12.153. A circle of radius $r$ is inscribed in a sector of radius $R$. Find the perimeter of the sector.
## Solution.

Let $D$ be the point of tangency of the circle inscribed in the sector $A O B$ (Fig. 12.26), with the center $O_{1}$ located on the bisector $O K$ of the angle $A O B$.

Let $\angle A O B=\alpha$, then $\angle D O K=\frac{\alpha}{2}$ and from $\triangle O D O_{1}\left(\angle O D O_{1}=\frac{\pi}{2}\right)$ $\sin \frac{\alpha}{2}=\frac{O_{1} D}{O O_{1}}=\frac{r}{R-r}$. Therefore, $\alpha=2 \arcsin \frac{r}{R-r}$, the length of the arc $A K B$ is $\alpha R=2 R \arcsin \frac{r}{R-r}$ and the perimeter of the sector is $2 R\left(1+\arcsin \frac{r}{R-r}\right)$. Answer: $2 R\left(1+\arcsin \frac{r}{R-r}\right)$.