11.107. The volume of a regular octagonal prism is $8 \mathrm{~m}^{3}$, and its height is 2.2 m. Find the lateral surface area of the prism.
Solution.

Let $O$ be the center of the regular octagon $A B C D E F K N$, which is the base of the given regular prism (Fig. 11.4). The volume of the prism $V=8 \mathrm{M}^{3}$, height $H=2.2$ m, then the area of the base of the prism $S_{\text {base}}=\frac{8}{2.2}=\frac{40}{11}\left(\mathrm{m}^{2}\right)$. Let $A N=a, O Q$ be the height of $\triangle A O N$.

Then $A Q=\frac{a}{2} ; \angle A O N=\frac{360^{\circ}}{8}=45^{\circ}, \angle A O Q=\frac{1}{2} \angle A O N=22.5^{\circ}$. From

$$
\begin{aligned}
& \triangle A Q O\left(\angle A Q O=90^{\circ}\right): O Q=A Q \operatorname{ctg} 22.5^{\circ}=\frac{a}{2} \cdot \frac{1+\cos 45^{\circ}}{\sin 45^{\circ}}=\frac{a(\sqrt{2}+1)}{2} \\
& S_{\text {base}}=8 S_{A O N}=8 \cdot \frac{1}{2} A N \cdot O Q=4 a^{2} \cdot \frac{\sqrt{2}+1}{2}=2 a^{2}(\sqrt{2}+1)
\end{aligned}
$$

$$
\text { Then } 2 a^{2}(\sqrt{2}+1)=\frac{40}{11} ; a^{2}=\frac{20}{11(\sqrt{2}+1)}=\frac{20(\sqrt{2}-1)}{11} ;
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0823.jpg?height=760&width=352&top_left_y=60&top_left_x=264)

Fig. 11.5

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0823.jpg?height=434&width=454&top_left_y=340&top_left_x=758)

Fig. 11.6

$a=\sqrt{\frac{20(\sqrt{2}-1)}{11}}=\frac{2}{11} \sqrt{55(\sqrt{2}-1)}$ (m). The lateral surface area of the prism $S=8 a H=8 \cdot \frac{2}{11} \sqrt{55(\sqrt{2}-1)} \cdot 2.2=16 \sqrt{2.2(\sqrt{2}-1)}\left(\mathrm{M}^{2}\right)$.

Answer: $16 \sqrt{2.2(\sqrt{2}-1)}\left(\mathrm{M}^{2}\right)$.