11.209 A plane is drawn through the side of the base of a regular quadrilateral pyramid, which cuts off a triangle with an area of $4 \mathrm{~cm}^{2}$ from the opposite face. Find the lateral surface of the pyramid that is cut off by the drawn plane from the given pyramid, if the lateral surface of the given pyramid is $25 \mathrm{~cm}^{2}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-265.jpg?height=312&width=322&top_left_y=937&top_left_x=156)

Let the section $C D M N$ of the given regular pyramid $F A B C D$ cut off a lateral face $\triangle F M N$ with an area of 4. $C D\|A B \Rightarrow M N\| A B$ (Theorem 14) $\Rightarrow$ $\Rightarrow \triangle F M N \sim \triangle F A B \Rightarrow\left(\frac{F M}{F A}\right)^{2}=\frac{S_{F M N}}{S_{F A B}}=\frac{16}{25}$ (Theorem 7) $\Rightarrow \frac{4}{5}=\frac{F M}{F A}=\frac{S_{F D M}}{S_{F D A}}$ (Theorem 10) $\Rightarrow$

$\Rightarrow S_{F D M}=5$ and the area of the lateral surface of the pyramid $F M N C D$ is

$S_{F M N}+2 S_{F D M}+S_{F C D}=4+10+\frac{25}{4}=20.25$.

Answer: $20.25 \mathrm{~cm}^{2}$.