10.120. Two circles of radius $R$ with centers at points $O_{1}$ and $O_{2}$ touch each other. They are intersected by a line at points $A, B, C$ and $D$ such that $A B=B C=C D$. Find the area of the quadrilateral $O_{1} A D O_{2}$.
Solution.

$\triangle A O_{1} B, \triangle O_{1} B O, \triangle B O C, \triangle O C O_{2}, \triangle C O_{2} D$ - equilateral with side $R$ (Fig. 10.112). The area of $\triangle A O_{1} B \quad S_{\triangle A O, B}=\frac{R^{2} \sqrt{3}}{4}$. The area of the desired quadrilateral $O_{1} A D O_{2} \quad S=5 S_{\triangle A O_{1} B}=\frac{5 R^{2} \sqrt{3}}{4}$.

Answer: $\frac{5 R^{2} \sqrt{3}}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-613.jpg?height=352&width=562&top_left_y=96&top_left_x=127)

Fig. 10.113