jmc

First, we note that the radius of the inscribed circle is 2 cm (since $\pi r^2=4\pi$ implies that $r=2$, given that $r$ is nonnegative).

Let $X$ be the midpoint of side $BC$. Thus segment $OX$ is a radius of the inscribed circle: Then $COX$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle, so the sides $OX$, $CX$, and $CO$ are in the ratio $1:\sqrt{3}:2$. Since $OX=2$, we have $CX=2\sqrt{3}$.

Triangle $ACX$ is also a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle, so the sides $CX$, $AX$, and $AC$ are in the ratio $1:\sqrt{3}:2$. Thus, $AX=(2\sqrt{3})(\sqrt{3})=6$.

Triangle $ABC$ has base $BC=2(XC)=4\sqrt{3}$ and corresponding height $AX=6$, so its area is $\frac{1}{2}(4\sqrt{3})(6)=\boxed{12\sqrt{3}}$.