2008 AMC 12B Problem 25

Because $AB \parallel CD,$ $\angle A + \angle D = 180^\circ,$ so the bisectors of $\angle A$ and $\angle D$ meet at right angles, $\angle APD = 90^\circ.$ Then the midpoint $M$ of $\overline{AD}$ is the circumcenter of right triangle $APD,$ giving $MP = MA = MD.$ Thus $\angle MPA = \angle PAM = \angle PAB,$ where the last equality uses the angle bisector at $A.$ Therefore $MP \parallel AB.$ The same argument gives $QN \parallel AB$ for the midpoint $N$ of $\overline{BC}.$ Hence $M, P, Q, N$ are collinear on the midline. The midline has length $\tfrac{AB + CD}{2} = 15,$ while $MP = \tfrac{AD}{2} = \tfrac72$ and $QN = \tfrac{BC}{2} = \tfrac52.$ Hence $PQ = 15 - \tfrac72 - \tfrac52 = 9.$ Drawing $AE \parallel BC$ with $E$ on $\overline{CD}$ gives $AE = 5$ and $DE = CD - AB = 8.$ In $\triangle ADE,$ $\cos(\angle AED) = \tfrac{8^2 + 5^2 - 7^2}{2 \cdot 8 \cdot 5} = \tfrac12,$ so $\angle AED = 60^\circ$ and the trapezoid’s height is $AF = 5\sin 60^\circ = \tfrac{5\sqrt3}{2}.$ The segment $PQ$ sits at half the height, so the hexagon splits into two trapezoids and $$\begin{aligned} &[ABQCDP] \\ &= \frac{AF}{4}\bigl(AB + CD + 2\,PQ\bigr) \\ &= \frac{\frac{5\sqrt3}{2}}{4}(11 + 19 + 18) \\ &= 30\sqrt3. \end{aligned}$$ Thus, the correct answer is B.