1954 AHSME Problems/Problem 8

Problem 8

The base of a triangle is twice as long as a side of a square and their areas are the same. Then the ratio of the altitude of the triangle to the side of the square is:

$\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 4$

Solution

Let the base and altitude of the triangle be $b, h$ , the common area $A$ , and the side of the square $s$ . Then $\frac{2sh}{2}=s^2\implies sh=s^2\implies s=h$ , so $\fbox{C}$

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1954 AHSME Problems/Problem 8

Problem 8

Solution