10.340. A circle of radius $R$ is inscribed in a rhombus. Find the area of the rhombus if its larger diagonal is 4 times the radius of the inscribed circle.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0803.jpg?height=422&width=523&top_left_y=80&top_left_x=94)

Fig. 10.125
## Solution.

Let $O$ be the center of the circle inscribed in the rhombus $ABCD$ (Fig. 10.125), $M$ be the point of tangency of the circle with the side of the rhombus. Then $OM=R, AO=2R$, $\angle AMO=90^{\circ}$. Therefore, $\angle OAM=30^{\circ}, \angle BAD=60^{\circ}$. Let $BE$ be the height of the rhombus. Then $BE=2R$ and from $\triangle AEB$ $\left(\angle AEB=90^{\circ}\right):$

$AB=\frac{BE}{\sin \angle BAD}=\frac{4R}{\sqrt{3}}=\frac{4R\sqrt{3}}{3} \cdot$ The area of the rhombus $S=AD \cdot BE=\frac{8R^2\sqrt{3}}{3}$. Answer: $\frac{8R^2\sqrt{3}}{3}$.