2.34. 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 $d_{1}>d_{2}$.
2.34. The desired volume $V=\frac{1}{3} S_{\text {osn }} \cdot S A$, where $S_{\text {osn }}=\frac{1}{2} d_{1} d_{2}$ (Fig.

2.29). In $\triangle S A O$ we have $S A=\sqrt{S O^{2}-A O^{2}}$, and $A O=\frac{d_{1}}{2}$, while

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-473.jpg?height=413&width=481&top_left_y=1074&top_left_x=350)

Fig. 2.29

$S O=\frac{2 Q}{d_{2}}$, since by the condition $\frac{1}{2} d_{2} \cdot S O=Q$. Therefore,

$$
S A=\sqrt{\frac{4 Q^{2}}{d_{2}^{2}}-\frac{d_{1}^{2}}{4}}=\frac{\sqrt{16 Q^{2}-d_{1}^{2} d_{2}^{2}}}{2 d_{2}}
$$

Finally, we get

$$
V=\frac{1}{3} \cdot \frac{1}{2} d_{1} d_{2} \cdot \frac{\sqrt{16 Q^{2}-d_{1}^{2} d_{2}^{2}}}{2 d_{2}}=\frac{d_{1}}{12} \sqrt{16 Q^{2}-d_{1}^{2} d_{2}^{2}}
$$

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