PROPOSITION XXX. THEOREM 659. Every section of a circular cone made by a plane parallel to the base is a circle. Hyp. A'B'C'D' is a section of cone V-ABCD made by a plane || to ABCD, O the center of the base, O' the point where axis VO intersects the section. OA, OD, and O'A', O'D' are intersections of a plane through V and any element VA with Proof. O'A' and O'D' are || to OA and OD respectively. 660. COR. 1. The axis of a circular cone passes through the center of every section which is parallel to the base, or The locus of the centers of the sections of a circular cone made by planes parallel to the base is the axis of the cone. 661. COR. 2. Sections made by planes parallel to the bases of a circular cone are to each other as the squares of their radii, or as the squares of their distances from the vertex of the cone. PROPOSITION XXXI. THEOREM 662. The lateral area of a cone of revolution is equal to half the product of the slant height by the circumference of the base. Hyp. S is lateral area, C the circumference of the base, and L the slant height of the cone; S' the lateral area, P the perimeter of the regular polygon forming the base of a circumscribed pyramid. HINT. - Circumscribe a pyramid; its slant height is L. Use Theorem of Limits. 663. COR. If S is the lateral area, T the total area, H the altitude, L the slant height, R the radius of the base, of a cone of revolution, S = RL. T = R(L+R). Ex. 1130. Find the lateral area of a cone of revolution if the hypotenuse of the generating triangle be 10 inches and the acute angles be 45° each. PROPOSITION XXXII. THEOREM 664. The volume of a cone is equal to one-third the product of its base by its altitude. Hyp. V is the volume, B the base, and H the altitude of the cone; V' the volume, B' the regular polygon forming the base of an inscribed pyramid. To prove V = B x H. V' = B' × H. HINT. - Use Theorem of Limits. (594) 665. COR. If the cone is a cone of revolution, with Ras radius of B, then V=RH. Ex. 1131. Find the lateral area of a cone of revolution whose radius is 4 and whose altitude is 3. Ex. 1132. Find the volume of a cone of revolution with radius 6 and altitude 2. Ex. 1133. Find the volume of a cone of revolution whose radius is 5 and whose slant height is 13. Ex. 1134. Find the lateral surface of a cone of revolution if its volume is 314 and its altitude is 2. PROPOSITION XXXIII. THEOREM 666. The lateral areas, or the total areas, of two similar cones of revolution are to each other as the squares of their altitudes, as the squares of their radii, or as the squares of their slant heights; and their volumes are to each other as the cubes of their altitudes, as the cubes of their radii, or as the cubes of their slant heights. A A L R H LH Hyp. S, S' are the lateral areas, T, T' the total areas, V, V' the volumes, H, H' the altitudes, R, R' the radii, L, L,' the slant heights of two similar cones of revolution. To prove and S : S' = T: T' = H2 : H2=R2 : R2=L2 : L2 13 13 V: V' = H3 : H3 = R3 : R'3 = L3 : L'3 The proof is similar to that of Prop. XXVIII. Ex. 1135. Find the ratio of the total areas of two similar cones of revolution whose altitudes are (a) 12 inches and 24 inches respectively. (b) 5 inches and d inches respectively. Ex. 1136. The volume of a cone of revolution is 343 cubic inches and its altitude is 5 inches. Find volume of a similar cone of altitude (a) 7 inches. (b) 15 inches. PROPOSITION XXXIV. THEOREM 667. The lateral area of a frustum of a cone of revolution is equal to one-half the sum of the circumferences of its bases multiplied by its slant height. Hyp. S is the lateral area, C and C" the circumferences of the bases, R and R' their radii, and L the slant height of the frustum; S' the lateral area, P and P' the perimeters of the bases, of a frustum of a regular pyramid circumscribed about the cone. Proof. The slant height of the frustum of the circumscribed pyramid = L, (Why?) (Why?) [To be completed by the student. HINT. - Use Theorem of Limits.] NOTE. It can be shown that the limit of the sum of a finite number of variables is equal to the sum of their respective limits. 668. COR. 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. For C=2TR, and C2R', and 2. 2 is the circumference of a section equidistant from the bases. Y |