A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder. $\textbf{(A)}\ \frac{8}3\qquad$ $\textbf{(B)}\ \frac{30}{11}\qquad$ $\textbf{(C)}\ 3\qquad$ $\textbf{(D)}\ \frac{25}{8}\qquad$ $\textbf{(E)}\ \frac{7}{2}$
Solution
Let $r$ be the cylinder's radius, so its height is $2r$. The cone has radius $5$ and height $12$.
At height $2r$ from the base, the cone's radius is $5\left(1-\frac{2r}{12}\right)$, which must equal $r$.
Solving $r=5-\frac{5r}6$ gives $r=\frac{30}{11}$. The answer is (\textbf{B}).
A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $12$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
$\textbf{(A)}\ \frac{8}3\qquad$ $\textbf{(B)}\ \frac{30}{11}\qquad$ $\textbf{(C)}\ 3\qquad$ $\textbf{(D)}\ \frac{25}{8}\qquad$ $\textbf{(E)}\ \frac{7}{2}$ 
