10.311. The lengths of the diagonals of a rhombus are in the ratio $3: 4$. How many times larger is the area of the rhombus compared to the area of the circle inscribed in it?
## Solution.

Let point $O$ be the intersection of the diagonals and the center of the inscribed circle of radius $r$ of the rhombus $ABCD$, with $BD:AC=3:4$ (Fig. 10.100). $\frac{BO}{CO}=\frac{BD}{AC}=\frac{3}{4}$. Let $BO=3x$, then $CO=4x, BC=5x$ (from $\left.\triangle BOC\right)$. Let $P$ be the point of tangency of the circle with side $BC$. Then $OP \perp BC, OP=r$. $S_{\triangle BOC}=\frac{1}{2} BC \cdot OP=\frac{1}{2} OC \cdot OB . OP=r=\frac{OC \cdot OB}{BC}=\frac{12x}{5}$. The area of the circle $S_{1}=\pi r^{2}=\frac{144 \pi x^{2}}{25}$. The area of the rhombus $S_{2}=\frac{1}{2} AC \cdot BD=2 BO \cdot OC=24 x^{2}$. $\frac{S_{2}}{S_{1}}=\frac{24 x^{2} \cdot 25}{144 \pi x^{2}}=\frac{25}{6 \pi}$

Answer: $\frac{25}{6 \pi}$.