Calculating area of a shape with circular boundaries with elementary methods Question. $\square ABCD$ is a square with $AB = 10$. Circle $O$ inscribes the $\square ABCD$. The center of the arc is $A$. What is the area of the colored area?

Explanation: This problem can be solved by integral. But, this problem is from an elementary math book, which means that integral is not a good solution for this.
Our teacher solved this problem with integral, but she was not able to solve it with elementary math. She asked my school students about this problem, but none of us were able to solve it. How can we solve the problem using only elementary math?
Korean elementary math does not contain:


*

*irrationals and imaginary numbers

*functions

*Of course, angle functions (such as $sin$, $cos$, $tan$, etc)

*inscribed angle

*non-linear equations

*similarity and congruence


If you are unsure that you can use certain formulas or something else, comment me and I'll answer you.
Edit 1. Actually, we do not study $\pi$ in elementary school. We just assimilate it to $3.14$. But it's not so important, so let's use $\pi$.
 A: Hint
The two yellow areas are equal.
You can easily calculate $$2 \mbox{yellow}+\mbox{blue}+\mbox{purple}=\mbox{Area} ABCD-\frac{1}{4} \mbox{Area} C(A, AB)$$
Next
$$4\mbox{purple}=\mbox{Area} ABCD- \mbox{Area} C(0, OA)$$
Finally
$$\mbox{Blue}=\mbox{Area} (sector \, OEF) -\mbox{Area} (sector \, AEF) +\mbox{Area}(AEOF)$$
A: 
Evaluate the areas of the circular sectors ABP and OPE, as well as the quadrilateral APOF and then subtract them from the half square to get the shaded area. (A unit square assumed below.)
With A being the origin, the two circles are
$$r=1; \space \space r(\sin\theta + \cos\theta)=r^2+\frac{1}{4}.$$ 
Eliminate $r$ to obtain,
$$\sin\beta = \frac{5+\sqrt{7}}{8},\>\>\>\>\>
\sin\alpha = \frac{0.5-\cos\beta}{0.5}=\frac{\sqrt{7}-1}{4}$$
Then, the area of the quadrilateral AFOP which is made of two triangles AFP and FPO, is
$$A_q = \frac{1}{4}\sin\beta +  \frac{1}{4}\left(\frac{1}{2}-\cos\beta \right)=\frac{2+\sqrt{7}}{16}
$$
And the area of the two sectors is
$$A_c=\frac{1}{2}\left( \frac{\pi}{2}-\beta\right) + \frac{\alpha}{8}
=\frac{\pi}{8}-\frac{1}{8}\sin^{-1} \left(\frac{23\sqrt{7}+67}{128}\right). $$
Therefore, the shaded area is
$$A_s=\frac{1}{2}-A_c-A_q=\frac{6-\sqrt{7}-2\pi}{16}+\frac{1}{8}\sin^{-1} \left(\frac{23\sqrt{7}+67}{128}\right). $$
