Frustum of a square pyramid

Text

A frustum of a right pyramid has square bases of edge $10$ cm and $4$ cm and apothem (slant height of a lateral face) $5$ cm. Find the lateral surface, the total surface, the height and the volume.

Solution

The difference between the base apothems is $\dfrac{10-4}{2}=3$; the frustum height is $h=\sqrt{5^2-3^2}=4$ cm. Lateral surface: $S_\ell=\dfrac12(P_1+P_2)\,a=\dfrac12(40+16)\cdot5=140$ cm$^2$. Total surface: $S_t=140+100+16=256$ cm$^2$. Volume: $V=\dfrac h3\left(\mathcal A_1+\mathcal A_2+\sqrt{\mathcal A_1\mathcal A_2}\right)=\dfrac43(100+16+40)=\dfrac43\cdot156=208$ cm$^3$. $\boxed{S_\ell=140;\ S_t=256\ \text{cm}^2;\ h=4\ \text{cm};\ V=208\ \text{cm}^3}$