jmc

If one angle of the parallelogram is 120 degrees, then another angle between adjacent sides has measure $180^{\circ }-120^{\circ }=60^{\circ }$. Shown in the diagram above, let the parallelogram be $ABCD$, with $\overline{AD}$ one of the short sides. Drawing the altitude from $D$ to $\overline{AB}$ gives us 30-60-90 triangle $AXD$, from which we find $XD=(AD/2)\sqrt{3}=4\sqrt{3}$, so the area of $ABCD$ is $(AB)(XD)=\boxed{60\sqrt{3}}$.