...and the solidity of the cylinder will be rR*xH, or »/?*//. MMMA. 520. The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude. Demonstration. This surface is equal to the sum of the rectangles AFGB, BGHC, CHID, &c. (j%. 25S),...

...prism ; their bases AB, BC, CD, &c. taken together, make up the perimeter of the prism's base. Hence the sum of these rectangles, or the convex surface...perimeter of its base, multiplied by its altitude. Cor. If two right prisms have the same altitude, their convex surfaces will be to each other as the...

...prism is equal to the perimeter of its base multiplied by its altitude. Now the altitudes ^F, .BG, C7/, &c., of these rectangles are each equal to the altitude...The convex surface of a cylinder is greater than the conves surface of any inscribed prism, and less than the convex surface of any circumscribed prism....

...(fig. 252), which compose it. Fig. 2S2. Now the altitudes AF, BG, CH, &c., of these rectangles ate each equal to the altitude of the prism. Therefore...surfaces of these prisms will be to each other as th« perimeters of the bases. LEMMA. 522. The convex surface of a cylinder is greater than the convex...

...and the solidity of the cylinder will be n R' x H, or nR'H. LEMMA. 520. The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude. Now the altitudes AF, BG, CH, &c., of these rectangles are each equal to the altitude of the prism....

...solidity of the cylinder will be n R2 x H, or nR*H. LEMMA. •* 520. The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude. Now the altitudes ^F, BG, CH, &c., of these rectangles arc each equal to the altitude of the prism....

...prism ; their bases AB, BC, CD, &c. taken together, make up the perimeter of the prism's base. Hence the sum of these rectangles, or the convex surface...perimeter of its base, multiplied by its altitude. 521. Cor. If two right prisms have the same altitude, their convex surfaces will be to each other as the...

...its altitude. Now it has already been demonstrated, lemma 520, that ' the convex surface of a right prism, is equal to the perimeter of its base multiplied by its altitude.' Admitting, then, our principle, the convex surface of a cylinder will consist of an infinite number...

...its altitude. Now it has already been demonstrated, lemma 520, that ' the convex surface of a right prism, is equal to the perimeter of its base multiplied by its altitude.' Admitting, then, our principle, the convex surface of a cylinder will consist of an infinite number...

...Now, the altitudes AF, BG, CH, &c. of the rectangles, are equal to the altitude of the prism. Hence, the sum of these rectangles, or the convex surface of the prism, isequa!to BC AF ; that is, to the perimeter of the base of the prism multiplied by its altitude. Cor....