6. In trapezoid $\mathrm{PQRS}$, it is known that $\angle \mathrm{PQR}=90^{\circ}, \angle Q R S>90^{\circ}$, diagonal $S Q$ is 24 and is the bisector of angle $S$, and the distance from vertex $R$ to line $Q S$ is 5. Find the area of trapezoid PQRS.
Answer: $\frac{27420}{169}$.

Solution $\angle \mathrm{RQS}=\angle \mathrm{PSQ}$ as alternate interior angles, so triangle $\mathrm{RQS}$ is isosceles. Its height $\mathrm{RH}$ is also the median and equals 5. Then $S R=\sqrt{R H^{2}+S H^{2}}=13$. Triangles SRH and SQP are similar by two angles, the similarity coefficient is $\frac{S R}{S Q}=\frac{13}{24}$. Therefore, $P Q=\frac{24}{13} R H=\frac{120}{13}$, $P S=\frac{24}{13} S H=\frac{288}{13} \cdot$ The area of the trapezoid is $P Q \cdot \frac{Q R+P S}{2}=\frac{27420}{169}$.