smc

First, drop perpendiculars from points $A$ and $B$ to segment $EC$. Since $\angle EAB=120^{\circ }$, $\angle EAF=30^{\circ }$. This implies that $△EAF$ is a 30-60-90 Triangle, so $EF=1$ and $AF=\sqrt{3}$. Similarly, $GB=\sqrt{3}$ and $GC=1$. Since $FABG$ is a rectangle, $FG=AB=2$. Now, notice that since $EC=EF+FG+GC=1+2+1=4$, triangle $DEC$ is equilateral. Thus, $[ABCDE]=[EABC]+[DCE]=\frac{2+4}{2}(\sqrt{3})+\frac{4^2\cdot \sqrt{3}}{4}=3\sqrt{3}+4\sqrt{3}=\boxed{7\sqrt{3}}$