surface of the other, and be equal to it. But the convex surface of the prism is equal to its perimeter multiplied by its altitude AF; (5.7;) therefore the convex surface of the cylinder is equal to the circumference of its base multiplied by its altitude; that is, equal to circ. RCX H. Hence, The convex surface of a cylinder is equal, &c. The solidity of a cylinder is equal to the area of its base, multiplied by its altitude. Let H be the altitude of the given cylinder, RC the radius of its base, and the area of its base be represented by area RC. It is to be shown that the solidity of the cylinder is equal to area RCXH. Inscribe in the cylinder a right prism of an indefinite number of sides, and of equal altitude with the cylinder, as in the last proposition. Now since the number of sides of the prism's base is indefinitely increased, it will coincide with the base of the cylinder and be equal to it. (9. 5. Cor. 2.) And since they are both of the same altitude, the convex surface of the one will coincide F M R B K H with the convex surface of the other and be equal to it; hence the solidity of the prism must be equal to that of the cylinder. (Ax. 13.) But the solidity of a prism of any number of sides, is equal to the product of its base by its altitude; (12.7;) therefore the solidity of the cylinder, which is of equal base and altitude with the prism, must be equal to the product of its base by its altitude; that is, equal to area RCXH. Hence, The solidity of a cylinder is equal, &c. Cor. 1. Cylinders of the same altitude are to each other as their bases; and cylinders of the same base are to each other as their altitudes. Cor. 2. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of their bases. For the bases are as the squares of their diameters; and the cylinders being similar, the diameters of their bases are to each other as the altitudes; (Def. 4. 8;) hence the bases are as the squares of the altitudes; therefore the bases, multiplied by the altitudes, or the cylinders themselves, are as the cubes of their altitudes, or of their diameters. Scholium. Let R be the radius of a cylinder's base; H the altitude: the area of the base will be я.R2; (11. 5. Cor. 2.;) and the solidity of the cylinder will be The convex surface of a cone is equal to the circumference of its base, multiplied by half its side. Let RA be the radius of the base of the given cone, S its vertex, and SD its side; then will its convex surface be equal to circ. RAXISD. Inscribe in the cone a regular pyramid of the same altitude with the cone, and having S for its vertex; (Def. 7. 8;) from H, the middle point of the side BC, draw SH perpendicular to BC; SH will be the slant-height of the pyramid. (Def. 16. 7.) Now, if the number of sides of the regular polygon ABCDEF, which forms the base of the pyramid, be indefinitely increased, the perimeter of the pyramid will coincide with the circumference of the cone; hence A SH, the slant-height of the pyramid will terminate in the circumference S R H of the cone's base, and consequently be equal to SD the side of the cone; and since the pyramid and cone are of the same altitude and their bases coincide, their convex surfaces must coincide and be equal to each other. But the convex surfaces of the pyramid is equal to the perimeter of its base multiplied by half of SH, its slant-height; (15. 7;) therefore the convex surface of the cone must be equal to the circumference of its base multiplied by half its side; that is, equal to circ. RAX SD. Hence, The convex surface of a cone is equal, &c. Scholium. Let L be the side of a cone, R the radius of its base; the circumference of this base will be 2.R, and the surface of the cone will be 2 R ×L, or πRL. The solidity of a cone is equal to the area of its base multiplied by a third of its altitude. Let SR be the altitude of the given cone, RA the radius of its base, and area RA, represent the surface of its base; then will the solidity of the cone be equal to area RA× SR. Inscribe in the cone a regular pyramid of an indefinite number of sides, as in the last proposition, having its vertex S, the same as the vertex of the cone. B R Now since the number of sides of the pyramid's base is increased indefinitely, its base will coincide with the base of the cone and be equal to it; and since both have the same vertex, the convex surface of the one must coincide with that of the other and be equal to it; consequently the solidity of the one must be equal to the solidity of the other. (Ax. 13.) But the solidity of the pyramid is equal to the area of its base into a third of its altitude; (18.7;) hence the solidity of the cone must be equal to the area of its base multiplied by a third of its altitude; that is, equal to Hence, The solidity of a cone is equal, &c. Cor. A cone is a third of a cylinder having the same base and the same altitude; whence it follows, 1. That cones of equal altitudes are to each other as their bases; 2. That cones of equal bases are to each other as their altitudes; 3. That similar cones are as the cubes of the diameters of their bases, or as the cubes of their altitudes. Scholium. Let R be the radius of a cone's base, H its altitude; the solidity of the cone will be «R3×H, or RaH. PROPOSITION V. THEOREM. The convex surface of the frustum of a cone AGBCED, is equal to its side AD multiplied by half the sum of the circumferences of its two bases; that is, AD × (circ. OB+circ. HC). D E H For, if we inscribe in the truncated cone a frustum of a regular pyramid, having the same altitude, and the number of its lateral faces be indefinitely increased, the bases of the frustum of M the pyramid will ultimately coincide with the bases of the truncated cone, and its slant-height will be equal to the side of the truncated cone; consequently the lateral surface of the that of the other and be equal to it. of the frustum of a pyramid is equal to the product of its slantheight into half the sum of the perimeters of its bases; (17. 7;) therefore the convex surface of a truncated cone must be equal to G one will coincide with But the convex surface ADX(circ. OB+circ. HC). Hence, The convex surface of a frustum of a cone is equal to its side multiplied by half the sum of the circumferences of its two bases. Cor. The convex surface of a truncated cone, is equal to the product of its side into the circumference of a section at equal distances from the two bases. For, bisect AD in M, and draw ML parallel to AB: the lateral surface of the lower part of the frustum is equal to AM × (AB+ML), that of the upper part is MD or AMX (DC+ML); hence the 'ateral surface of the whole frustum is equal to AM×(AB -DC+ML); but AM× (AB+DC) equals half the late |