6. The base of the pyramid $T A B C D$ is a trapezoid $A B C D (B C \| A D)$. The distances from points $A$ and $B$ to the plane $T C D$ are $r_{1}$ and $r_{2}$, respectively. The area of triangle $T C D$ is $S$. Find the volume of the pyramid $T A B C D$.
Solution: The volume of the pyramid $T A B C D$ is equal to the sum of the volumes of the pyramids $T B C D$ and $T A B D: V_{T A B C D}=V_{T B C D}+V_{T A B D}$. Moreover, $V_{T A B D}=V_{T A C D}$, because the pyramids $T A B D$ and $T A C D$ have a common height (from vertex $T$), and the areas of their bases are equal: $S_{A B D}=S_{A C D}$ (these triangles have a common base $B C$ and equal heights drawn from vertices $B$ and $C$, since $A B C D$ is a trapezoid by condition). Therefore, $V_{T A B C D}=V_{T B C D}+V_{T A C D}=\frac{1}{3} \cdot S \cdot r_{2}+\frac{1}{3} \cdot S \cdot r_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c242b26e7e2e4fc7d1ccg-4.jpg?height=377&width=300&top_left_y=288&top_left_x=1643)

Answer: $\frac{S\left(r_{1}+r_{2}\right)}{3}$.