11.131. At the base of an oblique prism lies an equilateral triangle with a side equal to $a$. One of the lateral faces of the prism is perpendicular to the base plane and is a rhombus, the diagonal of which is equal to $b$. Find the volume of the prism.
Solution.

By condition

the face $B B_{1} C_{1} C$ is a rhombus

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0842.jpg?height=533&width=684&top_left_y=757&top_left_x=562)

Fig. 11.28

(Fig. 11.28) and is perpendicular to the base. Then the height $B_{1} D$ of the rhombus is the height of the prism. Let $B B_{1}=a=B C, B_{1} C=b$.

The perimeter $P_{\triangle B B_{1} C}=2 a+b$, by Heron's formula the area of the triangle $BB C \quad S=\sqrt{\left(a+\frac{b}{2}\right)\left(a+\frac{b}{2}-a\right)^{2}\left(a+\frac{b}{2}-b\right)}=\frac{b \sqrt{4 a^{2}-b^{2}}}{4} \cdot$ On the other hand $S=\frac{1}{2} B C \cdot B_{1} D=\frac{a}{2} B_{1} D$. Therefore, $B_{1} D=\frac{2 S}{a}=\frac{b \sqrt{4 a^{2}-b^{2}}}{2 a}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0843.jpg?height=611&width=601&top_left_y=62&top_left_x=95)

Fig. 11.29

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0843.jpg?height=583&width=535&top_left_y=80&top_left_x=738)

Fig. 11.30

The volume of the prism $V=S_{\triangle A B C} \cdot B_{1} D=\frac{a^{2} \sqrt{3}}{4} \cdot \frac{b \sqrt{4 a^{2}-b^{2}}}{2 a}=\frac{a b \sqrt{12 a^{2}-3 b^{2}}}{8}$. Answer: $\frac{a b \sqrt{12 a^{2}-3 b^{2}}}{8}$.