# Difference between revisions of "1962 AHSME Problems/Problem 6"

## Problem

A square and an equilateral triangle have equal perimeters. The area of the triangle is $9 \sqrt{3}$ square inches. Expressed in inches the diagonal of the square is:

$\textbf{(A)}\ \frac{9}{2}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ \frac{9\sqrt{2}}{2}\qquad\textbf{(E)}\ \text{none of these}$

## Solution

To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is $\dfrac{x^2\sqrt{3}}{4}$. This has to be equal to $9 \sqrt{3}$, which means that $x=6$, or the side length of the triangle is $6$. Thus, the triangle (and the square) have a perimeter of $18$. It follows that each side of the square is $\dfrac{18}{4}=4.5$. If we draw the diagonal, we create a 45-45-90 triangle, whose hypotenuse (also the diagonal of the square) is $\boxed{\text{(D)} \dfrac{9\sqrt{2}}{2}}$.