A truncated cone has horizontal bases with radii $18$ and $2$. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere? $\mathrm{(A)}\ 6 \qquad$ $\mathrm{(B)}\ 4\sqrt{5} \qquad$ $\mathrm{(C)}\ 9 \qquad$ $\mathrm{(D)}\ 10 \qquad$ $\mathrm{(E)}\ 6\sqrt{3}$
Solution
In the axial cross-section, the sphere's great circle is inscribed in an isosceles trapezoid with parallel sides $2R$ and $2r$. By equal tangent lengths, each slanted side is $R+r$, so the height satisfies $h^2=(R+r)^2-(R-r)^2=4Rr$. The sphere's diameter is the height, so its radius is $\frac h2=\sqrt{Rr}$.
With $R=18,r=2$: sphere radius $=\sqrt{36}=6$. The answer is (\textbf{A}).
A truncated cone has horizontal bases with radii $25$ and $4$. A sphere is tangent to the top, the bottom, and the lateral surface of the truncated cone. What is the radius of the sphere?
$\mathrm{(A)}\ 6 \qquad$ $\mathrm{(B)}\ 4\sqrt{5} \qquad$ $\mathrm{(C)}\ 9 \qquad$ $\mathrm{(D)}\ 10 \qquad$ $\mathrm{(E)}\ 6\sqrt{3}$ 