E', EF = E'F', etc. Z GCD = z. G'C'D', Z CDE = Z.C'D'E', etc. .*. Polygon CF^ polygon (484). (?) (124). (499). (150). Q.E.D. Ex. 1. All right sections of a prism are equal. Ex. 2. Any section of a parallelepiped made by a plane cutting two pairs of opposite faces is a parallelogram. Ex. 3. How many edges has a cube? how many vertices? how many dihedral angles? how many trihedral angles? Ex. 4. In the figure of 566, if DE is the altitude of one of the lateral faces, how could one express the area of that face ? Ex.6. In 566, is CF' a prism ? Why? Is CB a prism? Why? Is A F a prism ? Why ? What name is given CB ? 818 BOOK VII. SOLID GEOMETRY PROPOSITION II. THEOREM 667. The opposite faces of a parallelepiped are congruent and parallel. H G Given: (?). To Prove: Face ^.F^and II to face DG. Proof : Faces AF and DG are ZI7 AB = DC, AE = DH AB is II to DC and AE is II to DH .'. face AF^ face DG Also face AF is II to face DG PROPOSITION III. THEOREM (124). (?) (499). (133). (499). Q.E.D. 568. The lateral area of a prism is equal to the product of a lateral edge by the perimeter of a right section. Given : Prism RU' ; edge = E ; right section AD. PRISMS 319 To Prove : Lateral area of EU 1 = E x perimeter of AD. Proof : AB is J_ to RR f , EC is _L to S8 1 , etc. (473). Area O #,s' = ^ . AB (359). Area O sr' = E EC (?). Area O ru 7 = E CD (?). etc. etc. Adding, The lateral area = E (AE + EC + CD+ etc.) (Ax. 2). = E perimeter of rt. sect. (Ax. 6). Q.E.D. 569. COROLLARY. The lateral area of a right prism is equal to the product of its altitude and the perimeter of its base. L = H - Pr. (Where L = lateral area, H = altitude, and Pr = perimeter of base, of a right prism.) 570. COROLLARY. T = L + 2 B. (Where T = total area, and E = area of base.) Ex. 1. Any section of a parallelepiped made by a plane parallel to any edge is a parallelogram. Ex. 2. The sum of the face angles at all the vertices of any parallele- piped is equal to 24 right angles. Ex. 3. The sum of the plane angles of all the dihedral angles of any parallelepiped is equal to 12 right angles. Proof : Pass three planes JL to three intersecting edges. Prove these sections i7 whose A are the plane angles of the dihedral angles, etc. Ex. 4. Find the lateral area of a right prism whose altitude is 8 ft. and each side of whase triangular base is 5 ft. Ex. 5. Find the total area of a regular prism whose base is a regular hexagon, 10 in. on a side, if the altitude of the prism is 15 in. Ex. 6. Find the lateral area of a prism whose edge is 12 in. and whose right section is a pentagon, the sides of which are 3 in., 5 in., 6 in., 9 in., and 11 in. 320 BOOK VII. SOLID GEOMETRY PROPOSITION IV. THEOREM 571. Two prisms are congruent if three faces including a trihedral angle of one are congruent, respectively, to three faces including a trihedral angle of the other, and similarly placed. Given: Prisms AO and A'o'-, face AM ^ face A f M r ; face AP ^ face A'P'\ face AD 3* face A'D'. To Prove : Prism AO ^ prism A r O f . Proof : The three face A at A are respectively = to the three face A at A f (27). . . trih. Z A = trih. Z. A f (546). Superpose prism AO upon prism A'o', making the equal trihedral A A and A! coincide. Face AD coincides with face A f D f , face AM with A'M', face AP with A'P f (They are ^ by hyp.). That is, point L falls on L 1 ; M on M 1 '; and P on P'. .*. the plane O falls upon the plane L'O' (477). Polygon LO^ polygon L r O f (Ax. 1). .-. these bases coincide (?). Similarly face BN coincides with B f N f , CO with c'o', etc. .. the prisms are ^ (Def. 556). Q.E.D. 572. COROLLARY. Two right prisms are congruent if they have congruent bases and equal altitudes. (Explain.) 573. COROLLARY. Two truncated prisms are congruent if three faces including a trihedral angle of one are congruent, respectively, to three faces including a trihedral angle of the other, and are similarly placed. (Explain.) PRISMS PROPOSITION V. THEOREM 321 574. An oblique prism is equal to a right prism whose base is a right section of the oblique prism, and whose altitude is equal to the lateral edge of the oblique prism. Given: Oblique prism AC 1 ; right prism PN r whose base is PN, a right section of AC 1 ', and whose altitude PP' = edge EE f . To Prove: Oblique prism AC f = right prism PN r . Proof : Edge EE f = PP r (Hyp.). Subtract PE r from each, and EP = E f P f (Ax. 2). Likewise AL = A r L r , BM = B'M', CN ^ C f N f , etc. (1) Face AC^ face A'c' (553). (2) In faces AP and A'P f , EP = E f P r , AL= A r L f (Ax. 2). Also AE=A'E', PL = P'L' (124). That is, face AP and face A'P' are mutually equilateral. Also Z EAL = Z E'A'L', Z P.^1 = Z z ^LLP = z A'L'P', z ^PL = Z That is, faces AP and ^'p' are mutually equiangular. . . face AP ^ face A'P' (150). (3) Similarly, face AM ^face A'M'. .-. truncated prism AN ^ truncated prism A' N ? (573). Now, add, solid PC' = solid PC 1 (Iden.). Oblique prism AC 1 = right prism PN 1 (Ax. 2). Q.E.D. Ex. Prove, in the figure of 574, that truncated prism A N is congruent to truncated prism A' N', by the method of superposition. (67). 322 BOOK VII. SOLID GEOMETRY PROPOSITION VI. THEOREM 675. The plane containing two diagonally opposite edges of a parallelepiped divides the parallelepiped into two equal tri- angular prisms. Given: Parallelepiped BH and plane AG containing the opposite edges AE and CG. To Prove : Prism ABC- F= prism ADC - H. Proof: Pass a right section RSTV intersecting the given plane in RT. Face AF is II to face DG (?). .-. RS is II to VT (484). Also RV is II to 8T (?). .-. RSTV is a O (?). .'. ARST^A EVT (126)- Prism ABC-F=& right prism whose base is RST and whose altitude = EA (574). Prism ADC-H = a right prism whose base is RVT and whose altitude = EA (?). But these imaginary right prisms are congruent (572). .-. prism ABC-F= prism ADC-H (Ax. 1). Q.E.D. Ex. 1. The section of a parallelepiped made by a plane containing two diagonally opposite edges, is a parallelogram. Ex. 2. Are the two triangular prisms in the diagram of 575 congruent ? Why? PARALLELEPIPEDS PROPOSITION VII. THEOREM 323 576. Two rectangular parallelepipeds having congruent bases are to each other as their altitudes. Given : Rectangular parallelepipeds P and Q, having ^ bases, and their altitudes AB and CD, respectively. To Prove : P : Q = AB : CD. Proof : I. If the altitudes are commensurable. (Consult 524. ) II. If the altitudes are incommensurable. There does not exist a common unit (225). Suppose AB divided into equal parts. Apply one of these as a unit of measure to CD. There will be a remainder, DX Pass a plane XT, through X and II to the base. Now P:CY=AB:CX Indefinitely increase, etc. as in 524. ROBBINGS NEW SOLID OEOM. 6 (?). 324 BOOK VII. SOLID GEOMETRY 577. COROLLARY. Two rectangular parallelepipeds having two dimensions of the one equal respectively to two dimen- sions of the other, are to each other as their third dimension. The faces having the sides of one equal to the sides of the other, re- spectively, may be considered the bases and the third dimensions the altitudes. Thus this statement is the same as 576. PROPOSITION VIII. THEOREM 678. Two rectangular parallelepipeds having equal altitudes are to each other as their bases. Hi Given: Rectangular parallelepipeds R and 5, having the same altitude H ; and other dimensions a, 5, and b' .H' (579). (Ax. 3). Q.E.D. Ex. 1. In the above diagram, if a = 6 in., b = 6 in., and H = 7 in., find the length of the diagonal of L. Ex. 2. Find the length of the diagonal of a room a ft. long, b ft. wide, and c ft. high. Ex. 3. Show that the four diagonals of a rectangular parallelepiped are all equal. Ex. 4. Find the diagonal of a cube whose edge is 4 in. Find the edge of a cube whose diagonal is 10 in. 326 BOOK VII. SOLID GEOMETRY PROPOSITION X. THEOREM 581. The volume of a rectangular parallelepiped is equal to the product of its three dimensions. Given: (?). To Prove: (?). Proof : Let U be a unit of vol. P = a - b - H U~ 1-1.1 = a b H But - = vol. of P U vol. of P = a b - H (580). (556). (Ax. 1). Q.E.D. 582. COROLLARY. The volume of a rectangular parallele- piped is equal to the product of its base by its altitude. V = B H. (See 581.) 583. COROLLARY. The volume of a cube is equal to the cube of its edge. Ex. 1. A rectangular tank is 56 in. long, 44 in. wide, and 60 in. deep (inside). How many gallons will it hold? [231 cu. in. = 1 gal.] Ex. 2. Three persons measured the above tank inaccurately. A found it to be 57 in. x 44 in. x 60 in. ; B, 56 in. x 45 in. x 60 in. ; and C, 56 in. x 44 in. x 61 in., each recording two dimensions correctly, and one, 1 in. too great. Whose error was most serious, judging by the capacity of the tank? Which dimensions should be most carefully measured? Ex. 3. What is the volume of a cube 6 in. on each edge? What is the edge of a cube having double the volume? What happens to the volume of a cube if we double each edge ? if we halve each edge ? PARALLELEPIPEDS 327 PROPOSITION XI. THEOREM 584. The volume of any parallelepiped is equal to the prod- uct of its base by its altitude. Given : Parallelepiped #, whose base = B and alt. = H. To Prove : Volume of E B - H. Proof : Prolong the edge AD and all edges II to AD. On the prolongation of AD, take EF= to AD. Through E and F pass planes EG and .FT, _L to EF, form- ing the right parallelepiped S. Again, prolong FJ and all the edges II to FJ. On the prolongation of FJ, take KL = to FJ. Through K and L pass planes KM and LN, _L to KL, form- ing the rectangular parallelepiped T. Consider FI the base of S, and EF its altitude. Then R=s (574). Also B = B' (360). Consider FP the base of S, KM the base of T, and KL its altitude. Then s = T; also B' ^ C (574, 134). Hence E = T and B = C (Ax. 1). And the altitude of T is H (509). But the volume of T = C H (582). .-. Vof^ = B-H (Ax. 6). Q.E.D. 328 BOOK VII. SOLID GEOMETRY 585. COROLLARY. Two parallelepipeds having equal alti- tudes and equal bases are equal. (Ax. i.) 586. COROLLARY. Two parallelepipeds having equal altitudes are to each other as their bases. Proof : Q = B n and E = B r - H. .. = (584, Ax. 3). R it' 587. COROLLARY. Two parallelepipeds having equal bases are to each other as their altitudes. (?.) 588. COROLLARY. Any two parallelepipeds are to each other as the products of their bases by their altitudes. (?.) PROPOSITION XII. THEOREM 589. The volume of a triangular prism is equal to the product of its base by its altitude. Given : Triangular prism ACD-F \ base = B ; alt. = H. To Prove : Volume of ACD-F = B - n. Proof: Construct parallelepiped AS having as three of its lateral edges AE, CF, DG. Vol. AS = AGED - H (584). Hence J volume of AS = \ AGED - H (Ax. 3). But \ volume of AS= volume of prism ACD-F (575). And \ ACRD= B (126). .-. volume of ACD-F = B H (Ax. 6). Q.E.D. PRISMS 329 PROPOSITION XIII. THEOREM 590. The volume of any prism is equal to the product of its base by its altitude. Given : Prism AD ; base = to B ; altitude = to H. To Prove : Vol. of AD = B H. Proof: Through any lateral edge, AC, and other lateral edges not adjoining AC, pass planes cutting the prism into triangular prisms I, II, III, having bases R, 8, T, respectively. Vol. of prism I = R H ' (589). Vol. of prism II = s H Vol. of prism III = T H Adding, Vol. of prism AD = (# -f s + r) // = B H (Ax. 2). Q.E.D. 591. COROLLARY. Two prisms having equal altitudes and equal bases are equal. 592. COROLLARY. Two prisms having equal altitudes are to each other as their bases. 593. COROLLARY. Two prisms having equal bases are to each other as their altitudes. 594. COROLLARY. Any two prisms are to each other as the products of their bases by their altitudes. 330 BOOK VII. SOLID GEOMETRY ORIGINAL EXERCISES 1. Which rectangular parallelepiped contains the greater volume, one whose edges are 5 in., 7 in., in., or one whose edges are 4 in., 6 in., 13 in.? 2. The base of a prism is a right triangle whose legs are 8 m. and 12 m. and the altitude of the prism is 20 m. Find its volume. 3. During a rain, half an inch of water fell. How many gallons fell on a level ten-acre park, allowing 7 gal. to the cubic foot ? 4. Counting 38 cu. ft. of coal to a ton, how many tons will a coal bin 18 ft. long, 6 ft. wide, and 9| ft. deep contain, when even full? 6. How many faces has a parallelepiped? edges? vertices? How many faces has a hexagonal prism? edges? vertices? 6. Every lateral face of a prism is parallel to the lateral edges not in that face. 7. Every lateral edge of a prism is parallel to the faces that do not contain it. 8. Every plane containing one and only one lateral edge of a prism is parallel to all the other lateral edges. 9. Any lateral face of a prism is less than the sum of the other lateral faces. 10. The diagonals of a rectangular parallelepiped are equal. Proof: Pass the plane ACGE. This is a rectangle (?), etc. 11. The four diagonals of a parallelepiped bisect each other. [First prove that one pair bisect each other; thus prove that any pair bisect each other, etc.] 12. Two triangular prisms are equal if their lateral faces are equal each to each. 13. Any prism is equal to the parallelepiped hav- ing the same altitude and an equal base. 14. The square of the diagonal of a rectangular parallelepiped is equal to the sum of the squares of its three dimensions. To Prove : AC* = AE* + ED* + DC*. Proof: AD is the hypotenuse of rt. &AED, and AC, of rt. A A CD. 16. The diagonal of a cube is equal to the edge multiplied by V3. A E ORIGINAL EXERCISES 331 16. The volume of a triangular prism is equal to half the product of the area of any lateral face by the perpendicular drawn to that face from any point in the opposite edge. 17. Every section of a prism made by a plane parallel to a lateral edge is a parallelogram. To Prove : LMRN a O. Proof: LM is II to NR (?). LN and MR are each II to any edge. (Explain.) 18. Every polyhedron has an even number of face angles. Proof: Consider the faces as separate polygons. The number of sides of these polygons = double the number of edges of the polyhedron. (Explain.) But the number of sides of these polygons = the number of their angles, that is, the number of face angles. .*. the number of face angles = double the number of edges = an even number (V). 19. There is no polyhedron having fewer than 6 edges. 20. Find the contents and total area of a room 7 m. x 5 m. x 3 in. 21. Find the volume, lateral area, and total area of an 8-inch cube. 22. A right prism whose height is 12 ft. has for its base a right tri- angle whose legs are 6 ft. and 8 ft. Find the volume, lateral area, and total area of the prism. 23. Find the altitude of a rectangular parallelepiped whose base is 21 in. x 30 in., equivalent to a rectangular parallelepiped whose dimen- sions are 27 in. x 28 in. x 35 in. 24. A cube and a rectangular parallelepiped whose edges are 6 in., 16 in., and 18 in., have the same volumes. Find the edge of the cube. 25. Find the volume of a rectangular parallelepiped whose total area is 620 sq. in. and whose base is 14 in. x 9 in. 26. How many bricks each 8 in. x 2| in. x 2 in. will be required to build a wall 22 ft. x 3 ft. x 2 ft. (not allowing for mortar) ? 27. If a triangular prism is 20 in. high and each side of its base is 8 in., how many cubic inches does it contain ? 28. Find the lateral area, total area, and volume of a regular hexag- onal prism each side of whose base is 10 in. and whose altitude is 15 in. 29. A box is 12 in. x 9 in. x 8 in. What is the length of its diagonal ? 30. Each edge of a cube is 8 in. Find its diagonal. 31. The diagonal of a cube is 10 V3 in. Find its edge, volume, and total area. 332 BOOK VII. SOLID GEOMETRY 32. A trench is 180 ft. long and 12 ft. deep, 7 ft. wide at the top and 4 ft. at the bottom. How many cubic yards of earth have been removed? 33. A metallic tank, open at the top, is made of iron 2 in. thick; the internal dimensions of the tank are, 4 ft. 8 in. long, 3 ft. 6 in. wide, 4 ft. 4 in. deep. Find the weight of the tank if empty; if full of water. [Water weighs 62^ Ib. to the cubic foot and iron is 7.2 times as heavy as water.] 34. The base of a right parallelepiped is a rhombus whose sides are each 25 in., and the shorter diagonal is 14 in. The height of the parallel- epiped is 40 in. Find its volume and total surface. 35. If the diagonal of a cube is 12 ft., find its surface. 36. If the total surface of a cube is 54 sq. ft., find its volume. 37. A right prism whose altitude is 25 in. has for its base a triangle whose sides are 11 in., 13 in., 20 in. Find its lateral area, total area, and volume. PYRAMIDS 595. A pyramid is a polyhedron, one of whose faces is a polygon and whose other faces are all triangles having a common vertex. The lateral faces of a pyramid are the triangles. The lateral edges of a pyramid are the intersections of the lateral faces. The vertex of a pyramid is the common vertex of all the lateral faces. The base of a pyramid is the face opposite the vertex. The lateral area of a pyramid is the sum of the areas of the lateral faces. The total area of a pyramid is the sum of the lateral area and the area of the base. The altitude of a pyramid is the perpendicular distance from the vertex to the plane of the base. A triangular pyramid is a pyramid whose base is a triangle. It is called also a tetrahedron. (See 551.) 596. A regular pyramid is a pyramid whose base is a regu- lar polygon and whose altitude, from the vertex, meets the base at its center. PYRAMIDS 333 PYRAMIDS RKGULAR TRUNCATKD FRUSTUM OP PYRAMIDS PYRAMID A PVRAMID The slant height of a regular pyramid is the line drawn in a lateral face, from the vertex perpendicular to the base of the triangular face. It is the altitude of any lateral face. 597. The frustum of a pyramid is the part of a pyramid included between the base and a plane parallel to the base. The altitude of a frustum of a pyramid is the perpendicular distance between the planes of its bases. The slant height of the frustum of a regular pyramid is the perpendicular distance, in a face, between the bases of that face. A truncated pyramid is the part of a pyramid included between the base and a plane cutting all the lateral edges. PRELIMINARY THEOREMS 598. THEOREM. The lateral edges of a regular pyramid are all equal. (504, II.) 599. THEOREM. The lateral faces of a regular pyramid are congruent isosceles triangles. (598, 78.) 600. THEOREM. The lateral edges of the frustum of a regu- lar pyramid are all equal. (Ax. 2.) 601. THEOREM. The lateral faces of the frustum of a regu- lar pyramid are congruent isosceles trapezoids. (484.) 602. THEOREM. The lateral faces of the frustum of any pyramid are trapezoids. (?.) 603. THEOREM. The slant height of a regular pyramid is the same length in all the lateral faces. 334 BOOK VII. SOLID GEOMETRY THEOREMS AND DEMONSTRATIONS PROPOSITION XIV. THEOREM 604. The lateral area of a regular pyramid is equal to half the product of the perimeter of the base by the slant height. 0, Given : Regular pyramid O-ABCDE ; lateral area = to L ; perimeter of base = to P; slant height OH= to s. To Prove : L = \ P s. Proof : Area A AOB = AB s j Area A BOC = ^ BC s\ etc. etc. Adding, Lateral area = | AB s + | BC - s + etc. (Ax. 2). That is, L = \(AB + C + etc.) . s, or, Lateral area, L = \ P s (Ax. 6). Q.E.D. Ex. 1. Prove that the bases of any frustum of a pyramid are mutually equiangular. Ex. 2. The foot of the altitude of a regular pyramid drawn from the vertex, coincides with the center of the circles inscribed in, and circum- scribed about, the base. Ex. 3. The sum of the medians of the lateral faces of the frustum of a pyramid is equal to half the sum of the perimeters of the bases. Ex. 4. To what rectangle is the lateral area of a regular pyramid equal? the total area? PYRAMIDS PROPOSITION XV. THEOREM 335 605. The lateral area of the frustum of a regular pyramid is equal to half the sum of the perimeters of the bases multiplied by the slant height. Given: (?). To Prove : L = |( p +^) 8 - Proof : Area trapezoid C I = J(CD + HI) s Area trapezoid BH = \(BC + GH) s Area trapezoid AG = \(AB + FG) 8 Adding, Lateral area, L |(J + p) . s (?), (?). (?) etc. Explain. Q.E.D. Ex. 1. Using L for lateral area and B for area of base express the formula for the total area, T, of a regular pyramid. Ex. 2. The slant height of a regular pyramid whose base is a square, of which each side is 8 ft., is 15 ft. Find the lateral area; the total area. Ex. 3. A regular pyramid stands on a hexagonal base 16 in. on a side, and the slant height is 2 ft. Find the lateral and total areas. Ex. 4. The slant height of the frustum of a regular pyramid is 12 in., and the bases are squares, 10 in. and 6 in. on a side, respectively. Find the lateral area and the total area. Ex. 6. How many square feet of tin will be required to line a vat in the form of the frustum of a regular pyramid, having inside measure- ments as follows : the slant height is 14 ft. ; the bases are regular hexa- gons whose sides are 7 ft. and 6 ft. respectively. Ex. 6. The lateral edge of a pyramidal church spire is 61 ft. Each side of its octagonal base is 22 ft. What will be the cost of painting the spire at 2^ a square foot ? 336 BOOK VII. SOLID GEOMETRY PROPOSITION XVI. THEOREM 606. If a pyramid is cut by a plane parallel to the base : I. The lateral edges and altitude are divided proportionally. n. The section is a polygon similar to the base. III. The area of the section is to the area of the base as the square of its distance from the vertex is to the square of the altitude of the pyramid. Given : Pyr. O-ABCDE-, plane Fi OL. to the base ; altitude To Prove II. III. OF_ = OG s= OH == = 03f OA OB OC OL Section FI is similar to the base AD. section FI _ OM 2 base AD QL 2 ' Proof: I. Imagine a plane through o II to plane AD. This plane is J_ to OL And II to plane FI OF _ OG _ OH _ _ OM ' OA~~ OB~ OC~ 7 OL II. FG is II to AB, GH is II to BC, etc. .-. Z FGH= Z ABC ; Z GUI Z BCD; etc. That is, the polygons are mutually equiangular. A OFG is similar to A OAB\ A OGH, to A OBC; etc. (305). (496). (489). (484). (499). PYRAMIDS 337 ... =()= ^ = (25?)= *U etc. (313). AH \OBJ BC \OCj CD .'. section FI is similar to base AD (301). Q.E.D. III. Section FI is similar to base AD (II, above), section FI 1G 2 ^OT*\ : - (OlO). base AD 2^ 2 Now A OFG is similar to A OAB (305). .-. = : (313). AB OA r> 4 OF OM , T T N But -= - (I, above). OA OL FG OM .'