1955 AHSME Problems/Problem 31

An equilateral triangle whose side is $2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is:

$\textbf{(A)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ 2+\sqrt{2}\qquad\textbf{(D)}\ \frac{2+\sqrt{2}}{2}\qquad\textbf{(E)}\ \frac{2\sqrt{3}-\sqrt{6}}{2}$

Solution

The area of the large equilateral triangle is $2^2*\frac{\sqrt{3}}{4}=\sqrt{3}$ square units, so the smaller triangle is $\frac{\sqrt{3}}{2}$ . The side length of the smaller length $s$ is $\sqrt{\frac{4}{\sqrt{3}}*\frac{\sqrt{3}}{2}} = \sqrt{2}$ . The median of the trapezoid can be determined by the average of the two bases of the trapezoids, so it works out to be $\boxed{\textbf{(D)}\frac{2+\sqrt{2}}{2}}$ units long.

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1955 AHSME Problems/Problem 31

Solution

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