Volume of a Frustum of a PyramidDate: 06/03/2002 at 13:36:25 From: Zizza Subject: the frustum of an pyramid Hi Dr. Math. I don't know how to prove that the formula V = h/3 * (B1 + sqrt(B1*B2) + B2) is correct for a frustum of a pyramid. Date: 06/03/2002 at 22:40:14 From: Doctor Peterson Subject: Re: the frustum of an pyramid Hi, Zizza. Here is one way to derive the formula: If we reconstruct the entire pyramid, the top part, with base area B2, is similar to the whole pyramid, with base area B1. Their heights must therefore be in the ratio sqrt(B1):sqrt(B2). The height of the whole pyramid is therefore sqrt(B1)/(sqrt(B1)-sqrt(B2)) h, and its volume is Vwhole = 1/3 B1 * sqrt(B1)/(sqrt(B1)-sqrt(B2)) h while the volume of the removed top part is Vremoved = 1/3 B2 * sqrt(B2)/(sqrt(B1)-sqrt(B2)) h Subtracting, we get Vfrustum = Vwhole - Vremoved = h/3 * [B1*sqrt(B1) - B2*sqrt(B2)]/[sqrt(B1)-sqrt(B2)] = h/3 * [sqrt(B1)^3 - sqrt(B2)^3]/[sqrt(B1)-sqrt(B2)] But the difference of cubes can be factored: a^3 - b^3 = (a - b)(a^2 + ab + b^2) so we get Vfrustum = h/3 * (B1 + sqrt(B1*B2) + B2) which is just what we wanted. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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