A [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle) with diagonal length $x$ is twice as long as it is wide. What is the area of the rectangle?
$(\mathrm {A}) \ \frac 14x^2 \qquad (\mathrm {B}) \ \frac 25x^2 \qquad (\mathrm {C})\ \frac 12x^2 \qquad (\mathrm {D}) \ x^2 \qquad (\mathrm {E})\ \frac 32x^2$
Let $w$ be the width, so the length is $2w$. By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $w^2 + 4w^2 = x^2 \Longrightarrow \frac{x}{\sqrt{5}} = w$. The area of the rectangle is $(w)(2w) = 2w^2 = 2\left(\frac{x}{\sqrt{5}}\right)^2 = \frac{2}{5}x^2 \ \mathrm{(B)}$.