2005 AMC 10A Problems/Problem 4

Revision as of 20:14, 12 October 2016 by Bluespruce (talk | contribs) (Solution)


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

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


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

Using the Pythagorean Theorem:



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


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