7.6 On a horizontal line, points $A$ and $B$ are marked, the distance between which is 4. Above the line, two semicircles with a radius of 2 are constructed, centered at points

A and B. Additionally, one circle, also with a radius of 2, is constructed, for which the point of intersection of these semicircles is the lowest point - see the figure. Find the area of the figure obtained by subtracting from the area of the circle the parts common to it and the two semicircles (the shaded area in the figure). Justify your answer.
Solution. Method 1. Let the midpoint of segment $A B$ be $E$, the center of the circle be $O$, and the second points of intersection of the circle and the semicircles be $C$ and $D$ (see figure). Since the radii of the semicircles and the radius of the circle are equal, segments $A D, E O$, and $B C$ will be equal to each other. Then points $D, O, C$ lie on the same line (parallel to line $A B$), and quadrilateral $A B C D$ will be a rectangle with sides 4 and 2. The area of this rectangle is 8. We will show that this rectangle has the same area as the desired figure. Indeed, the figures marked with the Roman numeral I (and also with the number II) are equal due to symmetry. $D C$ is a diameter, which divides the area of the circle in half. The figure in question is a semicircle plus figure I plus figure II. The rectangle consists of these same figures, only arranged differently.
![](https://cdn.mathpix.com/cropped/2024_05_06_6f59ebd1a852c759e945g-8.jpg?height=318&width=605&top_left_y=709&top_left_x=1248)

To the condition of problem 7.6, the area of the rectangle is 8. We will show that this rectangle has the same area as the desired figure. Indeed, the figures marked with the Roman numeral I (and also with the number II) are equal due to symmetry. $D C$ is a diameter, which divides the area of the circle in half. The figure in question is a semicircle plus figure I plus figure II. The rectangle consists of these same figures, only arranged differently.
![](https://cdn.mathpix.com/cropped/2024_05_06_6f59ebd1a852c759e945g-8.jpg?height=328&width=1286&top_left_y=1432&top_left_x=409)

To the solution of problem 7.6.

Method 1.

Method 2.

Method 2. Consider the square $A B C D$ lying on the same side of line $A B$ as the figure of interest. Draw semicircles with centers at points $C$ and $D$ with a radius of 2, as shown in the figure. The points of tangency of the semicircles lie exactly on the drawn circle and divide it into 4 equal parts. If we now connect these points sequentially, a square inscribed in the circle is formed. Our figure will result from this square if we replace two segments shaded in a "striped" pattern with two such segments shaded in a "checkerboard" pattern, so its area is equal to the area of the square. The diagonal of this square is equal to the length of $A B$, that is, equal to 4. Then the area of the square (and the area of our figure) is 8.

Answer: 8.

| THERE IS IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| There are arithmetic errors in the correct solution, possibly leading to an incorrect answer | 6 points |
| There is an idea to find a figure with the same composition, the area of which is easy to find | 3 points |
| The correct approach is indicated (for example, subtract the areas of the common parts of the circle and semicircles from the area of the circle), but this plan is not implemented | 1 point |
| Reasoning and calculations from which the solution approach is not clear | 0 points |