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

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

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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