11.024. The bases of a regular truncated pyramid are squares with sides $a$ and $b(a>b)$. The lateral edges are inclined to the plane of the base at an angle of $45^{\circ}$. Determine the volume of the truncated pyramid.
Solution.

Let $\quad A D=D C=B C=A B=a, \quad A_{1} B_{1}=B_{1} C_{1}=C_{1} D_{1}=D_{1} A_{1}=h$ (Fig. 11.19), $\angle D_{1} D H=45^{\circ}$, where $D_{1} H$ is the height of the truncated pyramid.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-664.jpg?height=554&width=386&top_left_y=63&top_left_x=191)

Fig. 11.18

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-664.jpg?height=492&width=550&top_left_y=126&top_left_x=653)

Fig. 11.19

$O O_{1} \| D_{1} H, O H=O_{1} D_{1} . O_{1} D_{1}=\frac{1}{2} B_{1} D_{1}=\frac{1}{2} b \sqrt{2}$. Similarly, $O D=\frac{1}{2} a \sqrt{2}$. Therefore, $H D=O D-O H=\frac{\sqrt{2}}{2}(a-b) . H D_{1}=H D \operatorname{tg} 45^{\circ}=\frac{\sqrt{2}}{2}(a-b)$; $S_{A B C D}=a^{2}, S_{A_{1} B_{1} C_{1} D_{1}}=b^{2}$. The volume of the truncated pyramid

$$
\begin{aligned}
V & =\frac{1}{3} H D_{1}\left(S_{A B C D}+\sqrt{S_{A B C D} S_{A_{1} B_{1} C_{1} D_{1}}}+S_{A_{1} B_{1} C_{1} D_{1}}\right)=\frac{1}{3} \frac{\sqrt{2}}{2}(a-b)\left(a^{2}+a b+b^{2}\right)= \\
= & \frac{\sqrt{2}}{6}\left(a^{3}-b^{3}\right) \\
& \text { Answer: } \frac{\sqrt{2}}{6}\left(a^{3}-b^{3}\right)
\end{aligned}
$$