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

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+ | == Problem == | ||

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+ | 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? | ||

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+ | <math> \textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200 </math> | ||

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+ | ==Solution== | ||

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Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is <math>5</math>, the diameter is <math>2\cdot 5 = 10</math>. | Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is <math>5</math>, the diameter is <math>2\cdot 5 = 10</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>. | 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{E}</math>. | Therefore the area is <math>20\cdot 10 = 200</math> or <math>\boxed{E}</math>. |

## Revision as of 18:59, 25 February 2012

## 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?

## Solution

Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is , the diameter is . Since the sides of the rectangle are in a ratio, the longer side has length . Therefore the area is or .