11.135. Determine the volume of a regular truncated quadrilateral pyramid if the side of the larger base is $a$, the side of the smaller base is $b$, and the acute angle of the lateral face is $60^{\circ}$.
Solution.

Let $D_{1} T$ be the height of the isosceles trapezoid $D D_{1} C_{1} C$, which is a lateral face of the regular truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.32). Then $D T=\frac{C D-C_{1} D_{1}}{2}=\frac{a-b}{2} ; \quad D_{1} T=D T \operatorname{tg} \angle D_{1} D T=\frac{a-b}{2} \operatorname{tg} 60^{\circ}=\frac{(a-b) \sqrt{3}}{2}$. $L_{1}$ is the midpoint of $C_{1} D_{1}$, and $L$ is the midpoint of $C D$. Then $L L_{1}=D_{1} T=\frac{(a-b) \sqrt{3}}{2}$. Drop the perpendicular $L_{1} K$ to the plane $A B C D$. Then $K$ lies on the segment $O L$, where $O$ is the center of the lower base, $K L=\frac{a-b}{2}$. From $\Delta L_{1} K L\left(\angle L_{1} K L=90^{\circ}\right): L_{1} K=\sqrt{L L_{1}{ }^{2}-K L^{2}}=\sqrt{\frac{3(a-b)^{2}}{4}-\frac{(a-b)^{2}}{4}}=\frac{a-b}{\sqrt{2}}$. The volume of the truncated pyramid is calculated by the formula:

$$
V=\frac{H}{3}\left(S+S_{1}+\sqrt{S S_{1}}\right) \text {, where } H-\text { height of the pyramid, } S \text { and } S_{1} \text { - areas of the }
$$

lower and upper bases, respectively.

Then $V=\frac{a-b}{3 \sqrt{2}}\left(a^{2}+b^{2}+a b\right)=\frac{a^{3}-b^{3}}{3 \sqrt{2}}=\frac{\left(a^{3}-b^{3}\right) \sqrt{2}}{6}$.

Answer: $\frac{\left(a^{3}-b^{3}\right) \sqrt{2}}{6}$.