# Difference between revisions of "2005 AMC 10A Problems/Problem 4"

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==Problem== | ==Problem== | ||

− | A rectangle with a diagonal of length <math>x</math> is twice as long as it is wide. What is the area of the rectangle? | + | A rectangle with a [[diagonal]] of length <math>x</math> is twice as long as it is wide. What is the area of the rectangle? |

<math> \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 </math> | <math> \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 </math> | ||

==Solution== | ==Solution== | ||

− | Let the width of the rectangle be <math>w</math>. | + | Let the width of the rectangle be <math>w</math>. Then the length is <math>2w</math> |

− | |||

− | Then the length is <math>2w</math> | ||

Using the [[Pythagorean Theorem]]: | Using the [[Pythagorean Theorem]]: | ||

− | <math> | + | <math>x^{2}=w^{2}+(2w)^{2}</math> |

<math>x^{2}=5w^{2}</math> | <math>x^{2}=5w^{2}</math> | ||

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<math>2w=\frac{2x}{\sqrt{5}}</math> | <math>2w=\frac{2x}{\sqrt{5}}</math> | ||

− | So the area of the rectangle is <math> w \cdot 2w = \frac{x}{\sqrt{5}} \cdot \frac{2x}{\sqrt{5}} = \frac{2}{5}x^{2} \ | + | So the [[area]] of the [[rectangle]] is <math> w \cdot 2w = \frac{x}{\sqrt{5}} \cdot \frac{2x}{\sqrt{5}} = \frac{2}{5}x^{2} \Longrightarrow \mathrm{(B)}</math> |

==See Also== | ==See Also== | ||

*[[2005 AMC 10A Problems]] | *[[2005 AMC 10A Problems]] | ||

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*[[2005 AMC 10A Problems/Problem 5|Next Problem]] | *[[2005 AMC 10A Problems/Problem 5|Next Problem]] | ||

+ | |||

+ | [[Category:Introductory Geometry Problems]] |

## Revision as of 10:37, 2 August 2006

## Problem

A rectangle with a diagonal of length is twice as long as it is wide. What is the area of the rectangle?

## Solution

Let the width of the rectangle be . Then the length is

Using the Pythagorean Theorem:

So the area of the rectangle is