jmc

Let $A$, $B$, and $C$ be the vertices of the right triangle corresponding to the 60, 90, and 30 degree angles, respectively. Also, let $F$ be the foot of the perpendicular from $B$ to the hypotenuse $AC$. Notice that $△BAF$ is a 30-60-90 triangle. The longer leg of a 30-60-90 is $\sqrt{3}$ times the shorter leg, so $AF=3/\sqrt{3}=\sqrt{3}$ units. The hypotenuse of a 30-60-90 triangle is twice the shorter leg, so $AB=2\sqrt{3}$ units. Since $△CAB$ is a 30-60-90 triangle as well, $BC=2\sqrt{3}\cdot \sqrt{3}=6$ units. The area of $ABC$ is $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(2\sqrt{3})(6)=\boxed{6\sqrt{3}}$ square units.