Difference between revisions of "2012 AMC 12B Problems/Problem 2"
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== Problem== | == Problem== | ||
A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle? | A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle? | ||
| − | + | <asy>draw((0,0)--(0,10)--(20,10)--(20,0)--cycle); | |
| + | draw(circle((10,5),5));</asy> | ||
<math>\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200</math> | <math>\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200</math> | ||
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==Solution== | ==Solution== | ||
Revision as of 23:09, 25 January 2013
Problem
A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?
![[asy]draw((0,0)--(0,10)--(20,10)--(20,0)--cycle); draw(circle((10,5),5));[/asy]](http://latex.artofproblemsolving.com/2/6/a/26acbad4cb3f70e1ecbadac04ad375e66a2a06c5.png)
Solution
If the radius is 
, then the width is 
, hence the length is 
. 
See Also
| 2012 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 12 Problems and Solutions | |
