 | J. M. Scribner - 1849 - 264 pages
...the Igth. of the base. 90~x 22x8-6=2660 cubic ft. PROBLEM Via To find the Solidity of a Rectangular Prismoid. Rule. — To the sum of the areas of the two ends, abc, d ef, add four tunes the area of a section, gh, parallel to and equally distant from the parallel... | |
 | Oliver Byrne - 1851 - 310 pages
...inches. And 7140 -r- 1728 = 4-1319 solid feet, the content. To find the solidity of a prismoid. — To the sum of the areas of the two ends add four times the area of a section parallel to and equally distant from both ends, and this last sum multiplied by ^ of the height will give the solidity.... | |
 | Thomas Tate - 1855 - 286 pages
...ABPE and DCHF, rectangles. A frustum is a cone or pyramid having its top cut off. GENERAL RTJLE. To the sum of the areas of the two ends, add four times the area of the middle or mean section parallel to the ends, multiply this sum by the height, and one-sixth the... | |
 | Charles Haslett - 1856 - 530 pages
...47 '75 gallons nearly, as before. PROBLEM VI. To find the solid content of the frustum of a pyramid. RULE. To the sum of the areas of the two ends add the square root of their product ; multiply this sum by the perpendicular height, and \ of the product... | |
 | Anthony Nesbit - 1859 - 494 pages
...of a square pyramid, or a cylindroid ; to find its content in imperial gallons and bushels. KULE. To the sum of the areas of the two ends, add four times the area of the middle section parallel to them ; multiply this sum by the perpendicular depth, and \ of the product... | |
 | Charles Haslett - 1859 - 572 pages
...47-75 gallons nearly, as before. B PROBLEM VI. To find the solid content of the frustum of a pyramid. RULE. To the sum of the areas of the two ends add the square root of their product ; multiply this sum by the perpendicular height, and £ of the product... | |
 | Alfred Newsom Niblett - 1861 - 204 pages
...pflrnllelnfrrams. being parallel but not similar to eacli other; and its sides four plane trapezoids. RULES. To the sum of the areas of the two ends add four times the area of a section parallel to and equally distant from bo.h ends; multiply this sum by the perpendicular height, and i of the product... | |
 | Oliver Byrne - 1863 - 600 pages
...inches. And 7140 -5- 1728 = 4-1319 solid feet, the content. To find the solidity of a prismoid. — To the sum of the areas of the two ends add four times the area of a section parallel to and equally distant from both ends, and this last sum multiplied by J of the height will give the solidity.... | |
 | Ellwood Morris - 1872 - 206 pages
...means of initial prismoids, which, by a little development, can be made quite useful. Hutton's General Rule. " To the sum of the areas of the two ends add four times the area of a section parallel to, and equally distant from, both ends, multiply the last sum by the hight, and J of the product will... | |
 | Ellwood Morris - 1872 - 206 pages
...means of initial prismoids, which, by a little development, can be made quite useful. Hutton's General Rule. " To the sum of the areas of the two ends add four times the area of a section parallel to, and equally distant from, both ends, multiply the last sum by the bight, and i of the product will... | |
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RULE.* To the sum of the areas of the two ends add four times the area of a section parallel to and equally distant from both ends, and this last sum multiplied by £ of the height will give the solidity. 
