Fig.150. Fig.147. Fig.151. Fig.149. ing triangle DEF of the other ; and we say—Two sym- will have for its measure ; it. [(AM*) — (BN*)] × OD. Fig.152. This expression may take another form. From I, the middle of AB, draw IK perpendicular to OD, and through B draw BP parallel to OD; we shall have AM -- BN = 2 IK, and AM – BN = AP ; consequently (AM + BN) × (AM – BN) = (AM)” — (BN)*, (176), - 2 IK X AP. The volume of the body generated by ABO, may therefore be expressed by # or X IK X AP × OD. But if we draw OI, as OAB is an isosceles triangle, OI will be perpendicular to AB; and the two triangles APB, OID, are similar, and give . OI OD the proportion AP of AB hence AP × OD = OIX AB; and substituting OIX AB for its value in the above expression we have ; tı X IK X AB. × OI. But the similar triangles ABP, OIK, give the proportion A : - #. wherefore AB X IK = OIX BP = OI X MN; and substituting this value of AB × IK in the expression for the volume of the body in question, we have # or X OI2 × MN. The area of a circle whose radius is OI is expressed by it X (OI)”. We therefore say— The volume of a body generated by the revolution of an s isosceles triangle about a straight line which meets it only in its summit, and is in the same plane, has for its measure fu'o thirds of the area of a circle whose radius is the height of the triangle, multiplied by the part of the avis intercepted by perpendiculars drawn to it from the vertices at the base of the triangle. 302. [If the axis of revolution is parallel to the base of the triangle, MN will equal AB, the surface generated by AB will be a cylindrical surface, and the volume of the body generated by the triangle will be two thirds of the volume of a cylinder the radius of whose base is OI, the height of the triangle, and whose height is AB, the base of the triangle. It follows, therefore, that the volume of a body generated by the revolution of the same triangle about its base as an axis, will be equivalent to one third of the cylinder of the above dimensions.] 303. Returning to figure 149, it will be readily seen that the portion of the polygon a b c d, inscribed in the quadrant ad is composed of isosceles triangles whose common height is the radius of the inscribed circle. The volume of the body generated by a Ob, will have for its 10% Fig.149. Fig.148. measure # It X (Oh)" × a e ; that generated by b Oc remain the same, the limit for the product is ; it x (O a)2 × a p. height is equal to the diameter of the sphere (fig. 148), and whose common base is equal to a great circle of the Fig.148. sphere; we may express these convex surfaces as sollows: that of the cone (5): 7t. R2 ; sphere 4 it. R2 ; cylinder = 4 at . R2. The volumes of these bodies have the following expressions: Remark. The sphere is therefore two thirds of the circumscribed cylinder. The cone is one third of the circumscribed cylinder. The area of the sphere is equivalent to that of the convex surface of the circumscribed cylinder. 305. The bodies of revolution which have now been discussed, are said to be similar when the figures which generate them are similar. Two cones generated by similar triangles revolving about homologous sides, have their heights in proportion to the radii of their bases. Let A represent the area of a come whose height is H and the radius of whose base is R, and whose side is S; we shall have the equation d = 2 it. B. × 4 S, (272). For another similar cone we shall have a = 2 it. r X , s. If we compare these expressions, cancelling the common . A may write - = - We therefore say— (s. 7.2 T s 2 T /;2 The areas of similar comes are to each other as the seeond powers of their corresponding dimensions. 306. Two similar cylinders are generated by similar rectangles (fig. 144). Their areas will have for their ex- Fig.144. pressions, A = 2 it. It X H, a = 2 it. r X. h. If we compare these neglecting the common factors in the terms of H R2 - 2 the second ratio, we shall have #= R × H_ _II*. r x sh T r2 T h” ” that is—The areas of similar cylinders are to each other as the second powers of their homologous measures. 307. All spheres are similar, because every sphere has for its generatrix a semicircle, and all semicircles are similar. The areas of two spheres have for their expressions, A = 4 it. R*, a = 4 it. r2 ; comparing these and cancelling the common factors in the second ratio, 2 ... . A R g will give a = z: ; that is—The areas of two spheres are to each other as the second powers of their radii. 308. If we compare the volumes of two similar cones, which are expressed as follows; V = } it. R* x H, and V R2 × H R 3 V = } t. ro X / ; we shall have — = sco × 11 io — * Q) r2 × h 7.3 = 7; as, r = 7 : that is—The volumes of similar cones are to each other as the third powers of their homologous lines. 309. The volumes of two similar cylinders have for their expressions, V' = r. R2 × H, v' = r. ro X. h. If volumes of similar cylinders are to each other as the third powers of their homologous measures. 310. Two spheres, being always similar bodies, and having for the expressions of their volumes, V" = # It. R*, and !!! - 3 or”, oy be readily compared; 3 : and will give v. ~ R* – D". That is—The volQ) 7° 3 als umes of two spheres are to each other as the third powers of their radii, or as the third powers of their diameters. 311. It was observed (227) that a regular polyedron has all its laces equal regular polygons, all its diedral angles equal, and all its polyedral angles equal. Let us inquire into the number of regular polyedrons which may be constructed. in the first place, it is manifest that the faces of these regular bodies must be equilateral triangles, squares or regular pentagons; for the angles of a regular hexagon are equal to four right angles; no polyedral angle therefore, can be formed with regular hexagons, or with regular polygons of a greater number of sides. Of those bodies contained by equilateral triangles, the |