10.131. A circle with an area of $Q$ is inscribed in a rhombus with an acute angle of $30^{\circ}$. Find the area of the rhombus.
## Solution.

Draw radii $O K, O L, O M, O N$ to the points of tangency (Fig. 10.118). $\angle A B C=180^{\circ}-\angle B A D=180^{\circ}-30^{\circ}=150^{\circ}$. Since the diagonals of a rhombus are angle bisectors, $\angle A B O=\frac{150^{\circ}}{2}=75^{\circ}$. Therefore, $B O=\frac{K O}{\sin 75^{\circ}}=\frac{K O}{\cos 15^{\circ}}$. Then $A B=\frac{B O}{\sin \angle O A B}=\frac{K O}{\sin 15^{\circ} \cos 15^{\circ}}=\frac{2 K O}{\sin 30^{\circ}}=$ $=4 K O=4 r$. For the rhombus, $A B=B C=C D=A D=4 r$. The area of the rhombus $S=A B \cdot A D \sin 30^{\circ}=16 r^{2} \sin 30^{\circ}$. The area of the circle $\dot{Q}=\pi r^{2}$, hence $r^{2}=\frac{Q}{\pi} \cdot$ Therefore, finally $S=\frac{8 Q}{\pi}$.

Answer: $\frac{8 Q}{\pi}$.