Difference between revisions of "2001 AMC 10 Problems/Problem 21"
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==Solution== | ==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); | ||
| + | </asy> | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | |||
Let the diameter of the cylinder be <math> 2r </math>. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, <math> \frac{12-2r}{12}=\frac{2r}{10} </math> which we solve to find <math> r=\frac{30}{11} </math>. Our answer is <math> \boxed{\textbf{(B)}\ \frac{30}{11}} </math>. | Let the diameter of the cylinder be <math> 2r </math>. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, <math> \frac{12-2r}{12}=\frac{2r}{10} </math> which we solve to find <math> r=\frac{30}{11} </math>. Our answer is <math> \boxed{\textbf{(B)}\ \frac{30}{11}} </math>. | ||
Revision as of 17:10, 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 
and altitude 
, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
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); [/asy]](http://latex.artofproblemsolving.com/a/5/b/a5b814fc56d5c6fdb6bc02cd936c2db77dcf2135.png)
Let the diameter of the cylinder be 
. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, 
which we solve to find 
. Our answer is 
.
See Also
| 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 | ||