. -- = - (Ax. 1). AB OL And ^ = ^ (287). AB" OL Hence section f/ = o^ K ^s 5Z 2 Ex. 1. The bases of the frustum of a regular pyramid are equilateral triangles whose sides are 12 in. and 20 in., respectively. The slant height is 40 in. Find the lateral area ; the total area. Ex. 2. The bases of the frustum of a regular pyramid are regular hexa- gons whose sides are 8 in. and 18 in., respectively. The slant height is 25 in. Find the lateral area and total area. Ex. 3. A pyramid whose altitude is 10 in. and whose base contains 80 sq. in. is cut by a plane bisecting the altitude and parallel to the base. Find the area of the section of the pyramid made by this plane. Ex. 4. Cutting a pyramid whose altitude is 16 ft. is a plane parallel to the base and 6 ft. from it. The area of the base is 192 sq. ft. What is the area of the section ? Ex. 5. If a plane parallel to the base of a pyramid bisects the altitude, how does the area of the section compare with the area of the base ? Ex. 6. Two planes, parallel to the base of a pyramid, trisect the alti- tude. How do the areas of the sections compare with the area of the base ? 338 BOOK VII. SOLID GEOMETRY PROPOSITION XVII. THEOREM 607. If two pyramids have equal altitudes and equal bases, sections made by planes parallel to the bases and at equal distances from the vertices are equal. Given: Pyramids O-A BCD E and o'-PQRS-, alt. OL = o' L f ; base AD = base PR ; planes of the sections FI and TV II to the bases; OM = O'M'. To Prove : Section FI = section TV. :2 Proof: But section FI OM" T section TV and O'M base AD QL 2 base PR OM=0'M' and OL = O r L 1 section FI _ section TV base AD base PR Now base AD = base PR Multiplying, section FI= section TV 5 (806, III). (Hyp-)- (Ax. 1). (Hyp.). (Ax. 3). Q.E.D. 608. If a plane is passed parallel to the base of a pyramid, intersect- ing all the lateral edges, and upon this section, as a base, a prism is con- structed wholly inside the pyramid, but having one lateral edge in a lateral edge of the pyramid, this prism is called an inscribed prism. If upon this section as a base a prism is constructed partly outside the pyramid, having one lateral edge in one of the lateral edges of the pyra- mid, this prism is called a circumscribed prism. PYRAMIDS 339 PROPOSITION XVIII. THEOREM 609. The volume of a triangular pyramid is the limit of the sum of the volumes of a series of inscribed or circumscribed prisms, having equal altitudes, if the number of prisms is in- definitely increased. Given : Triangular pyramid O-ABC, having a series of prisms inscribed in it, and another series circumscribed about it, all the prisms having equal altitudes. To Prove : O-ABC is the limit of the sum of each series of prisms as their number is indefinitely increased. Proof: Denote the volume of the pyramid by F, the sum of the volumes of the series of inscribed prisms by 8& and the sum of the volumes of the series of circumscribed prisms by S c . The uppermost circumscribed prism = the uppermost in- scribed prism (591). The second pair of prisms also are equal (?). And so on, until the last circumscribed prism, D-ABC, re- mains, for which there is no equivalent inscribed prism. Hence it is evident that 8 C 8 t = D-ABC, the lowest circumscribed prism. Now by indefinitely increasing the number of the prisms, the altitude of D-ABC becomes indefinitely small, and hence the volume of D-ABC approaches zero as a limit. The altitude can never actually equal zero, nor can the volume equal zero. Hence S c S { can be made less than any mentionable quantity, but cannot equal zero. ROBBINS'S NEW SOLID GEOM. 7 340 BOOK VII. SOLID GEOMETRY Now s c = s c and F > 8 t (Ax. 5). .*. S c V < S c Si (Ax. 9). v < s c (Ax. 5). and Si = .'. V-8 t < S e -8 { (Ax. 7). That is, 8 e V and F S t are each less than S c 8# which itself approaches zero. Hence 8 e F approaches zero and F S { approaches zero. .-. s c approaches F as a limit, and S { approaches F as a limit (227). (See note on p. xviii.) Q.E.D. Ex. 1. If a plane is passed parallel to the base of a pyramid, cutting the lateral edges, the section is to the base as the square of the lateral edge of the pyramid cut away by this plane is to the square of the lateral edge of the original pyramid. [See proof of 60C, III.] Ex. 2. If two pyramids have equal altitudes and are cut by planes parallel to the bases and at equal distances from the vertices, the sec- tions formed will be to each other as the bases of the pyramids. Ex. 3. In the figure of 609 prove that the planes of the faces of the prisms that are opposite OC are parallel to OC and AB. Ex. 4. State the theorems leading up to the theorem of 575. Ex. 5. State the theorems leading up to the theorem of 584. Ex. 6. State the theorems leading up to the theorem of 590. Ex. 7. The base of a pyramid is 180 sq. in. and its altitude is 15 in. What is the area of the section made by a plane parallel to the base, and 5 in. from the vertex? Ex. 8. The base of a pyramid is 200 eq. in., and its altitude is 12 in. At what distance from the vertex must a plane be passed so that the section shall contain half the area of the base ? Ex. 9. The base of a pyramid is B sq. in., and the altitude is h in. How far from the vertex must a plane parallel to the base be passed so that the area of the section shall contain I B sq. in. ? \ B sq. in. ? Find the area of the section if the plane is passed \ h in. from the vertex ; if it is passed \ h in. from the vertex. Ex. 10. A pyramid having an altitude of 2 ft. and a base which is an equilateral triangle of 8 in. on a side, is cut by a plane parallel to the base and 18 in. from it. Find the area of the section. PYRAMIDS 341 PROPOSITION XIX. THEOREM 610. Two triangular pyramids having equal altitudes and equal bases are equal. Given: Triangular pyramids O-ABC and o f -A f B'c ( having equal altitudes and base ABC = base A f B f C f . To Prove : O-ABC = O'-A'B'C'. Proof : Divide the altitude of each pyramid into any number of equal parts. Through these points of division pass planes II to the bases, forming triangular sections. Upon these sections as bases construct inscribed prisms. Denote the volumes of the pyramids by v and V 1 ', and the sums of the volumes of these series of prisms by S and 8 f . The corresponding sections are equal (607). .*. the corresponding prisms are equal (591). Hence 8 = s f (Ax. 2). By indefinitely increasing the number of equal parts into which the altitudes are divided, the number of prisms be- comes indefinitely great. .-. S approaches F as a limit (609). And s' approaches V 1 as a limit (?). .-. F= V' (229). That is, O-ABC = O'-A'B'C'. Q.E.D. NOTE. As in plane geometry, A ABC is the same as ABAC, so in solid geometry the pyramid O-ABC is the same as the pyramid A-BCO or B-ACOor C-ABO. 342 BOOK VII. SOLID GEOMETRY PROPOSITION XX. THEOREM 611. The volume of a triangular pyramid is equal to one third the product of its base by its altitude. Given : Triangular pyramid O-AMC, whose base = B and altitude = H. To Prove : Volume O-AMC = B H. Proof : Construct a prism AMC-DOE, having AMC as its base, and OM as one of its lateral edges. Pass a plane through DO and o<7, cutting the face AE in line CD. The prism is now divided into three triangular pyramids. In pyramids O-AMC and C-ODE, the altitudes are = (509). The bases AMC and ODE are ^ (553). .-. pyramid O-AMC = pyramid C-ODE (610). In pyramids C-AMO and C-AOD, the altitudes are the same line, u _L from C to plane DM (491). The bases AMO and AOD are ^ (126). .*. pyramid C-AMO = pyramid C-AOD (610). Hence O-AMC = C-ODE = C-AOD (Ax. 1). That is, O-AMC = J the prism. But the volume of the prism = B H (?). .-. volume of pyramid O-AMC =/*// (Ax. 6). Q.E.D. Ex. 1. In the figure of 611 prove pyramid 0-ACD = O-CDE. Ex. 2. The area of the base of a triangular pyramid is 30 sq. in., and its altitude is 20 in. Find the volume. Find the volume of the prism having the same base and altitude. PYRAMIDS 343 PROPOSITION XXI. THEOREM 612. The volume of any pyramid is equal to one third the product of its base by its altitude. O, Given : Pyramid o-CDEFG, whose base = B and altitude = H. To Prove : Volume of o-CDEFG = % n H. Proof : Through any lateral edge, OC, and lateral edges not adjoining OC, pass planes dividing the pyramid into tri- angular pyramids. Vol. of O-CFG = 1 CFG - H Vol. of 0-CDE = 1 CDE H Vol. of 0-CEF = 1 CEF- H (611). Adding, Vol. of O-CDEFG = J B H (Ax. 2, 4). Q.E.D. 613. COROLLARY. Any two pyramids having equal altitudes and equal bases are equal. (Ax. 1.) 614. COROLLARY. Two pyramids having equal altitudes are to each other as their bases. (Prove.) 615. COROLLARY. Two pyramids having equal bases are to each other as their altitudes. (Prove.) 616. COROLLARY. Any two pyramids are to each other as the products of their bases by their altitudes. (Prove.) Ex. The altitude of a pyramid is 15 ft., and each side of its square base is 8 ft. Find the volume. What is the volume of a prism having the same base and altitude ? 344 BOOK VII. SOLID GEOMETRY PROPOSITION XXII. THEOREM 617. The volume of the frustum of a triangular pyramid is equal to one third the altitude multiplied by the sum of the lower base, the upper base and a mean proportional between the bases of the frustum. Given : The frustum ED of a triangular pyramid whose lower base = B ; upper base = b ; altitude = H. To Prove : Volume of ED = J H IB + b + V# b~\. Proof : Pass a plane through edge CE and vertex s, and another through edge RS and vertex E, dividing the frustum into three triangular pyramids, S-CDE, E-RST, E-CRS. I. S-CDE = I H-.B II. E-RST = %H b III. We shall now prove E-CRS = E-CSD A .E-CflS " A C&D^ A CRS E-CSD E-CRS ' (368). (Ax. 1). Likewise ^ = ^ (?). 8-ERT AERT PYRAMIDS 345 And But Hence That is, E-CRS (Substituting from I and II). .-. E-CRS = V n B J H - b = i // VB b (289). .-. Volume of the frustum = #[5 + 6+ Vfl.fr] (Ax. 2). Q.E.D. NOTE. The theorem of 617 is sometimes stated thus: The frustum of a triangular pyramid is equal to the sum of three pyramids whose altitudes are the same as the altitude of the frustum and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum. A CER CE (?) (?). (606, II). .(?) (Ax. 1). A ERT RT S-ERT RT CDE and RST are similar CD_CE RS~ RT E-CSD S-CER or E-CRS E-CRS S-ERT J H B _ E-CRS E-CRS I II I Ex. 1. Find the volume of the pyramid whose altitude is 18 in. and whose base is 10 in. square. Ex. 2. Find the volume of the frustum of a triangular pyramid whose altitude is 20 in. and the areas of whose bases are 18 sq. in. and 32 sq. in. Ex. 3. If, in the formula for the volume of the frustum of a triangu- lar pyramid, b = B, show that the formula becomes the correct formula for the volume of a prism. And, again, if b.= 0, show that the formula of 617 becomes the correct formula for the volume of a pyramid. Ex. 4. What is the locus of the vertices of all pyramids upon the same base and having the same volume? Historical Note. The Greeks, as early as the fourth century B.C., knew that the pyramid is the third part of the prism having the same base and altitude. Eudoxus, an Athenian mathematician, is given credit for the discovery and proof of this truth. 346 BOOK VII. SOLID GEOMETRY PROPOSITION XXIII. THEOREM 618. The volume of the frustum of any pyramid is equal to one third the altitude multiplied by the sum of the lower base, the upper base, and a mean proportional between the bases of the frustum. o P Given : Pyr. O-ADEFG ; frustum A r F, whose lower base = U, upper base = #, altitude = H. To Prove : Volume of frustum = i H [B + b + VlTT| . Proof : Construct a A QRS = to polygon AF (395). Upon A QRS as a base, construct a pyramid whose altitude = the altitude of O-ADEFG. Pass a plane Q'R'S' II to QRS and at a distance from QRS= to H. Vol. of Q'R = % H [A QRS + A Q'R'S' -f- VA QRS - A Q'R'S'~\ (617). The alt. of P-Q'R'S' = alt. of O-A'D'E'F'G' (Ax. 2). Also QRS = B (Const.); and Q'R'S' = b (607). .'. vol. of O-ADEFG = vol. of P-QRS (613). And vol. of O-A'D'E'F'G' = vol. of P-Q'R'S' (?). Subtracting, vol. of frustum A'F vol. of frustum Q'R (Ax. 2). Vol. of frustum A'F = ^ H[fl + 6+VB-&] (Ax. 6). Q.E.D. Ex. The bases of the frustum of a pyramid are regular hexagons whose sides are 10 in. and 6 in., respectively. The altitude of the frus- tum is 2 ft. Find its volume. PYRAMIDS 347 PROPOSITION XXIV. THEOREM 619. A truncated triangular prism is equal to three triangu- lar pyramids whose bases are the base of the prism and whose vertices are the three vertices of the face opposite the base (the inclined section). HI Given: The truncated triangular prism ABC-RST, whose base is ABC and whoso opposite vertices are JR, S, T. Let it be divided by the planes ACS, ABT, BCR. To Prove : ABC-liST = R-ABC + S-ABC -f- T-ABC (HI). Proof : In Fig. I, S-ABC is obviously one of these pyramids. In Fig. II, A-CST = A-BCT (613). That is, A-CST= T-ABC. In Fig. Ill, T-ARS = T-ABR (613). T-ABR = C-ABR (613). .'. T-ARS = R-ABC (Ax. 1). Now ABC-RST= T-ARS + S-ABC + A-CST (Ax. 4). Hence ABC-RST = R-ABC -f S-ABC + T-ABC (Ax. 6). Q.E.D. Ex. There are approximately 1 cu. ft. in a bushel. Find the capacity, in bushels, of a grain elevator, 30 ft. high, in the shape of the frustum of a square pyramid, and having bases 24 ft. square and 16 ft. square. 348 BOOK VII. SOLID GEOMETRY 620. COROLLARY. The volume of a trun- cated triangular prism is equal to the prod- uct of the base by one third the sum of the three altitudes drawn to the base from the three vertices opposite the base. 621. COROLLARY. The volume of a trun- cated right triangular prism is equal to the product of the base by one third the sum of its lateral edges. 622. COROLLARY. The volume of any trun- cated triangular prism is equal to the product of its right section by one third the sum of its lateral edges. Proof : The right section divides the solid into two truncated right prisms. Hence the volume = the right section x ^ lateral edges. the sum of the (621.) Q.E.D. Ex. 1. A pyramid whose volume is V and whose altitude is h, is bisected by a plane parallel to the base. Find the distance of this plane from the vertex. Ex. 2. The altitude of a square pyramid each side of whose base is 6 ft., is 10 ft. Parallel to the base and 2 ft. from the vertex a plane is passed. Find the area of the section. Find the volumes of the two pyramids concerned, and hence find the volume of the frustum. Ex. 3. The base of a pyramid is a rhombus whose diagonals are 7 m. and 10 m. Find the volume if the altitude is 15 m. Ex. 4. The areas of the bases of the frustum of a pyramid are 3 sq. ft. and 27 sq. ft. The volume is 104 cu. ft. Find the altitude. Ex. 6. State what distances must be known in order to find the volume of a truncated right triangular prism, and of a truncated oblique triangular prism. Now explain how to use these lengths to find the volume. PYRAMIDS 349 PROPOSITION XXV. THEOREM 623. Two triangular pyramids (tetrahedrons) having a tri- hedral angle of one equal to a trihedral angle of the other are to each other as the products of the three edges including the equal trihedral angles. Given : Triangular pyramids S-ABC, S-PQR ; having the trill. A at S equal ; and their volumes V and v' . To Prove: Z = Ml^l^. V 1 SP SQ SB Proof : Place the pyramids so that the equal trihedral A coincide. Draw the altitudes AX and'PF and the projection SXY, of the edge PS, in plane SQR. A SAX is similar to A SPY V A SBC -AX A SBC AX (304). (616). (313). V f ASQR-PY A SQR PY A SBC _ SB - SC (374X and AX _ SA ASQR~SQ-SR PY SP But Hence, by substituting in the first equation above, V SB SC SA SA SB SC V f SQ SR SP SP SQ SR (Ax. 6). Q.E.D. Ex. 1. To what is the ratio of A SAB to A SPQ equal? Ex. 2. Reduce the formula of 623 if SA = SB = SC and SP = SQ = SR. 350 BOOK VII. SOLID GEOMETRY PROPOSITION XXVI. THEOREM 624. In any polyhedron the number of edges increased by two is equal to the number of vertices increased by the num- ber of faces. Given : A polyhedron ; E = number of edges ; F = num- ber of faces ; F = number of vertices. To Prove : E + 2 = v + F. Proof: Suppose the surface of the polyhedron is put to- gether, face by face. For one face, E = V (145). (Begin with the base.) By attaching an adjoining face, the number of edges is one greater than the number of vertices. That is, for two faces, E = V + 1. For three faces, E=V+2. For four faces, E=V+S. For five faces, E = V + 4. For n faces, E=V + (nl). For F-l faces, E= F + (F-2). By attaching the last face, neither the number of edges nor the number of vertices is increased. That is, for F faces, E = V + F 2. .-. for the complete solid, ^ + 2 = V+F (Ax. 2). Q.E.D. 625. COROLLARY. In any polyhedron the difference between the number of edges and the number of faces is two less than the number of vertices, that is, E - F = V - 2. POLYHEDRONS 351 PROPOSITION XXVII. THEOREM 626. The sum of all the face angles of any polyhedron is equal to 4 right angles multiplied by two less than the number of vertices, that is, S A ( F 2) 4 rt. A = ( V 2) 360. i J Given: A polyhedron ; E = number of edges; F= number of faces ; F = number of vertices. To Prove : Sum of all the face A = (F - 2) 4 rt. A. Or, 8A= (r-2) 360. Proof : If the faces are considered as separate polygons, it is obvious that each edge is a side of two polygons, that is, the number of sides of the several faces = 2 E. .*. the number of vertices of all the polygons = 2 E (145). Suppose an exterior Z formed at each of these 2 E vertices. Then the sum of the exterior A of each face =4 rt. A (?). Sum of int. and ext. A at each vertex = 2 rt. A (?). . -. the int. A + ext. A at all the 2 E vertices = 4 E rt. A But the ext. A of all the F faces = 4 Frt. A (?). . *. the int. A of these polygons = 4 E rt. A 4 F rt. A (?). = (E - F) 4 rt. A. But E - F=V -2 (625). .-. SA = (V-V) 4rt. Zs = (F-2)360 (Ax. 6). Q.E.D. 627. Prismatoid. A prismatoid is a polyhedron two of whose faces, called bases, are polygons in parallel planes, and 352 BOOK VII. SOLID GEOMETRY whose other faces are triangles, trapezoids, or parallelo- grams. The altitude of a prismatoid is the distance between the planes of the bases. The mid-section of a prismatoid is the section made by a plane parallel to the bases and bisecting the altitude. A prismoid is a prismatoid whose lateral faces are either trapezoids or parallelograms. PROPOSITION XXVIII. THEOREM 628. The volume of a prismatoid is equal to the product of one sixth of its altitude by the combined sum of the two bases and four times the mid-section. Given: The prismatoid ACD-EFGKT, with bases b and .B, mid-section Jf, altitude H, and volume F. To Prove : V = % H\b + B + M]. Proof : Through any point, P, in the mid-section, pass planes, each containing an edge of the solid. These planes will divide the prismatoid into pyramids : I. The pyramid P-ACD, whose vertex is P, base 6, alti- tude \ H. Of this pyramid, v 1 = J b H (612). II. The pyramid P-EFGKT, whose vertex is P, base .B, altitude J H. Of this pyramid, v z = J BH. POLYHEDRONS 353 III. Several pyramids, like P-AEF, whose combined vol- ume = | M - //, as we shall now prove. AE = 2 AR (?). .'. AAEF=4AARS (376). .*. pyramid P-AEF = 4 - pyramid P-ARS (614). But pyramid P-ARS = ^ A PRS | J/= J A PRS - II (611). That is, pyramid P-AEF = | A PRS - II. Hence, for the sum of all such pyramids in the prismatoid, V S = M-H (Ax. 4). By addition, F = bH + 7*# + 3f // (Ax. 2). Or, V=\H[b + ^ + 43f]. Q.E.D. Ex. 1. A prismatoid has an upper base 5 sq. in., a lower base 11 sq. in., a mid-section 8 sq. in., and an altitude 9 in. Find the volume. Ex. 2. How does the mid-section of a prism compare with the base ? the mid-section of a pyramid? of a cube-? Ex. 3. Will the formula for the volume of a prismatoid give the vol- ume of a cube? Will it give the formula for the volume of a prism? Ex. 4. Reduce the prismatoid formula to a formula for the volume of a parallelepiped. Ex. 6. Derive the formula for the volume of a pyramid from the prismatoid formula. Ex. 6. Is a prism a prismatoid ? is a pyramid ? is a truncated prism? is the frustum of a pyramid? is a parallelepiped? Historical Note. The prismatoid formula was discovered by a German, E. F. August, in the latter part of the nineteenth century. Its importance in the mensuration of polyhedrons was recognized at once by mathema- ticians throughout the world. REGULAR AND SIMILAR POLYHEDRONS 629. A regular polyhedron is a polyhedron whose faces are equal regular polygons and whose polyhedral angles are all equal. Similar polyhedrons are polyhedrons which have the same number of faces similar each to each and similarly placed, and which have their homologous polyhedral angles equal. 354 BOOK VII. SOLID GEOMETRY PROPOSITION XXIX. THEOREM 630. There can exist no more than five kinds of regular polyhedrons. Proof: The faces must be equilateral A, squares, regular pentagons, or some other regular polygons. (629.) There must be at least three faces at each vertex. (565.) The sum of the face A at each vertex is < 360. (549.) I. Each Z of an equilateral A = 60 (?). Hence we may form a polyhedral Z by placing 3, 4, or 5 equilateral A at a vertex, but not 6 (?). That is, only three regular poly- hedrons can be formed having equilateral triangles for faces. II. Each Z of a square = 90 (?). Hence we may form a polyhedral Z by placing 3 squares at a vertex ; but not 4. That is, only one regular polyhedron can be formed having squares for faces. III. Each Z of a regular pentagon = 108 (155). Hence we may form a polyhedral Z by placing 3, but not 4 regular pentagons at a vertex. That is, only one regular polyhedron can be formed having regular pentagons for faces. IV. Each Z of a regular hexagon = 120 (?). .. no polyhedral Z can be formed by hexagons (?). Consequently there can be no more than five kinds of regular polyhedrons, three kinds bounded by triangles, one kind by squares, and one by pentagons. Q.E.D. 631. The names of the regular polyhedrons. TOTAL NUMBKR OF NAMES NUMBER FACRB AT KINDS OF FACES OK FACES EACH VERTEX Regular tetrahedron 4 3 Equilateral triangles Regular hexahedron (cube) 6 3 Squares Regular octahedron 8 4 Equilateral triangles Regular dodecahedron 12 3 Regular pentagons Regular icosahedron 20 5 Equilateral triangles POLYHEDRONS 355 KKGUL.AR CUHK RKGUL.AR RKUULAR KKGULAK TETRAHEDRON OCTAHEDRON DODECAHEDRON ICOSAHEDRON DIRECTIONS FOR CONSTRUCTION. Mark on cardboard larger figures similar to the drawings. Cut the dotted lines half through and the solid lines entirely through. Fold along the dotted lines, closing the solids up and forming the figures. Paste strips of paper along the edges. Historical Note. Pythagoras knew about the existence of all the regu- lar polyhedrons except the dodecahedron. This was discovered it) 470 B.C. by Hippasus, who having boasted of his discovery was drowned by the other Pythagoreans. The regular polyhedrons were supposed to have certain magical properties and their study was greatly emphasized. ROBBINS'S NEW SOLID GEOM 8 356 BOOK VII. SOLID GEOMETRY PROPOSITION XXX. THEOREM 632. In two similar polyhedrons : I. Homologous edges are proportional. n. Homologous faces are to each other as the squares of any two homologous edges. III. Total surfaces are to each other as the squares of any two homologous edges. Proof : I. Homologous faces are similar (629). AB EC CD DH AE BF A'B' B'C' C'D' D'H' A'E' B'F* Face DG im* face AH ~AE' Face D'G 1 .z/jj' 2 face A'a ~ Face DG face AH face GE Face D'(?' " face A'H' face G r E f Total surface of A G face DG DH AE _ ^ ~ Total surface of A'Q' face D' G 1 vTa A'W 2 Or, T:T' = e2;e'2. Q.E.D. 633. COROLLARY. If a pyramid is cut by a plane parallel to the base, the pyramid cut away is similar to the original pyramid. _ (629.) Ex. 1. Show that the theorem of 624 holds true in the case of a cube. Ex. 2. Show that the theorem of 626 holds true in the case of a cube. Ex. 3. Show that 624 and 626 are true of a regular octahedron. POLYHEDRONS 357 PROPOSITION XXXI. THEOREM 634. Two similar tetrahedrons are to each other as the cubes of any two homologous edges. O' Given : Similar tetrahedrons O-ABC and O'-A'B'C' ; whose volumes = V and V'. To Prove : v : V 1 = 2s 3 : J 7 ^ 73 = etc. Proof : Trihedral Z A = trihedral Z ^ (629). . F AB.AC.AO (623) (632, I). (Ax. 6). etc. Q.E.D. ' V' A'B' .A'C' ^A'O' AB AC AO A'B' A'C' A'O' AB AC AO A' B' A'C' A'O 1 p V AB AB AB ' V' A'B' A'B' A'B' That is, F : V' = AB B : A'B' & = AC 3 : A'C' Or, S:T r ' = e* : e> 3. Ex. 1. If, in the figure of 631, AO= 4 in. and A'0' = l in., what is the ratio of the total surfaces of the tetrahedrons ? of their volumes ? Ex. 2. Two homologous edges of two similar polyhedrons are 2 in. and 3 in. What is the ratio of their total surfaces ? Ex. 3. Two homologous edges of two similar tetrahedrons are 2 in. and 5 in. The total surface of the less is 28 sq. in., and its volume is 40 cu. in. Find the area of the total surface and the volume of the other. Ex. 4. Show that the theorems of 624 and 626 are true in the cases of regular dodecahedrons and regular icosaliedrous. 