jmc

Drawing an altitude of an equilateral triangle splits it into two 30-60-90 right triangles:

The altitude is the longer leg of each 30-60-90 triangle, so $AM=\sqrt{3}\cdot BM$ in the diagram above. Since $AM=\sqrt{6}$, we have $$BM=\frac{AM}{\sqrt{3}}=\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{\frac{6}{3}}=\sqrt{2}.$$ Therefore, we have $BC=2BM=2\sqrt{2}$, so the area of the triangle is $(BC)(AM)/2=(2\sqrt{2})(\sqrt{6})/2=\sqrt{12}=\boxed{2\sqrt{3}}$ square units.