2.30. Find the volume of a regular triangular pyramid, where the dihedral angle at the vertex is $90^{\circ}$, and the distance between a lateral edge and the opposite side of the base is $d$.
2.30. According to the condition, $B S \perp S A$ and $B S \perp S C$ (Fig. 2.25), i.e., $B S$ is perpendicular to the face $S A C$ and $S D=d$. Therefore, the required volume $V=\frac{1}{3} S_{\triangle A C S} \cdot B S$. In $\triangle S A D$ we have $\angle S D A=90^{\circ}, \angle A S D=45^{\circ}$, from which $A D=S D=d$ and $S_{\triangle A C S}=d^{2}$. Further, in $\triangle B S D$ we have $\angle B S D=90^{\circ}, B D=2 d \cdot \frac{\sqrt{3}}{2}=d \sqrt{3}$, from which

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-470.jpg?height=502&width=492&top_left_y=1028&top_left_x=660)

Fig. 2.25 $B S=\sqrt{B D^{2}-S D^{2}}=\sqrt{3 d^{2}-d^{2}}=d \sqrt{2}$. Finally, we find

$$
V=\frac{1}{3} d^{2} \cdot d \sqrt{2}=\frac{1}{3} d^{3} \sqrt{2}
$$

Answer: $\frac{d^{3} \sqrt{2}}{3}$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-471.jpg?height=459&width=400&top_left_y=109&top_left_x=46)

Fig. 2.26