Equilateral $\triangle ABC$ has side length $1$, and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$? $\textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad$ $\textbf{(B)}\ \dfrac92\qquad$ $\textbf{(C)}\ 3+\sqrt3\qquad$ $\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad$ $\textbf{(E)}\ 6$
Solution
The equilateral triangle has area $\frac{\sqrt3}4$.
The three squares have total area $3$.
The remaining $3$ pieces are isosceles triangles with two sides $1$ and apex angle $360^\circ-60^\circ-90^\circ-90^\circ=120^\circ$, with total area $3\times \frac12\cdot 1\cdot 1\cdot \sin120^\circ=\frac{3\sqrt3}4$.
The total area is $3+\frac{\sqrt3}4+\frac{3\sqrt3}4=3+\sqrt3$.
$\textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad$ $\textbf{(B)}\ \dfrac92\qquad$ $\textbf{(C)}\ 3+\sqrt3\qquad$ $\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad$ $\textbf{(E)}\ 6$ 
