12.448 At the base of a quadrilateral pyramid lies an isosceles trapezoid with bases $a$ and $b(a>2 b)$ and an angle $\varphi$ between the equal segments of its diagonals. The vertex of the pyramid is projected onto the point of intersection of the diagonals of the base. The angles that the lateral faces passing through the bases of the trapezoid form with the base plane are in the ratio $1: 2$. Find the volume of the pyramid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-317.jpg?height=353&width=1076&top_left_y=909&top_left_x=223)

Since $P N = F P \cdot \operatorname{ctg} x$ and $P M = F P \cdot \operatorname{ctg} 2 x$, then

$\frac{P M}{P N} = \frac{b}{a} = (\Delta B C P \sim \triangle A D P) = \frac{\operatorname{tg} x}{\operatorname{tg} 2 x} = \frac{1 - \operatorname{tg}^2 x}{2} \Rightarrow \operatorname{tg} x = \sqrt{\frac{a - 2 b}{a}} \Rightarrow$

$\Rightarrow F P = P N \cdot \operatorname{tg} x = \frac{a}{2} \cdot \operatorname{tg} \frac{\varphi}{2} \cdot \sqrt{\frac{a - 2 b}{a}}$.

Answer: $\quad V_{F A B C D} = \frac{1}{3} S_{A B C D} \cdot F P = \frac{(a + b)^2}{24} \cdot \operatorname{tg}^2 \frac{\varphi}{2} \cdot \sqrt{a(a - 2 b)}$.