11.025. The lateral edges of a regular truncated triangular pyramid are inclined to the base plane at an angle of $60^{\circ}$. The sides of the lower and upper bases are equal to $a$ and $b(a>b)$, respectively. Find the volume of the truncated pyramid.
Solution.

Let $A B=B C=A C=a, A_{1} B_{1}=B_{1} C_{1}=C_{1} A_{1}=b, \angle A_{1} A H=60^{\circ}$ (Fig. 11.20). Since $\triangle A B C$ and $A_{1} B_{1} C_{1}$ are equilateral, $O$ and $O_{1}$ are the centers

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-665.jpg?height=416&width=440&top_left_y=172&top_left_x=76)

Fig. 11.20

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-665.jpg?height=510&width=628&top_left_y=79&top_left_x=556)

Fig. 11.21

of the circumcircles around them. This means that $A_{1} O_{1}=R_{1}=\frac{b \sqrt{3}}{3}$, $A O=R=\frac{a \sqrt{3}}{3} . A_{1} H \| O O_{1}$. Therefore, $A H=A O-H O=R-R_{1}=\frac{\sqrt{3}}{3}(a-b)$. The height $A_{1} H$ can be found: $A_{1} H=A H \operatorname{tg} 60^{\circ}=\frac{\sqrt{3}}{3}(a-b) \cdot \sqrt{3}=(a-b)$. The area $S_{A B C}=\frac{a^{2} \sqrt{3}}{4}, S_{A_{1} B_{1} C_{1}}=\frac{b^{2} \sqrt{3}}{4}$. Then the volume

$$
\begin{aligned}
& V=\frac{1}{3} A_{1} H\left(S_{A B C}+\sqrt{S_{A B C} S_{A_{1} B_{1} C_{1}}}+S_{A_{1} B_{1} C_{1}}\right)=\frac{1}{3}(a-b)\left(\frac{a^{2} \sqrt{3}}{4}+\frac{\sqrt{3}}{4} a b+\frac{b^{2} \sqrt{3}}{4}\right)= \\
& =\frac{\sqrt{3}}{12}\left(a^{3}-b^{3}\right)
\end{aligned}
$$

Answer: $\frac{\sqrt{3}}{12}\left(a^{3}-b^{3}\right)$.