1951 AHSME Problems/Problem 25

Problem

The apothem of a square having its area numerically equal to its perimeter is compared with the apothem of an equilateral triangle having its area numerically equal to its perimeter. The first apothem will be:

$\textbf{(A)}\ \text{equal to the second}\qquad\textbf{(B)}\ \frac{4}{3}\text{ times the second}\qquad\textbf{(C)}\ \frac{2}{\sqrt{3}}\text{ times the second}\\ \textbf{(D)}\ \frac{\sqrt{2}}{\sqrt{3}}\text{ times the second}\qquad\textbf{(E)}\ \text{indeterminately related to the second}$

Solution

First we try to find the size of the square. Let $s$ be the side length of the square. It states that $s^2=4s$ , therefore $s=0,4$ . We cross out the trivial case $s=0$ , so the side length of the square is $4$ . The apothem of the square is simply half its side length, or $2$ .

Let the side length of the equilateral triangle be $t$ . The problem states that $3t=\frac{t^2\sqrt{3}}{4}$ , so $12t=t^2\sqrt{3}$ , therefore $t=0, \frac{12}{\sqrt{3}}$ . Again we cross out the trivial case $t=0$ , so we have the side length of the triangle as $\frac{12}{\sqrt{3}}$ . The apothem of the triangle is $\frac{1}{3}$ of the height. The height of the triangle is $\frac{\sqrt{3}}{2}\cdot \frac{12}{\sqrt{3}} = 6$ , so the apothem is $2$ Therefore, the answer is $\textbf{(A)}\ \text{equal to the second}$

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1951 AHSME Problems/Problem 25

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