Problem 11.8. In the quadrilateral pyramid $S A B C D$

- the lateral faces $S A B, S B C, S C D, S D A$ have areas 9, 9, 27, 27 respectively;
- the dihedral angles at the edges $A B, B C, C D, D A$ are equal;
- the quadrilateral $A B C D$ is inscribed, and its area is 36.

Find the volume of the pyramid $S A B C D$.
Answer: 54.

Solution. Let the angle between the lateral face and the base of the pyramid be denoted as $\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8af0c885427e3e323cf9g-45.jpg?height=438&width=711&top_left_y=483&top_left_x=371)

Fig. 10: to the solution of problem 11.8

Draw the height $S H$ to the base of the pyramid, and denote its length as $h$ (Fig. 10). Also, draw the heights $S H_{a}, S H_{b}, S H_{c}, S H_{d}$ to the edges $D A, A B, B C, C D$ respectively. Since $S H_{a} \perp D A$ and (by the theorem of three perpendiculars) $H H_{a} \perp D A$, we have $\angle S H_{a} H=\alpha$. From the right triangle $S H_{a} H$, we get $S H_{a}=S H / \sin (\alpha)$.

Similarly, we find that the heights $S H_{b}, S H_{c}$, and $S H_{d}$ also have the same value. Denote their length as $h_{1}=h / \sin (\alpha)$. Using the areas of the lateral faces, we get $D A=27 x, A B=9 x, B C=9 x, C D=27 x$, where $x=2 / h_{1}$.

Since triangles $B A D$ and $B C D$ are equal by three sides, the quadrilateral $A B C D$ is a kite (symmetric with respect to the diagonal $B D$). In the inscribed quadrilateral $A B C D$, the angles $D A B$ and $D C B$ are equal and supplement each other to $180^{\circ}$, so they are each $90^{\circ}$. The area of the right triangle $D A B$ is thus $\frac{1}{2} \cdot 27 x \cdot 9 x=\frac{1}{2} 243 x^{2}$; the area of the entire base is twice this, and equals $243 x^{2}$. According to the problem, it is equal to 36, from which we find $x=\sqrt{36 / 243}=2 /(3 \sqrt{3})$, so $h_{1}=2 / x=3 \sqrt{3}$.

Now we find $\alpha$. For this, consider how the areas of the lateral faces change when projected onto $A B C D$. The heights of the lateral faces, drawn from the vertex $S$, are multiplied by $\cos (\alpha)$, while the corresponding bases of the triangles remain unchanged; thus, the areas are multiplied by $\cos (\alpha)$. On the other hand, the union of the projections of the lateral faces is exactly the quadrilateral $A B C D$, from which we get $36=(9+9+27+27) \cdot \cos (\alpha)$. Therefore, $\cos (\alpha)=\frac{1}{2}$, so $\alpha=\pi / 3$.

Finally, we find $h=h_{1} \sin (\alpha)=4.5$ and the volume of the pyramid $\frac{1}{3} \cdot S_{A B C D} \cdot h=\frac{1}{3} \cdot 36 \cdot 4.5=54$.