2.2. The base of the pyramid is a triangle with sides $a, a$ and $b$. All lateral edges are inclined to the base plane at an angle of $60^{\circ}$. Determine the volume of the pyramid.
2.2. According to the condition, $A B = B C = a$ (Fig. 2.2), therefore

$$
B H = a^{2} - \frac{b^{2}}{4} = \frac{1}{2} \sqrt{4 a^{2} - b^{2}}
$$

From this

$$
S_{\mathrm{OCH}} = \frac{1}{2} A C \cdot B H = \frac{1}{4} b \sqrt{4 a^{2} - b^{2}}
$$

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

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-456.jpg?height=629&width=498&top_left_y=766&top_left_x=655)

Fig. 2.2 of this circle be $R$. Then $O B = R = \frac{a^{2} b}{4 S_{\text {base }}} = \frac{a^{2}}{\sqrt{4 a^{2} - b^{2}}}$. In $\triangle S O B$ we have $O B = \frac{S B}{2}$; hence, $S O = \sqrt{S B^{2} - O B^{2}} = \sqrt{4 O B^{2} - O B^{2}} = O B \sqrt{3} = \frac{a^{2} \sqrt{3}}{\sqrt{4 a^{2} - b^{2}}}$. Therefore,

$$
V = \frac{1}{3} S_{\mathrm{OCH}} \cdot S O = \frac{1}{3} \cdot \frac{b \sqrt{4 a^{2} - b^{2}}}{4} \cdot \frac{a^{2} \sqrt{3}}{\sqrt{4 a^{2} - b^{2}}} = \frac{a^{2} b \sqrt{3}}{12}
$$

Answer: $\frac{a^{2} b \sqrt{3}}{12}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-457.jpg?height=410&width=512&top_left_y=445&top_left_x=46)

Fig. 2.3