358 BOOK VII. SOLID GEOMETRY PROPOSITION XXXII. THEOREM 635. Two similar polyhedrons can be decomposed into the same number of tetrahedrons similar each to each and sim- ilarly placed. Given : (?). To Prove : (?). Proof: Suppose diagonals drawn in every face of AT^ ex- cept the faces containing vertex A, dividing the faces into A. (The figure shows only 8V.) Suppose lines drawn from A to the several vertices of these A. (The figure shows only AS, AV.) Obviously, this process divides the solid (by planes) into tetrahedrons, each of which has a vertex at A. Then construct homologous lines in solid A'T*. There will evidently be as many lines in A f T f as in AT and as many tetrahedrons, and these will be similarly placed. Now, in the tetrahedrons A-SVR and A f -S r V r R f , A AYR is similar to A A'V'R' ; A ASR is similar to A A'S'R' ; ' A SVR is similar to A S f V r R f (318). AV VR VS RS AS A1 Also AV = VS A'V' ~ V'S 1 POLYHEDRONS 359 Hence A ASV is similar to A A' 8 1 V 1 (308). Also the trihedral A R and R f are equal ; 8 and S f are equal ; V and V r are equal, etc. (546). .-. the two tetrahedrons are similar (Def. 629). Furthermore, after removing these tetrahedrons, the re- maining polyhedrons are similar. (Def. 629.) By the same process other pairs of tetrahedrons may be removed and proved similar, and the process may be con- tinued until the polyhedrons are completely decomposed into tetrahedrons similar each to each and similarly placed. Q.E.D. PROPOSITION XXXIII. THEOREM 636. The volumes of two similar polyhedrons are to each other as the cubes of any two homologous edges. Given: Similar polyhedrons AT and A'T' ; volumes V and V'\ AB and A'R 1 , any two homologous edges. (See figure of 635.) To Prove : (?). Proof: These solids may be decomposed, etc. (635). Denote the volumes of tetrahedrons of AT by w, x, y, z, etc.; of A'T' by w', x f , y', z' , etc. rpi w Alf x AR 3 y AI? /i?oi\ Ihen = = r^; -7= ; - = ^; etc. (b34). v* A'R'* *' A'R' 3 y' A'R' 3 (Ax. 1). (291) (Ax. 6). Q.E.D. A'R' 637. COROLLARY. The volumes of two similar pyramids are to each other as the cubes of their altitudes. (Explain.) Hence Therefore w + x w' x' y' -h?/4-etc. _w w' + x> That is. ' +y' H- etc. w 1 3 360 BOOK VII. SOLID GEOMETRY ORIGINAL EXERCISES 1. What plane through the vertex of a given tetrahedron will divide it into two equal parts? Prove. 2. The area of the base of any pyramid is less than the sum of the lateral faces. [Draw the altitudes of the lateral faces and the projections of the altitudes upon the base.] 3. Three of the edges of a parallelepiped that meet in a point are also the lateral edges of a pyramid. What part of the parallelepiped is this pyramid? 4. A plane is passed containing one vertex of a parallelepiped and a diagonal of a face not contain- ing that vertex. What part of the volume of the parallelepiped is the pyramid thus cut off ? 6. Any section of a tetrahedron made by a plane parallel to two opposite edges, is a parallelogram. Given : Section DEFG II to OA and BC. To Prove : DEFG is a O. Proof: EF is II to OA, DG is II to OA (?). Also DE is II to BC and GF is II to BC (?), etc. 6. The three lines that join the midpoints of the opposite edges of a tetrahedron meet in a point and bisect one another. Given : LM, PQ, RS, three lines, etc. To Prove : (?). Proof : Join PS, SQ, QR, PR. PS is II to and = BC ; RQ is II to and = \ BC. (Explain.) .-.fig. PSQRis&O (?). Similarly, discuss LM and SR. 7. A pyramid having one of the faces of a cube for its base and the center of the cube for its vertex, contains one sixth of the volume of the cube. 8. A plane containing an edge of a regular tetra- hedron and the midpoint of the opposite edge : (a) contains the medians of two faces ; (b) is perpendicular to the opposite edge ; (c) is perpendicular to these two faces ; M (d) contains two altitudes of the tetrahedron. 9. The altitude of a regular tetrahedron meets the base at the point of intersection of the medians of the base. ORIGINAL EXERCISES 361 10. The altitude of a regular tetrahedron = \/6 times the edge. 11. The altitudes of a regular tetrahedron meet at a point. 12. The lines joining the vertices of any tetrahedron to the point of intersection of the medians of the opposite face meet in a point that divides each line into segments in the ratio 3 : 1. Given : OM, CR, two such lines. To Prove : The four such lines meet, etc. Proof: OM and CR lie in the plane determined by OC and point D, the midpoint of AB. .: OM and CR intersect. Draw RM. DR =\RO 1 ( ^ . PR _ RO DM=\ MC \ ' DM~ MC .-.^M is II to OC B .-. A DMR and DCO are similar (?) ; and DR:DO = RM: OC. (?) Thus RM = OC. (Explain.) Also & PRM and OPC are similar (?). Hence OP : PM = OC : RM = 3:1.) And CP:PR=OC:RM = 3:l.\ (Explain.) Q.E.D. NOTE. This point P is called the center of gravity of the tetrahedron. 13. There can be no polyhedron having seven edges and only seven. 14. The planes bisecting the dihedral angles of any tetrahedron meet in a point that is equally distant from the faces. 16. The lines perpendicular to the faces of any tetrahedron, at the centers of the circles circumscribed about the faces, meet in a point that is equally distant from the vertices. Proof : RX and SY are loci of points, etc. (511). Plane MN, _L to AB at M, the midpoint of AB, is the locus of points, etc. .-. RX and SY lie in MN and intersect at 0, etc. 16. If a plane is passed through the midpoints of the three edges of a parallelepiped that meet at a vertex, what part of the whole solid is the pyramid thus cut off? 17. The plane bisecting a. dihedral angle of a tetrahedron divides the opposite edge into two segments proportional to the areas of the faces that form the dihedral angle. 362 BOOK VII. SOLID GEOMETRY 18. Two tetrahedrons are similar if a dihedral angle of one equals a dihedral angle of the other and the faces forming these dihedral angles are respectively similar. 19. If from any point within a regular tetrahedron perpendiculars to the four faces are drawn, their sum is constant and equal to the altitude of the tetrahedron. |E 20. Construct a regular tetrahedron upon a given edge. Construction: Upon AB, construct an equilateral &ABC. Erect ED _L to plane of &ABC, at Z>, the center of circumscribed O. Take V on ED such that AV = BV = CV = AB,etc. 21. Construct a regular hexahedron upon a given Construction: Upon AB construct a square ABCD. At the vertices erect _ls = to AB and join the extremities, etc. 22. Construct a regular octahedron upon a given . edge. \ Construction: Upon AB construct a square D ABCD. At M, the center of the square, erect XX' _L to plane of ABCD. On XX' take MV = MV = MD. Draw the edges from V and V. Statement : VV is a regular octahedron. Proof: The right A DM V, DMC, DMV are equal. (Explain.) A Thus the 12 edges are equal and the 8 faces are equal. (Explain.) Figures AVCV, DVBV, ABCD are equal squares (Explain.) Then, pyramids V-ABCD, D-AVCV, etc., are equal and the 6 polyhedral angles are equal. (Explain.) . 23. Pass a plane through a cube so that the section will be hexagon. 24. Pass planes through three given lines in space, no two of which are parallel, which shall inclose a parallelepiped. 25 Find the lateral area and the total area of a regular pyramid whose slant height is 20 in. and whose base is a square, 1 ft. on a side. ORIGINAL EXERCISES 363 26. Find the volume of a pyramid whose altitude is 18 in. and whose base is an equilateral triangle each side of which is 8 in. 27. A regular hexagonal pyramid has an altitude 1 of 9 ft. and each edge of the base is 6 ft. Find the volume. 28. The base of a pyramid is an isosceles triangle whose sides in inches are 14, 25, 25, and the altitude of the pyramid is 12 in. Find its volume. 29. The altitude of the frustum of a pyramid is 25 in., and the bases are squares whose sides are 4 in. and 10 in., respectively. Find the volume of the frustum. 30. The frustum of a regular pyramid has hexagons for bases whose sides are 5 in. and 9 in., respectively. The slant height of the frustum is 14 in. Find its lateral area. Find its total area. Q 31. The altitude of a regular pyramid is 15 in., and each side of its square base is 16 in. Find the slant height, the lateral edge, the total area, and the volume. OA* = O& + DA 2 = (15) 2 + (8) 2 = 289. .-.A0 = 17. OC 2 = OA 2 + A C* = 289 + 64 = 353. .-. OC = v/353 = 18.78+. 32. The slant height of a regular pyramid is 39 ft., the altitude is 36 ft., and the base is a square. Find the lateral area and the volume. 33. The lateral edge of a regular pyramid is 37 in. and each side of the hexagonal base is 12 in. Find the slant height, the lateral area, the total area, and the volume. In rt. A A CD, CD = 12, A C = 6, .-. AD = 6\/3. Inrt. AvlCO, CO = 37, AC = 6, .-. .40 = V1333. In rt. A CDO, CO = 37, CD = 12, .-. OD = 35, etc. 34. Find the total area and volume of a regular tetrahedron whose edge is 6 in. The four faces are equal equilateral &. .-. AO=AC = 3\/3 in.; .-. AD = V3 in. and CD = 2V3 in. Hence OD = 2 V6 in. Area of any face = 9V3 sq. in., etc. 36. Find the total area and the volume of a regular tetrahedron whose edge is 10 in. 36. Find the total area and the volume of a regular hexahedron whose edge is 8 in. 864 BOOK VII. SOLID GEOMETRY 37. Find the total area and the volume of a regular octahedron whose edge is 16 in. The 8 faces are equal equilateral A. ^40 = 8V3. In A ADO, one finds OD = 8V2. The volume of the octa- hedron = the volume of two pyramids, etc. 38. Find the total area and the volume of a regular octahedron whose edge is 18 in. 39. The altitude of a regular pyramid is 16 in. and each side of the square base is 24 in. Find the lateral area and the volume. 40. The slant height of a regular pyramid is 16 in. and each side is an equilateral triangle whose side is 20 \/3 in. Find the total area and the volume. 41. The altitude of a regular pyramid is 29 in. and its base is a regu- lar hexagon whose side is 10 in. Find the total area and the volume. 42. Find the total area and the volume of a regular tetrahedron whose edge is 18 in. 43. Find the total area and the volume of a regular octahedron whose edge is 20 in. 44. If the edge of a regular tetrahedron is e in., show that the total area is e 2 \/3 in. and the volume is ^ e 3 V2 cu. in. 45. If the edge of a regular octahedron is e in., show that the total area is 2 e 2 v"3 sq. in. and the volume is \ e 3 V2 cu. in. 46. A pyramid whose base is a square 9 in. on a side, contains 360 cu. in. Find its height. 47. A pyramid has for its base a hexagon whose side is 7 units and the pyramid contains 675 cubic units. Find the altitude. 48. The volume of a regular tetrahedron is 144 V2 cu. in. Find its edge. 49. The volume of a regular octahedron is 243 V2 cu. in. Find its 60. The volume of a square pyramid is 676 cu. in. and the altitude is 1 ft. Find the side of the base. Find the lateral area. 61. The altitude of the Great Pyramid is 480 ft. and its base is 764 ft. square. It is said to have cost $ 10 a cubic yard and f 3 more for each square yard of lateral surface (considered as planes). What was the cost? 62. The total surface of a regular tetrahedron is 324\/3sq. in. Find its volume. 63. The base of a pyramid is an isosceles right triangle whose hypot- enuse is 8 in. The altitude of the pyramid is 15 in. Find the volume. ORIGINAL EXERCISES 365 64. Find the area of the section of a triangular pyramid, each side of whose base is 8 in. and whose altitude is 18 in., made by a plane parallel to the base and 1 ft. from the vertex. 66. The altitude of a frustum of a pyramid is 6 in., and the areas of the bases are 20 sq. in. and 45 sq. in. Find the altitude of the complete pyramid. Find the volume of this frustum by two distinct methods. 66. A granite inonument in the form of a frustum of a pyramid, having rectangular bases one of which is 8 ft. wide and 12 ft. long, and the other 6 ft. wide, is 30 ft. high. It is surmounted by a granite pyramid having the same base as the less base of the frustum, and 10 ft. in height. Find the entire volume and the weight. [1 cu. ft. of water weighs 62| Ib. and granite is 3 times as heavy as water.] 67. If a square pyramid contains 40 cu. in. and its altitude is 15 in., find the side of its base. 68. A church spire in the form of a regular hexagonal pyramid whose base edge is 8 ft. and whose altitude is 75 ft. is to be painted at the rate of 18 ^ per square yard. Find the cost. 69. Find the edge of a cube whose volume is equal to the volumes of two cubes whose edges are 4 in. and 6 in. 60. The base of a certain pyramid is an isosceles trapezoid whose parallel sides are 20 ft. and 30 ft. and the equal sides each 13 ft. Find the volume of the pyramid if its altitude is 12 yd. 61. The lateral edge of the frustum of a regular square pyramid is 53 in. and the sides of the bases are 10 in. and 66 in. Find the altitude, the slant height, the lateral area, and the volume. 62. The sides of the base of a triangular pyramid in inches are 33, 34, 65, and the altitude of the pyramid is 80. Find its volume. 63. The sides of the base of a tetrahedron in inches are 17, 25, 26, and its altitude is 90. Find its volume. 64. If there are 1 cu. ft. in a bushel, what is the capacity (in bushels) of a hopper in the shape of an inverted pyramid, 12 ft. deep and 8 ft. square at the top ? 66. In the corner of a cellar is a pyramidal heap of coal. The base of the heap is an isosceles right triangle whose hypotenuse is 20 ft. and the altitude of the heap is 7 ft. If there are 35 cu. ft. in a ton of coal, how many tons are there in this heap? 66. How many cubic yards of earth must be removed in digging an artificial lake 15 ft. deep, whose base is a rectangle 180 ft. x 20 ft. and whose top is a rectangle 216 ft. x 24 ft. ? [The frustum' of a pyramid.] 366 BOOK VII. SOLID GEOMETRY 67. One pair of homologous edges of two similar tetrahedrons are 3 ft. and 5 ft. Find the ratio of their surfaces ; of their volumes. 68. A pair of homologous edges of two similar polyhedrons are 5 in. and 7 in. Find the ratio of their surfaces ; of their volumes. 69. The edge of a cube is 3 in. What is the edge of a cube twice as large ? four times as large ? half as large ? 70. An edge of a tetrahedron is 6 ft. What is tjie edge of a similar tetrahedron three times as large? eight times as large? nine times as large ? one third as large ? 71. An edge of a regular icosahedron is 3 in. What is the edge of a similar solid five times as large? ten times as large? fifty times as large? a thousand times as large? 72. The edges of a trunk are 2 ft., 3 ft., 5 ft. Another trunk is twice as long (the other edges 2 ft. x 3 ft.). How do their volumes compare? A third trunk has each dimension double those of the first. How does its volume compare with the first ? How do their surfaces compare ? 73. If the altitude of a certain regular pyramid is doubled, but the base remains unchanged, how is the volume affected? If each edge of the base is doubled, but the altitude is unchanged, how is the volume affected ? If the altitude and each edge of the base are all doubled, how is the volume affected ? 74. A contractor agrees to build a dam 60 ft. long, 15 ft. high, 11 ft. wide at the bottom and 7 ft. wide at the top for $ 8.25 a cubic yard. Find his profit if it costs him only $2000. 75. A pyramid is cut by a plane parallel to the base and bisecting the altitude. What part of the entire pyramid is the less pyramid cut away by this plane ? 76. The volume of a certain pyramid, one of whose edges is 7 in., is 686 cu. in. Find the volume of a similar pyramid whose homologous edge is 8 in. 77. A certain polyhedron whose shortest edge is 2 in. weighs 40 Ib. What is the weight of a similar polyhedron whose shortest edge is 5 in.? 78. An edge of a polyhedron is 5 in. and the homologous edge of a similar polyhedron is 7 in. The entire surface of the first is 250 sq. in. and its volume is 375 cu. in. Find the entire surface and volume of the second. 79. A berry box, sold to contain a quart of berries, is in the form of the frustum of a pyramid 5 in. square at the top, 4| in. square at the bottom, and 2 in. deep. If a U.S. dry quart contains 67.2 cu. in., does this box contain more or less than a quart? BOOK VIII CYLINDERS AND CONES CYLINDERS 638. A cylindrical surface is a surface generated by a moving straight line which continually intersects a given curved line in a plane, and which is always parallel to a given straight line not in the plane of the curve. The generating line is the generatrix. The directing curve is the directrix. An element of a cylindrical surface is the generating line in any position. CYLINDRICAL SURFACE RIGHT CIRCULAR OBLIQUE CYLINDER CYLINDER CYLINDER OF REVOLUTION 639. A cylinder is a solid bounded by a cylindrical sur- face and two parallel planes. The bases of a cylinder are the parallel plane sections. The lateral area of a cylinder is the area of the cylindrical surface included between the planes of the bases. The total area of a cylinder is the sum of the lateral area and the areas of the bases. The altitude of a cylinder is the perpendicular distance between the planes of the bases. 367 368 BOOK VIII. SOLID GEOMETRY 640. A right cylinder is a cylinder whose elements are per- pendicular to the planes of the bases. A circular cylinder is a cylinder whose base is a circle. An oblique cylinder is a cylinder whose elements are not perpendicular to the planes of the bases. A right circular cylinder is a right cylinder whose base is a circle. A cylinder of revolution is a cylinder generated by the revolution of a rectangle about one of its sides as an axis. Similar cylinders of revolution are cylinders generated by similar rectangles revolving on homologous sides. 641. A right section of a cylinder is a section made by a plane perpendicular to all the elements. A plane is tangent to a cylinder if. it contains one element of the cylindrical surface and only one, however far it may be extended. A prism is inscribed in a cylinder if its lateral edges are elements of the cylinder and the bases of the prism are in- scribed in the bases of the cylinder. A prism is circumscribed about a cylinder if its lateral faces are tangent to the cylinder and the bases of the prism are circumscribed about the bases of the cylinder. PRELIMINARY THEOREMS 642. THEOREM. Any two elements of a cylinder are paral- lel and equal. (495 and 509.) 643. THEOREM. A line drawn through any point in a cylin- drical surface, parallel to an element, is itself an element. (Ax. 13.) 644. THEOREM. A right circular cylinder is a cylinder of revolution. Ex. If a plane is defined as a surface generated by a moving straight line, what is the directrix ? CYLINDERS 369 THEOREMS AND DEMONSTRATIONS PROPOSITION I. THEOREM 645. Every section of a cylinder made by a plane contain- ing an element is a parallelogram. " Given : Cylinder AB ; plane CE containing element CD. To Prove : CE is a O. Proof : At E draw EF II to CD in plane CE. Also, EF is an element of the cylinder. (643.) .-. EF is the intersection of the plane and the cylindrical surface. (466.) Also OF is II to DE (-184). .-. CDEFis&CJ (120). PROPOSITION II. THEOREM 646. The bases of a cylinder are congruent. Given: (?). To Prove : (?). Proof : Suppose 2?, 5, T three points in the perim. of base AC. (642). 370 BOOK VIII. SOLID GEOMETRY Draw elements RR r , SS r < Tf . Also draw RS, ST, RT, R'S', S'T', R'T' RR f = and is II to 88 r RR 1 = and is II to Tl* 88 r = and is II to TT f .-. RS f is a O, RT 1 is a O, ST 1 is a O (129). .'.RS = R'S'; 8T=8 r T 1 -, RT=R'T' (124). \' / * .. base AC may be placed upon base BD so that R, S, and T coincide with R f , s', and r', respectively. But S is any point on the boundary ; hence every point on the boundary of AC will coincide with a corresponding point on the boundary of BD. .-. base AC ^ base BD (Def. 26). Q.E.D. 647. COROLLARY. Parallel plane sections of a cylinder (cutting all the elements) are congruent. 648. COROLLARY. Every section of a right cylinder made by a plane containing an element is a rectangle. 649. COROLLARY. The line joining the centers of the bases of a circular cylinder is equal and parallel to an element. 650. COROLLARY. All sections of a circular cylinder parallel to its bases are equal circles, and the straight line joining the centers of the bases passes through the centers of all the parallel sections. To Prove : HF = any other II section and C is its center. Proof : Pass a plane AE cutting the three planes in AG, CF, and BE. DF is a cylinder (Def. 639). That is, HF is a O and equal to any other section CYLINDERS 371 Now AG is II to CF (484). And A K is II to GE (649). .