# 1964 AHSME Problems/Problem 10

## Problem

Given a square side of length $s$. On a diagonal as base a triangle with three unequal sides is constructed so that its area equals that of the square. The length of the altitude drawn to the base is: $\textbf{(A)}\ s\sqrt{2} \qquad \textbf{(B)}\ s/\sqrt{2} \qquad \textbf{(C)}\ 2s \qquad \textbf{(D)}\ 2\sqrt{s} \qquad \textbf{(E)}\ 2/\sqrt{s}$

## Solution

The area of the square is $s^2$. The diagonal of a square with side $s$ bisects the square into two $45-45-90$ right triangles, so the diagonal has length $s\sqrt{2}$.

The area of the triangle is $\frac{1}{2}bh$. The base $b$ of the triangle is the diagonal of the square, which is $b = s\sqrt{2}$. If the area of the triangle is equal to the area of the square, we have: $s^2 = \frac{1}{2}bh$ $s^2 = \frac{1}{2}s\sqrt{2}\cdot h$ $s = \frac{\sqrt{2}}{2}h$ $h = \frac{2}{\sqrt{2}}s$ $h = s\sqrt{2}$

This is option $\boxed{\textbf{(A)}}$

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