1956 AHSME Problems/Problem 30

Revision as of 19:00, 12 February 2021 by Justinlee2017 (talk | contribs) (Solution)

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

\[\frac{(2\sqrt 2)^2 \cdot \sqrt{3}}{4} = \frac{8 \sqrt{3}}{4} = 2 \sqrt {3}}\] (Error compiling LaTeX. Unknown error_msg)

$\boxed{B}$

~JustinLee2017