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

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==Solution==
 
==Solution==
If the radius is 5, then the width is 10, hence the length is 20. 10x20=200, E
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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=200</math>, <math>\boxed{\text{E}}</math>

Revision as of 15:02, 24 February 2012

Problem

A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to it's width is 2:1. What is the area of the rectangle


Solution

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