HMMT February

Let the midpoint of $P Q$ be $M$; note that $M$ lies on the midline of $A B C D$. Let $B^{\prime} C^{\prime}$ be a translate of $B C$ (parallel to $A B$ and $C D$) so that $M$ is the midpoint of $B^{\prime}$ and $C^{\prime}$. Since $M B^{\prime}=M C^{\prime}=25 / 2=M P=M Q, B^{\prime}$ and $C^{\prime}$ are one of the four intersections of the circle with diameter $P Q$ and the sides $A B$ and $C D$. We may also define $A^{\prime}$ and $D^{\prime}$ similarly and get that they are also among the four points. It follows that the desired quadrilateral is $B^{\prime} D^{\prime} C^{\prime} A^{\prime}$, which is a rectangle with height equal to the height of $A B C D$ (which is 24), and width equal to $\frac{1}{2}(31-17)=7$. Thus the area is $24 \cdot 7=168$.