11.158. In a triangular pyramid, each of the lateral edges of which is equal to $a$, one of the planar angles at the vertex of the pyramid is right, and each of the others is $60^{\circ}$. Calculate the volume of the pyramid.
Solution.

Let $D$ be the vertex of the pyramid $DABC$ (Fig. 11.53). According to the problem, $DA=DB=DC=a, \angle ADC=\angle BDC=60^{\circ}, \angle ADB=90^{\circ}$.

Then $\triangle ADC$ and $\triangle BDC$ are equilateral, $AC=BC=a, AB=a\sqrt{2}$. Since $AC^{2}+BC^{2}=AB^{2}$, then $\angle ACB=90^{\circ}$ and $S_{\triangle ACB}=\frac{a^{2}}{2}$. DO is the height of the pyramid. Since the lateral edges of the pyramid are equal, point $O$ is the center of the circle circumscribed around the base of the pyramid. Since $\triangle ABC$ is a right triangle, point $O$ is the midpoint of the hypotenuse $AB, AO=\frac{a\sqrt{2}}{2}$. From

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0864.jpg?height=632&width=553&top_left_y=76&top_left_x=95)

Fig. 11.53

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0864.jpg?height=563&width=555&top_left_y=142&top_left_x=687)

Fig. 11.54

$\triangle AOD\left(\angle AOD=90^{\circ}\right): DO=\sqrt{AD^{2}-AO^{2}}=\sqrt{a^{2}-\frac{a^{2}}{2}}=\frac{a\sqrt{2}}{2}$. The volume of the pyramid $V=\frac{1}{3} S_{\triangle ACB} \cdot DO=\frac{a^{3}\sqrt{2}}{12}$.

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