12.462 The base of the pyramid is an isosceles trapezoid with an acute angle $\alpha$. This trapezoid is circumscribed around the base circle of a cone. The vertex of the pyramid lies on one of the generatrices of the cone, and its projection onto the base plane coincides with the point of intersection of the diagonals of the trapezoid. Find the volume of the pyramid if the generatrix of the cone is equal to $l$ and forms an angle $\beta$ with the height.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-326.jpg?height=259&width=344&top_left_y=1189&top_left_x=95)

Let the isosceles trapezoid $ABCD$ $(BC \| AD, BC < AD)$ be the base of the given pyramid $FABCD$, $O$ be the center of the base of the given cone, $K$ be the vertex of the cone, $P$ be the projection of $F$ onto the base $ABCD$, and $\alpha = \angle BAD$. Draw the height $MN$ of the trapezoid $ABCD$ passing through points $P$ and $O$.

Since $O$ is the center of the circle inscribed in $ABCD$, then $MN = 2 \cdot MO = 2l \cdot \sin \beta$, $S_{ABCD} = MN(BM + AN)$, $AO$ and $BO$ are the angle bisectors of $\angle A$ and $\angle B$, $AN = ON \cdot \operatorname{ctg} \angle OAN = l \cdot \sin \beta \cdot \operatorname{ctg} \frac{\alpha}{2}$, $BM = OM \cdot \operatorname{tg} \angle OBM = l \cdot \sin \beta \cdot \operatorname{tg} \frac{\alpha}{2}$.

Since $KM$ is the generatrix of the cone, then $KOM \perp ABCD \Rightarrow$ the perpendicular to the plane $ABCD$ through point $P$ lies in $\triangle KOM \Rightarrow$ the vertex $F$ of the pyramid lies on the generatrix $KM$. Therefore,
$FP = MP \cdot \operatorname{ctg} \beta \left(\angle FMO = 90^\circ - \beta\right)$. Additionally,

$\triangle PBC \sim \triangle PAD \Rightarrow \frac{MP}{PN + MP} = \frac{BM}{AN + BM} \Rightarrow MP = \frac{BM \cdot MN}{BM + AN}$.

In the end, we have

Answer: $V_{FABCD} = \frac{1}{3} S_{ABCD} \cdot FP = \frac{\operatorname{ctg} \beta}{3} MN^2 \cdot BM = \frac{2}{3} l^3 \operatorname{tg} \frac{\alpha}{2} \cdot \sin \beta \cdot \sin 2\beta$.