 CD, &c., taken together, make up the perimeter of the prism's base : hence the sum of these rectangles, or the convex surface of the prism, is equal to the perimeter of its base multiplied by its altitude.  Elements of Geometry - Pagina 180door Adrien Marie Legendre - 1825 - 224 pagina’sVolledige weergave - Over dit boek
 | Adrien Marie Legendre - 1822 - 394 pagina’s
...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... | |
 | Adrien Marie Legendre - 1825 - 276 pagina’s
...(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... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 294 pagina’s
...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.... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 280 pagina’s
...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.... | |
 | Adrien Marie Legendre - 1828 - 346 pagina’s
...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... | |
 | Jared Sparks, Edward Everett, James Russell Lowell, Henry Cabot Lodge - 1828 - 598 pagina’s
...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... | |
 | Jared Sparks, Edward Everett, James Russell Lowell, Henry Cabot Lodge - 1828 - 598 pagina’s
...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... | |
 | Adrien Marie Legendre - 1836 - 394 pagina’s
...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.... | |
 | Adrien Marie Legendre - 1839 - 372 pagina’s
...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, is equal to (AB + BC + CD + DE + EA) x AF ; that is, to the perimeter of the base of the prism mult! plied by its... | |
 | Adrien Marie Legendre - 1841 - 288 pagina’s
...the solidity of the cylinder will be * W x If, or * IPH. LEMMA. 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. (fig. 252),... | |
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