A rectangle with a [diagonal](https://artofproblemsolving.com/wiki/index.php/Diagonal) of length $x$ is twice as long as it is wide. What is the area of the rectangle? 
$\textbf{(A) } \frac{1}{4}x^2\qquad \textbf{(B) } \frac{2}{5}x^2\qquad \textbf{(C) } \frac{1}{2}x^2\qquad \textbf{(D) } x^2\qquad \textbf{(E) } \frac{3}{2}x^2$
Let's set our length to $2$ and our width to $1$.
We have our area as $2*1 = 2$ and our diagonal: $x$ as $\sqrt{1^2+2^2} = \sqrt{5}$ (Pythagoras Theorem)
Now we can plug this value into the answer choices and test which one will give our desired area of $2$.

All of the answer choices have our $x$ value squared, so keep in mind that $\sqrt{5}^2 = 5$
Through testing, we see that ${2/5}*\sqrt{5}^2 = 2$
So our correct answer choice is $\boxed{\textbf{(B) }\frac{2}{5}x^2}$
-JinhoK