2005 AMC 10A Problem 4

The diagonal cuts the rectangle into two right triangles, so the Pythagorean theorem connects the diagonal to the sides.

1. Name the sides. Let the width be $w$. The length is twice as long, so it is $2w$.
2. Use the right triangle. The legs are $w$ and $2w$, and the hypotenuse is the diagonal $x$:
$$x^2 = w^2 + (2w)^2 = w^2 + 4w^2 = 5w^2.$$
So $w^2 = \frac{x^2}{5}$.
3. Find the area. Area is length times width:
$$2w \cdot w = 2w^2 = 2 \cdot \frac{x^2}{5} = \frac{2}{5}x^2.$$

The answer is $\boxed{\textbf{(B)}\ \frac{2}{5}x^2}$.

A rectangle's width, length and diagonal always make a right triangle. Once both sides are written using one letter, the diagonal squared and the area are both "some number times $w^2$," and comparing those two numbers ($5$ and $2$) gives the answer.

Pick easy numbers. Let the width be $1$ and the length $2$. Then the area is $2$ and the diagonal squared is $1 + 4 = 5$. Look for the choice that turns $x^2 = 5$ into $2$: only $\frac{2}{5} \cdot 5 = 2$ works.

Let the width be $w$, so the length is $2w$. The diagonal is the hypotenuse of a right triangle with legs $w$ and $2w$:
$$x^2 = w^2 + (2w)^2 = 5w^2, \qquad\text{so}\qquad w^2 = \frac{x^2}{5}.$$
The area is length times width:
$$2w \cdot w = 2w^2 = \frac{2}{5}x^2.$$

The answer is $\boxed{\textbf{(B)}\ \frac{2}{5}x^2}$.