Solid Geometry Advanced – Problem 15: Find the surface area of that sphere

A regular square frustum has bottom base side length $5$, top base side length $4$, and height $5$. Assume all 8 vertices lie on a common sphere.

Find the surface area of that sphere.

(1) Place coordinates so the frustum axis is the $z$-axis. Put the bottom square in plane $z=0$ centered at the origin with half-side $\frac52$, and the top square in plane $z=5$ centered at the origin with half-side $2$.

Bottom vertex can be taken as $B\left(\frac52,\frac52,0\right)$. Top vertex can be taken as $T(2,2,5)$.

(2) By symmetry, the sphere center is $(0,0,z_0)$. Equate squared distances: $$\left(\frac52\right)^2+\left(\frac52\right)^2+z_0^2=2^2+2^2+(5-z_0)^2.$$ Compute: $$\frac{25}{4}+\frac{25}{4}+z_0^2=8+25-10z_0+z_0^2$$ $$\frac{25}{2}=33-10z_0\Rightarrow 10z_0=\frac{41}{2}\Rightarrow z_0=\frac{41}{20}.$$ (3) Then $$R^2=\frac{25}{2}+\left(\frac{41}{20}\right)^2=\frac{25}{2}+\frac{1681}{400}=\frac{6681}{400}.$$ So sphere surface area is $$S=4\pi R^2=4\pi\cdot\frac{6681}{400}=\frac{6681\pi}{100}.$$

The sphere surface area is $\dfrac{6681\pi}{100}$.