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 $ABQCDP$?
In this isosceles trapezoid, $AB=10$, $CD=4$, and each leg is $5$. The height is $\sqrt{5^2-3^2}=4$.
The angle-bisector intersections are $P=(4,2)$ and $Q=(6,2)$ when $A=(0,0)$, $B=(10,0)$, $D=(3,4)$, and $C=(7,4)$.
The trapezoid's area is $\frac{10+4}{2}\cdot4=28$. The two corner triangles $ADP$ and $BCQ$ each have area $5$, so the hexagon area is $28-5-5=\boxed{18}$.
