10.398 Four equal circles are inscribed in a circle of radius $R$, each touching the given circle and two adjacent ones. Calculate the area of the figure bounded by these four circles.
Solution: Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the centers of four circles of radius $r$, each touching two adjacent ones, and inscribed in a circle of radius $R$ with center $P$. Clearly, $P$ coincides with the center of the square $O_{1} O_{2} O_{3} O_{4}$ with side $2 r$. Moreover, $P O_{1}=R-r$ (the line $P O_{1}$ passes through the point of tangency of the circles $P$ and $O_{1}$). Hence, $2 r=O_{1} O_{2}=P O_{1} \cdot \sqrt{2}=(R-r) \sqrt{2} \Rightarrow r=R(\sqrt{2}-1)$.

The area of the figure $\Phi$, bounded by the circles $O_{i}, i=1,2,3,4$, is equal to the area of the square $O_{1} O_{2} O_{3} O_{4}$ minus the areas of four $90^{\circ}$ sectors of radius $r$, i.e.,

$$
S_{\Phi}=4 r^{2}-4 \frac{\pi r^{2}}{4}=R^{2}(4-\pi)(3-2 \sqrt{2})
$$

Answer: $\quad R^{2}(4-\pi)(3-2 \sqrt{2})$.