6. In trapezoid $P Q R S$, it is known that $\angle P Q R=90^{\circ}, \angle Q R S<90^{\circ}$, diagonal $S Q$ is 24 and is the bisector of angle $S$, and the distance from vertex $R$ to line $Q S$ is 16. Find the area of trapezoid PQRS.

Answer. $\frac{8256}{25}$.
Given $\angle R Q S=\angle P S Q$ as alternate interior angles, therefore triangle $R Q S$ is isosceles. Its height $R H$ is also a median and equals 16. Then $S R=\sqrt{R H^{2}+S H^{2}}=20$. Triangles $S R H$ and $S Q P$ are similar by two angles, the similarity coefficient is $\frac{S R}{S Q}=\frac{5}{6}$. Therefore, $P Q=\frac{6}{5} R H=\frac{96}{5}$, $P S=\frac{6}{5} S H=\frac{72}{5} \cdot$ The area of the trapezoid is $P Q \cdot \frac{Q R+P S}{2}=\frac{8256}{25}$.