Advanced Number Theory · Challenge problem

Square $ABCD$ has side length $10$. Point $P$ lies on $\overline{AB}$ with $AP=4$, and point $Q$ lies on $\overline{BC}$ with $BQ=3$. Segments $\overline{DP}$ and $\overline{AQ}$ intersect at point $R$. Then the area of $\triangle APR$ equals $\dfrac{p}{q}$ for relatively prime positive integers $p,q$. Find $p+q$.