imc

We find that the area of the triangle is $\frac{6\times 2\sqrt{3}}{2}=6\sqrt{3}$. By the Pythagorean Theorem, we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3}{)}^2+6^2}=4\sqrt{3}$. Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way. Let $h$ be the third height of the triangle. We have $\frac{4\sqrt{3}h}{2}=6\sqrt{3}⟹h=\frac{6\cdot 2}{4}⟹h=\boxed{3}$ Note: The third altitude of a right triangle is always the product of the lengths of the two legs divided by the hypotenuse.