Arched area

the side of small square is $a$ . $\begin{aligned} A_{square} &= a^2 = 2^2 + 2^2 -2 \times 2 \times 2 \times \cos {30\degree} \\ &= 8 - 4\sqrt 3 \\ A_{circular segment} &= \dfrac {\pi \times 2^2}{12} - \dfrac 12 \times 2 \times 2 \times \sin {30\degree} \\ &= \dfrac {\pi}3 - 1 \\ A &= A_{square} + 4 \times A_{circular segment} \\ &= \dfrac {4\pi}{3} + 4- 4\sqrt 3 \\ &\approx \boxed{1.26} \end{aligned}$

The 1st step is to realize that the area we're after can be divided into 4 equal shapes: The 2nd step is to figure out the area of one of these shapes:

Label the corners of the square with grid points and set up the equations for the two (quarter) circles: Calculate where the two quarter circles intersect:

x²+y²-4=(x-2)²+y²-4 gives x=1 and y=√3. The other intersection point then is (√3,1): We can now calculate some angles: Sin α= $\frac{√3}{2}$ so α=60° and then β=90°-60°=30°. For the other intersection point this is the same but mirrored: If we now combine the angles we've found we can see we're finally getting somewhere: Together with part of the quarter circle we now have a 30° sector of a circle with a radius of 2. The area of this sector can be calculated with

$\frac{30}{360}$ * π* r²= $\frac{1}{12}$ * π* 4= $\frac{π}{3}$ : To make our shape we only need a small part of this sector, which is the top segment. So we'll need to deduct the area of the triangular shaped bit which can be divided up into two identical triangles: Sin 15°= $\frac{b}{2}$ so b=2* sin 15°. Cos 15°= $\frac{h}{2}$ so h=cos 15°* 2. The area of each the two identical triangles then is $\frac{1}{2}$ * 2* sin 15°* cos 15°* 2=(\frac{1}{2}). The area of the two together then is 2* (\frac{1}{2})=1. So the area of the little segment then is $\frac{π}{3}$ -1.

If we now draw two lines from each end of the segment to the middle point of the square we get a 90°-45°-45° triangle with two sides with length a and a hypotenuse with length 2b: We know 2b=2* 2* sin 15°=4* sin 15° so 2a²=(4* sin 15°)² which means a²=8* (sin 15°)². With this we can calculate the area of this triangle:

$\frac{1}{2}$ * 8* (sin 15°)²=4* ( $\frac{1}{2}$ - $\frac{√3}{4}$ )=2-√3. This added to the area of the small segment gives: 2-√3+ $\frac{π}{3}$ -1= $\frac{π}{3}$ +1-√3. We have 4 of these shapes so the total shaded area is:

4* ( $\frac{π}{3}$ +1-√3)

which approximates to 1,26.