1956 AHSME Problems/Problem 30

Solution

Drawing the altitude, we see that is opposite the angle in a triangle. Thus, we find that the shortest leg of that triangle is , and the side of the equilateral triangle is then . 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)

~JustinLee2017