741 The lateral area of a frustum of a cone of revolution is equal to half the sum of the circumferences of its bases multiplied by its slant height. HYPOTHESIS. S is the lateral area, C and C' are the circumferences of the bases, R and R' their radii, and L is the slant height. CONCLUSION. S = { (C+C') × L. PROOF Circumscribe about the frustum of the cone a frustum of a regular pyramid, and denote its lateral area by S', the perimeters of its bases by P and P'. Its slant height is L. § 730 If the number of lateral faces of the frustum of the pyramid is indefinitely increased, P+P' approaches C + C' as a limit. · · · 1⁄2 (P + P') × L approaches (C + C') x L as a limit. Also, S' approaches S as a limit. 742 COROLLARY. § 455 §§ 268, 267 § 735 Proved § 264 Q. E. D. The lateral area of a frustum of a cone of revolution is equal to the circumference of a section equidistant from its bases multiplied by its slant height. 743 The volume of a frustum of a circular cone is equivalent to the sum of the volumes of three cones whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and the mean proportional between the bases of the frustum. HYPOTHESIS. V is the volume of a frustum of a circular cone, B its lower base, B' its upper base, and H its altitude. CONCLUSION. V=H (B+B' + √B × B'). PROOF Let V' denote the volume of an inscribed frustum of a pyramid whose bases are regular polygons, b its lower base, and b' its upper base. The altitude of the frustum is H. § 548 § 674 frustum is § 455 § 272 .. √b × b' approaches VB x B' as a limit. § 271 · · · { H (b + b' + √b x b') approaches 3 } H (B + B' + VB X B') as a limit. §§ 268, 267, 272 Also, V' approaches V as a limit. ... V = ¦ H (B + B' + √B × B'). 744 COROLLARY. If R and R' are the radii of the a frustum of a circular cone, then B = = T πR2, B' = πR'2. ·· √B x B' = √TR2 X TR 12 = TRR'. .. V = } πH (R2 + R'2 + RR'). § 735 Proved § 264 Q. E. D. bases of § 464 PROBLEMS OF COMPUTATION 1243 Compute the lateral area of a cone of revolution of which the radius of the base is 3 inches and the altitude is 4 inches. 1244 Compute the total area of a right circular cone of which the radius of the base is 7 feet and the slant height is 23 feet. 1245 Find the volume of a cone of revolution, the radius of its base being 48 inches, and its slant height being 73 inches. 1246 Find the volume of a circular cone whose altitude is 10 feet, and the diameter of whose base is 10 feet. 1247 The radii of the bases of a frustum of a cone of revolution are 6 inches and 4 inches, and the slant height is 5 inches. Find the lateral area. 1248 The diameters of the bases of a frustum of a circular cone are 18 feet and 12 feet, and the altitude is 16 feet. Find the volume. 1249 The radii of the bases of a frustum of a right circular cone are 7 feet and 4 feet, and its altitude is 5 feet. Find the altitude of the cone from which the frustum is cut off. 1250 The altitudes of two similar cones of revolution are as 2:1. What is the ratio of their total areas? of their volumes ? SOLUTION. T: T′ = H2 : H/2 = (2)2: (1)2 = 4 : 1. V: V' = H3: H/3 = (2)3: (1)3 = 8:1. 1251 The volumes of two similar cones are 216 cu. ft. and 512 cu. ft. What is the height of the first, if the height of the second is 12 ft. ? 1252 The altitude of a cone is 36 inches. How far from the vertex must it be cut by a plane so that the frustum shall be equivalent to half the cone ? 1253 How many square inches of tin are required to make a funnel, if the slant height is 9 inches, and the diameters of its bases are 6 inches and 1 inch, allowing inch for seam? 1254 The volumes of two similar cones of revolution are as 27: 125. Compare their convex surfaces. 1255 Of two cylinders of revolution, the altitude of one is three times the altitude of the other. Compare their convex surfaces and their volumes. 1256 Find the edge of a cube equivalent to a right circular cylinder whose diameter is 6 ft. and whose altitude is 10 ft. BOOK VIII THE SPHERE DEFINITIONS 745 A sphere is a solid bounded by a surface all points of which are equidistant from a point within called the center of the sphere. A sphere may be generated by the revolution of a semicircle about its diameter as an axis. 746 The radius of a sphere is a straight line drawn from the center to any point of the surface. The diameter of a sphere is a straight line drawn through the center and terminated by the surface. 747 A line or a plane is tangent to a sphere when it has one point only in common with the surface of the sphere. Two spheres are tangent to each other when their surfaces have one point only in common. 748 COROLLARY. All radii of a sphere are equal. All diameters of a sphere are equal. Spheres having equal radii are equal, and conversely. A point is within a sphere, on its surface, or without a sphere, when its distance from the center is less than, equal to, or greater than, the radius. PROPOSITION I. THEOREM 749 Every section of a sphere made by a plane is a circle. B D A HYPOTHESIS. O is the center of a sphere, and ABC is any section made by a plane. CONCLUSION. The section ABC is a circle. PROOF Draw OD to the section, and join O to any two points in the perimeter of the section, as A and B. Then OA OB... DA § 748 That is, all points in the perimeter of the section ABC are equidistant from the point D. ... the section ABC is a circle. Q. E. D. 750 DEFINITION. A circle of a sphere is any section of the sphere made by a plane. 751 COROLLARY 1. The line joining the center of a sphere and the center of a circle of the sphere is perpendicular to the plane of the circle. 752 COROLLARY 2. Circles of a sphere equidistant from the center of the sphere are equal, and conversely. 753 COROLLARY 3. Of two circles unequally distant from the center of a sphere, the more remote is the smaller, and conversely. |