Solid geometry

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1902 edition. Excerpt: ...be indefinitely increased. The proof is exactly the same as that of Prop. XXXII, if we substitute cone for cylinder and pyramids for prisms. 721. Cor. The volume of a frustum of a cone is the limit of the volumes of the frustums of the inscribed and circumscribed pyramids, if the number of lateral faces is indefinitely increased, and the lateral area of the frustum of a cone is the limit of the lateral areas of the frustums of the inscribed and circumscribed pyramids. Proposition XXXIX. Theorem. 722. The lateral area of a cone of revolution is equal to half the product of the slant height by the circumference of the base. Let S denote the lateral area, C the circumference of the base, and L the slant height, of the given cone. To prove that S = iC X L. Proof. Circumscribe about the cone a regular pyramid. Denote the perimeter of its base by P, and its lateral area by 8. Then 8 = J P X L. §643 If the number of the lateral faces of the circumscribed pyramid is indefinitely increased, S' approaches S as a limit, and P approaches C as a limit..-. P X L approaches £ C X L as a limit. But 8 = iP X L, always..-.S = iCxL. Proposition XL. Theorem. 724. The volume of a circular cone is equal to one third the product of its base by its altitude. Let V denote the volume, B the base, and H the altitude of the given cone. To prove that V = i-B X H. Proof. Inscribe in the cone a pyramid with a regular polygon for its base. Denote its volume by V, and its base by £'. Then V' = iB'xH. §652 If the number of the lateral faces of the inscribed pyramid is indefinitely increased, V approaches V as a limit, § 720 and B' approaches B as a limit. § 454.-. B' X H approaches i B X H as a limit. § 279 But V = iB' X H, always. § 652.-....