9. (10 points) As shown in the figure, the area of trapezoid $ABCD$ is 117 square centimeters, $AD \parallel BC$, $EF=13$ centimeters, $MN$ $=4$ centimeters, and it is also known that $EF \perp MN$ at $O$. Therefore, the total area of the shaded parts is $\qquad$ square centimeters.
【Analysis】According to the graph, the areas of $\triangle A B M$ and $\triangle E F M$ are equal, the areas of $\triangle C D N$ and $\triangle E F N$ are equal, and the sum of the areas of $\triangle E F M$ and $\triangle E F N$ is $E F \times(M O+N O) \div 2$. Then, subtracting twice the sum of the areas of $\triangle E F M$ and $\triangle E F N$ from the area of the parallelogram gives the area of the shaded part. According to this, solve the problem.

【Solution】Solution: According to the above analysis, the area of the shaded part is:
$$
\begin{array}{l}
117-(S \triangle E F M+\triangle E F N) \times 2, \\
=117-[E F \times(M O+N O) \div 2] \times 2, \\
=117-[13 \times 4 \div 2] \times 2, \\
=117-26 \div 2 \times 2, \\
=117-52, \\
=65 \text { (square centimeters). }
\end{array}
$$

Therefore, the answer is: 65.