One tricky question from geometry, finding area of shaded region This is a question which has been giving me all sorts of goosebumps. I am losing confidence due to this question. I have constructed the hypotenuse which is $10\sqrt2$, but I later realized that the two small tiny parts are making it hard.
Please guide me.

 A: 
This is a partial answer using geometry. Draw a chord $GH$. The area in question is the area of circular segment $GHB$ minus circular segment $GH$ of the larger circle. The area of circular segment is the area of sector minus area of isosceles triangle. All we need to find is the angles $GEH$ and $GAH$. Note that we know all sides in $\triangle GAE$ therefore we can use law of cosine to find $\cos \angle GEA=\frac{100+50-25}{100\sqrt 2}=\frac{5}{4\sqrt 2}$. Thus, $\angle GEH=2 \arccos \frac{5}{4\sqrt 2}$, $GH=\frac{5\sqrt 7}{\sqrt 2}$, $\angle GAH=2 \arcsin \frac{\sqrt 7}{2\sqrt 2}$. I'll leave the remaining calculations to you.
A: I have thought of another way to calulate the shaded area, without using any calculus.
First let's draw lines from the bottom right corner to the intersection points, lines from the center of the square to the intersection points, and finally a line connecting the center of the square to the bottom right corner.

We obtained two similar triangles, and we actually know all of their sidelengths, since the greater circle has radius $10$ cm, the smaller circle has radius $5$ cm, and the center of the two circles are $5 \sqrt{2}$ distance away from each other (using simple Pyhegoras).
We obtain this triangle twice:

And we can calculate $\alpha$ from the Law of cosines:
$$10^2 = 5^2 + (5\sqrt{2})^2 - 2\times 5\sqrt{2}\times 5 \cos(\alpha) \\
100 = 25 + 50 - 50\sqrt{2}cos(\alpha) \\
50\sqrt{2}cos(\alpha)= -25 \\
cos(\alpha) = -\frac{1}{2\sqrt{2}} \\
\alpha \approx 110.705°$$
Since from construction, $360° = \theta + 2\alpha$,
$$\theta = 360° - 2\alpha \approx 138.59°$$
We can also calculate $\beta$ using the Law of consines again:
$$5^2 = (5\sqrt{2})^2 + 10^2 - 2 \times (5\sqrt{2}) \times 10 \times \cos(\beta) \\
25 = 50 + 100 - 100\sqrt{2}\cos(\beta) \\
100\sqrt{2}\cos(\beta) = 125 \\
\cos(\beta) = \frac{5}{4\sqrt{2}} \\
\beta \approx 27.89°$$
And therefore, $2\beta \approx 55.77°$.
So we have:

So to calculate the area of the shaded region, we simply need to calculate the $139.59°$ sector of the smaller circle and twice the area of the mentioned triangle, and subtract the $55.77°$ sector of the larger circle.
$$\text{Shaded area} = 5^2 \times \pi \times \frac{138.59°}{360°} + 2\times \text{Area of the triangle} -  10^2 \times \pi \times \frac{55.77°}{360°} \approx \\ \approx  30.24 + 2\times 21.28 - 48.67 = 24.13.$$
So the area of the shaded region is approximately $24.13$ $cm^2$. Let me know if I made a mistake anywhere. You could probably generalize this to a square with a different sidelength as well.
A: I'm not sure if you are familiar with definite integrals, but this sort of problem can easily be solved with them.
This is how I would go about it:
First place the construction into the $(x,y)$ plane, maybe like this (you can place the origin elsewhere, that would change the calculation slightly):

Then define the curves $f_1$, $f_2$, $f_3$ and $f_4$, and calculate the $x$-coordinate of their intersections: $x_2$ and $x_4$. I also denoted the $x$-coordiante of the left side of the square by $x_1$; and the $x$ value where the lower curve intersects with the $x$ axis splitting it into the $f_4$ and $f_2$ curves by $x_3$. ($f_4$ and $f_2$ are basically the same curves, but this notation will be useful for the integrals).
Depending on the scale, you'll need to write up the equations of the two circles and find their intersections. Their $x$-coordinate will give $x_2$ and $x_4$. You'll also need to solve
$$\text{equation of the larger circle} = 0$$
to find $x_3$. Finally, $x_1$ simply comes from construction.
Then the area in question will be the sum of the green and the blue areas, which is:
$$\text{Area} = \text{Green Area} + \text{Blue Area} = \\ = -\left(\int_{x_1}^{x_2} f_3(x)dx + \int_{x_2}^{x_3} f_4(x)dx\right) + \left(\int_{x_1}^{x_4} f_1(x)dx - \int_{x_3}^{x_4} f_2(x)dx\right)$$
Note that for the blue area, we calculated a difference of areas described under the $f_1$ and the $f_2$ curves, respectively, while the green area is simply a sum of two areas.
If you'd like any more specifics, I can edit my comment to show the exact calculation. However I encourage you to try this on your own.
