HMMT November

Solution:

Let $Q$ be the circumcenter of $\triangle B P C$. Thus, $\angle Q B C = \angle Q C B = 90^{\circ} - \angle B P C$, and so $\triangle P A D$ and $\triangle Q B C$ are congruent. This means that $P Q$, $A B$, and $C D$ share the common perpendicular bisector.

We now find the area by determining the altitude. Note that we have all four side lengths of isosceles trapezoids $P Q A B$ and $P Q C D$. Thus, one can compute their altitudes via Pythagorean theorem:

$$\begin{aligned} & \operatorname{distance}(P, A B) = \sqrt{14^{2} - \left(\frac{18-14}{2}\right)^{2}} = 8 \sqrt{3} \\ & \operatorname{distance}(P, C D) = \sqrt{14^{2} - \left(\frac{28-14}{2}\right)^{2}} = 7 \sqrt{3} \end{aligned}$$

so the altitude of trapezoid $A B C D$ is $15 \sqrt{3}$, so the final answer is
$$\frac{1}{2} \cdot (14 + 18) \cdot 15 \sqrt{3} = 345 \sqrt{3}.$$