Circles with centers $O$ and $P$ have radii $4$ and $9$, respectively, and are externally tangent. Points $A$ and $B$ on the circle with center $O$ and points $C$ and $D$ on the circle with center $P$ are such that $AD$ and $BC$ are common external tangents to the circles. What is the area of hexagon $AOBCPD$?

1. The circles are externally tangent, so $OP=4+9=13$. Join $OA$ and $PD$: a tangent is perpendicular to the radius at the point of tangency, so $OA\perp AD$ and $PD\perp AD$, with $OA=4$ and $PD=9$.
2. Draw $OE\perp PD$ with $E$ on $PD$. Then $AOED$ is a rectangle, so $DE=OA=4$, $EP=9-4=5$, and $OE=AD$.
3. In the right triangle $OEP$, $OE=\sqrt{13^2-5^2}=12$, so $AD=12$.
4. The quadrilateral $AOPD$ is a right trapezoid with parallel sides $OA=4$, $PD=9$ and height $AD=12$, so $[AOPD]=\dfrac12(4+9)\cdot12=78$.
5. By symmetry about line $OP$, $[BOPC]=78$ as well, and the hexagon $AOBCPD$ is made of these two trapezoids, so its area is $2\cdot78=\boxed{156}$.