Frustum of a pyramid with a square base

Statement

A frustum of a pyramid has square bases with sides $4$ and $2$ and the frustum’s height is $6$. Calculate its volume.

Solution

Method by difference. Similarity ratio $k = 2/4 = 1/2$. The heights of the complete pyramid and of the small one are in the ratio $h/h' = 1/k = 2$, with $h = h'+6$: from which $h' = 6$, $h = 12$. $V_{\text{grande}} = \tfrac{1}{3}\cdot 16\cdot 12 = 64, \qquad V_{\text{piccola}} = \tfrac{1}{3}\cdot 4\cdot 6 = 8.$ $V_{\text{tronco}} = 64 - 8 = \boxed{56}.$

Check with the frustum formula $V=\tfrac{h}{3}\bigl(A_1+A_2+\sqrt{A_1A_2}\bigr)$, with $A_1=16$, $A_2=4$: $V = \tfrac{6}{3}\bigl(16+4+\sqrt{64}\bigr) = 2\cdot 28 = 56. \quad ✓$

Links

Topics: Synthetic geometry in space
Concepts: Similarity of solids · Frustum of a pyramid · Volume
Skills: Calculating · Using formulae
Exercise types: Geometric problem