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

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== Problem==
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== 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);  
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draw(circle((10,5),5));</asy>
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<asy>
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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==
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== 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>
 
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>
  

Latest revision as of 01:07, 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]

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

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
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

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