If the altitude of an equilateral triangle is $\sqrt {6}$, then the area is: 
$\textbf{(A)}\ 2\sqrt{2}\qquad\textbf{(B)}\ 2\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 6\sqrt{2}\qquad\textbf{(E)}\ 12$
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