Two opposite lateral faces of the quadrilateral pyramid $SABCD$ are perpendicular to the base, the height of the pyramid is $\sqrt{5}$. At the base of the pyramid lies an isosceles trapezoid $ABCD$ ($AD = BC$), circumscribed around a circle, such that $AB = 6$, $\angle BAD = \frac{\pi}{3}$. Find the distance from point $D$ to the plane $SAB$. Inside the pyramid, a cone is located such that the circle of its base is inscribed in the triangle $SCD$, and the vertex lies on the face $SAB$. Find the volume of the cone.
Since the planes of the opposite lateral faces SAD and SBC (Fig.1) are perpendicular to the base plane of the pyramid, the line $l$ of their intersection is also perpendicular to the base plane and passes through the common point $S$ of the planes $S A D$ and $S B C$. Since the point $H$ of intersection of the lines $A D$ and $B C$ is also a common point of these planes, the line $l$ coincides with the line $S H$, meaning $S H$ is the height of the pyramid. The circle with center $O$, inscribed in the isosceles trapezoid $A B C D$, is also the inscribed circle of the equilateral triangle $A H B$ with side $A B=6$. If $r$ is the radius of the circle, and $M$ and $N$ are the midpoints of $A B$ and $C D$ respectively, then

$$
M N=2 r=2 \cdot \frac{1}{3} H M=2 \cdot \frac{1}{3} \cdot \frac{A B \sqrt{3}}{2}=2 \cdot \frac{1}{3} \cdot \frac{6 \sqrt{3}}{2}=2 \sqrt{3}
$$

The line $C D$ is parallel to the line $A B$, which lies in the plane $A B S$, so the line $C D$ is parallel to this plane. Therefore, the distance from point $D$ to the plane $A B S$ is equal to the distance from this plane to any point on the line $C D$, in particular, from point $N$. Let $P$ be the foot of the perpendicular dropped from point $N$ to the line $S M$. Since the line $A B$ is perpendicular to the intersecting lines $H M$ and $S M$ in the plane $S M H$, the line $A B$ is perpendicular to the plane $S N M$, meaning $P N \perp A B$ and $N P$ is the perpendicular to the plane $A S B$. Therefore, the distance from point $D$ to the plane $S A B$ is equal to the length of the segment $N P$. Consider the right triangle $S M H$ (Fig.3). Let $\angle S M H=\alpha$. Then

$$
\operatorname{ctg} \alpha=\frac{M H}{S H}=\frac{3 \sqrt{3}}{\sqrt{5}}, \sin \alpha=\frac{1}{\sqrt{1+\alpha g^{2} q^{2}}}=\frac{1}{\sqrt{1+\frac{37}{5}}}=\frac{\sqrt{5}}{4 \sqrt{2}} \text {. }
$$

Therefore,

$$
P N=M N \sin \alpha=2 \sqrt{3} \cdot \frac{\sqrt{5}}{4 \sqrt{2}}=\frac{\sqrt{30}}{4}
$$

From the similarity of triangles $D H C$ and $A H B$, we find that

$$
C D=A B \cdot \frac{H N}{H M}=6 \cdot \frac{1}{3}=2
$$

Then

$$
H C=H D=C D=2, S D=S C=\sqrt{S H^{2}+H C^{2}}=\sqrt{5+4}=3
$$

From the right triangles $S H M$ and $S H N$, we find that

$$
S M=\sqrt{S H^{2}+M H^{2}}=\sqrt{5+27}=4 \sqrt{2}, S N=\sqrt{S H^{2}+N H^{2}}=\sqrt{5+3}=2 \sqrt{2}
$$

Let $Q$ be the center of the base circle of the cone mentioned in the problem, $R$ be the radius of this circle, and $p$ be the semiperimeter of triangle $C S D$. Since the base circle of the cone is the inscribed circle of triangle $C S D$,

![](https://cdn.mathpix.com/cropped/2024_05_06_1f91526468b1b4d32730g-29.jpg?height=103&width=595&top_left_y=0&top_left_x=734)

Let the perpendicular from point $Q$ to the segment $S N$ intersect the segment $S M$ at point $E$. Then $E Q$ is the height of the cone. Let $\angle M S N=\beta$ (Fig.3). By the cosine rule,

$$
\cos \beta=\frac{S M^{2}+S N^{2}-M N^{2}}{2 S \cdot S M N}=\frac{(4 \sqrt{2})^{2}+(2 \sqrt{2})^{2}-(2 \sqrt{3})^{2}}{2 \cdot 4 \sqrt{2} \cdot 2 \sqrt{2}}=\frac{7}{8}
$$

Then

$$
\operatorname{tg} \beta=\sqrt{\frac{1}{\cos ^{2} \beta}-1}=\sqrt{\frac{64}{49}-1}=\frac{\sqrt{15}}{7}
$$

From the right triangle $S Q E$, we find that

$$
E Q=S Q \operatorname{tg} \beta=\frac{3 \sqrt{2}}{2} \cdot \frac{\sqrt{15}}{7}=\frac{3 \sqrt{30}}{14}
$$

Let $V$ be the volume of the cone. Then

$$
V=\frac{1}{3} \pi R 2 \cdot E Q=\frac{1}{3} \pi \cdot\left(\frac{\sqrt{2}}{2}\right) 2 \cdot \frac{3 \sqrt{30}}{14}=\frac{\pi \sqrt{30}}{28}
$$

## Answer

$\sqrt{30}, \pi \sqrt{30}$