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Thread: Frustum of Cone & Pyramid

  1. #1
    Jan 2010

    Frustum of Cone & Pyramid

    1.) The inside dimensions of the bases of a liquid container are 50m and 75m in diameter (frustum of a cone). If it's height is 35m. Find it's capacity in gallons.

    2.) If the water pail is in the form of a frustum of a regular pyramid with square bases whose sides are equal to 65 cm and 86 cm, and a slant height of 45 cm, find it's capacity in liters; in gallons.
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  2. #2
    MHF Contributor

    Apr 2005
    The formula for volume of a cone, with base radius r and height h, is $\displaystyle \pi r^2 h$. Further, since the truncated cone goes from radius 75 to 50 m (losing 1/3 its value) in 35 meters, it will go to 0 in an additional 2(35)= 70 m. Find the volume of two cones, one of radius 75 and height 35+ 70= 105, and the other of radius 50 and height 70, and subtract.

    The volume of a square pyramid, with has length s and height h, is $\displaystyle \frac{1}{3}s^2h$. The truncated pyramid goes from edge length 86 to 65, losing 86- 65= 21 cm (losing 21/86 of its value) in slant height 45 cm. If the slant height of the whole pyramid were h, we would have h/86= 45/21 or h= (86)(45)/21= (86)(15)/7= 1290/7= 184.24 cm. A pyramid with slant height 184.24 and base length 86 would have, by the Pythagorean theorem, height $\displaystyle \sqrt{(184.24)^2- 2(86)^2}$ = 138.5 cm. A pyramid with slant height 184.24- 45= 139.25 and base length 65 would have height $\displaystyle \sqrt{(139.25)^2- 2(65)^2}$= 104.6. Find the volumes of those two pyramids and subtract.
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