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Difference between revisions of "2012 AMC 10B Problems/Problem 2"

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Since the sides of the rectangle are in a <math>2:1</math> ratio, the longer side has length <math>2\cdot 10 = 20</math>.
 
Since the sides of the rectangle are in a <math>2:1</math> ratio, the longer side has length <math>2\cdot 10 = 20</math>.
 
Therefore the area is <math>20\cdot 10 = 200</math> or <math>\boxed{\textbf{(E)}\ 200}</math>
 
Therefore the area is <math>20\cdot 10 = 200</math> or <math>\boxed{\textbf{(E)}\ 200}</math>
 +
{{MAA Notice}}

Revision as of 12:14, 4 July 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

Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is $5$, the diameter is $2\cdot 5 = 10$. Since the sides of the rectangle are in a $2:1$ ratio, the longer side has length $2\cdot 10 = 20$. Therefore the area is $20\cdot 10 = 200$ or $\boxed{\textbf{(E)}\ 200}$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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