-. ^FisaO (120). And AG = CF (124). But JIF is a O and F is any point on it. Hence C is equally distant from all points on HF, and is, therefore, the center. (179.) Q.E.D. PROPOSITION III. THEOREM 651. THEOREM. If a regular prism is inscribed in, or circum- scribed about, a right circular cylinder and the number of sides of the base is indefinitely increased, the lateral area of the cylinder is the limit of the lateral area of the prism. Given : A regular prism inscribed in and a regular prism circumscribed about a right circular cylinder ; the lateral area of the cylinder = L, and of the prisms, L t and L c , respectively. To Prove : That as the number of sides of the bases of the prisms is indefinitely increased, L is the limit of both Li and L c . Proof : If the number of sides of the bases of the prisms is indefinitely increased, their perimeters will approach the circumference of the base of the cylinder as a limit. (?.) Hence, it is obvious that the lateral area of the cylinder is the limit of the lateral area of either prism. Q.E.D. ROBBINS'S NEW SOLID GEOM. 9 372 BOOK VIII. SOLID GEOMETRY PROPOSITION IV. THEOREM 652. If a prism having a regular polygon for a base is in- scribed in, or circumscribed about, any circular cylinder and the number of the sides of the base of the prism is indefinitely increased, the volume of the cylinder is the limit of the volume of the prism. Given: (?). To Prove: (?). Proof : If the number of sides of the base of either prism is indefinitely increased, the area of the base of the prism approaches the area of the base of the cylinder. (424, II.) .*. it is obvious that the volume of the cylinder is the limit of the volume of either prism. Q.E.D. Ex. 1. If the cylindrical surface of a cylinder is cut along an element, and this surface is placed in coincidence with a plane, what plane geometrical figure will it become 9 Ex. 2. What two lines determine the size of a right circular cylinder? Ex. 3. What is the locus of all points 2 in. from a circular cylindrical surface? Is the answer to this question affected by the radius of the given surface? If so, explain. Ex. 4. From a log 36 in. in diameter at its less end, is to be cut the largest prismatic piece of timber possible, having square ends. Find the side of this square. Ex. 6. A lead pencil whose ends are regular hexagons was cut from a cylindrical piece of wood, with the least waste of wood. If the original piece was 8 in. long and \ in. in diameter, find the volume of the pencil. CYLINDERS 373 PROPOSITION V. THEOREM 653. The lateral area of a right circular cylinder is equal to the product of the circumference of the base by an element. Given : A right circular cylinder, the circumference of whose base = O, and whose element = E. To Prove : Lateral area L = C-E. Proof : Inscribe in the cylinder a regular prism, the perimeter of whose base is P, whose lateral edge is E, and whose lateral area is L r . Then L f = P-E (?). If the number of sides of the base of the prism is indefi- nitely increased, L f approaches L as a limit (?). P approaches C as a limit (?). .-. L = C-E (?). Q.E.D. 654. COROLLARY. Area of a right circular cylinder: (Where L = lateral area, H = altitude, E T = total area.) radius of base, and NOTE. The lateral area of an oblique circular cylinder equals the product of the perimeter of a right section of the cylinder by an element. The right section of an oblique cir- cular cylinder is not a circle. The right section of an inscribed prism, having a regular polygon for a base, is not a regular polygon. Hence the proof of this theorem is omitted. 374 BOOK VIII. SOLID GEOMETRY PROPOSITION VI. THEOREM 655. The volume of a circular cylinder is equal to the product of its base by its altitude. Given : A circular cylinder whose base = #, altitude = IT, and volume = F. To Prove : v = B H. Proof : Inscribe a prism having a regular polygon for its base, whose base = B 1 and volume = V 1 . Then V 1 = B r - H (?). If the number of sides of the base of the prism is indefi- nitely increased, v' approaches F as a limit (?). B 1 approaches B as a limit (?). B 1 - H approaches B H as a limit. .-. F = B .H (229). Q.E.D. 656. COROLLARY. Volume of a circular cylinder: r= irR 2 jy. (Where F = volume, H altitude, and R = radius of base.) Ex. 1. Find the lateral area and the total area of a cylinder whose altitude is 15 in. and radius 14 in. Ex. 2. How many square inches of tin are required to make a cylin- drical pipe 10 J in. in diameter and 8 ft. long? Ex. 3. What is the capacity of a cylindrical pail 1 ft. high and 9 in. in diameter? CYLINDERS 375 PROPOSITION VII. THEOREM 657. Of two similar cylinders of revolution: I. The lateral areas are to each other as the squares of their altitudes or as the squares of the radii of their bases. II. The total areas are to each other as the squares of their altitudes or as the squares of the radii of their bases. III. The volumes are to each other as the cubes of their altitudes or as the cubes of the radii of their bases. Given : Two similar cylinders of revolution whose lateral areas = L and I ; total areas = T and t ; volumes = V and v ; altitudes = H and A, and radii are R and r. To Prove : I. L : 1= H 2 : A 2 = R 2 : r 2 . II. T: t=H 2 : h 2 = R 2 : r*. III. F : v= H s : A 3 = # 3 : r 3 . Proof : The generating rectangles are similar. .-. H: h=R : r Hence H+R_H_R h -\- r h U ff TT TT rr2 7?2 7 = (Def. 640.) (291). 2 Trrh RH rh TT r n R H+R r r H A = A 2 r 2 __ ( . fi . ' ~~ 7rR 2 H 91 irr 2 h R 2 Ff 97 r z h TTT ___ _ __ - ~ - * " -- ~ * - - ~~~ 9 r 2 H T" h H 19 - - * h 2 h h 6 r 3 Q.E.D. 376 BOOK VIII. SOLID GEOMETRY ORIGINAL EXERCISES (NUMERICAL) TT = 3f. 1 bu. = 2150.42 cu. in. 1 gal. = 231 cu. in. In a cylinder of revolution : 1. If 72 = 5 in., // = 14 in., find L ; T-, V. 2. If R = 7 m., If = 10 in., find L ; T- V. 3. If 72 = 4f ft., 77 = 18 ft., find ; T; V. 4. If 72 = 6 in., L = 792 sq. in., find //; T- V. 6. If R = 4 ft., T = 352 sq. ft., find //; Z; V. 6. If 72 = 2 in., F = 22 cu. in., find 77; L ; T 7 . 7. If 77 = 5.6 in., Z = 352 sq. in., find R ; T 7 ; F. 8. If 77 = 9 in., T = 440 sq. in., find R ; L ; F. 9. If 77 = 9 in., F = 66 cu. in., find R ; L ; 7 1 . 10. How many square inches of tin will be required to make a cylin- drical pail 10 in. in diameter and 1 ft. in height, without any lid? How many gallons will it contain ? 11. The diameter of a cylindrical well is 5| ft. and the water is 14 ft. deep. How many gallons of water does the well hold ? 12. In a cylinder of revolution generated by a rectangle 30 in. x 14 in. revolving about its shorter side as an axis, find L ; 7*; V. 13. In a cylinder of revolution generated by the rectangle of No. 12, revolving about its longer side as an axis, find L ; T] V. 14. A cylindrical vessel 9 in. high, closed at one end, required 361f sq. in. of tin in its construction. Find its radius. 15. A cylindrical pail 12 in. high holds exactly 2 gal. Find 72. 16. How many cubic feet of metal are there in a hollow cylindrical tube 42 ft. long, whose outer and inner diameters are 10 in. and 6 in.? 17. A tunnel whose cross section is a semicircle 18 ft. high is 1 mi. long. How many cubic yards of material were removed in the excavation ? 18. An irregular stone is placed in a cylindrical vessel a in. in diame- ter and partly full of water. The water rises b in. Find volume of stone. 19. A rod of copper 18 ft. long and 2 in. square at the end is melted and formed into a wire in. in diameter. Find the length of the wire. 20. Plow many miles of platinum wira ^ z in. in diameter can be made from a cubic foot of platinum? 21. If a cubic foot of copper weighs 550 lb., what is the weight of a copper wire in. in diameter and 5 mi. long? CONES 377 CONES 658. A conical surface is a surface generated by a moving straight line that continually intersects a given curve in a plane, and passes through a fixed point not in this plane. The generating line is the generatrix. The directing curve is the directrix. The fixed point is the vertex of the conical surface. An element of a conical surface is the generating line in any position. CONICAL SURFACE RIGHT CIRCULAR CONK FRUSTUM OF A CONE CONE OF REVOLUTION OBLIQUE CONE 659. A cone is a solid bounded by a conical surface and a plane cutting all the elements. The base of a cone is its plane surface. The lateral area of a cone is the area of the conical surface. The total area of a cone is the sum of the lateral area and the area of the base. The altitude of a cone is the perpendicular distance from the vertex to the plane of the base. 660. A circular cone is a cone whose base is a circle. The axis of a circular cone is the line drawn from the vertex to the center of the base. A right circular cone is a circular cone whose axis is per- pendicular to the plane of the base. An oblique circular cone is one whose axis is oblique to the plane of the base. 378 BOOK VIII. SOLID GEOMETRY A cone of revolution is a cone generated by the revolution of a right triangle about one of the legs as an axis. Similar cones of revolution are cones generated by the revolution of similar right triangles revolving about homol- ogous sides. The slant height of a cone of revolution is any one of its elements. 661. A frustum of a cone is the portion of a cone between the base and a plane parallel to the base. The altitude of a frustum of a cone is the perpendicular distance between the planes of its bases. The slant height of a frustum of a cone is the portion of an element included between the bases. The lateral area of a frustum is the area of its curved surface. The total area of a frustum is the sum of the lateral area and the area of the bases. The mid- section of a frustum is the section made by a plane parallel to the bases and bisecting the altitude and the slant height. 662. A plane is tangent to a cone if it contains one element of the conical surface and only one, however far it may be extended. A pyramid is inscribed in a cone if its base is inscribed in the base of the cone, and its vertex is the vertex of the cone. A pyramid is circumscribed about a cone if its base is cir- cumscribed about the base of the cone, and its vertex is the vertex of the cone. The frustum of a pyramid is inscribed in, or circumscribed about, the frustum of a cone if the bases of -the pyramid are inscribed in, or circumscribed about, the bases of the cone. Ex. 1. Find the slant height of a right circular cone whose altitude is 8 in. and whose radius is 6 in. Ex. 2. What is the locus of all points 3 in. from a conical surface ? Ex. 3. What is the locus of all lines forming a given angle with a given line at a given point in the line? CONES 379 PRELIMINARY THEOREMS 663. THEOREM. The elements of a right circular cone are all equal. (504,11.) 664. THEOREM. A right circular cone is a cone of revo- lution. 665. THEOREM. The altitude of a cone of revolution is the axis of the cone. 666. THEOREM. A straight line drawn from the vertex of a cone to any point in the perimeter of the base is an element. (39.) 667. THEOREM. The lateral edges of a pyramid inscribed in a cone are elements of the cone. 668. THEOREM. The lateral faces of a pyramid circum- scribed about a cone are tangent to the conical surface. 669. THEOREM. The slant height of a regular pyramid cir- cumscribed about a right circular cone is the same as the slant height of the cone. 670. THEOREM. The slant height of the frustum of a regu- lar pyramid circumscribed about the frustum of a right circular cone is the same as the slant height of the frustum of the cone. 671. THEOREM. The radius of the mid-section of a frustum of a right circular cone is equal to half the sum of the radii of the bases. Proof : The radius of the mid-section is the median of a trapezoid whose bases are the radii of the bases of the frustum. That is, m = J(B + r). (139.) Q.E.D. Ex. If a conical surface is cut along an element and the surface there placed in coincidence with a plane, what geometrical figure does the sur- face become ? 380 BOOK VIII. SOLID GEOMETRY 672. THEOREM. If a regular pyramid is inscribed in, or circumscribed about, a right circular cone and the number of sides of the base is indefinitely increased, the lateral area of the cone is the limit of the lateral area of the pyramid. (See Fig. A.) Demonstration is similar to that of 651. FIG. A FIG. B 673. THEOREM. If a pyramid having a regular polygon for a base is inscribed in, or circumscribed about, any circular cone and the number of sides of its base is indefinitely in- creased, the volume of the cone is the limit of the volume of the pyramid. (See Fig. B.) Demonstration is similar to that of 652. 674. THEOREM. If a frustum of a regular pyramid is in- scribed in, or circumscribed about, the frustum of a right circular cone and the number of sides of the bases is in- definitely increased, the lateral area of the frustum of the cone is the limit of the lateral area of the frustum of the pyramid. 675. THEOREM. If the frustum of a pyramid having regular polygons for its bases is inscribed in, or circumscribed about, a frustum of any circular cone and the number of sides of the bases of the frustum is indefinitely increased, the volume of the frustum of the cone is the limit of the volume of the frustum of the pyramid. CONES 881 THEOREMS AND DEMONSTRATIONS PROPOSITION VIII. THEOREM 676. Any section of a cone made by a plane passing through the vertex is a triangle. Given : Cone O-AB ; plane OCD. To Prove : Section OCD is a A. Proof : Draw straight lines OC, OD, in plane OCD. They are elements (666). .*. OC and OD compose the intersection of the plane and the conical surface. (466.) Also CD is a straight line (?). .-. OCD is a A (23). Q.E.D. Ex. 1. Can the plane section of a cone, containing the vertex, ever be a right triangle? Explain. Can it be an isosceles triangle? Explain. What kind of triangle is this section, in general? Ex. 2. What is the locus of all straight lines making a given angle with a given plane at a given point in the plane ? Ex. 3. Does every cone have a slant height, or only certain kinds of cones ? Ex. 4. If a circular disk is held between a source of light and a wall, and parallel to the wall, what is the shape of the shadow on the wall? What is the shape of the shadow region between the light and the wall? (Consider the source of light a point.) 382 BOOK VIII. SOLID GEOMETRY PROPOSITION IX. THEOREM 677. Any section of a circular cone made by a plane parallel to the base is a circle, whose center is the intersec- tion of the plane with the axis. Given : Cone O-AB ; circle C its base ; section A f B 1 II to base, and axis OC. To Prove : A'B' also a O, whose center is C r . Proof : Pass planes OCD, OCE intersecting the base in CD, CE respectively, and the section in C'D', C'E'. In A OCD and OCE, D r C r is II to DC ; C'E' is II to CE (484). .-. A OC'D' is similar to A OCD; A OC'E' is similar to A OCE OC But .. by multiplying, CD C'D' OC c f E f CE CD CE CD = CE C'D' (305). (313). (Ax. 1). (187). (Ax. 3). That is, all points on the boundary of A'B' are equally distant from C 1 . .-. A'B' is a O whose center is C (179). Q.E.D. Ex. Any section of a circular cone parallel to the base is to the base as the square of its distance from the vertex is to the altitude of the cone. Proof : A'B' : AB = C^D' 2 : CD* = 0~C' 2 : OC 2 . (Explain.) CONES 383 PROPOSITION X. THEOREM 678. The lateral area of a right circular cone is equal to half the product of the circumference of the base by the slant height. Given: Right circular cone O-J.D, the circumference of whose base = C and whose slant height = s. To Prove : Lateral area = ^ C s. Proof : Circumscribe a regular pyramid and denote the lateral area by L r and the perimeter of the base by P. Slant height OA = s. (663.) Then L f = P . s (598). Now indefinitely increase the number of sides of the base of the pyramid and L r approaches L as a limit (672), P approaches C as a limit (424, I), P s approaches ^ C 8 as a limit (?). Hence L = -J- C s (229). Q.E.D. 679. COROLLARY. Area of a right circular cone : L = i (2 TrR)s = trRs. T = TrRs + irJB 2 = irR(s + .R). (Where L = lateral area, T= total area, * = slant height, and R = radius of base.) Ex. 1. Of a right circular cone whose slant height is 15 in. and radius is 9 in., find the lateral area, the total area, and the altitude. Ex. 2. If the radius of a right circular cone is 8 in. and the altitude is 15 in., find the slant height, the lateral area, and the total area. 384 BOOK VIII. SOLID GEOMETRY PROPOSITION XI. THEOREM 680. The lateral area of the frustum of a right circular cone is equal to half the sum of the circumferences of the bases multiplied by the slant height. Given : Frustum of right circular cone, whose lateral area is L ; whose slant height is s ; and the circumferences of whose bases are C and c. To Prove : L = J(c + = ir(JB + r)s. L = 7r(2 rri)s = 2 irms. T= TT(R 4- r)s + 7r/i 2 + -rrr 2 . .-. T=Tr[(,R + r> + jR 2 + r 2 ]. (Where , T, K, r and denote magnitudes as before, and m = the radius of the midsection.) CONES 385 PROPOSITION XII. THEOREM 682. The volume of a circular cone is equal to one third the product of the area of the base by the altitude. Given: Circular cone O-AV, whose volume = F; area of whose base = B ; altitude = OE = H. To Prove : F = J B - H . Proof : Circumscribe (or inscribe) a pyramid having a regular polygon for its base. Denote the volume of the pyramid by F', its base by B r . Its altitude = OE = H (491). Indefinitely increase the number of the sides, etc. Then V' approaches F as a limit (?). B' approaches B as a limit (424, II). \ B f H approaches ^ B H as a limit. Hence V=^B-H (229). Q.E.D. 683. COROLLARY. Volume of a circular cone: F=*TrR 2 JJ. (Where F= volume, H = altitude, and It = radius of base.) Ex. 1. Find the volume of a circular cone whose altitude is 14 in. and the radius of the base, 6 in. Ex. 2. A right triangle, whose legs are 15 in. and 20 in., is revolved about the lesser leg as an axis, forming a cone of revolution. Find the lateral area, the total area, and the volume. Find the lateral area, the total area, and the volume of the cone formed by revolving this triangle about the greater leg as an axis. 380 BOOK VIII. SOLID GEOMETRY PROPOSITION XIII. THEOREM 684. The volume of the frustum of a circular cone is equal to one third the product of the altitude by the sum of the lower base, the upper base, and a mean proportional between the bases of the frustum. Given : A frustum of any circular cone, whose volume = F, whose bases are B and >, whose altitude is H. To Prove: v= ^ H\_B + b + V# >]. Proof : Inscribe (or circumscribe) a frustum of a pyramid having regular polygons for bases. Denote its volume by F', bases by B f and >', and altitude by H. Then V = ^H [B f +b f + V#' . b'] (618). Indefinitely increase, etc. V f approaches F as a limit (?), B f and b' approach B and b respectively as limits (?), V.B' b f approaches V^ b as a limit. %H[B f + b' + ^/B f - >'] approaches ^H[s+b + V# &]. Hence v=^H[B + l + Vz? 6] (229). Q.E.D. 685. COROLLARY. Volume of the frustum of a circular cone, -f ^^2 _|_ V7TJ2 2 TTr 2 ] (Where F= volume, etc.) CONES 387 PROPOSITION XIV. THEOREM 686. Of two similar cones of revolution : I. The lateral areas are to each other as the squares of thek altitudes, or as the squares of their radii, or as the squares of their slant heights. II. The total areas are to each other as the squares of their altitudes, or as the squares of their radii, or as the squares of their slant heights. III. The volumes are to each other as the cubes of their altitudes, or as the cubes of their radii, or as the cubes of their slant heights. Given: Two similar cones of revolution, whose respec- tive lateral areas are L and Z, total areas are T and , vol- umes are V and t;, altitudes are H and h, radii are R and r, slant heights are -S and s. 'T rr2 t.2 To Prove : I O O r 2 s 2 ii. *...* O O r 2 s 2 III. -1 = ^=^ = ^-. TJ* -ry O Proof: Generating A are similar and =-=- h r s Hence L ~ I 7TT8 R+S_R_S_H r -h s r s h 7TRS = R S^H H = H 2 = B?. r s h h h 2 r 2 (291). (Ax. 6). III. v - = T_7TR(R+S) == R R 2 H H # 3 I? 3 (?). Q.E.D. ROBBINS'S NEW SOLID GEOM. 10 388 BOOK VIII. SOLID GEOMETRY ORIGINAL EXERCISES In a cone of revolution : 1. If //= 12 in., * = 13in., find/?. 2. If H = 15 ft., R = 8 ft., find S. 3. If R = 18 cm., s = 30 cm., find H. 4. If // = 6 in., s = 10 in., find R ; L ; T; V. 6. If // = 20 ft, R = 21 ft., find s ; L ; 7 1 ; F. 6. If # = 7 m., * = 25 m., find //; Z; T; F. 7. If Z = 4070 sq.,in., .s = 37 in., find R ; //; T 7 ; F. 8. If Z = 46.64 sq. in., 72 = 2.8 in., find s; H; T; V. 9. If Z = 400 sq. ft, T = 500 .q. ft., find s ; H; H ; F. 10. If T = 80 TT sq. in., R = 5 in., finds; #; Z; F. 11. If T = 10 TT sq. ft, s = 3 ft, find R; //; Z; F. 12. If F = 462 cu. in., R = 21 in., find //; s ; Z ; T. 13. If F=8 T f y cu. ft, 7/=3ft., find 72; *; Z; T. 14. What would be the cost at 10 ^ a square foot of painting a conical church steeple, 112 ft. high and 30 ft. in diameter at the base? 16. The sides of an equilateral triangle are each 12 in. Find the lateral surface, the total surface, and the volume of the solid generated by revolving this triangle about an altitude as an axis. 16. An isosceles right triangle whose legs are each 8 in. is revolved about the hypotenuse as an axis. Find the total surface and the volume of the solid generated. 17. The sides of an equilateral triangle are each 10 in. Find the total surface and the volume of the solid generated by revolving this triangle about one of its sides as an axis. 18. Find the volume of a cone of revolution whose slant height is 16 in. and whose lateral area is 192 TT sq. in. 19. Find the lateral area of a cone of revolution whose altitude is 20 in. and whose volume is 240 TT cu. in. 20. How many bushels are there in a conical heap of grain whose base is a circle 35 ft. in diameter, and whose height is 25 ft. ? 21. A regular hexagon whose side is 6 in. revolves about one of the longer diagonals. Find the surface and the volume of solid generated. 22. Find the volumes of the right circular cones inscribed in and circumscribed about a regular tetrahedron whose edge is a m. ORIGINAL EXERCISES 889 23. A right triangle whose legs are 15 in. and 20 in. is revolved about the hypotenuse as an axis. Find the surface and the volume of the solid generated. In the frustum of a right circular cone : 24. If // = 8 in., R = 10 in., r = 4 in., find s ; L ; T 7 ; V. 25. If H = 30 cm., s = 34 cm., r = 5 cm., find R ; L ; T; V. 26. If s = 19 ft., 72 = 10i ft., r = 3 ft., find #; Z; T; F. 27. How many square feet of tin are required to make a funnel 2 ft. long, if the diameters of the ends are 20 in. and 56 in., respectively? 28. A chimney 150 ft. high has a cylindrical flue 3 ft. in diameter. The bases of the chimney are circles whose diameters are 28 ft. and 7 ft. Find the number of cubic yards of masonry in the chimney. 