# 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 $\boxed{\textbf{(A)}}$

## See Also

 1964 AHSC (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions

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