10,11 In the quadrilateral pyramid $O A B C D$, the base is a trapezoid $A B C D$, and the lateral faces $O A D$ and $O B C$ are perpendicular to the base. The area of the face $O A B$ is 9, the area of the face $O C D$ is 20, the edge $A B$ is 3, and the edge $C D$ is 5. Find the volume of the pyramid.
Suppose that $A D \| B C$. Since the planes of the lateral faces $O A D$ and $O B C$ are perpendicular to the base plane, each of these planes contains a line that is perpendicular to the base plane. These two lines are parallel because both are perpendicular to the same plane. Then, two intersecting lines in the plane $O A D$ are respectively parallel to two intersecting lines in the plane $O B C$. Therefore, the planes $O A D$ and $O B C$ are parallel, which contradicts the condition ( $O$ - the common point of these planes). Hence, $A B$ and $C D$ are the bases of the trapezoid $A B C D$, and $A D$ and $B C$ are its lateral sides. Since each of the planes $O A D$ and $O B C$ is perpendicular to the plane $A B C D$, the line of intersection of the planes $O A D$ and $O B C$ is perpendicular to the plane $A B C D$. Therefore, if $H$ is the point of intersection of this line with the plane $A B C D$, then $O H$ is the height of the given pyramid. Let $O M$ be the height of the triangle $A O B$. Then, by the theorem of three perpendiculars, $H M \perp A B$, i.e., $H M$ is the height of the triangle $H A B$, and since $C D \| A B$, then $H M \perp C D$. Let the lines $H M$ and $C D$ intersect at point $N$. Then $H N \perp C D$ and $O N \perp C D$, i.e., $O N$ is the height of the triangle $O C D$.

The triangles $H A B$ and $H D C$ are similar with a ratio of $\frac{3}{5}$. Let $H M=3 x$. Then $H N=5 x, M N=H N-H M=2 x$, and $M N$ is the height of the trapezoid $A B C D$. In the right triangle $O H N$, it is known that

$$
\begin{gathered}
O N=2 \cdot \frac{S_{S O A N}^{C D}}{C D}=2 \cdot \frac{2 M}{5}=8, O M=2 \cdot \frac{S_{\triangle O A B}}{A B}=2 \cdot \frac{9}{3}=6 \\
H N=5 x, H M=3 x, O N 2-H N 2=O H 2=O M 2-H M 2
\end{gathered}
$$

Therefore, $64-25 x 2=36-9 \times 2$, from which $x=\frac{\sqrt{7}}{2}$. Thus,

$$
\begin{gathered}
O H=\sqrt{36-9 x^{2}}=\sqrt{36-\frac{93}{1}}=\sqrt{\frac{81}{1}}=\frac{9}{2} \\
S_{A B C D}=\frac{1}{2}(A B+C D) \cdot M N=4 \cdot 2 x=8 x=4 \sqrt{7}
\end{gathered}
$$

Therefore,

$$
V_{O A B C D}=\frac{1}{3} S_{A B C D} \cdot O H=\frac{1}{3} \cdot 4 \sqrt{7} \cdot \frac{9}{2}=6 \sqrt{7}
$$

## Answer

$6 \sqrt{7}$.