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2005 AMC 10A Problems/Problem 4: Difference between revisions

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The [[area]] of the [[rectangle]] is <math>2w^2=\frac{2}{5}x^2</math>
The [[area]] of the [[rectangle]] is <math>2w^2=\frac{2}{5}x^2</math>


<math>B)</math>
<math>(B)</math>


==See Also==
==See Also==

Revision as of 19:14, 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 $2w^2=\frac{2}{5}x^2$

$(B)$

See Also

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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