## Task B-2.5.

From each vertex of the square, a circular arc of radius $a$ has been drawn, where $a$ is the length of the side of the square. Determine the perimeter and area of the shaded figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_d18d2bbdee068dba95dfg-10.jpg?height=420&width=422&top_left_y=430&top_left_x=820)
## First Solution.

Notice that the area of the square is equal to the sum of four areas of the shaded part $P_{x}$ and the desired area. First, we will calculate the area $P_{x}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_d18d2bbdee068dba95dfg-10.jpg?height=611&width=642&top_left_y=1228&top_left_x=707)

The area $P_{x}$ is obtained by subtracting the area of the circular segment of the circle with center $D$, bounded by the chord $\overline{A F}$ and the arc $\overparen{A F}$, from the area of the circular sector $A B F$. Triangle $A F D$ is an equilateral triangle (its sides are equal to the radii of the drawn arcs). Therefore, $\varangle F A B=30^{\circ}$, $\varangle F D A=60^{\circ}$. It follows that

$$
P_{x}=a^{2} \pi \cdot \frac{30^{\circ}}{360^{\circ}}-\left(a^{2} \pi \cdot \frac{60^{\circ}}{360^{\circ}}-\frac{a^{2} \sqrt{3}}{4}\right)=\frac{a^{2} \sqrt{3}}{4}-\frac{a^{2} \pi}{12}
$$

The desired area is equal to

$$
P=a^{2}-4 P_{x}=a^{2}\left(1-\sqrt{3}+\frac{\pi}{3}\right)
$$

The desired perimeter is equal to four times the length of the arc $\overparen{E F}$, which corresponds to a central angle of $30^{\circ}$. Therefore, $o=4 \cdot \frac{1}{12} \cdot 2 a \pi=\frac{2 a \pi}{3}$.