Difference between revisions of "2001 AMC 10 Problems/Problem 21"

Revision as of 17:29, 26 August 2011

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

$[asy] draw((5,0)--(-5,0)--(0,12)--cycle); unitsize(.75cm); draw((-30/11,0)--(-30/11,60/11)); draw((-30/11,60/11)--(30/11,60/11)); draw((30/11,60/11)--(30/11,0)); draw((0,0)--(0,12)); label("$2r$",(0,30/11),E); label("$12-2r$",(0,80/11),E); label("$2r$",(0,60/11),S); label("$10$",(0,0),S); label("$A$",(0,12),N); label("$B$",(-5,0),SW); label("$C$",(5,0),SE); label("$D$",(-30/11,60/11),W); label("$E$",(30/11,60/11),E); [/asy]$

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)
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Difference between revisions of "2001 AMC 10 Problems/Problem 21"

Revision as of 17:29, 26 August 2011

Problem

Solution

See Also

@@ Line 18: / Line 18: @@
 label("$2r$",(0,60/11),S);
 label("$10$",(0,0),S);
+label("$A$",(0,12),N);
+label("$B$",(-5,0),SW);
+label("$C$",(5,0),SE);
+label("$D$",(-30/11,60/11),W);
+label("$E$",(30/11,60/11),E);
 </asy>