29. A plane is passed parallel to the base of a right circular cone and | the distance from the vertex to the base. Find the ratio of the smaller cone thus formed to the original cone. Compare the volume of the less cone with the frustum formed. Original cone = 5J ? , ^ Less cone 2 3 Hence Original cone - Less cone = 125 - 8 (? ^ Less cone 8 30. The altitude of a cone of revolution is 12 in. What is the alti- tude of the frustum of this cone that shall contain one fourth the volume of the whole cone ? 31. The altitudes of two similar cylinders of revolution are 3 in. and 5 in. What is the ratio of their lateral areas? of their total areas? of their volumes? 32. The altitudes of two similar cylinders of revolution are 5 in. and 6 in., and the lateral area of the first is 200 sq. in. Find the lateral area of the second. If the volume of the first is 500 cu. in., what is the volume of the second? 33. The total areas of two similar cones of revolution are 24 TT sq. in. and 216 TT sq. in. and the radius of the first is 3 in. Find the radius of the second. The slant height of the first is 5 in. Find the lateral area of the second. Find the altitude of the first and the volume of the second. 34. The volumes of two similar cones of revolution are 27 IT cu. in. and 343 TT cu. in. The altitude of the first is 9 in. Find the altitude of the second. Find the radius of the base of each. 390 BOOK VIII. SOLID GEOMETRY 35. A cone of revolution whose radius is 10 in. and altitude 20 in., has the same volume as a cylinder of revolution whose radius is 15 in. Find the altitude of the cylinder. 36. A cylinder of revolution whose radius is 8 in. and altitude 30 in., is formed into a cone of revolution whose altitude is 40 in. Find the radius of its base. 37. The heights of two equivalent cylinders of revolution are in the ratio of 4 : 9. If the diameter of the first is 92 ft., what is the diameter of the second? 38. A cylinder of revolution 8 ft. in diameter is equivalent to a cone of revolution 7 ft. in diameter. If the height of the cone is 16 ft., what is the height of the cylinder? 39. Two circular cylinders having equa-1 altitudes are to each other as their bases. 40. Two circular cylinders having equal bases are to each other as their altitudes. 41. Two circular cylinders having equal bases and equal altitudes are equal. 42. Two circular cones having equal altitudes are to each other as their bases. 43. Two circular cones having equal bases are to each other as their altitudes. 44. Two circular cones having equal bases and equal altitudes are equal. 45. If the altitude of a right circular cylinder is equal to the radius of the base, the lateral area is half the total area. 46. If the altitude of a right circular cylinder is half the radius of the base, the lateral area is equal to the area of the base. 47. If the slant height of a right circular cone is equal to the diameter of the base, the lateral area is double the area of the base. 48. The lateral area of a cone of revolution is equal to the area of a circle whose radius is a mean proportional between the slant height and the radius of the base. 49. The lateral area of a cylinder of revolution is equal to the area of a circle whose radius is a mean proportional between the altitude of the cylinder and the diameter of its base. 60. What relation does the section of a circular cone made by a plane parallel to the base have to the base ? Prove. ORIGINAL EXERCISES 391 61. At what distance from the vertex of a right circular cone whose altitude is h must a plane parallel to the base be passed, so as to bisect the lateral area? At what distance must it be passed so as to bisect the volume? 52. What does the volume Tof a right circular cone become, if the altitude is doubled and the base undisturbed? Prove. What does the volume V become if the radius of the base is doubled but the altitude undisturbed ? Prove. What does the volume become if both radius of base and altitude are doubled ? Prove. 63. The intersection of two planes tangent to a cylinder is a line parallel to an element. 64. The intersection of two planes tangent to a cone is a line through the vertex. 56. One straight line can be drawn upon a cylindrical surface through a given point, and only one. 66. If two cylinders of revolution have equivalent lateral areas, their volumes are to each other as their radii. 57. If a rectangle is revolved about its unequal sides as axes, the vol- umes of the two solids generated are inversely proportional to the axes, and directly proportional to the radii of the bases. 68. Show that the formula for the volume of a circular cone can be derived from the formula for the volume of a frustum of a circular cone if one base of the frustum becomes a point. 59. Reduce the formula for the volume of a frustum of a circular cone if the radius of one base is double the radius of the other. 60. Could you prove the theorem of 680 by inscribing a frustum of a pyramid? Could you prove the theorem of 684 by circumscribing a frustum of a pyramid? Give reasons for your answer. 61. Could you prove the theorem of 678 by inscribing a pyramid? Could you prove the theorem of 682 by inscribing a pyramid? Give reasons. 62. Pass a plane tangent to a circular cylinder and containing a given element. Construction : Draw a line in plane of base tangent to the base at the end of the given element, etc. 63. Pass a plane tangent to a circular cone and containing a given ele- ment. 64 . Divide the lateral surface of a cone of revolution into two equiva- lent parts by a plane parallel to the base. 392 BOOK VIII. SOLID GEOMETRY 65. Pass a plane tangent to a circular cylinder and through a given point without it. Construction: From the point draw a line II to an element, meeting the plane of the base. From this point of intersection draw a line tan- gent to the base of the cylinder. Through the point of contact draw an element, etc. 66. Pass a plane tangent to a circular cone through a given point without it. Construction : Connect this point with the vertex of the cone and pro- long this line to meet the plane of the base, etc. 67. Find the locus of points at a given distance from a given straight line. 68. Find the locus of points equally distant from two given points and at a given distance from a straight line. Discuss. 69. Find the locus of points at a given distance from a given plane and at a given distance from a given line. Discuss. 70. Find the locus of points at a given distance from a given cylin- drical surface whose generatrix is a circle, and whose elements are per- pendicular to the plane of the circle. 71. Find the locus of a point at a given distance from a given line and equally distant from two given planes. Discuss. 72. Find a point A', at a given distance from a given line and equally distant from three given points. Discuss. 73. A grain elevator in the form of a frustum of a right circular cone is 30 ft. high, and the radii of its bases are 12 ft. and 8 ft. respectively. If a bushel contains approximately 1 ^ cu. ft., how many bushels of wheat will this elevator hold? 74. A certain coffee pot is 7 in. deep, 3 in. in diameter at the top, and 5 in. at the bottom. If there are 5 cups in a quart, how many cups of coffee will this coffee pot hold? (Disregard fractional parts of a cup.) 76. Find the radius of a circle having the same area as the lateral area of a cone of revolution whose radius is 4 and slant height is 9. 76. The frustum of a circular cone is 15 in. high and the bases are circles whose radii are 3 in. and 5 in. Find the edge of an equivalent cube. 77. The frustum of a right circular cone has a slant height of 9 ft. and the radii are 5 ft. and 7 ft. Find the lateral area and the total area. What is the length of the altitude of this frustum ? Find the altitude of the cone that was removed to leave this frustum. Now find in two ways the volume of the frustum. BOOK IX THE SPHERE 687. A sphere is a solid bounded by a surface, all points of which are equalty distant from a point within, called the center. The surface of a sphere is called a spherical surface. A radius of a sphere is a straight line drawn from the center to any point of the surface. A diameter of a sphere is a straight line that contains the center and has its extremities in the surface. Ex. 1. What is the locus of all points 2 in. from the surface of a sphere whose radius is 10 in. ? Ex. 2. Consider the center of a sphere at one of the vertices of a rec- tangular box or room. What part of the sphere is within the box or room? Ex. 3. Name several familiar objects that are usually regarded as spheres. Ex. 4. What is the shape of the celestial bodies ? Ex. 5. Why do you believe the earth to be spherical ? Ex. 6. Why do you believe the moon to be spherical ? Ex. 7. Describe fully the locus of points 3 in. from a line 8 in. long. Historical Note. Archimedes was the discoverer of the formulas for the surface and volume of the sphere. Menelaus (100 A.D.) gave us the proper- ties of spherical triangles. V; 394 BOOK IX. SOLID GEOMETRY PROPOSITION I. THEOREM 688. Every plane section of a sphere is a circle. M Given : Sphere whose center is O ; plane MN intersecting sphere in AB. To Prove : The figure AB is a O. Proof : Draw OD _L to plane JOT, meeting the plane at D. Take P and Q, any two points on the perimeter of the sec- tion, and draw DP, DQ, OP, OQ. A ODP and ODQ are rt. A (?). In right A ODP and ODQ, OD = OD (Iden.). OP = OQ OA (504, I). .-. X lies without the sphere (704). Hence every point of plane MN, except A, is without the sphere ; that is, plane MN is tangent to the sphere. (690.) Q.E.D. PROPOSITION III. THEOREM 707. A plane tangent to a sphere is perpendicular to the radius drawn to the point of contact. [Converse.] Given: Plane MN tangent to sphere O at A; radius OA. To Prove : OA is _L to plane MN. Proof : Every point in MN, except A, is without the sphere (690). Take any point X in MN and draw OX. Now ox is > OA (704). That is, OA is the shortest line from O to MN. .'. OA is i. to MN (504, I). Q.E.D. 398 BOOK IX. SOLID GEOMETRY PROPOSITION IV. THEOREM 708. All points in the circumference of a circle of a sphere are equally distant from either pole; that is, the polar dis- tances of all points in the circumference of a circle are equal. Given : P and L, the poles of O C on sphere O, and great PAL, PEL. To Prove : arc PA = arc PB ; arc AL = arc BL. Proof : Draw the axis PL meeting plane of O C at C. Draw AC AP BC BP PC is _L to plane DAB (Def. of axis, 689). .-. chord PA = chord PB (?). Hence arc PA = arc PB (?). Likewise arc AL == arc BL. Q.E.D. 709. COROLLARY. The polar distance of a great circle is a quadrant. Ex. 1. If the radius of a sphere is 26 in. and a plane is passed, 10 in. from the center, find the radius of the circular section. Ex. 2. What geographical circles on the earth's surface are great circles ? Which are small circles ? Ex. 3. Can two circles on the surface of a sphere intersect in more than two points ? Why ? Ex. 4. The area of a section of a sphere 45 in. from the center is 784 ir sq. in. Find the radius of the sphere. Ex. 6: The area of a section of a sphere 7 in. from the center is 576 TT sq. in. Find the area of a section 6 in. from the center. THE SPHERE 399 PROPOSITION V. THEOREM 710. If a point on the surface of a sphere is at the distance of a quadrant from two other points on the surface, not the ends of a diameter, it is the pole of the great circle containing these two points. Given : P, a point, and R and 2V, two other points (not the ends of a diameter), all on the surface of sphere O ; arcs PR and P2V, quadrants ; great circle ARNB. To Prove : P is the pole of O ARNB. Proof : Draw the radii OP, OR, ON. A PON and FOR are rt. A (232). .-. PO is _L to plane AB (485). .-. PO is the axis of O ARNB (Def. 689). .-. P is the pole of O ARNB (Def. 689). Q.E.D. Ex. 1. Considering the earth as a true sphere, what is the pole of the equator? If a city has a latitude of 43, what is its polar distance? What is the polar distance of a place upon the equator ? Ex. ( 2. In the diagram above, is a trihedral angle. What arcs are the measures of its face angles? What is an isosceles trihedral angle? Explain by this diagram that the name is consistent with the etymology of the word. Ex. 3. Prove that two lines tangent to a sphere at a point determine a plane tangent to the sphere at that point. Ex. 4. Find the volume of a cube circumscribed about a sphere whose radius is 6 in. Find the volume of the cube inscribed in this sphere. 400 BOOK IX. SOLID GEOMETRY PROPOSITION VI. THEOREM 711. A spherical angle is measured by the arc of a great circle having the vertex of the angle as a pole and intersected by the sides of the angle. Given : Spherical Z.AVB\ arc AB of great O whose pole is F, on sphere O. To Prove : Z.AVB is measured by arc AB. Proof: Draw radii OJ, O5, OF, and at F draw FC tangent to O VA, and VD tangent to O VB. VB is a quadrant. (709.) OF is -L to VD (203), OF is -L to (232)r / OF is -L to FC and to OA (?). .*. FZ> is II to OB, and FC is II to OA (62). ... ^ CVD=Z.AOB (499). Z CVD is the spherical AVB (693). /. AOB is measured by arc AB (232). .-. /. CVD is measured by arc AB (Ax. 6). That is, AVB is measured by arc AB (Ax. 6). Q.E.D. 712. COROLLARY. All arcs of great circles containing the pole of a great circle are perpendicular to the great circle. 713. COROLLARY. A spherical angle is equal to the plane angle of the dihedral angle formed by the planes of its sides. 714. COROLLARY. If two great circles are perpendicular to each other, each contains the pole of the other. (528; 689.) THE SPHERE 401 PROPOSITION VII. THEOREM 715. A sphere may be inscribed in any tetrahedron. Given: Tetrahedron A-BCD. To Prove : (?). Proof: Pass plane OAB bisecting dih. Z AB, and plane OBC bisecting dih. Z BC, and plane OCD bisecting dih. Z CD, the three planes meeting at point O. Point O, in plane OAB, is equally distant from faces ABC and ABD. (?.) Point O, in plane OBC, is equally distant from faces ABC and BCD. (?.) Point O, in plane OCD, is equally distant from faces BCD and ACD. (?.) .*. O is equally distant from all four faces (Ax. 1). Hence the sphere constructed with O as a center and OB as a radius, is tangent to each of the four faces. .-. that sphere is inscribed in the tetrahedron (691). Q.E.D. 716. COROLLARY. The six planes bisecting the six dihedral angles of any tetrahedron meet in a point. Ex. 1. The volume of any polyhedron circumscribed about a sphere, is equal to one third the product of the surface of the polyhedron and the radius of the sphere. PROOF : Pass planes each containing the center of the sphere and two vertices of the polyhedron. These form pyramids whose altitude . . . etc. Ex. 2. What is true of the point at which the six planes meet in 716 ? Ex. 3. Explain, so that a blind boy could understand it, the process of inscribing a sphere in a tetrahedron. 402 BOOK IX. SOLID GEOMETRY PROPOSITION VIII. THEOKEM 717. A sphere may be circumscribed about any tetrahedron, Given : (?). To Prove : (?). Proof : Take E and F, the centers of circles circumscribed about the faces ACD and BCD, respectively. Erect EG and FH _L to these faces. Find Jf, the midpoint of edge CD. EG is the locus of all points equally distant from points A, D, and C. (511.) FH is the locus of all points equally distant from points B, C, and D. (?.) That is, all points in EG and FH are equally distant from C and D. (Ax. 1.) But all points equally distant from C and D are in a plane J- to CD at M. (510.) .-. EG and FH are in this plane and are not parallel. (Not J_ to the same plane.) That is, EG and FH must intersect at o. Hence O is equally distant from A, J5, C, and D. (Ax. 1.) That is, using O as a center and OA, or OB, or OC, or OD, as a radius, a sphere may be circumscribed about the tetra- hedron A-BCD. (691.) Q.E.D. 718. COROLLARY. Through any four points not in the same plane a sphere may be described. 719. COROLLARY. The six planes perpendicular to the edges of any tetrahedron at their midpoints meet in a point. THE SPHERE 403 PROPOSITION IX. THEOREM 720. The intersection of two spherical surfaces is a circle whose plane is perpendicular to the line which joins the cen- ters of the spheres and whose center is in that line. Given: Two intersecting circles O and o'; common chord CD ; line of centers XY, intersecting CD at M. To Prove : The spherical surfaces generated by the revo- lution of these (D intersect in a circle. Proof : If these are revolved upon XY as an axis, they will generate spheres. (701.) CM=MD (219). Point c, common to both , the arc of a great circle (by 723). Statement: Arc PAE is _L to circle BEDC. Q.E.F. Proof : ED, EP, and PD are quadrants (709). .-. E is the pole of arc PD Hence Z. PED is measured by quadrant PD .-. /- PED is a right angle (232). That is, arc PAE is _L to circle BEDC. Q.E.D. Ex. 1. Construct a plane tangent to a sphere at a given point on the surface. Ex. 2. Construct a plane tangent to a sphere from a given point with- out the sphere. How many such planes are there ? 408 BOOK IX. SOLID GEOMETRY SPHERICAL TRIANGLES 725. A spherical triangle is a portion of the surface of a sphere bounded by three arcs of great circles. The bounding arcs are the sides of the triangle. The intersections of the sides are the vertices of the triangle. The angles formed by the sides are the angles of the tri- angle. Spherical triangles are equilateral, equiangular, isosceles, scalene, acute, right, obtuse, under the same conditions as in plane triangles. 726. A birectangular spherical triangle is a spherical tri- angle two of whose angles are right angles. 4\\ : A f / \ A trirectangular spherical triangle is a spherical triangle all of whose angles are right angles. The unit usually employed in measuring the sides of a spherical triangle is the degree. It is obvious that three great circles (not meeting at a point) divide the surface of a sphere into eight spherical triangles. 727. Two spherical triangles are mutually equilateral if the sides of the triangles are equal each to each ; and they are mutually equiangular if their angles are equal each to each. SPHERICAL TRIANGLES 409 MUTUALLY EQUILATERAL SPHERICAL TRIANGLES POLAR TRIANGLES MUTUALLY EQUIANGULAR SPHERICAL TRIANGLES 728. If three great circles are described, having as their poles the vertices of a spherical triangle, one of the eight triangles thus formed is the polar triangle of the first. The polar triangle is the one whose vertices are nearest the vertices of the original triangle. 729. Symmetrical spherical triangles are triangles that have their parts equal but arranged in reverse order. They correspond to symmetrical trihedral angles. Vertical spherical triangles correspond to vertical trihe- dral angles. If the diameters of a sphere are drawn to the vertices of a spherical triangle, the original triangle and the triangle whose vertices are the opposite ends of these diameters are vertical spherical triangles. 730. A spherical polygon is a portion of the sphere bounded by three or more arcs of great circles. Two spherical polygons are congruent if they can be made to coincide. The diagonal of a spherical polygon is the arc of a great circle connecting two vertices not in the same side. Only convex spherical polygons are considered in this book. 410 BOOK IX. SOLID GEOMETRY PRELIMINARY THEOREMS 731. THEOREM. The planes of the sides of a spherical tri- angle form a trihedral angle: I. Whose vertex is the center of the sphere. n. Each of whose face angles is measured by the inter- cepted side of the triangle. (232.) III. Each of whose dihedral angles is equal to the corre- sponding angle of the triangle. (713.) 732. THEOREM. Two symmetrical spherical triangles are mutually equilateral and mutually equiangular. (51, 193, 522, 713.) 733. THEOREM. The homologous parts of two symmetrical spherical triangles are arranged in reverse order. Proof : If the eye is at the center of the sphere, the order of the vertices J., .B, C is the same in direction as. the motion of the hands of a clock. But the order of A 1 , B' , cf is in the opposite direction. (See 541, Note.) Hence the parts are arranged in reverse order. Q.E.D. 734. THEOREM. The homologous parts of two symmetrical spherical triangles are equal. (732.) 735. THEOREM. Two symmetrical isosceles spherical triangles are con- gruent. Proof : The method of superposi- tion, as in the case of plane triangles. Historical Note. It was not until the seventeenth century that polar triangles were invented by Gerard of Holland. It was he also who found the formulas for the area of a spherical triangle and of a spherical polygon. SPHERICAL TRIANGLES 411 THEOREMS AND DEMONSTRATIONS PROPOSITION XIV. THEOREM 736. One side of a spherical triangle is less than the sum of the other two. Given : (?). To Prove : AB < AC + BC. Proof : Draw radii OA, OB, OC. In the trihedral Z o, Z AOB < Z AOC + Z BOG Z AOB is measured by arc AB, etc. .-. arc AB < arc AC -f- arc BC PROPOSITION XV. THEOREM (548). CO: (Ax. 6). Q.E.D. 737. In a birectangular spherical triangle the sides opposite the right angles are quadrants, and the third angle is meas- ured by the third side. Given : Birectangular A ABC', Z B and Z c, right A. To Prove : I. AB and AC quadrants. II. Z A is measured by arc BC. 412 BOOK IX. SOLID GEOMETRY Proof : I. Draw radii OA, OB, OC. Arc AB is -L to arc BC and arc AC is J_ to arc BC (Hyp.). .-. A is the pole of arc BC (714)- .-. AB and AC are quadrants (709). II. Z A is measured by arc BC. (711). Q.E.D. 738. COROLLARY. The three sides of a trirectangular spher- ical triangle are quadrants. Ex. 1. If two sides of a spherical triangle are quadrants, the triangle is birectangular. (710, 712.) Ex. 2. If all sides of a spherical triangle are quadrants, the triangle is trirectangular. PROPOSITION XVI. THEOREM 739. The sum of the sides of any spherical polygon is less than 330. Given: (?). To Prove : (?). Proof : Draw radii to the several vertices of the polygon, forming the polyhedral Z O. Then Z AOB + Z BOG + Z COD + Z AOD < 360 (549). But Z AOB is measured by arc AB, etc. (?). .-. arcs AB + BC+CD + AD < 360 (Ax. 6). Q.E.D. SPHERICAL TRIANGLES 413 740. COROLLARY. The sum of the sides of any spherical polygon is less than the circumference of a great circle. PROPOSITION XVII. THEOREM 741. If one spherical triangle is the polar of a second tri- angle, then the second is the polar of the first. Given : Spherical A ABC and its polar A A'B'C ! . To Prove : A ABC is the polar A of A A'B'C'. Proof : A is the pole of arc B r C f (Hyp.). .-. B' is the distance of a quadrant from A (709). C is the pole of arc A'B' (?). .-. B f is the distance of a quadrant from C (?). Hence B r is the pole .of arc AC (710). Also A r is the pole of BC, and (f is the pole of AB. .