| 1801 - 426 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, William Hawney - 1817
...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 - 434 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
...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 - 252 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
...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
...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 - 428 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 - 208 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 - 264 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... | |
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