The convex pentagon $ABCDE$ has $\angle{A}=\angle{B}=120^\circ,EA=AB=BC=2$ and $CD=DE=4$. What is the area of ABCDE?
$\text{(A) } 10\quad \text{(B) } 7\sqrt{3}\quad \text{(C) } 15\quad \text{(D) } 9\sqrt{3}\quad \text{(E) } 12\sqrt{5}$
First, drop perpendiculars from points $A$ and $B$ to segment $EC$.
Since $\angle EAB = 120^{\circ}$, $\angle EAF = 30^{\circ}$.
This implies that $\triangle EAF$ is a 30-60-90 Triangle, so $EF = 1$ and $AF = \sqrt3$.
Similarly, $GB = \sqrt3$ 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, $[ABCDE] = [EABC]+[DCE] = \frac{2+4}{2}(\sqrt3)+\frac{4^2\cdot\sqrt3}{4} = 3\sqrt3+4\sqrt3=\boxed{7\sqrt3 (B)}$
-AOPS81619