Solid Geometry Advanced – Problem 14: Find the surface area of this sphere

A lampshade is a right circular frustum (a truncated cone). The radii of the top and bottom circular rims are $5\,\text{cm}$ and $12\,\text{cm}$, and the vertical height is $17\,\text{cm}$.

Assume the frustum has a circumscribed sphere (a sphere passing through both circular rims). Find the surface area of this sphere.

(1) Consider the axial cross-section. It is an isosceles trapezoid with vertices $$(12,0),\ (-12,0),\ (5,17),\ (-5,17)$$ in the $(x,z)$-plane, where $z$ is the axis direction.

(2) By symmetry, the circumcenter of this trapezoid lies on the $z$-axis, say at $(0,k)$. Equate distances to two vertices: $$12^2+k^2=5^2+(17-k)^2.$$ Compute: $$144+k^2=25+289-34k+k^2\Rightarrow 144=314-34k\Rightarrow k=5.$$ (3) The circumradius is $$R^2=12^2+5^2=169\Rightarrow R=13\,\text{cm}.$$ (4) Rotating the circumcircle around the axis gives the circumscribed sphere. Its surface area is $$S=4\pi R^2=4\pi\cdot 169=676\pi\,\text{cm}^2.$$

The circumscribed sphere has radius $13\,\text{cm}$, so its surface area is $676\pi\,\text{cm}^2$.