Difference between revisions of "2012 AMC 12B Problems/Problem 2"

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==Solution==
 
==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=200</math>, <math>\boxed{\text{E}}</math>
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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>
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2012|ab=B|num-b=1|num-a=3}}
 
{{AMC12 box|year=2012|ab=B|num-b=1|num-a=3}}

Revision as of 22:25, 12 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?

$\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200$


Solution

If the radius is $5$, then the width is $10$, hence the length is $20$. $10\times20= \boxed{\textbf{(E)}\ 200}.$

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 12 Problems and Solutions
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