2001 AMC 10 Problems/Problem 21

Problem

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 the diameter of the cylinder be $2r$ . Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, $\frac{12-2r}{12}=\frac{2r}{10}$ which we solve to find $r=\frac{30}{11}$ . Our answer is $\boxed{\textbf{(B)}\ \frac{30}{11}}$ .

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search

2001 AMC 10 Problems/Problem 21

Problem

Solution

See Also