5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle $\alpha=45^{\circ} .(15$ points $)$
![](https://cdn.mathpix.com/cropped/2024_05_06_66ff5449eae3e24affe1g-19.jpg?height=608&width=1530&top_left_y=1101&top_left_x=294)
Solution. Let the area of the semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$, and the sector $A C B=a$. Then, if the area of the segment $A C$ is $b$, we have: $x+b=S_{0}=b+a \Rightarrow x=a$.

The entire shaded area is equal to $x+y=a+y$, but $a+y$ is the area of the circular sector $A B B_{1}$ with radius $2 R$. Therefore, the area of $A B B_{1}=\frac{1}{2}(2 R)^{2} \frac{\pi}{4}=\frac{\pi R^{2}}{2}$.

Answer. $\frac{\pi R^{2}}{2}$.

## VARIANT 7