-. ABC is the polar A of A A'B'C' (728). Q.E.D. NOTE. Many properties of a trihedral angle are common to the cor- responding spherical triangle. The polyhedral angle is similarly related to the spherical polygon. Sometimes it is advantageous to employ one, sometimes the other. The spherical triangle is perhaps simpler and more suggestive of properties than the corresponding trihedral angle, when the plane angles of its dihedral angles appear in the same diagram. It is a most instructive and helpful exercise for the student to draw spherical triangles and their polars, etc., on a material sphere, such as a slate globe, a large apple, a ball, or other spherical object. A great many geometrical truths can be fixed in the mind by an orange and three long needles. 414 BOOK IX. SOLID GEOMETRY PROPOSITION XVIII. THEOREM 742. In two polar spherical triangles each angle of one and the opposite side of the other are supplementary. Given : Polar A ABC and A'B'C'. To Prove: Z A + a' = 180 ; Zvl' + a Z V + V = 180; Z B' + 5 = 180; Zc+ 180 ; II. Z.A + Z B + Z. c < 540. Proof : I. Construct A A'H'C', the polar A of A ABC. ^A + a' = 180, Z B + b r = 180, Z C + c' = 180 (742). Adding, ^A + Z.B + ^.C + a 1 + & + > = 540 (Ax. 2). But Subtracting, V + b' + c' < 360 (739). Z^+Z^+ZC > 180 (Ax. 9). Q.E.D. '.A + ZB + Z.C+a'+b' + c' = 540 (Ax. 2). a' + V + (725). But Subtracting, Z^ + Z7i + ZC < 540 (Ax. 9). Q.E.D. 746. COROLLARY. The sum of the angles of a spherical tri- angle is greater than two, and less than six, right angles. 746. COROLLARY. A spherical triangle may have one, two, or three obtuse angles. 416 BOOK IX. SOLID GEOMETRY PROPOSITION XX. THEOREM 747. Two symmetrical spherical triangles are equal. B Given : Two symmetrical spherical A ABC and A'B'C'. To Prove : A ABC = A A'B'C'. Proof : Suppose P is the pole of the O containing A, B, C. Draw the diameters AA f , BB', cc', PP\ and the arcs of great , PA, PB, PC, P'A', P'B', P'C'. (193). (708). (Ax. 1). (735). .-. arc PA = arc P'A' Also arc PB = arc P'B' and arc PC= arc P'C'. But PA = PB = PC .-. P'A' = P'B' = P'C' Hence A APB ^ A A' P'B' A AGP =* A A ^PC ^ A Adding, A ABC = A ' (Ax. 2)- Q.E.D. NOTE. If the pole P should be without the A ABC, one of the pairs of equal A would be without the original & and would be subtracted from the sum of the others to obtain & ABC and A'B'C'. 748. COROLLARY. Vertical spherical triangles are symmet- rical and equal. Ex. 1. Are symmetrical spherical triangles ever congruent? Ex. 2. What unit is used in measuring the sides and angles of a spherical triangle? SPHERICAL TRIANGLES 417 PROPOSITION XXI. THEOREM 749. Provided two spherical triangles on the same sphere (or on equal spheres) have their parts arranged in the same order, they are congruent : I. If two sides and the included angle of one are equal respectively to two sides and the included angle of the other. II. If a side and the two angles adjoining it of one are equal respectively to a side and the two angles adjoining it in the other. III. If three sides of the one are equal respectively to three sides of the other; that is, if they are mutually equilateral. On Equal Spheres. Given: (?). To Prove : A ABC ^ A RST. Proof : I and II. Superposition as in plane A. III. Draw radii of the sphere to all the vertices of the A. The face A of the trih. Z O = the face A of the trih. Z N, respectively. (?.) Hence trih. Z o = trih. Z N (546). .-. dih. Z OA = dih. Z NR ; dih. /.OB dih. Z N8 ; etc. .-. the A are mutually equiangular (731, III). Hence the A can be made to coincide. .'. A ABC ^ A RST (26). Q.E.D. 418 BOOK IX. SOLID GEOMETRY PROPOSITION XXII. THEOREM 750. Provided two spherical triangles on the same sphere (or on equal spheres) have their parts arranged in reverse order, they are symmetrical : I. If two sides and the included angle of one are equal respectively to two sides and the included angle of the other. II. If a side and the two angles adjoining it of one are equal respectively to a side and the two angles adjoining it of the other. III. If three sides of one are equal respectively to three sides of the other; that is, if they are mutually equilateral. Given : (?). To Prove : (?). Proof : In each of these cases construct a third spherical A R'S'T*, symmetrical to the A RST. Then AR'S'T' will have its parts equal to the parts of A ABC and arranged in the same order. .-. AR'S'T'^AABC (749). Hence A RST is symmetrical to A ABC (Ax. 6). Q E.D 751. COROLLARY. Two mutually equilateral spherical tri- angles are mutually equiangular and are congruent or sym- metrical. When are they congruent? When are they symmetrical? Ex. Is the corresponding theorem about plane triangles true? SPHERICAL TRIANGLES 419 PROPOSITION XXIII. THEOREM 752. Two mutually equiangular spherical triangles on the same sphere (or on equal spheres) are mutually equilateral, and are congruent or symmetrical. Given : A A and A 1 , mutually equiangular. To Prove : A A and A r mutually equilateral, and congruent or symmetrical. Proof : Construct A E and E 1 , the polar A of A and A'. The sides of E are supplements of the A of A. (742V The sides of E 1 are supplements of theziof A' . j But the A of A are = respectively to the A of A f (Hyp.). .-. &E and E 1 are mutually equilateral (49). Hence A E and E f are mutually equiangular (751). Again A A and A 1 are the polar A of E and E f (741). .. the sides of A are supplements of the A of E ) , ? , the sides of A' are supplements of the A of E' \ Hence A A and A 1 are mutually equilateral (?). . *. they are congruent (when ?) ; or symmetrical (when ?). Q.E.D. Ex. Ts the corresponding theorem about plane triangles true ? Is there any theorem concerning the congruence of plane triangles that is not true of spherical triangles? If the parts of two plane triangles were arranged in reverse order and they were kept in a plane, could they be made to coincide ? ROBBINS'S NEW SOLID GEOM. 12 420 BOOK IX. SOLID GEOMETRY PROPOSITION XXIV. THEOREM 753. The angles opposite the equal sides of an isosceles spherical triangle are equal. Given: (?). To Prove : Z B = Z C. Proof : Suppose X the midpoint of BC. Draw AX, the arc of a great Q. Now the two spherical A ABX and AXC are mutually equilateral. (Explain.) .-. they are mutually equiangular and symmetrical (751). .-. Z = ZC (734). Q.E.D. 754. COROLLARY. The arc of a great circle drawn from the vertex of an isosceles spherical triangle to the midpoint of the base bisects the vertex angle and is perpendicular to the base. PROPOSITION XXV. THEOREM 755. If two angles of a spherical triangle are equal, the sides opposite are equal. Given: (?). To Prove: (?). SPHERICAL TRIANGLES 421 Proof : Construct A A'B'C', the polar A of A ABC. Then A' B 1 is the supplement of Z C. And A'C' is the supplement of Z 73. .-. Z7/=Z C' Again A ABC is the polar A of A A'B'C' AB is the supplement of Z (/, and ^C of Z .-. ^7i = 4C PROPOSITION XXVI. THEOREM (49). (753). (741). CO- (49). Q.E.D. 756. If two angles of a spherical triangle are unequal, the sides opposite are unequal and the greater side is opposite the greater angle. Given: A ABC', Z ABC > Z C. To Prove : AC > AB. Proof : Suppose BR drawn, the arc of a great Q, making Z CBR = Z C and meeting AC at R. Now ^72 + 7*>47* (736). But BR = CR (755). .-. ^#4- CR > .47? (Ax. 6). That is, AC > AB Q.E.D. Ex. Compare Propositions XXIV, XXV, XXVI with the correspond- ing theorems about plane triangles. 422 BOOK IX. SOLID GEOMETRY PROPOSITION XXVII. THEOREM 757. If two sides of a spherical triangle are unequal, the angles opposite are unequal and the greater angle is opposite the greater side. [Converse.] Given: (?). To Prove : Z ABC > Zc. Proof : Z ABC is either < Z. C or = Z C or > Z C. Continue by method of exclusion (90). PROPOSITION XXVIII. THEOREM 768. If two circles on a sphere contain a point on the arc of a great circle that joins their poles, they have no other point hi common. Given : Point P on the arc AB of a great O of a sphere, and P common to two circles whose poles are A and B. To Prove : P is the only point common to these . Proof : Suppose X is another common point. Draw arcs of great AX and EX. Then AX-}- BX > AP + J5P (736). But AX = AP (708). Subtracting, BX > BP (Ax. 7). That is, X is without the O B and cannot be in both the circles. Q.E.D. SPHERICAL TRIANGLES 428 PROPOSITION XXIX. THEOREM 759. The shortest line that can be drawn on the surface of a sphere, between two points on the surface, is the less arc of the great circle containing the two points. Given : Points A and B, and AB the arc of a great O join- ing them ; line ADEB, any other line on the surface of the sphere, between A and B. To Prove : Arc AB < line ADEB. Proof : Take on arc AB any point C, and describe two cir- cles through C, having A and B as their poles, and intersect- ing ADEB at D and E. Point C is the only point common to these two . (758.) No matter what kind of line AD is, a line of equal length can be drawn from A to C, on the surface ; and a line can be drawn from B to C equal in length to BE. [Imagine AD revolved on the surface of the sphere, using A as a pivot, and D will move along the O to point C. Similarly with BE.~\ There is now a line from A to B, through C, < ADEB. That is, whatever the nature of ADEB, there is a shorter line from A to B, which contains C, any point of arc AB. Thus the shortest line contains all the points of AB and therefore is the line AB. Q.E.D. NOTE. This theorem justifies the definition of the "distance" be- tween two points, etc., in 692. 424 BOOK IX. SOLID GEOMETRY ORIGINAL EXERCISES 1. Vertical spherical angles are equal. 2. If two spherical triangles, on the same or equal spheres, are mutu- ally equilateral, their polar triangles are mutually equiangular. 3. The polar triangle of an isosceles spherical triangle is isosceles. 4. The polar triangle of a birectangular spherical triangle is birec- tangular. 6. If two dihedral angles of a trihedral angle are equal, the opposite face angles also are equal. Proof: Construct a sphere having the vertex as center, etc. 6. If two face angles of a trihedral angle are equal, the opposite di- hedral angles also are equal. 7. A trirectangular spherical triangle is its own polar triangle, 8. Two symmetrical spherical polygons are equal. 9. Any side of a spherical polygon is less than the sum of the other sides. [Draw diagonals from a vertex.] 10. If the three face angles of a trihedral angle are equal, the three dihedral angles also are equal. 11. State and prove the converse of No. 10. 12. A straight line cannot meet a spherical surface in more than two points. 13. If two dihedral angles of a trihedral angle are unequal, the oppo- site face angles are unequal, and the greater face angle is opposite the greater dihedral angle. 14. State and prove the converse of No. 13. 15. All the tangent lines drawn to a sphere from an external point are equal. 16. The volume of any tetrahedron is equal to one third the product of its total surface by the radius of the inscribed sphere. 17. Every point of a great circle that is perpendicular to an arc at its midpoint is equally distant from the ends of the arc. A 18. The points of contact of all lines tangent to a sphere from an external point lie in the cir- cumference of a circle. 19. Any point in the arc of a great circle that X. bisects a spherical angle is equally distant from ( the sides of the angle. B ""T C ORIGINAL EXERCISES 425 20. If the opposite sides of a spherical quadrilateral are equal, the opposite angles are equal. 21. If the opposite sides of a spherical quadrilateral are equal, the diagonals bisect each other. 22. If the diagonals of a spherical quadrilateral bisect each other, the opposite sides are equal. 23. The exterior angle of a spherical triangle is less than the sum of the opposite interior angles. 24. The sum of the angles of a spherical quadrilateral is more than four right angles. 25. If two spheres are tangent to each other, the straight line joining their centers passes through the point of contact. 26. The sum of the angles of a spherical polygon is more than 2 n 4 right angles and less than 2 n right angles. 27. The arcs of great circles bisecting the angles of a spherical tri- angle meet in a point. 28. If a tangent line and a secant are drawn to a sphere from an ex- ternal point, the tangent is a mean proportional between the whole secant and the external segment. 29. The product of any secant that can be drawn to a sphere from an external point, by its external segment, is constant for all secants drawn through the same point. 30. If two spherical surfaces intersect and a plane is passed contain- ing their intersection, tangents from any point in this plane to the two spherical surfaces are equal. 31. Find the distance from the center of a sphere whose radius is 15 in. to the plane of a small circle whose radius is 8 in. 32. The polar distance of a small circle is 60 arid the radius of the sphere is 12 in. Find the radius of the circle. 33. The total surface of a tetrahedron is 90 sq. m., and the radius of the inscribed sphere is 4 m. Find the volume of the tetrahedron. 34. Find the radius of the sphere inscribed in a tetrahedron whose volume is 250 cu. in. and total surface is 150 sq. in. 35. Find the total surface of a tetrahedron whose volume is 320 cu. in., if the radius of the inscribed sphere is 8 in. 426 BOOK IX. SOLID GEOMETKY 36. Find the radius of the sphere inscribed in a regular tetrahedron whose edges are each 10 in. 37. Find the radius of the sphere circumscribed about a regular tetrahedron whose edges are each 18 in. 38. Find the radii of the spheres inscribed in and circumscribed about a cube whose edges are each 10 in. 39. The sides of a spherical triangle are 60, 80, 110. Find the angles of its polar triangle. 40. The angles of a spherical triangle are 74, 119, 87. Find the sides of its polar triangle. 41. The chord of the polar distance of the circle of a sphere is 12 m., and the radius of the sphere is 9 m. Find the radius of the circle. 42. The polar distance of a circle is 60 and the diameter of the circle is 8 ft. Find the diameter of the sphere. [Denote by R, each side of an equilateral triangle whose altitude is 4 ft.] 43. The radii of two spherical surfaces are 11 in. and 13 in., and their centers are 20 in. apart. Find the radius of the circle of their intersec- tion. Find also the distances from the centers of the spheres to the center of this circle. 44. The radii of two spherical surfaces are 20 m. and 37 m., and the distance between their centers is 19m. What is the length of the diame- ter of their intersection ? 45. Bisect an arc of a great circle. 46. Draw an arc of a great circle perpendicular to a given arc of a great circle through a given point in the arc. 47. Bisect a spherical angle. 48. Bisect an arc of a small circle. 49. Circumscribe a circle about a given spherical triangle. 60. Construct a spherical angle equal to a given spherical angle at a given point on the same sphere. 61. Construct a spherical triangle having the three sides given. 62. Construct a spherical triangle having the three angles given. 63. Construct a plane tangent to a sphere at a given point on the surface. 54. Construct a spherical surface having the radius given and contain- ing three given points. SPHERICAL AREAS AND VOLUMES 427 65. Construct a spherical surface that shall have a given radius, touch a given plane, and contain two given points. 66. Construct a spherical surface that shall have a given radius, shall be tangent to a given sphere, and contain two given points. 57. Construct a spherical surface that shall contain four given points. 58. Construct a plane that shall contain a given line and be tangent to a given sphere. 69. Construct a plane tangent to a given sphere and parallel to a given plane. 60. What is the locus of points on the surface of a sphere : (a) Equally distant from two given points on the surface? (b) Equally distant from two given points not on the surface ? 61. What is the locus of the centers of those spherical surfaces that pass through two given points ? 62. What is the locus of the centers of the spherical surfaces of given radius that contain two given points ? 63. What is the locus of the centers of the spherical surfaces that pass through three given points ? SPHERICAL AREAS AND VOLUMES 760. A lune is a portion of the surface of a sphere bounded by two great semicircles. The points of intersection of the sides of a lune are the vertices of the lune. The angles made at the vertices by the sides are the angles of the lune. LUNE (a) SPHERICAL SECTOR (6) SPHERICAL PYRAMID ZONE SPHERICAL CONE SPHERICAL SEGMENT 428 BOOK IX. SOLID GEOMETRY 761. A zone is a portion of the surface of a sphere bounded by two circles whose planes are parallel. The bases of a zone are the circles bounding it. The altitude of a zone is the perpendicular distance be- tween the planes of its bases. If one of the planes is tangent to the sphere, the zone is a zone of one base. 762. A spherical degree is y ^ of the surface of a sphere. If the surface of a sphere is divided into 720 equal parts, each part is a spherical degree. The size of a spherical degree depends on the size of the sphere. It may be easily conceived to be half a lune whose angle is 1 degree, that is, a birectangular spherical triangle whose third angle is l c . How many spherical degrees are there in a trirectangular spherical triangle ? 763. The spherical excess of a spherical triangle is the sum of its angles less 180. That is, E= A+ B+ C 180. 764. A spherical pyramid is a portion of a sphere bounded by a spherical polygon and the planes of its sides. The vertex of a spherical pyramid is the center of the sphere. The base of a spherical pyramid is the spherical polygon. 765. A spherical sector is the solid generated by the revo- lution of the sector of a circle about any diameter of the circle as an axis. The base of the spherical sector is the zone generated by the arc of the circular sector. A spherical cone is a spherical sector whose base is a zone of one base. 766. A spherical segment is a portion of a sphere in- cluded between two parallel planes that intersect the sphere. SPHERICAL AREAS AND VOLUMES 429 The bases of a spherical segment are the circular sections made by the parallel planes. The altitude of a spherical segment is the perpendicular distance between the bases. A spherical segment of one base is a segment one of whose bounding planes is tangent to the sphere. A hemisphere is a spherical segment of one base, which base is a great circle. A spherical wedge is a portion of a sphere bounded by a lune and the planes of its sides. Ex. 1. What is the spherical excess of a spherical triangle whose angles are 60, 70, and 100? Ex. 2. Distinguish between a zone and a spherical segment. Ex. 3. Find the area of a spherical degree on a sphere whose surface is 3600 sq.in. Ex. 4. Find the area of a spherical triangle containing 80 spherical degrees, on a sphere whose surface is 450 sq. ft. Ex. 6. Find the area of a spherical polygon containing 152 spherical degrees on a sphere whose surface is 630 sq. yd. Ex. 6. A spherical triangle containing 128 spherical degrees has an area of 72 sq. in. What is the area of the spherical surface ? PRELIMINARY THEOREMS 767. THEOREM. Either angle of a lune is measured by the arc of a great circle described with the vertex of the lune as a pole, and included between the sides of the lune. (711.) 768. THEOREM. The angles of a lune are equal. 769. THEOREM. Every great circle of a sphere divides the sphere into two equal hemispheres, and the surface into two equal zones. 770. THEOREM. The spherical excess of a spherical n-gon is equal to the sum of its angles less (n 2) 180. Proof: (?). 430 BOOK IX. SOLID GEOMETRY 771. THEOREM. If a regular polygon having an even num- ber of sides is inscribed in, or circumscribed about, a circle, and the figure is made to revolve about one of the longest diagonals of the polygon, the surface generated by the perim- eter of the polygon approaches the surface of the sphere gene- rated by the circle, as a limit, if the number of sides of the polygon is indefinitely increased. 772. THEOREM. If a polyhedron is circumscribed about a sphere and the number of its faces is indefinitely increased, the surface of the polyhedron approaches the surface of the sphere as a limit, and the volume of the polyhedron approaches the volume of the sphere as a limit. NOTE. If a regular polygon having an even number of sides is in- scribed in, or circumscribed about, a circle, and the figure is made to revolve about one of the longest diagonals of the polygon, the surface generated by the polygon is composed of the surfaces of cones, a cylin- der, and frustums, and the surface generated by the circle is a spherical surface. Ex. 1. Find the spherical excess of a polygon whose angles are 80, 110, 140, 130, 160. Ex. 2. The spherical excess of a spherical polygon is the difference between the sum of its angles and the sum of the angles of a plane poly- gon having the same number of sides. Ex. 3. The sum of the angles of a spherical quadrilateral is less than eight right angles. Ex. 4. Find the spherical excess of a spherical hexagon if each of its angles equals 128. If each angle equals 155, find the excess. Ex. 6. If the opposite angles of a spherical quadrilateral are equal, the opposite sides are also equal. Proof: Prolong one pair of opposite sides in both directions until they meet. Now prove two triangles congruent. AREAS OF SPHERES 431 THEOREMS AND DEMONSTRATIONS PROPOSITION XXX. THEOREM 773. The area of the surface generated by a straight line revolving about an axis in its plane is equal to the product of the projection of the line upon the axis by the circumference of a circle whose radius is the line perpendicular, to the revolv- ing line at its midpoint, and terminating in the axis. Given : Line AB revolving about axis XX' ; CD = projection of AB on XX 1 '; MP = a = _L erected at mid- point of AB and terminating in XX 1 ; MO = radius of mid-section. To Prove : Surface generated by AB = CD 2 Tra. Proof : I. The surface generated by A B is the surface of the frustum of a right circular cone whose bases are gene- rated by AC and BD, and the mid-section, by MO. Area of surface = 2 wMO - AB (681). Now A ABH and MPO are similar (310). .'. MO : AH = MP : AB (?). Hence MO AB = AH MP = CD a . (?). . . area of surface = 2 TTCD a = CD 2 tra (Ax. 6). II. If AB is II to XX* r , the surface is cylindrical and equals CD - 2 Tra (654). III. If AB meets XX 1 at C, the entire surface is conical and equals TrBD - AB (680). Now BD=2MO (136). And MO - AB = CD. a. (?.) . '. TrBD AB = TT 2 MO - AB = TT 2 CD - a = CD 2 Tra (Ax. 6). .