# Difference between revisions of "1960 AHSME Problems/Problem 21"

## Problem

The diagonal of square $I$ is $a+b$. The perimeter of square $II$ with twice the area of $I$ is:

$\textbf{(A)}\ (a+b)^2\qquad \textbf{(B)}\ \sqrt{2}(a+b)^2\qquad \textbf{(C)}\ 2(a+b)\qquad \textbf{(D)}\ \sqrt{8}(a+b) \qquad \textbf{(E)}\ 4(a+b)$

## Solution

Since the diagonal of square $I$ is $a+b$ units long, the side length of square $I$ is $\frac{a+b}{\sqrt{2}}$, so the area of square $I$ is $\frac{(a+b)^2}{2}$.

The area of square $II$ is twice as much as the area of square $I$, so the area of square $II$ is $(a+b)^2$. That means the side length of square $II$ is $a+b$, so the perimeter of square $II$ is $4(a+b)$, or answer choice $\boxed{\textbf{(E)}}$.