[Circles](https://artofproblemsolving.com/wiki/index.php/Circle) with centers $O$ and $P$ have radii $2$ and $4$, 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 the [concave](https://artofproblemsolving.com/wiki/index.php/Concave) [hexagon](https://artofproblemsolving.com/wiki/index.php/Hexagon) $AOBCPD$?

$\mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2}$
When we see this problem, it practically screams similar triangles at us. Extend $OP$ to the left until it intersects lines $AD$ and $BC$ at point $E$. Triangles $EBC$ and $ECP$ are similar, and by symmetry, so are triangles $EAC$ and $EDP$. Then, the area of kite $EAOB$ is just $4 \sqrt{2} \times 2$ and the area of kite $EDPC$ is $8 \sqrt{2} \times 4$ (using our similarity ratios). The difference of these yields the area of hexagon $AOBCPD$ or $24\sqrt{2} \Longrightarrow \boxed{\mathrm{(B)}}$.
~icecreamrolls8