 | Adrien Marie Legendre - Geometry - 1819 - 574 pages
...by the half of its side should be the measure of the surface of a less cone. We conclude then that the convex surface of a cone is equal to the circumference of the base multiplied by half of its side. 529. Scholium. Let L be the side of a cone, and R the radius... | |
 | Adrien Marie Legendre - Geometry - 1822 - 394 pages
...multiplied by half the side cannot be the measure of the surface of a smaller cone. Hence finally, the convex surface of a cone is equal to the circumference of its base multiplied by half its side. Scholium. Let L be the side of a cone, B, the radius of its base ; the... | |
 | Adrien Marie Legendre - Geometry - 1825 - 276 pages
...the frustum of a cone has for its measure iOPx (7iX-dO + 7ixDP+nxAOxDP) (422), or \n : THEOREM. 528. The convex surface of a cone is equal to the circumference of its base multiplied by half its side. Kg. 259. Demonstration. Let AO (Jig. 259), be the radius of .the base... | |
 | Adrien Marie Legendre, John Farrar - Geometry - 1825 - 280 pages
...measure i OP x (nx AO + nx DP + nx AO x DP) (422), or i 7i x OP x (AO*+ PD + AO x DP). THEOREM. . 528. The convex surface of a cone is equal to the circumference of its base multiplied by half its side. Fig. 259. Demonstration. Let AO (fig. 259), be the radius of the base... | |
 | Adrien Marie Legendre - 1825 - 570 pages
...cone has for its measure i OP x (re x AO + 71 x DP + re x AO x DP) (422), or THEOREM. 528. TVie conrex surface of a cone is equal to the circumference of its base multiplied by half its side. . 259. Demonstration. Let AO (fig. 259), be the radius of the base of... | |
 | Adrien Marie Legendre, John Farrar - Geometry - 1825 - 294 pages
...by the half of its side should be the measure of the surface of a less cone. We conclude then, that the convex surface of a cone is equal to the circumference of the base multiplied by half of its side. 529. Scholium. Let L be the side of a cone, and R the radius... | |
 | Adrien Marie Legendre - Geometry - 1836 - 394 pages
...surface of the pyramid is equal to the perimtter of the base multiplied by half the slant height : hence the convex surface of a cone is equal to the circumference of the base multiplied by half the side. Scholium. Let L be the side of a cone, R the radius of its base... | |
 | Charles Davies - Geometrical drawing - 1840 - 264 pages
...S—CDB, the cone S—FKH be taken away, the remaining part is called the frustum of a cone. M? 8. The convex surface of a cone is equal to the circumference of the base multiplied by half the slant height. Thus, the convex surface of the cone C — AED is equal... | |
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The convex surface of a cone is equal to the circumference of the base multiplied by half the slant height. 
