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?

1. In an axial cross-section we get an isosceles trapezoid with parallel sides $50$ and $8$ and an inscribed circle of radius $\rho$; the height is $2\rho$.
2. The tangent lengths from the vertices to the circle give each leg the length $25+4=29$.
3. A leg is also the hypotenuse of a right triangle with legs $25-4=21$ and $2\rho$: $21^2+4\rho^2=29^2$, so $4\rho^2=400$.
4. $\rho=\boxed{10}$.