10,11

In a triangular pyramid, each lateral edge is equal to 1, and the lateral faces are equal in area. Find the volume of the pyramid, given that one of the dihedral angles at the base is a right angle.
Let $S H$ be the height of the given triangular pyramid $S A B C$ with vertex $S$, and the dihedral angle at edge $B C$ is $90^{\circ}$.

Since the lateral edges of the pyramid are equal, the height of the pyramid passes through the center of the circle circumscribed around the base, and since the lateral face $S B C$ is perpendicular to the plane of the base, the height of the pyramid lies in this face, meaning that point $H$ lies on edge $B C$. Furthermore, triangle $S B C$ is isosceles, so $H$ is the midpoint of $B C$.

The center $H$ of the circumscribed circle of triangle $A B C$ lies on side $B C$, so triangle $A B C$ is a right triangle, and its hypotenuse $B C$ is equal to the diameter of the circumscribed circle.

The lateral faces of the pyramid are equal in area, so

$$
\frac{1}{2} S A \cdot S B \sin \angle A S B=\frac{1}{2} S B \cdot S C \sin \angle B S C=\frac{1}{2} S A \cdot S C \sin \angle A S C
$$

from which we find that $\sin \angle A S B=\sin \angle B S C=\sin \angle A S C$, and since $B S C$ is the largest of the planar angles at the vertex of the pyramid ($B C$ is the largest side of triangle $A B C$), then $\angle B S C>90^{\circ}$, and $\angle A S B=\angle A S C=180^{\circ}-\angle B S C$. Let $R$ be the radius of the circumscribed circle of the right triangle $A B C$. Then $B C=2 R$, and since $A B=A C$, we have $A B=A C=R \sqrt{2}$. By the cosine theorem,

$$
\begin{aligned}
& \cos \angle B S C=\frac{S B^{2}+S C^{2}-B C^{2}}{2 S B \cdot S C}=\frac{1+1-4 R^{2}}{2 \cdot 1 \cdot 1}=2-4 R^2 \\
& \cos \angle A S B=\frac{S A^{2}+S B^{2}-A B^{2}}{2 S A \cdot S B}=\frac{1+1-2 R^{2}}{2 \cdot 1 \cdot 1}=2-2 R^2
\end{aligned}
$$

and since $\angle B S C=180^{\circ}-\angle A S B$, then $\cos \angle B S C=-\cos \angle A S B$, or $2-4 R^2=2 R^2-2$, from which we find that $R=\sqrt{\frac{2}{3}}$. Therefore,

$$
\begin{aligned}
V_{S A B C} & =\frac{1}{3} S_{\triangle A B C} \cdot S H=\frac{1}{3} \cdot \frac{1}{2} R \sqrt{2} \cdot R \sqrt{2} \cdot \sqrt{1-K^{2}}= \\
& =\frac{1}{3} R^2 \sqrt{1-K^{2}}=\frac{1}{3} \cdot \frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{9 \sqrt{3}}
\end{aligned}
$$

## Answer

$\frac{2}{9 \sqrt{3}} \cdot$