1956 AHSME Problems/Problem 30

Solution

Drawing the altitude, we see that is opposite the $60^{\circ}$ angle in a $30-60-90$ triangle. Thus, we find that the shortest leg of that triangle is $\sqrt{\frac{6}{3}} = \sqrt{2}$ , and the side of the equilateral triangle is then $2\sqrt{2}$ . Thus, the area is $\frac{(2\sqrt 2)^2 \cdot \sqrt{3}}{4}$ $= \frac{8 \sqrt{3}}{4}$ $= 2 \sqrt {3}$ $\boxed{B}$

~JustinLee2017

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1956 AHSME Problems/Problem 30

Solution