 | 1801 - 658 sider
...base 8 feet 6 inches. Ans, PROBLEM VI. fo find tin convex surface of the frustum »fa right com. 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. EXAMPLES. I. If the circumferences... | |
 | Thomas Keith - 1817 - 306 sider
...41), and this product by the height, for the solidity. Nate. To find the superficies, multiply half the sum of the perimeters of the two ends by the slant height, and the produce will be the suiface of the sides ; to which add the areas of the cuds, and the sum will... | |
 | Anthony Nesbit - 1824 - 476 sider
...will be the solidity. Nett. The surface of the frustum of a pyramid may be found thus : 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 which add the areas of the ends, and the sum will... | |
 | John Nicholson (civil engineer.) - 1825 - 1008 sider
...125664 15708 2 ) 282744 141-372 Ansr. Pnl. 6. To find the Convex Surfaceof the Frnstum of a Right Cone. 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. Ex. If the circumferences of the... | |
 | John Bonnycastle - 1829 - 256 sider
...surface? Ans. 98.09375. PROBLEM VII. To find the convex surface of the frustum of a right cone, ' RULE.* 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. 2. What is the convex surface of... | |
 | Thomas Curtis - 1829 - 810 sider
...XXV.* To find the convex surface of the frustum of a right cone or pyramid. Rule. — Multiply half the sum of the perimeters of the two ends by the slant height of the frustum, and the product will be the area. Example. — The circumferences of the ends of the... | |
 | William Templeton (engineer.) - 1833 - 224 sider
...28=2932.]6cubifiincheS) and MENSURATION PROBLEM III. To find the Surface of the Frustum 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.... | |
 | William Galbraith - 1834 - 454 sider
...area of the base by one-third the perpendicular height. 3. Frustum of a. Pyramid. (1.) Multiply half the sum of the perimeters of the two ends by the slant height. To this add the areas of the two ends, the sum will be the whole surface. (2.) Capacity. Add a diameter... | |
 | James Gallier - 1836 - 228 sider
...'will be the surface. To find the Convex Surface of a Frustum 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 Haynes Haswell - 1844 - 298 sider
...To find the Convex Surface of a Frustrum of a Right Cone or Pyramid — figs. 32 and 34. RULE. — Multiply the sum of the perimeters of the two ends by the slant height or side, and half the product will be the surface. OF SPHERES. To find the Convex Surface of a Sphere... | |
| |
Pyramid, RULE. — Multiply the sum of the perimeters of the two ends by the slant height, and... 
