Confusing problem determining the exact formula for area of 'complex' shape Draw a square with length n on each side, and then draw arcs of 1/4 circle along the DIAGONAL LINES inside this square. So we have $4$ arcs of 1/4 circle inside this square. My question is, What is the exact formula for area of the region at the center of this square?. I don't know any methods to find this, and I gave up. I want to draw it here using latex but I don't know how to do it, I'm new here. I will be amazed if someone can solve this problem quickly.I wish the formula only contain n, constant pi, rational numbers,sin/cos/tan/log/in, roots of any degree.
 A: In case this is your diagram:


Think of the area in the middle as:


You have the twice the total area of the areas of two sectors, $FEC$ and $BEG$, less areas of $\triangle PGB$ and $\triangle FQC$, less the area of $\triangle PEQ$. Since  the polygons are symmetric, you just have to solve the area of one each:
  $$\frac12 A=2 A_{FEC}-2A_{\triangle FQC}-A_{\triangle PEQ}\tag{1}$$


WLOG, consider your square to lie on the $xy$ -plane, centered at $(0,0)$. Thus, your vertices will be a combination of the points $\left(\pm\frac n2,\pm\frac n2\right)$.
Consider the intersection between the two circles centered at $B,C$. Using the formula for circles, with $r=n$, we know they intersect at:
$$E=\left(0,\frac{1}{2}\left(\sqrt{3}n-n\right)\right)$$
The circles intersect the $x$-axis at the following points:
$$F=\left(-\frac{1}{2}\left(\sqrt{3}n-n\right),0\right)\\
G=\left(\frac{1}{2}\left(\sqrt{3}n-n\right),0\right)$$
For $O=(0,0)$, it's very clear that $FO=EO$, therefore $FE=FO\sqrt2$, and using the cosine law, we get that:
$$\left(\frac{n-\sqrt{3} n}{\sqrt{2}}\right)^2=-2 n^2 \cos (\angle FCE)+n^2+n^2\\
\angle FCE=\frac\pi6=30^\circ$$
Therefore, $$A_{FEC}=\pi n^2 \cdot \frac{30}{360}=\frac{\pi  n^2}{12}$$

To find $A_{FQC}$, consider the line $EC:=\frac{1}{2}\left(\sqrt{3}-1\right)n-\sqrt{3}x$, which intersects the $x$-axis at $Q=\left(\frac{\left(\sqrt{3}-1\right)n}{2\sqrt{3}},0\right)$. Since we already know $F$, then $FQ=\frac{n}{\sqrt{3}}$. Using the distance formula tells us that $QC=\frac{n}{\sqrt{3}}$ thus $A_ {FQC} $ is :
$$A_{FQC}=\frac12\cdot n\cdot \frac{n}{\sqrt{3}}\cdot \sin(30^\circ)=\frac{n^2}{4 \sqrt{3}}$$

To find $A_{\triangle PEQ}$, notice that $PQ=2QO=\frac{\left(\sqrt{3}-1\right) n}{\sqrt{3}}$. Solving for the length of $EQ$, tells us that $\triangle PEQ$ is equilateral, therefore:
$$A_{\triangle PEQ}=\frac12 \left(\frac{\left(\sqrt{3}-1\right) n}{\sqrt{3}}\right)^2\cdot \sin \frac \pi3=\frac{1}{6} \left(2 \sqrt{3}-3\right) n^2$$

Going back to $(1)$, we now get that the area of the figure in the middle is:

$$A=\frac{1}{3} \left(-3 \sqrt{3}+\pi +3\right) n^2 \tag{2}$$


Consider the circle centered at $B:=\frac{1}{2}\left(\sqrt{3n^2-4nx-4x^2}-n\right)$, for $n=2$, the area in question, for $G=\frac{1}{2} \left(2 \sqrt{3}-2\right)$, should be:
$$A=4\int_0^G \frac{1}{2} \left(\sqrt{-4 x^2-8 x+12}-2\right) \, dx$$
Which evaluates to:
$$A=4\cdot\left(\frac{1}{3} \left(-3 \sqrt{3}+\pi +3\right)\right)$$
which is exactly what we get from $(2)$
