11.160. The base of the pyramid is a rhombus with diagonals $d_{1}$ and $d_{2}$. The height of the pyramid passes through the vertex of the acute angle of the rhombus. The area of the diagonal section, passing through the smaller diagonal, is $Q$. Calculate the volume of the pyramid given that $d_{1}>d_{2}$.
Solution.

The required volume $V=\frac{1}{3} S_{\text{base}} \cdot S A$, where $S_{\text{base}}=\frac{1}{2} d_{1} d_{2}$ (Fig. 11.55). In $\triangle S A O$, we have $S A=\sqrt{S O^{2}-A O^{2}}$, where $A O=\frac{d_{1}}{2}$, and $S O=\frac{2 Q}{d_{2}}$, since by the condition $\frac{1}{2} d_{2} \cdot S O=Q$. Therefore, $S A=\sqrt{\frac{4 Q^{2}}{d_{2}^{2}}-\frac{d_{1}^{2}}{4}}=\frac{\sqrt{16 Q^{2}-d_{1}^{2} d_{2}^{2}}}{2 d_{2}}$. We obtain $V=\frac{1}{3} \cdot \frac{1}{2} d_{1} d_{2} \cdot \frac{\sqrt{16 Q^{2}-d_{1}^{2} d_{2}^{2}}}{2 d_{2}}=\frac{d_{1}}{12} \sqrt{16 Q^{2}-d_{1}^{2} d_{2}^{2}}$.

Answer: $\frac{d_{1}}{12} \sqrt{16 Q^{2}-d_{1}^{2} d_{2}^{2}}$.

11.161. In a triangular pyramid, two lateral faces are mutually perpendicular. The areas of these faces are $P$ and $Q$, and the length of their common edge is $a$. Determine the volume of the pyramid.

Solution.

Let in the pyramid $A B C S$ the faces $A B C$ and $A B S$ be mutually perpendicular, $\triangle A B C$ be the base of the pyramid, and point $S$ be its vertex (Fig. 11.56), $S_{\triangle A B C}=P, S_{\triangle A S B}=Q, A B=a$. In the plane $A B S$, drop a perpendicular $S K$ to $A B$. Then the segment $S K$ is perpendicular to the plane $A B C$ and is the height of the pyramid. From $\triangle A S B: S K=\frac{2 S_{\triangle A S B}}{A B}=\frac{2 Q}{a}$. The volume of the pyramid $V=\frac{1}{3} S_{\triangle A B C} \cdot S K=\frac{2 P Q}{3 a}$.

Answer: $\frac{2 P Q}{3 a}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0866.jpg?height=422&width=601&top_left_y=225&top_left_x=93)

Fig. 11.55

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0866.jpg?height=612&width=551&top_left_y=33&top_left_x=710)

Fig. 11.56