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Difference between revisions of "2005 AMC 10A Problems/Problem 4"
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<math>x^{2}=5w^{2}</math>
 
<math>x^{2}=5w^{2}</math>
  
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<math>w=\frac{x}{\sqrt{5}}</math>
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The [[area]] of the [[rectangle]] is <math>2w^2=\frac{5}{2}x^2</math>
  
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<math>2w=\frac{2x}{\sqrt{5}}</math>
 
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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]]

Revision as of 19:13, 12 October 2016

Problem

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

$\mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2$

Solution

Let the width of the rectangle be $w$. Then the length is $2w$

Using the Pythagorean Theorem:

$x^{2}=w^{2}+(2w)^{2}$

$x^{2}=5w^{2}$

The area of the rectangle is $2w^2=\frac{5}{2}x^2$

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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