1953 AHSME Problems/Problem 26

Problem 26

The base of a triangle is $15$ inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:

$\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad \textbf{(B)}\ 10\text{ inches} \qquad \textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad \textbf{(D)}\ 7.5\text{ inches}\\  \textbf{(4)}\ \text{none of these}$

Solution

Let the triangle be $\triangle ABC$ where $BC$ is the base. Then let the parallels be $MN$ and $PQ$, where $PQ$ is closer to our base $BC$.

It's obvious that $\triangle APQ \sim \triangle ABC$, where $|\triangle APQ|:|\triangle ABC|=2:3$, so $\frac{PQ}{BC}=\sqrt{\frac{2}{3}}$. Since we know $BC=15$, $PQ=\frac{15\sqrt{2}}{\sqrt{3}}=5\sqrt{6}$

Hence our answer is $\textbf{(A)}$

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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