11.018. The base of the pyramid is a triangle with sides equal to $a$, $a$, and $b$. All lateral edges are inclined to the base plane at an angle of $60^{\circ}$. Determine the volume of the pyramid.
Solution.

Given $A B=B C=a$ (Fig. 11.15); therefore, $B H=a^{2}-\frac{b^{2}}{4}=$ $=\frac{1}{2} \sqrt{4 a^{2}-b^{2}}$. Hence, $S_{\text {Base }}=\frac{1}{2} A C \cdot B H=\frac{1}{4} b \sqrt{4 a^{2}-b^{2}}$. Draw the height $SO$. Since all the edges of the pyramid are equally inclined to the base plane, point $O$ is the center of the circumscribed circle around $\triangle A B C$. Let the radius of this circle be $R$. Then

$$
\begin{aligned}
& O B=R=\frac{a^{2} b}{4 S_{\text{circ}}}=\frac{a^{2}}{\sqrt{4 a^{2}-b^{2}}} \text{ in } \triangle S O B \quad O B=\frac{S B}{2} ; \text{ therefore } \\
& S O=\sqrt{S B^{2}-O B^{2}}=\sqrt{4 O B^{2}-O B^{2}}=O B \sqrt{3}=\frac{a^{2} \sqrt{3}}{\sqrt{4 a^{2}-b^{2}}} \cdot \text{ We have } \\
& V=\frac{1}{3} S_{\text {base }} \cdot S O=\frac{1}{3} \cdot \frac{b \sqrt{4 a^{2}-b^{2}}}{4} \cdot \frac{a^{2} \sqrt{3}}{\sqrt{4 a^{2}-b^{2}}}=\frac{a^{2} b \sqrt{3}}{12} . \\
& \text { Answer: } \frac{a^{2} b \sqrt{3}}{12} .
\end{aligned}
$$