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

1964 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1964 AHSME Problems/Problem 10

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