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

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==Solution==
 
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Let the width of the rectangle be <math>w</math>.  Then the length is <math>2w</math>
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Let the width of the rectangle be <math>w</math>.  Then the length is <math>2w</math>.
  
 
Using the [[Pythagorean Theorem]]:
 
Using the [[Pythagorean Theorem]]:

Revision as of 10:25, 12 August 2018

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

$(B)$

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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