 | Scottish school-book assoc - 1845 - 444 sider
...base. The part cut off will be a pyramid or cone similar to the original one. RULE I. Multiply half the sum of the perimeters of the two ends by the slant height, and the product will be the surface of the sides; to this add the areas of the two ends, and the sum will... | |
 | Nathan Scholfield - 1845 - 894 sider
...PROBLEM V. To find the lateral surf ace of the frustum of a regular pyramid. RULE. — Multiply half the sum of the perimeters of the two ends by the slant height. For each side of a frustum of a regular pyramid is a trapezium, as AFDH, and the area of each of the... | |
 | John Bonnycastle - 1848 - 320 sider
...surface? Ans. 98.09375. f PROBLEM VII. To find the convex surface of thefrustrum of a right cone. RULE.* Multiply the sum of the perimeters of the two ends, by the slant height of the frustrum, and half the product witt be the surface required. * Demon. Let the perimeter of the... | |
 | Thomas Tate (mathematical master.) - 1848 - 284 sider
...product will be the solidity. Note. To find the convex surface of a frustum of a cone, multiply half the sum of the perimeters of the two ends by the slant height. The faces of a prismoid have, in general, the form of a trapezoid. EXAMPLES. the breadth AE or BP =... | |
 | D. M. Knapen - 1849 - 300 sider
...parts ? Ans. 1.80165. To find the surface of a frustum of a pyramid : — When the pyramid is regular, multiply the sum of the perimeters of the two ends by the lateral length, and to half the product add the areas of the two ends, and the sum will be the surface.... | |
 | John Radford Young - 1850 - 294 sider
...PROBLEM V. — To find the surface of the frustum of an upright regular pyramid, or circular cone. RULE. Multiply the sum of the perimeters of the two ends by the slant height, and half the product will be the slant surface, to which add the areas of the two ends, and the whole surface will be obtained.... | |
 | Oliver Byrne - 1851 - 310 sider
...feet, the convex 2 2 surface required. To find the convex surface of the frustum of a right cone. — Multiply the sum of the perimeters of the two ends, by the slant height of the frustum, and half the product will be the surface required. In the frustum ABDE, the circumferences... | |
 | Charles Haynes Haswell - 1851 - 346 sider
...the Convex Surface of a Fnistrum of a Right Cone or Pyramid — figs. 32 and 34. • . • RULE. — Multiply the sum of the perimeters of the two ends by the dant height or side, and half the product will be the surface. OF SPHERES. . . i To find the Convex... | |
 | Lucius D. Gould - 1853 - 234 sider
...will be the surface. To find the Convex Surface of a frustrum of -a Right Cone or Pyramid. Rule. — Multiply the sum of the perimeters of the two ends by the slant height or side of the frustum, and half the product will be the surface required. To find the Solidity of... | |
 | Charles W. Hackley - 1856 - 530 sider
...1728 = 1-697 cubic feet. PROBLEM lY. Tojind the surface of the fnt stum of a cone or pyramid. RULE. Multiply the sum of the perimeters of the two ends by the slant height, and half the product will be the slant surface ; to which add the areas of the two ends, and the product will be the whole surface.... | |
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Pyramid, RULE. — Multiply the sum of the perimeters of the two ends by the slant height, and... 
