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> | ||
| − | + | == Solution == | |
| − | + | If the radius is <math>5</math>, then the width is <math>10</math>, hence the length is <math>20</math>. <math>10\times20= \boxed{\textbf{(E)}\ 200}.</math> | |
| − | ==Solution== | ||
| − | If the radius is <math>5</math>, then the width is <math>10</math>, hence the length is <math>20</math>. <math>10\times20= | ||
== See Also == | == See Also == | ||
| − | {{AMC12 box|year=2012|ab=B|num-b= | + | {{AMC12 box|year=2012|ab=B|num-b=1|num-a=3}} |
| + | {{MAA Notice}} | ||
Latest revision as of 00:07, 19 October 2020
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/1/6/2/1627a134d8cbe234c9b8224ec84ebc3d479d85a8.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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
