[Theorem of Three Perpendiculars]

The base of the pyramid is a rhombus with an acute angle of $30^{\circ}$. The lateral faces are inclined to the base plane at an angle of $60^{\circ}$. Find the volume of the pyramid if the radius of the circle inscribed in the rhombus is $r$.
Prove that the height of the pyramid passes through the center of the rhombus.

## Solution

Let $H$ be the height of the pyramid $P A B C D$, the base of which is a rhombus $A B C D$ with an angle of $30^{\circ}$ at vertex $A$, and $P M$ is the perpendicular dropped to side $B C$. By the theorem of three perpendiculars, $H M \perp B C$. Therefore, $P M H$ is the linear angle of the dihedral angle between the lateral face $B C P$ and the plane of the base $A B C D$. Hence, $\angle P M H = 60^{\circ}$.

Dropping perpendiculars from vertex $P$ to the other sides of the rhombus and considering the resulting right triangles with the common leg $P H$ and the opposite angle equal to $60^{\circ}$, we can prove that point $H$ is equidistant from all four lines containing the sides of the rhombus $A B C D$. Therefore, $H$ is the center of the circle inscribed in this rhombus, i.e., the point of intersection of its diagonals.

Dropping the perpendicular $B F$ from the vertex of the rhombus to side $A D$, we have $B F = 2 r$. From the right triangle $A B F$, we find that $A B = 2 \cdot B F = 4 r$. Thus,

$$
S(A B C D) = A D \cdot B F \cdot \sin 30^{\circ} = A B \cdot B F \cdot \sin 30^{\circ} = 8 r^{2}
$$

From the right triangle $P M H$, we find that

$$
P H = H M \cdot \operatorname{tg} 60^{\circ} = r \sqrt{3}.
$$

Therefore,

$$
V(P A B C D) = \frac{1}{3} S(A B C D) \cdot P H = \frac{1}{3} 8 r^{2} \cdot r \sqrt{3} = \frac{8}{3} r^{3} \sqrt{3}
$$

## Answer

$$
\frac{8}{3} r^{3} \sqrt{3} .
$$

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