2. The base of a pyramid is a right trapezoid with an angle of $30^{\circ}$ and a longer leg of length 12. The lateral faces of the pyramid make equal angles with the base. Determine the height of the pyramid if the sum of the areas of the lateral faces is 90.
Solution. Let $ABCD$ be the base of the pyramid with apex $T$, where $AB \| CD$, $\overline{BC}=12$, $\measuredangle BAD=90^{\circ}$, $\measuredangle ABC=30^{\circ}$. It immediately follows that $\overline{AD}=\overline{BC} \sin 30^{\circ}=6$. Let $H$ be the height of the pyramid dropped from vertex $T$, and $E$ its foot. If the angle that each of the sides makes with the base is $\varphi$, then $\triangle ABT, \triangle BCT, \triangle CDT, \triangle DAT$ have equal heights dropped from vertex $T$, $h=\frac{H}{\sin \varphi}$. From the condition of the problem

$$
P_{ABT}+P_{BCT}+P_{CDT}+P_{DAT}=90
$$

it follows that

$$
\frac{h}{2}(\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA})=90
$$

On the other hand, the distance from point $E$ to each of the sides of $ABCD$ is $r=H \operatorname{ctg} \varphi$, so $ABCD$ is a tangential trapezoid with the center of the inscribed circle at $E$. Then $\overline{AB}+\overline{CD}=\overline{BC}+\overline{DA}=18$. Substituting in the above equation gives $h=5$ and we also conclude that $r=\frac{\overline{AD}}{2}=3$. Finally,

![](https://cdn.mathpix.com/cropped/2024_06_05_84a5520b68919fd51f4cg-07.jpg?height=469&width=413&top_left_y=477&top_left_x=1036)

$$
H=\sqrt{h^{2}-r^{2}}=4
$$