# 2005 AMC 10A Problems/Problem 4

## Problem

A rectangle with a diagonal of length $x$ is twice as long as it is wide. What is the area of the rectangle? $\textbf{(A) } \frac{1}{4}x^2\qquad \textbf{(B) } \frac{2}{5}x^2\qquad \textbf{(C) } \frac{1}{2}x^2\qquad \textbf{(D) } x^2\qquad \textbf{(E) } \frac{3}{2}x^2$

## Video Solution

CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M

~Charles3829

## Solution 1

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

## Solution 2

Call the length $2l$ and the width $l$.

The area of the rectangle is $2l*l = 2l^2$ $x$ is the hypotenuse of the right triangle with $2l$ and $l$ as legs. By the Pythagorean theorem, $(2l)^2+l^2 = x^2$ $4l^2 + l^2 = x^2$ $,$ $5l^2 = x^2$ $,$ and $l^2 = \frac{x^2}{5}$.

Therefore, the area is $\boxed{\textbf{(B) }\frac{2}{5}x^2}$

-mobius247

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 