A semicircle of radius $r$ is divided into 3 equal parts by points, and the points of division are connected by chords to one end of the diameter that spans this semicircle. Find the area of the figure bounded by the two chords and the arc between them.
Let $C D$ be a diameter, $O$ be the midpoint of $C D$, and $D A, A B$ and $B C$ be chords. Then the required area is equal to the area of

sector $A O B$,

## Solution

Let $C D$ be a diameter, $O$ be the midpoint of $C D$, and $D A, A B$ and $B C$ be chords. Triangles $A O D, A O B$ and $B O C$ are equilateral, $\angle A O D = \angle A O B = \angle B O C = 60^{\circ}$, so $A B \parallel D C$. Therefore, $S_{\triangle \text{ ADB }} = S_{\triangle \text{ AOB }}$. The problem requires finding the area of the figure composed of triangle $A D B$ and the segment $A B$ bounded by the arc $A B$ not containing point $D$. Note that the sector $A O B$ consists of triangle $A O B$ and this same segment, hence the required area is equal to the area of sector $A O B$, i.e., one-sixth of the area of the corresponding circle. Thus, the required area is $\frac{\pi r^{2}}{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-17.jpg?height=269&width=512&top_left_y=2538&top_left_x=773)

Answer

Problem