11.198 The base of a right prism is a triangle with sides 6, 8, and 10 cm. A certain plane section of this prism cuts off segments of 12 cm each from the lateral edges passing through the vertices of the larger and medium angles of the base, and a segment of 18 cm from the edge passing through the vertex of the smallest angle of the base. Find the volume and the area of the complete surface of the figure bounded by the plane of the base of the prism, the planes of the lateral faces, and the plane of the section.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-259.jpg?height=365&width=300&top_left_y=114&top_left_x=163)

Let $A B C A_{1} B_{1} C_{1}$ be the given right prism with base $A B C, A C=6, B C=8, A B=10$. Let $L M N$ be the given section, $A M=C N=12, B L=18$.

By the condition, $A B^{2}=100=A C^{2}+B C^{2} \Rightarrow \angle C$ is a right angle. Let the section $M L_{1} N \| A B C$. Then $L L_{1}=6$ and

$$
\begin{gathered}
V_{A B C M L N}=V_{A B C M L_{1} N}+V_{M L N L_{1}}= \\
=12 S_{A B C}+2 S_{M L_{1} N}=336
\end{gathered}
$$

$M N \perp N L_{1} \Rightarrow M N \perp N L$ (Theorem 12) $\Rightarrow$ the angle $\alpha=\angle L N L_{1}$ is the angle formed by the planes $M N L$ and $M N L_{1}$, and $\triangle M N L_{1}$ is the projection of $\triangle M N L \Rightarrow S_{M N L}=\frac{S_{M N L_{1}}}{\cos \alpha}$. But $\operatorname{tg} \alpha=\frac{3}{4}$. Therefore, $\cos \alpha=\frac{4}{5}$, and the area of the complete surface of the figure $A B C M L N$ is $S_{A B C}+\bar{S}_{M N L}+A M \cdot A C+B C \cdot \frac{C N+B L}{2}+A B \cdot \frac{A M+B L}{2}=396$. Answer: 336 cm, 396 cm.