11.108. The bases of the truncated pyramid are two regular octagons. The side of the lower base of the pyramid is 0.4 m, and the upper base is 0.3 m; the height of the truncated pyramid is 0.5 m. The truncated pyramid is completed to a full pyramid. Determine the volume of the complete pyramid.
Solution.

Let $AB$ be the side of the lower base, $A_{1}B_{1}$ be the corresponding side of the upper base of the given truncated octagonal pyramid, $CO$ be the height of the complete pyramid, $O_{1}$ be the intersection point of $CO$ and the plane of the upper base (Fig. 11.5). Then $OO_{1}$ is the height of the truncated pyramid, $OO_{1}=0.5 \, \text{m}, \triangle A_{1}O_{1}B_{1} \sim \triangle AOB$.

$$
\text{Then } \frac{O_{1}A_{1}}{OA}=\frac{A_{1}B_{1}}{AB}=\frac{0.3}{0.4}=\frac{3}{4} \cdot \Delta CO_{1}A_{1} \sim \Delta COA
$$

Then $\frac{CO_{1}}{CO}=\frac{O_{1}A}{OA}=\frac{3}{4}$. Let $CO=h$. Then $CO_{1}=h-0.5 ; \frac{h-0.5}{h}=\frac{3}{4}$; $h=2$ m: $P$ is the center of the regular octagon, which is the base of the pyramid (Fig. 11.6). Let $PA=PB=x, \angle APB=\frac{360^{\circ}}{8}=45^{\circ}$.

From $\triangle APB: AB^{2}=PA^{2}+PB-2PA \cdot PB \cos \angle APB$;

$$
\begin{aligned}
& 2x^{2}-2x^{2} \cos 45^{\circ}=0.4^{2} ; 2x^{2}(1-\cos 45^{\circ})=0.16 ; x^{2}(2-\sqrt{2})=0.16 \\
& x^{2}=\frac{0.16}{2-\sqrt{2}}=\frac{0.16(2+\sqrt{2})}{2}=0.08(2+\sqrt{2}) . S_{\triangle APB}=\frac{1}{2} AP^{2} \sin \angle APB= \\
& =\frac{1}{2} \cdot 0.08(2+\sqrt{2}) \sin 45^{\circ}=0.02 \sqrt{2}(2+\sqrt{2})=0.04(\sqrt{2}+1)=\frac{\sqrt{2}+1}{25}\left(m^{2}\right)
\end{aligned}
$$

The area of the octagon $S=8 S_{\triangle APB}=\frac{8}{25}(\sqrt{2}+1)$ m $^{2}$. The volume of the pyramid $V=\frac{1}{3} S h=\frac{16}{75}(\sqrt{2}+1) \mathbf{m}^{3}$.

Answer: $\frac{16}{75}(\sqrt{2}+1) \mathbf{m}^{3}$.