smc

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 $$\frac{(2\sqrt{2}{)}^2\cdot \sqrt{3}}{4}$$ $$=\frac{8\sqrt{3}}{4}$$ $$=2\sqrt{3}$$ $\boxed{2\sqrt{3}}$