Outside the right angle with vertex $C$, on the extension of its bisector, a point $O$ is taken, such that $O C=\sqrt{2}$. A circle with center at point $O$ and radius 2 is constructed. Find the area of the figure bounded by the sides of the angle and the arc of the circle enclosed between them.
Find the cosine of angle $A O B$

## Solution

Let $A$ and $B$ be the points of intersection of the circle with the sides of the angle. Denote $O A=O B=x$.

By the cosine theorem,

$$
A O^{2}=A C^{2}+O C^{2}-2 A C \cdot O C \cos 135^{\circ}
$$

or

$$
4=x^{2}+2+2 x \sqrt{2} \cdot \frac{\sqrt{2}}{2}, x^{2}+2 x-2=0
$$

From this equation, we find that $x=\sqrt{3}-1$. Then

$$
A B=x \sqrt{2}=\sqrt{2}(\sqrt{3}-1)
$$

By the cosine theorem from triangle $A O B$, we find that

$$
\cos \angle A O B=\frac{O A^{2}+O B^{2}-A B^{2}}{2 O A \cdot O B}=\frac{4+4-2(\sqrt{3}-1)^{2}}{8}=\frac{\sqrt{3}}{2}
$$

Therefore, $\angle A O B=30^{\circ}$ and the area of sector $A O B$ is $\frac{4 \pi}{12}=\frac{\pi}{3}$.

The desired area is the difference between the area of this sector and the sum of the areas of triangles $A C O$ and $B C O$, i.e.,

$$
\frac{\pi}{3}-2 \cdot \frac{1}{2} C A \cdot C O \sin 135^{\circ}=\frac{\pi}{3}-\sqrt{3}+1
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-11.jpg?height=480&width=560&top_left_y=1975&top_left_x=746)

Answer

$\frac{\pi}{3}-\sqrt{2}+1$

Submit a comment