| 1801 - 658 pages
...its 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... | |
| Charles Vyse - 1806 - 342 pages
...Frustum of a Pyramid •r right Cone. RULE. Multiply the Sum of the Perimeters or Circumferences of the Ends by the slant Height, and half the Product will be the Surface required. EXAMPLES. (13) How many square Feet are in the Surface of a Frustum of a Square Pyramid,... | |
| Samuel Webber - 1808 - 520 pages
...base 8 feet 6 inches. Ans. 66 7' 59. PROBLEM VI. Tofnd the convex surface of the f nation of a right cone. 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. NOTE, The same rule may be used to... | |
| Charles Vyse - 1815 - 340 pages
...frustum of a pyramid or right cone. , RULE. Multiply the sum of the perimeters or circumferences of the ends by the slant height, and half the product will be the surface required. EXAMPLES. (13) How many square feet are in the surface of a frustum of a square pyramid,... | |
| Thomas Keith - 1817 - 306 pages
...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 pages
...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 pages
...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 pages
...the convex 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 pages
...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... | |
| John Bonnycastle - 1833 - 310 pages
...find the surface of the frustrum of a right pyramid. Rule. Multiply the sum of the perimeters of the ends by the slant height, and half the product will be the surface required. 1. What is the convex surface of the frustrum of a right cone, the circumference... | |
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