-. the area of the surface = CD 2 Tra (Ax. 6). Q.E.D. 432 BOOK IX. SOLID GEOMETRY PROPOSITION XXXI. THEOREM 774. The surface of a sphere is equal in area to four great circles ; that is, to Given : Semicircle ACF ; diameter AF; S = surface of sphere generated by revolving the semicircle about AF as an axis ; R = radius of this sphere. To Prove : S = 4 irR 2 . Proof: Inscribe in this semicircle half of a regular poly- gon having an even number of sides. Draw the apothems, a. Draw the projections of the sides of the polygon on the diam- eter. Now, if the figure revolves on AF as an axis, the surface AB = AP 2 IT a the surface BC = PS - 2 ira (773). the surface CD = 8T 2 ira etc. Adding, the entire surface = (AP -f- PS+ ST + etc.) = AF 2 TTO. jra (Ax. 2). (Ax. 6). Now, if the number of sides of the polygon is indefinitely increased, the entire surface generated by the polygon ap- proaches S as a limit. (771.) a approaches R as a limit (422). Also AF 2 TTQ approaches AF 2 irR. c A v 9 -K> /^99CU . o ./I /* j-i 7T K \jUtiij j But AF=2R (?). (Ax. 6). Q.E.D. AREAS OF SPHERES 433 775. COROLLARY. The area of a spherical degree equals 4irR2 = T*^* sq. units. 720 180 776. COROLLARY. 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. Proof. s - ' S' PROPOSITION XXXII. THEOREM 777. The area of a zone is equal to the product of its alti- tude by the circumference of a great circle. Given : (The same as in 774). To Prove : The area of the zone generated by the arc EC = PS x 2 TTR. Proof: The area generated by chord BC= PS- 2 ira (773). If the number of sides of the inscribed polygon is indefi- nitely increased, the length of chord BC approaches arc BC and the surface generated by chord BC approaches the area of a zone. Also PS - 2 ira will approach PS 2 TTR. Hence Area of zone BC = PS 2 irB (229). Q.E.D. 778. COROLLARY. Area of a zone, Z = 2 (Where Z = area of the zone, H= its altitude, and R = radius of sphere.) Ex. 1. On a sphere whose radius is 6 in., find the area of a zone 2J in. in height. Ex. 2. What does the formula for the area of a zone become when the altitude is the diameter ? when the altitude is half the radius? 434 BOOK IX. SOLID GEOMETRY 779. COROLLARY. The area of a zone of one base is equal to the area of a circle whose radius is the chord of the generating arc. xl A D X' Given : Arc AB of semicircle ABC; diameter AC; chord AB. To Prove : Area of zone generated by arc AB = TrAB 2 . Proof : Area of zone AB = AD 2 TTR That is, area of zone AB = TT AD 2 B. Draw chord BC. A ABC is a rt. A .'. AD- AC=AB 2 That is, AD-2E = AB 2 Hence area of zone 41? = TrAB 2 That is, area of zone of one base = IT (chord) 2 . (777). (333). (Ax. 6). (Ax. 6). Q.E.D. Ex. 1. What is the area of a zone of one base whose chord is 7 in. in length? of one whose chord is 14 in. in length? Ex. 2. What does the formula for the area of a zone of one base be- come when the generating arc is a semicircle ? when the generating arc is a quadrant? Ex. 3. If the radius of the earth is approximately 4000 mi. and the altitude of the north temperate zone is 2080 mi., what is the area of the north temperate zone ? Ex. 4. Prove that on the same or equal spheres, zones having equal altitudes have equal areas. AREAS OF SPHERES 435 PROPOSITION XXXIII. THEOREM 780. The area of a lune is to the area of the surface of its sphere as the angle of the lune is to 360. Given : Lune ABCDA on sphere o ; L = area of lune ; 8 = area of sphere ; great O EB whose pole is A. To Prove : L : 8 = Z A : 360. Proof: I. If arc BD and the circumference of O .E^B'are commensurable. There exists a common unit of measure. Suppose this unit contained 5 times in BD; 32 times in the circumference. .-. arc BD : circumference = 5 : 32 (?). Arc BD measures Z A (711). .-. ZA : 360 = 5: 32 (Ax. 6). Pass great CD through the several points of division of circumference EB and vertex A, dividing the surface of the sphere into 32 equal lunes. Then L : S = 5 : 32 (Ax. 3). Hence L : s = /. A : 360 (Ax. 1). Q.E.D. II. If the arc and circumference are incommensurable. The proof is similar to that found in 293, 524. 781. COROLLARY. The number of spherical degrees in the area of a lune is double the number of degrees in its angle. Proof: Let L denote the area of the lune, expressed in spherical degrees. Then L : 720 = Z A : 360 (Subst. in 780). .-. Jy = 2Z4. Q.E.D. ROBBINS'S NEW SOLID GEOM. 13 436 BOOK IX. SOLID GEOMETRY 782. COROLLARY. The area of a lune expressed in square units is Q.E.D. 90 Proof : Substituting in 780, L : 4 TTR* = /. A : 360. L = ^A- 90 783. COROLLARY. Two lunes on the same or equal spheres are to each other as their angles. Proof : L : S = Z A : 360, and L r : S = Z A 1 : 360 (780). Dividing, L : L f = Z A : Z A 1 (Ax. 3). Q.E.D. PROPOSITION XXXIV. THEOREM 784. The number of spherical degrees in a spherical tri- angle is equal to the spherical excess of the triangle. Lune, Given : Spherical A ABC on sphere O ; spherical excess of the A = E. To Prove : Number of spherical de- grees in A ABC = E. Proof : Continue the sides of the A ABC to form the lunes ABA'cA, BAB'CB, CAC'BC\ draw diameters AA f , BB f , CC f . AABC' = AA'B'C (747). Lune CAC'BC = A ABC+A AC'B = A ABC+AA'B'C (Ax. 6). AREAS OF SPHERES 437 Now AABC + AA'B'c=lune CAC'BC. } And AABC + A A' BC = lune ABA'CA. And AABC + A AB'C = lune BAB'CB. . (Ax. 4.) Adding, A AB'C = lune A + lune B + lune C (Ax. 2). Now, first, 4 of these A compose a hemisphere and = 360 spherical degrees ; second, the 3 lunes = 2Z^ + 2Z + 2ZC (781). By substituting in the long equation above, 2 A ABC + 360 = 2Z4 + 2Z + 2ZC (Ax. 6). .-. AABC= Z4+Z+Z c-180 (Ax. 3). That is, A ABC = E (763). Q.E.D. 785. COROLLARY. The area of a spherical triangle expressed in square units is Proof : 1 spherical degree = ^ sq. units 180 . Area of a spherical A = E * R sq. units (Ax. 3). 180 Q.E.D. Ex. 1. How many spherical degrees are there in a lune whose angle equals 25? Ex. 2. On a sphere whose radius is 10 in., is a lune whose angle is 10. Find the area of the lune in square inches.. Ex. 3. Reduce the formula for the area of a lune in square units, if the angle is 90; if the angle is 180. Ex. 4. On a sphere whose radius is 9 ft. is a spherical triangle whose angles are 70, 145, and 60. Find the area of the triangle. 438 BOOK IX. SOLID GEOMETRY PROPOSITION XXXV. THEOREM 786. The number of spherical degrees in a spherical poly- gon is equal to its spherical excess. Given : A spherical n-gon. To Prove : The number of spherical degrees in this n-gon = the excess of the polygon. Proof : From any vertex draw diagonals, dividing the polygon into (n 2) A ; let the sums of the A of these A be denoted by s, s v 2 , etc. Now the number of spherical degrees in one A = s 180 (784). Number of spherical degrees in another A = 8 1 180 (?). Etc., for (n-2) A. Adding, the number of spherical degrees in the n-gon = the sum of its A-(n- 2) 180 (Ax. 2). The excess of w-gon = sum of its A - (n 2) 180 (770). .-. the number of spherical degrees in a spherical polygon = the excess of the polygon (Ax. 1). Q.E.D. Ex. 1. Find the area of a spherical triangle whose angles are 80, 125, and 95, on a sphere whose radius is 6.3 in. Ex. 2. Find the area of a spherical polygon whose angles are 135, 105, 85, 155, 120, on a sphere whose radius is 15 ft. Ex. 3. Find the area of a spherical triangle whose angles are 72, 97, and 101, on a sphere whose radius is 3 in. VOLUMES OF SPHERES 439 PROPOSITION XXXVI. THEOREM 4-rrJJ 8 787. The volume of a sphere = 3 Given : Sphere o ; radius = R ; surface = S ; volume = V. To Prove: F = i^- 3 . 3 Proof : Suppose a polyhedron circumscribed about the sphere, its surface denoted by S f , and its volume by V 1 . Suppose planes are passed through the edges of the poly- hedron and the center of the sphere, thus dividing the poly- hedron into pyramids whose vertices are all at the center, and whose common altitude is B. The volume of one such pyramid = ^ E its base (612). .*. volume of all pyramids = ^ E the sum of all the bases (Ax. 2). That is, V' = IR.8'. Indefinitely increase the number of faces of the polyhe- dron, thus indefinitely decreasing each face, and v' approaches V as a limit 1 f 772^ and s' approaches S as a limit I Hence R 8 f approaches J R 8 as a limit. .-. V=%R- 8 (229). But S = 47TE 2 (?). (Ax. 6). Q.E.D. 440 BOOK IX. SOLID GEOMETRY 788. COROLLARY. The volumes of two spheres are to each other as the cubes of their radii or as the cubes of their diameters. Proof: F _ 4 7T.R 3 4 TTR' 3 R 3 _ ( J _D) 3 __ D 3 , * n^ V 1 ~~ 3 3 ~ R' B (i D') 3 D' 3 Q.E.D. 789. COROLLARY. The volume of a spherical pyramid is equal to one third the product of the polygon that is its base, by the radius of the sphere. F= I (area of base)U. Proof : Similar to the proof of 787. 790. COROLLARY. The volume of a spherical wedge is to the volume of the sphere as the angle of its base is to 360. Proof : Similar to the proof of 780. 791. COROLLARY. Volume of a spherical wedge, 270 (Where A = the ^ of the lune, and R = the radius of sphere.) Pr.oof: F : wiz 8 = Z : 360 (790). 270 792. COROLLARY. The volume of a spherical sector is equal to one third the product of the zone that is its base by the radius of the sphere. Proof : Similar to the proof of 787. 793. COROLLARY. Volume of a spherical sector or a spher- ical cone, r =iz.*=|irl*ir. (Where F=the volume of the spherical sector or cone, H = the altitude of its base, JB = the radius of the sphere.) VOLUMES OF SPHERES 441 PROPOSITION XXXVII. PROBLEM 794. To derive a formula for the volume of a spherical segment. There are Three Cases 1. Spherical segment of one base. x x ^*-**^ ^ A, / Spherical Cone \ ( Con e . ^ ^ I - f-SpherrcdltieS'nent I (Generated by OAXJ ~~ I Generated by A CO I (Generated byACXJ Given : Spherical segment generated by the figure AGX\ semicircle XAY '; AG=r\ > radius of sphere = #; altitude = CX = H. Required : To find the volume of the spherical segment. Computation : Draw chords AX, AY, and radius AO. The right A AGO will generate a cone of revolution (Def. 660). The volume of spherical segment ACX = the volume of spherical cone OAX minus the volume of cone AGO. (793). (683). (331, II) . - #) (Ax. 6). Hence volume of spherical segment ACX = | 7TR 2 ri (| 7TR 2 H 7TRII 2 + J 7TH 3 ) (AX. 6). . . Volume of spherical segment of one base = ^ irlT 2 ( 3 H H) . Q.E.F. Volume of spherical cone OAX = f TrR 2 - H Volume of cone AGO = % Trr 2 - CO Now r 2 = CX CY = J/(2 B - H) Also CO = R H. .-. vol. AGO = 442 BOOK IX. SOLID GEOMETRY 2. Spherical segment not including the center. A ^ -^ A. tor I i / O?/7e Generated by ABOJ I I Generated by ACO Generated by BDO) (Generated by ACDB / Given : Spherical segment generated by figure . ACDB ; semicircle XABY; AC r\ BD = r r ; ra- g y dius of sphere = 12 ; altitude = CD = H. Required : To find the volume, F, of the sphc: - ical segment. Computation: The A AGO and BDO generate cones of revolution (Def. 660). Denote OD by d. The volume of spherical segment ACDB = the volume of spherical sector ABO plus the volume of cone AGO minus the volume of cone BDO. Now the volume of spherical sector ABO = % 7rR 2 H (792). And the volume of cone AGO = J irr\d + H) (683). And the volume of cone BDO = 4 7rr' 2 d (683). F= -f[* -r' 2 )] But in rt. A AGO, R 2 = r 2 + (d + tf) 2 and in rt. A BDO, R 2 = r' 2 + d 2 . . Subtracting and solving, d = Substituting in (3), ~4 I *.'4 i rr4 9 R 2 = T t- r f a. (1) (2)1 (3)1 (4) (334) ORIGINAL EXERCISES 443 Substituting (4) and (5) in (1), and simplifying, 3 _ = r rz H + fl 3 "! 2 J .-. V = TrH(r* + r'*) + TrH*. Q.E.F. 3. Spherical segment including the center. Given : Spherical segment generated by figure BDSR ; etc. Required : To find the volume, F, of the spherical segment. Computation: v = the volume of spherical sector BOB plus the volume of cone BDO plus the volume of cone R80. (Computation similar to that in 2, with same final formula.) ORIGINAL EXERCISES 1. Prove that the area of the surface of a sphere is equal to the square of the diameter multiplied by TT ; that is, 5 = TrZ) 2 . 2. Prove that the volume of a sphere is equal to one sixth the cube of the diameter multiplied by TT ; that is, V = $ TrZ) 8 . 3. The surface of a sphere is equal to the cylindrical surface of the circumscribed cylinder. 4. The total surface of a hemisphere is three fourths the surface of the sphere. 6. The 'volume of a sphere is two thirds the volume of the circum- scribed cylinder. 6. Upon the same circle as a base are constructed a hemisphere, a cylinder of revolution, and a cone of revolution, all having the same altitude. Prove that their total areas are 3 TrR 2 , 4 Tr# 2 , TrZZ 2 (l + \/2), respectively, and their volumes are f wR s , TrR 3 , j TrR 8 , respectively. 7. Two zones on the same sphere, or on equal spheres, are to each other as their altitudes. 8. The area of the surface of a sphere is equal to the area of the circle whose radius is the diameter of the sphere. 9. Show that the formula for the volume of a spherical segment of one base reduces to the correct formula for the volume of a hemisphere when the base of the segment is a great circle ; and to the correct for- mula for the volume of a sphere when the planes are both tangent. 444 BOOK IX. SOLID GEOMETRY a circle whose radius is R, there are in- square and an equilateral triangle having 10. In an equilateral triangle is inscribed a circle, and the figure is revolved about an altitude of the tri- angle as an axis. Prove : (a) That the surface generated by the circumfer- ence is two thirds the lateral surface generated by the triangle. (&) That the volume generated by the circle is four ninths the volume generated by the triangle. 11. Derive a formula for the surface of a sphere, containing only V and TT. 12. Derive a formula for the volume of a sphere, containing only S and TT. 13. In scribed a their bases parallel ; the whole figure is then revolved about the diameter perpendicular to the base of the triangle. Find, in terms of R : (a) The total areas of the three surfaces generated. (6) The volumes of the three solids generated. 14. If a cylinder of revolution having its altitude equal to the diam- eter of its base, and a cone of revolution having its slant height equal to the diameter of its base, are both inscribed in a sphere : (a) The total area of the cylinder is a mean proportional between the area of the surface of the sphere and the total area of the cone. (6) The volume of the cylinder is a mean proportional between the volume of the sphere and the volume of the cone. 16. About a circle whose radius is a there are circumscribed a square and an equilateral triangle having their bases in the same straight line. The whole figure is then revolved about an altitude of the triangle. Find, in terms of a : (a) The total areas of the three surfaces gen- erated. (&) The volumes of the three surfaces generated. 16. If a cylinder of revolution having its altitude equal to the diam- eter of its base, and a cone of revolution having its slant height equal to the diameter of its base, is circumscribed about a sphere : (a) The total area of the cylinder is a mean proportional between the area of the surface of the sphere and the total area of the cone. (7>) The volume of the cylinder is a mean proportional between the volume of the sphere and the volume of the cone. r-. A ORIGINAL EXERCISES 445 17. The line joining the centers of two intersecting spherical surfaces is perpendicular to the plane of the intersection at the center of the intersection. 18. A cube and a sphere have equal surfaces. Show that the sphere has the greater volume. 19. Prove that the parallel of latitude through a point having 30 north latitude bisects the surface of the northern hemisphere. 20. Prove that in order that the eye may observe one sixth of the surface of a sphere it must be at a distance from the center of the sphere equal to f of the radius. Proof: Zone TT = $ surface of sphere (Hyp.). .-. AB = $ diain. = | R. Hence BC - f R. In rt. A ETC, TC 2 = EC- BC (?) ; .-. R* = EC f R, or EC = f R. (Explain.) Q.E.D. 21. How many miles above- the surface of the earth (diameter of earth = 7960 mi.) must a person be in order that he may see one sixth of the earth's surface ? 22. If the area of a zone of one base is a mean proportional between the area of the remaining zone of the sphere and the area of the entire sphere, the altitude of the zone is R(Vo 1). 23. The area of a lune is to the area of a trirectangular spherical triangle as the. angle of the lune is to 45. 24. A cone, a sphere, and a cylinder have the same diameters and altitudes. Prove that their volumes are in arithmetical progression. 26. The surface of a sphere bears the same ratio to the total surface of the circumscribed cylinder of revolution as the volume of the sphere bears to the volume of the cylinder. 26. The smallest circle upon a sphere whose plane passes through a given point within the sphere, is the circle whose plane is perpendicular to the diameter through the given point. 27. What part of the surface of the earth could one see if he were at the distance of a diameter above the surface? 28. Prove that if any number of lines in space are drawn through a point, and from any other point perpendiculars to these lines are drawn, the feet of all of these perpendiculars lie on the surface of a sphere. 29. The volume of a sphere is to the volume of the circumscribed cube as TT : 6. The volume of a sphere is to the volume of the inscribed cube as TT : f \/3. 446 BOOK IX. SOLID GEOMETRY 30. There are five spheres that touch the four planes of the faces of a tetrahedron. 31. If two angles of a spherical triangle are supplementary, the sides of the polar triangle, opposite these angles, are supplementary. 32. A square, whose side is a, is revolved about a diagonal, and also about an axis bisecting two opposite sides. Which of these figures con- tains the greater volume? Which has the greater surface? 33. Find the area of the surface and the volume of a sphere whose radius is 6 in. 34. Find the area of a zone whose altitude is 4 in. on a sphere whose radius is 14 in. 35. Find the area of a lune whose angle is 30 on a sphere whose radius is 8 in. 36. Find the area of a spherical triangle whose angles are 110, 41, 92 on a sphere whose radius is 10 in. 37. Find the volume of a sphere whose radius is 5 m. 38. Find the volume of a spherical pyramid whose base is 35 sq. in. on a sphere whose radius is 12 in. 39. Find the area of a spherical polygon whose angles are 87, 108, 121, 128 on. a sphere whose radius is 25 cm. 40. What is the radius of a sphere whose surface is 1386 sq. yd. ? 41. What is the radius of a sphere whose volume is -- cu. in. ? 3 42. What is the- area of the surface of a sphere whose volume is 2887rcu. ft.? 43. What is the volume of a sphere the area of whose surface is 2464 sq. in. ? 44. Find the area of a zone whose altitude is 3 in. if the radius of the sphere is 7 in. 46. Find the volume of a spherical sector the altitude of whose base is 5 in. if the radius of the sphere is 6 in. 46. Find the diameter, the circumference of a great circle, and the volume of a sphere the area of whose surface is 25 TT sq. ft. 47. By how many cubic inches is a 9-in. cube greater than a 9-in. sphere ? 48. The radius of a sphere is 15 in., and the angles of the base of a spherical pyramid are 160, 127, 96, 145, and 117-. Find the volume of the pyramid. ORIGINAL EXERCISES 447 49. A cylindrical vessel 10 in. in diameter contains a liquid. A metal hall is immersed in the liquid and the surface rises in. What is the diameter of the ball? 60. If a sphere 3 ft. in diameter weighs 99 lb., how much will a sphere of the same material 4 ft. in diameter weigh ? 61. The radii of the bases of a frustum of a cone of revolution are 5 in. and 6 in., and the altitude of the frustum is 19| in. What is the diameter of an equal sphere ? 62. What is the radius of a sphere whose surface is equal to the total surface of a right circular cylinder having an altitude equal to 21 in. and radius of the base equal to 6 in. ? 53. Find the volume generated by the revolution of an equilateral triangle inscribed in a circle whose radius is 8 in. about an altitude of the triangle as an axis. (See Fig. of Ex. 55.) 64. In the figure of Ex. 55, find the volume of the segment generated by the figure A ED revolving about CD as an axis. 65. Find the area of the surface and the volume of the sphere generated by a circle that is circum- scribed about an equilateral triangle whose side is 10 in. 66. Circumscribing a sphere whose radius is 18 m. is a cylinder of revolution. Compare their total areas ; their volumes. 67. Circumscribing a cylinder of revolution whose altitude and diameter are each 6 in. is a sphere. Find the volume and area of the surface of the sphere. 58. Circumscribing a cylinder whose altitude is 4 in. and diameter is 3 in. is a sphere. Find the radius and volume of the sphere. 59. Each edge of a cube is 8 in. What is the area of the surface and the volume of the circumscribed sphere ? 60. Find the volume of one of the segments cut from a 10 in. sphere by the plane of one of the faces of the inscribed cube. 61. The volume of a certain sphere is 179f cu. ft. Find the radius of a sphere 8 times as large. Find the radius of a sphere 3 times as large. 62. The radius of a certain sphere is 5 in. What is the radius of a sphere twice as great ? half as great ? two thirds as great ? 448 BOOK IX. SOLID GEOMETRY 63. A hollow sphere lias an outer diameter of 20 in. and an inner diameter of 10 in. Find the volume of the metal in the shell. 64. Find the diameter of that sphere whose volume is, numerically, equal to the area of its surface. 66. A projectile consists of a right circular cylinder having a hemi- sphere at each end. If the cylinder is 9 in. long and 7 in. in diameter, what is the volume of one projectile ? 66. Inscribed in a regular tetrahedron whose edge is 4 in., and circum- scribed about it, are two spheres. Find their radii. 67. Find the radii of the spheres inscribed in and circumscribed about a regular hexahedron whose edge is 8 m. 68. Find the radii of the spheres inscribed in and circumscribed about a regular octahedron whose edge is 12 in. 69. How many spherical bullets \ in. in diameter can be made from a cube of lead 5 in. on each edge? 70. The area of a spherical triangle whose angles are 158, 77, 95 is 288| s q- ft. Find the radius of the sphere. 71. The area of a spherical triangle whose excess is 75 is 135 IT sq. in. Find the radius of the sphere. 72. If the radius of a sphere is 2.5 in., and the sides of a triangle on it are 104, 115, 101, find the area of the polar triangle. 73. In a trihedral angle the plane angles of the dihedral angles are 75, 85, 110. Find the number of degrees of surface of a sphere whose center is the vertex of the trihedral angle inclosed by the faces of this trihedral angle. 74. What is the area of a spherical hexagon, each of whose angles is 145, on a sphere whose radius is 15 m. ? 75. How many miles above the earth must a person be in order that he may see a third of its surface? one eighth of its surface? 