Let $ABCD$ be a trapezoid with $AB||CD$, $AB = 11$, $BC = 5$, $CD = 19$, and $DA = 7$. Bisectors of $\angle A$ and $\angle D$ meet at $P$, and bisectors of $\angle B$ and $\angle C$ meet at $Q$. What is the area of hexagon $ABQCDP$? $\textbf{(A)}\ 28\sqrt {3}\qquad$ $ \textbf{(B)}\ 30\sqrt {3}\qquad$ $ \textbf{(C)}\ 32\sqrt {3}\qquad$ $ \textbf{(D)}\ 35\sqrt {3}\qquad$ $ \textbf{(E)}\ 36\sqrt {3}$
Solution
Set $A=(0,0)$, $B=(11,0)$, $D=(-\frac{11}2,\frac{5\sqrt3}2)$, and $C=(\frac{27}2,\frac{5\sqrt3}2)$ (one checks $DA=7$, $BC=5$, $CD=19$). The height is $h=\frac{5\sqrt3}2$, and the trapezoid has area $\frac{11+19}2\cdot h=\frac{75\sqrt3}2$.
$P$ lies on the bisectors of $\angle A$ and $\angle D$, so it is equidistant from $AB$, $AD$ and $DC$; since $AB\parallel DC$, this distance is $\frac h2=\frac{5\sqrt3}4$. Hence $[ADP]=\frac12\cdot7\cdot\frac{5\sqrt3}4=\frac{35\sqrt3}8$, and similarly $[BCQ]=\frac12\cdot5\cdot\frac{5\sqrt3}4=\frac{25\sqrt3}8$.
Let $ABCD$ be a trapezoid with $AB\parallel CD$, $AB=10$, $BC=5$, $CD=4$, and $DA=5$. The bisectors of $\angle A$ and $\angle D$ meet at $P$, and the bisectors of $\angle B$ and $\angle C$ meet at $Q$. What is the area of hexagon $ABQCDP$?
$\textbf{(A)}\ 28\sqrt {3}\qquad$ $ \textbf{(B)}\ 30\sqrt {3}\qquad$ $ \textbf{(C)}\ 32\sqrt {3}\qquad$ $ \textbf{(D)}\ 35\sqrt {3}\qquad$ $ \textbf{(E)}\ 36\sqrt {3}$ 
