A Difficult Area Problem involving a Circle and a Square A few days ago, I encountered the following problem:

After a little bit of thinking, I managed to come up with the following solution:

*

*Rotate the square $90^\circ$ clockwise and let the new bottom left corner of the square be $(0,0)$.

*The circle inscribed in the square is hence centered at $(5,5)$ with a radius of $5$. The circle equation thus becomes $(x-5)^{2} + (y-5)^{2} = 25 \Rightarrow y = 5 + \sqrt{25 - (x-5)^{2}}$ in the first quadrant.

*Similarly for the quarter circle, the equation becomes $y = \sqrt{100-x^2}$.

The graph hence looks like this:

My intention is to find the shaded area in the above graph. To do so, first I find $X$ by equating $5 + \sqrt{25 - (x-5)^{2}} = \sqrt{100-x^2} \Rightarrow x=\frac{25 - 5\sqrt{7}}{4}$.
From this, I calculate the area of the shaded region as follows:
$$\text{Area} = (10 \cdot \frac{25 - 5\sqrt{7}}{4} - \int_0^\frac{25 - 5\sqrt{7}}{4} \sqrt{100-x^2} \,\mathrm{d}x) + (10 \cdot (5 - \frac{25 - 5\sqrt{7}}{4}) - \int_\frac{25 - 5\sqrt{7}}{4}^5 5 + \sqrt{25 - (x-5)^{2}} \,\mathrm{d}x) \approx 0.7285$$
Now, the diagram looks like this:

From here, I figured out the shaded area as follows:
$$\text{Area} \approx 10^{2} - \frac{\pi(10^{2})}{4} - (\frac{10^{2} - \pi(5^{2})}{4} + 2 \times 0.7285) \approx \boxed{14.6 \:  \text{cm}^{2}}$$
While I did figure out the correct solution, I find my approach to be rather lengthy. I was wondering if there is a quicker, simpler and more concise method (that probably does not require Calculus) that one can use and I would highly appreciate any answers pertaining to the same.
 A: Here is an alternate solution,

We assume right bottom vertex to be the origin then, equations of circles are
Circle S: $x^2+y^2 = 100$
Circle T: $(x+5)^2+(y-5)^2 = 25$
Solving both equations, intersection points are $A \left(\frac{5}{4}\left(\sqrt7-5\right), \frac{5}{4}\left(\sqrt7+5\right)\right)$ and $B\left(-\frac{5}{4}\left(\sqrt7+5\right), -\frac{5}{4}\left(\sqrt7-5\right)\right)$
So length of chord $AB$ at intersection is $\frac{5 \sqrt7}{\sqrt2}$.
We note that this chord $AB$ is common chord of both circles.
Angle subtended by chord at the center is given by,
At $P$, $\angle APB = \alpha = 2 \arcsin \left(\frac{\sqrt7}{2\sqrt2}\right)$
At $O$, $\angle AOB = \beta = 2 \arcsin \left(\frac{\sqrt7}{4\sqrt2}\right)$
Shaded area is difference of area of two circular segments of this chord, which are $ATB$ and $ASB$.
$ = \left(circular \ sector \ PATB - \triangle PAB\right) - \left(circular \ sector \ OASB - \triangle OAB\right)$
$ = \left(25 \times \frac{\alpha}{2} - \frac{25 \sqrt7}{8}\right) - \left(100 \times \frac{\beta}{2} - \frac{125 \sqrt7}{8}\right) \approx 14.638 $
EDIT: to find $AB$ without coordinate geometry, we know that $OP = 5 \sqrt2$. If perp from $P$ to $AB$ is $x$ then,
$\left(5\sqrt2 + x\right)^2 + \left(\frac{AB}{2}\right)^2 = 10^2$ and $x^2 + \left(\frac{AB}{2}\right)^2 = 5^2$. Solving them, we get value of $AB$.