76. Find the altitude of the zone whose area is equal to the area of a great circle of a sphere. 77. If the radius of a sphere is doubled, how is the amount of surface affected? the volume? the weight? 78. At a distance (= d) from the center of a sphere whose radius is r is an illuminating point. What is the altitude of the zone illuminated? 79. On a sphere having a radius of 5 in. is an equiangular spherical triangle whose area is 5 ir sq. in. Find the angles of the triangle. 80. Find the area of the surface of a sphere whose volume is 1 cu. yd. ORIGINAL EXERCISES 449 81. Find the volume of a sphere whose surface is 1 sq. yd. 82. If a circumference is described on the surface of a sphere by a pair of compasses whose points are 2f in. apart, what is the area of the /one bounded by this circumference ? 83. On a sphere the area of whose surface is 288 sq. ft. is a birectan- gular spherical triangle whose vertex angle is 100. Find the area of this triangle. 84. Five inches from the center of a sphere whose diameter is two feet a plane is passed. Find the areas of the two zones formed. Find the chords of their generating arcs. 85. The diameter of the moon is about 2000 mi. ; that of the earth, about 8000 mi. How do their surfaces compare? their volumes? 86. The radii of two concentric spheres are 12 in. and 13 in. A plane is tangent to the inner sphere. Find area of section of outer sphere. 87. If a solid sphere 4 ft. in diameter weighs 500 lb., what is the weight of a spherical shell, whose external diameter is 10 ft., if it is made of the same material and a foot thick? 88. The sun's diameter is about 109 times the diameter of the earth. How do the areas of their surfaces compare? their volumes? 89. How many quarter-inch spherical bullets can be made from a sphere of lead a foot in diameter? 90. A 12 -inch cube of lead is melted and cast in the form of a spheri- cal cannon ball. What is the radius of the cannon ball ? 91. Find the angles of an equiangular spherical triangle equal to the sum of three equiangular spherical triangles (upon the same sphere) whose angles are each 75. 92. What is the radius of a sphere equal to the sum of two spheres whose radii are 3 in. and 4 in. respectively? 93. What is the radius of a sphere equal to the difference of two spheres whose radii are 5 in. and 4 in. respectively? 94. The area of an equiangular spherical triangle is TT sq. in., and the radius of the sphere is 4 in. Find the angles of the triangle. 95. The volumes of two spheres are to each other as 64 : 343. What is the ratio of their surfaces ? 96. Find the volumes of the segments of a sphere whose radius is 12 in. formed by a plane whose distance from the center is 9 in. 450 BOOK IX. SOLID GEOMETRY 97. If the radius of a sphere is 20 in., find : (a) The area of its surface. (6) The area of a zone whose altitude is 2 in. (c) The edge of a cube inscribed in the sphere. (J) The area of a lune whose angle is 80. (e) The area of a spherical triangle whose angles are 75, 53, 72. (/) The area of a spherical polygon whose angles are 68, 119, 128, 147, 150. (#) The area of a birectangular spherical triangle whose vertex- angle is 54. (Ji) The area of a zone of one base whose altitude is 5 in. (i) The radius of a sphere whose surface is four times as large. (/) The volume of the sphere. () The volume of a wedge whose angle is 36. (/) The volume of a spherical pyramid whose base is the triangle of exercise (e). (in) The volume of the spherical sector whose base is the zone of exercise (6). (n) The volume of the spherical cone whose base is the zone of exercise (h). (0) The volume of a spherical segment of one base, whose altitude is 6 in. (jo) The volume of a spherical segment whose altitude is 4 in. and the radii of whose bases are 12 in. and 16 in. (q) The radius of a sphere whose volume is 64 times as large. 98. The angles of a spherical triangle are 80, 90, 100. Find the angle of an equal lune. 99. In a sphere whose radius is 26 in. two parallel planes are passed 34 in. apart. The radii of the two sections are 10 in. and 24 in. Find the volume of the spherical segment included between the planes. 100. In a certain refrigerating plant is a large tank of ice water. From this tank to a faucet is a pipe in. inside diameter, and 42 ft. long. The faucet is opened and 1 qt. of water runs out every 4 seconds. In what length of time will the cold water from the tank appear at the faucet? 101. A sphere 2 ft. in diameter is trisected by two concentric spherical surfaces. Find the radii of these surfaces, in inches. 102. If a cylindrical leaden bar, a ft. long and b in. in diameter, is melted and made into bullets, | in. in diameter, explain the successive steps necessary to be taken to ascertain the number of bullets there will be. SUMMARY OF FORMULAS OF SOLID GEOMETRY B area of base. m = radius of mid-section. b = area of upper base. P = perimeter of base. E = number of edges. P r = perimeter of right section, e, e' = homologous edges. p = perimeter of upper base. F = number of faces. R, r radius of base. H = altitude. s = slant height. L = lateral area. T = total area. M = mid-section. V = volume ; number of vertices. PRISMS AND PYRAMIDS Parallelepiped V = BH (584). Prism L = H P r (569). T=L + 2J3 (570). V=B H (590). Prismatoid V = \H(b + B + 4 M) (628). Regular Pyramid L = *! s (604). T=L + B (595). Pyramid V = \B H (612). Frustum of pyramid .... L = (P + p~)s (605). V = I H(B + b + VB^b) (618). Polyhedron. ..... JEJ + 2 = F 4- F (624). Sum of face A = (V- 2)360 (626). Similar polyhedrons . . T :T = e*.-.e'* (632). F:F' = e 3 :e' 3 (634). CYLINDERS AND CONES Right circular cylinder . . . L = 2TrRH (654). T=2TTR(H+R) (654). Circular cylinder V=irR*H (656). Right circular cone .... L = TTRs (679). T = irR(s + R) (679). Circular cone V = \TtR*H (683). Frustum of right circular cone L = TT(R + r)s (681), = 2irms (681). T = ir[(R + r)s + -R 2 + f 2 ] (681). V = l - irJff [ J? 2 + r 2 + B r] (685) . 3 EOBBINS'S NEW SOLID GEOM. 14 451 452 BOOK IX. SOLID GEOMETRY A = angle of lime. E spherical excess. H = altitude. L = area- of lime. R = radius of sphere. r, r' = radii of bases of spherical segment. S = area of spherical surface. V = volume. Z = area of zone. A = area of triangle. Spherical surface . THE SPHERE . . S = 4irR 2 . . Z - 2 TTllH (774). (778). Zone of one base . . . . Z - TT (chord) 2 . . L 2 A spherical degrees (779). (781). or L ' ~^ souare units <78^ Spherical triangle . . 90 . . A = E spherical degrees or A ~ Tf-R square units V. 1 ^>")- (784). (785) Spherical polygon . . Sphere ISO Polygon = E spherical degrees v 4irU 3 (780). (787). Spherical pyramid 3 . . F = |(base)U r _^A-TTR* (789). (79l\ Spherical sector . . . . V^ITTR^H . . F-firl? 2 JJ (793). (793). Spherical segment, one Any spherical segment base, r = lTTH^R-H) F = - TrH (r* + r' 2 ) + i IT IT 3 2 6 (794). (794). INDEX OF DEFINITIONS (The numbers refer to pages.) Abbreviations, vii. Adjacent dihedral angles, 290. Alternate interior dihedral angles, 291. Altitude, of cone, 377. of cylinder. 367. of frustum of cone, 378. of frustum of pyramid, 333. of prism, 314. of prismatoid, 352. of pyramid, 332. of spherical segment, 429. of zone, 428. Angle, between intersecting curves, 395. birectangular trihedral, 307. convex polyhedral, 305. dihedral, 290. isosceles trihedral, 307. plane angle of dihedral, 291. polyhedral, 305. rectangular trihedral, 307. spherical, 395. trihedral, 307. trirectangular trihedral, 307. Angles, adjacent dihedral, 290. complementary dihedral, 291. equal dihedral, 291. equal polyhedral, 306. face, of polyhedral, 305. of lune, 427. of spherical triangle, 408. right dihedral, 291. supplementary dihedral, 291. symmetrical polyhedral, 306. vertical dihedral, 291. vertical polyhedral, 306. Axis, of circle of sphere, 394. of circular cone, 377. Base, of cone, 377. of pyramid, 332. of spherical pyramid, 428. of spherical sector, 428. Bases, of cylinder, 367. of prism, 314. of spherical segment, 429. of zone, 428. Birectangular spherical triangle, 408. Birectangular trihedral angle, 307. Center of sphere, 393. Circles of sphere, 394. line tangent to, 395. Circular cone, 377. axis of, 377. right, 377. Circular cylinder, 368. right, 368. Circumscribed frustum of pyramid, 378. Circumscribed prism, 368. Circumscribed sphere about polyhe- dron, 395. Complementary dihedral angles, 291. Cone, 377. altitude of, 377. base of, 377. circular, 377. circular, axis of, 377. frustum of, 378. lateral area of, 377. lateral area of frustum of, 378. oblique circular, 377. 453 454 INDEX OF DEFINITIONS Cone (continued] of revolution, 378. of revolution, slant, height of, 378. plane tangent to, 378. right circular, 377. spherical, 428. total area of, 377. total area of frustum of, 378. Cones, 377. similar, of revolution, 378. Congruent solids, 315. Congruent spherical polygons, 400. Conical surface, 377. Convex polyhedral angle, 305. Convex polyhedron, 313. Corresponding dihedral angles, 201. Cube, 315, 354. Cylinder, 367. altitude of, 367. bases of, 367. circular, 368. lateral area of, 367. oblique, 368. of revolution, 368. right, 368. right circular, 368. right section of, 368. total area of, 367. Cylinders, 367. similar, of revolution, 368. Cylindrical surface, 367. Degree, spherical, 428. Determined, plane, 262. Diagonal, of polyhedron, 313. of spherical polygon, 400. Diameter of sphere, 303. Dihedral angle, 200. edge of, 200. faces of, 200. plane angle of, 201. Dihedral angles, 200. adjacent, 200. alternate interior, 201. complementary, 201. corresponding, 201. equal, 201. Dihedral angles (continued) right, 201. supplementary, 201. vertical, 201. Dimensions of parallelepiped, 315. Directrix, 367, 377. Distance, between points of surface of sphere, 305. from point to a plane, 286. Dodecahedron, 313. regular, 354. Edge of dihedral angle, 200. Edges, of polyhedral angle, 305. of polyhedron, 313. Element, of conical surface, 377. of cylindrical surface, 367. Equal dihedral angles, 201. Equal polyhedral angles, 306. Equal solids, 315. Equal spheres, 305. Equiangular spherical triangles, 408. Equilateral spherical triangles, 408. Excess, spherical, 428. Face angles of polyhedral angle, 305. Faces, of dihedral angle, 200. of polyhedral angle, 305. of polyhedron, 313. Foot of line, 261. Frustum of cone, 378,. altitude of, 378. lateral area of, 378. mid-section of, 378. slant height of, 378. total area of, 378. Frustum of pyramid, 333. altitude of, 333. circumscribed about frustum of cone, 378. inscribed in frustum of cone, 378. slant height of, 333. Generatrix, 367, 377. Geometrical solid, 316. Geometry, Solid, 261. Great circle of sphere, 394. INDEX OF DEFINITIONS 455 Great circle, axis of, 394. Hemisphere, 429. Hexahedron, 313. regular, 354. Historical notes : Archimedes, 393. August, 353. Eudoxus, 345. Gerard, 410. Hippasus, 355. Menelaus, 393. Pythagoras, 355. Icosahedron, 313. regular, 354. Inclination of line, 286. Inscribed frustum of pyramid, 378. Inscribed prism, 338. Inscribed prism in cylinder, 368. Inscribed sphere in polyhedron, 395. Intersection, 261. Isosceles trihedral angle, 307. Lateral area, of cone, 377. of cylinder, 367. of frustum of cone, 378. of prism, 314. of pyramid, 332. Lateral edges, of prism, 314. of pyramid, 332. Lateral faces, of prism, 314. of pyramid, 332. Line, foot of, 261. inclination of, 286. projection of, 262. straight, oblique to plane, 261. straight, parallel to plane, 261. straight, perpendicular to plane, 261. tangent to circle of sphere, 395. tangent to sphere, 395. Lune, 427. angles of, 427. vertices of, 427. Mid-section of frustum of cone, 378. of prismatoid, 352. Mutually equiangular spherical tri- angles, 408. Mutually equilateral spherical tri- angles, 408. Normal, 261. Oblique circular cone, 377. Oblique cylinder, 368. Oblique parallelepiped, 315. Oblique prism, 314. Octahedron, 313. regular, 354. Parallel planes, 261. Parallelepiped, 315. dimensions of, 316. oblique, 315. rectangular, 315. right, 315. Perpendicular planes, 291. Plane, 261. determined, 262. distance from point to, 286. straight line oblique to, 261. straight line parallel to, 261. straight line perpendicular to, 261. tangent to cone, 378. tangent to cylinder, 368. tangent to sphere, 395. Plane angle of dihedral angle, 291. Plane section of polyhedral angle, 305. Planes, parallel, 261. perpendicular, 291. Point, of contact, 395. of tangency, 395. projection of, 262. Polar distance, 395. Polar triangle, 409. Poles of circle of sphere, 394. Polygon, spherical, 409. Polyhedral angle, 305. convex, 305. edges of, 305. face angles of, 305. faces of, 305. 456 INDEX OF DEFINITIONS Polyhedral angle (continued) plane section of, 305. vertex of, 305. Polyhedral angles, 306. equal, 306. symmetrical, 306. vertical, 306. Polyhedron, 313. convex, 313. diagonal of, 313. edges of, 313. faces of, 313. inscribed sphere in, 395. regular, 353. vertices of, 313. Polyhedrons, 353. similar, 353. Prism, 314. altitude of, 314. bases of, 314. circumscribed about cylinder, 368. circumscribed about pyramid, 338. inscribed in cylinder, 368. inscribed in pyramid, 338. lateral area of, 314. lateral edges of, 314. lateral faces of, 314. oblique, 314. regular, 314. right, 314. right section of, 315. total area of, 314. triangular, 314. truncated, 314. Prismatoid, 351. altitude of, 352. mid-section of, 352. Prismoid, 352. Projection,'262. Pyramid, 332. altitude of, 332. altitude of frustum of, 333. base of, 332. circumscribed about cone, 378. circumscribed frustum of, 378. frustum of, 333. inscribed frustum of, 378. Pyramid (continued} inscribed in cone, 378. lateral area of, 332. lateral edges of, 332. lateral faces of, 332. regular, 332. slant height of frustum of regular, 333. slant height of regular, 333. spherical, 428. total area of, 332. triangular, 332. truncated, 333. vertex of, 332. Pyramids, 332. Quadrant, 395. Radius of sphere, 393. Rectangular parallelepiped, 315. Rectangular trihedral angle, 307. Regular dodecahedron, 354. Regular hexahedron, 354. Regular icosahedron, 354. Regular octahedron, 354. Regular polyhedron, 353, 354. Regular prism, 314. Regular pyramid, 332. slant height of, 333. Regular tetrahedron, 354. Revolution, cone of, 378. cylinder of, 368. similar cones of, 378. similar cylinders of, 368. Right circular cone, 377. Right circular cylinder, 368. Right cylinder, 368. Right dihedral angles, 21)1. Right parallelepiped, 315. Right prism, 314. Right section, of cylinder, 368. of prism, 315. Sector, base of spherical, 428. spherical, 428. Segment, base of spherical, 429. spherical, 428. INDEX OF DEFINITIONS 457 Sides of spherical triangle, 408. Similar cones of revolution, 378. Similar cylinders of revolution, 368. Similar polyhedrons, 353. Slant height, of cone of revolution, 378. of frustum of cone, 378. of frustum of pyramid, 333. of regular pyramid, 333. Small circle of sphere, 394. Solid, 261. geometrical, 316. volume of, 315. Solid geometry, 261. Solids, congruent, 315. equal, 315. Sphere, 393. axis of circle of, 394. center of, 393. circumscribed about polyhedron, 395. diameter of, 393. great circle of, 394. inscribed in polyhedron, 395. line tangent to, 395. line tangent to circle of, 395. plane tangent to, 395. small circle of, 394. Spheres, equal, 395. tangent, 395. Spherical angle, 395. Spherical cone, 428. Spherical degree, 428. Spherical excess, 428. Spherical polygon, 409. diagonal of, 409. Spherical polygons, congruent, 409. Spherical pyramid, 428. base of, 428. vertex of, 428. Spherical sector, 428. base of, 428. Spherical segment, 428. altitude of, 429. bases of, 429. of one base, 429. Spherical surface, 393. Spherical triangle, 408. angles of, 408. birectangular, 408. sides of, 408. symmetrical, 409. trirectangular, 408. unit of measure of, 408. vertices of, 408. Spherical triangles, 408. mutually equiangular, 408. mutually equilateral, 408. Spherical wedge, 429. Straight line, oblique to plane, 261. parallel to plane, 261. perpendicular to plane, 261. Supplementary dihedral angles, 291. Surface, conical, 377. cylindrical, 367. spherical, 393. Surfaces, 261. Symmetrical polyhedral angles, 306. Symmetrical spherical triangles, 409. Tangent spheres, 395. Tetrahedron, 313. regular, 354. Total area, of cone, 377. of cylinder, 367. of frustum of cone, 378. of prism, 314. of pyramid, 332. Triangle, spherical, 408. Triangular prisrn, 314. Triangular pyramid, 332. Trihedral angle, 307. birectangular, 307. isosceles, 307. rectangular, 307. trirectangular, 307. Trirectangular spherical triangle, 408. Trirectangular trihedral angle, 307. Truncated prism, 314. Truncated pyramid, 333. 458 INDEX OF DEFINITIONS Unit, of measure of spherical triangle, 408. of volume, 315. Vertex, of polyhedral angle, 305. of pyramid, 332. of spherical pyramid, 428. Vertical dihedral angles, 291. Vertical polyhedral angles, 306. Vertices, of lune, 427. Vertices, of polyhedron, 313. of spherical triangle, 408. Volume, of solid, 315. unit of, 315. Wedge, spherical, 429. Zone, 428. altitude of, 428. bases of, 428. of one base, 428. ADVERTISEMENTS ROBBINS'S PLANE TRIGONOMETRY By EDWARD R. ROBBINS, Senior Mathematical Mas- ter, William Perm Charter School, Philadelphia, Pa. $0.60 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. ^j 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. A knowledge of the principles of geometry needed in trigonometry is, as a rule, too freely taken for granted. The author gives at the beginning of the book a statement of the applied principles, with reference to the sections of his Geometry, where such theorems are proved in full. Cross references in the text of the Trigonometry to those theorems make it easy for the pupil to review or to supplement imperfect knowledge. ^[ Due emphasis is given to the theoretical as well as to the practical applications of the science. The number of ex- amples, both concrete and abstract, is far in excess of those in other books on the market. This book contains four times as many exercises as most books, and twice as many as that having the next lowest number. AMERICAN BOOK COMPANY (313) MILNE'S STANDARD ALGEBRA By WILLIAM J. MILNE, Ph.D., LL.D., President of the New York State Normal College, Albany, N. Y. $1.00 THE Standard Algebra conforms to the most recent courses of study. The inductive method of presentation is followed, but declarative statements and observations are used, instead of questions. Added to this kind of unfold- ing and development of the subject are illustrative problems and explanations to bring out specific points, the whole being driven home by varied and abundant practice, ^f The problems are fresh in character, and besides the tradi- tional problems include a large number drawn from physics, geometry, and commercial life. They are classified accord- ing to the nature of the equations involved, not according to subject matter. The statement of necessary definitions and of principles is clear and concise, but the proofs of principles, except some important ones, are left for the maturer years of the pupil. ^f Accuracy and self-reliance are encouraged by the use of numerous checks and tests, and by the requirement that re- sults be verified. The subject of graphs is treated after simple equations, introduced by some of their simple uses in repre- senting statistics, and in picturing two related quantities in the process of change, and again after quadratics. Later they are utilized in discussing the values of quadratic expressions. Factoring receives particular attention. Not only are the usual cases given fully and completely with plenty of practice, but the factor theorem is taught. ^f The helpful and frequent reviews are made up of pointed oral questions, abstract exercises, problems, and recent college entrance examination questions. The book is unusually handy in size and convenient for the pocket. The page size is small. AMERICAN BOOK COMPANY ADVANCED ARITHMETIC $0.75 By ELMER A. LYMAN, Professor of Mathematics, Michigan State Normal College, Ypsilanti THIS book meets the requirements of secondary and normal schools. In its preparation the author has aimed to make the work a study of the fundamental principles of arithmetic, and thereby emphasize the disciplinary value of the subject, and at the same time to apply these principles to the solution of practical business problems. To this end such methods as are used in the best commercial practice are em- phasized throughout the work, and obsolete methods and problems are carefully excluded. ^[ The exercises have been selected largely from actual busi- ness transactions, and nearly all of the problems in the appli- cation of percentage have been secured from business houses, or reviewed by representative business men. The chapters on banking, and stocks and bonds, give information of a prac- tical character which, though indispensable to a proper under- standing of the subject, is rarely found in text-books. ^j In order to economize time, pupils are encouraged to use every practical labor-saving device known to the science of arithmetic, but so-called short processes, which are compli- cated or cumbersome, have been carefully avoided. The use of checks is also strongly recommended, because it contributes greatly to accuracy in results, and cultivates a spirit of self- reliance. ^[ In addition to the special methods for solution given in connection with the various subjects, a chapter is devoted to the general method of approach to any problem. This offers pupils much helpful advice in attempting the solution of prob- lems of a miscellaneous character, such as are given in exam- inations. Attention is also called to the historical notes, to the treatment of graphical representations, and to the chapter on approximate results. AMERICAN BOOK COMPANY (58) PLANE SURVEYING FOR USE IN THE CLASSROOM AND FIELD $3.00 By WILLIAM G. RAYMOND, C.E., LL.D., Member American Society of Civil Engineers, Dean of the Col- lege of Applied Science, State University of Iowa. THIS standard textbook has now been completely re- vised, rewritten, rearranged, reset and remade. The new edition is distinguished by the convenience of its pocket form for field service ; the completeness of its text ; and the clearness of its tables. The book is light in weight, being printed on thin Bible paper, and is bound in flexible leather covers with rounded corners, so that it can be rolled without injury. The maps have been especially planned for hard service and therefore include no folded maps that are easily torn. ^[ The principles are carefully explained and the exercises are designed to show the student, by his own experience, not only the possibilities, but also the limitations of instru- ments, methods, and individuals. 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