Find the area of a circle part of which is in a square

Hint: You can use the image to get the intuition. Also remember that a circle is defined as $x^2+y^2=r^2$

Circle intersects square at $(\sqrt{11},5)$ and $(5,\sqrt{11})$, thus you have triangles with areas: $$A_{\triangle}=\frac12(5)(\sqrt{11})$$ The area of the sector is given by: $$A_{\text{sector}}=\pi r^2\cdot\frac{\theta}{360}=\pi(6^2)\cdot\frac{90-2\sin^{-1}\frac{\sqrt{11}}6}{360}$$ Thus you get the area you need: $$A=4(2\cdot A_{\triangle}+A_{\text{sector}})$$

Looking at the positive quadrant, you see two points of intersection of the circle and the square, at $x=5$ and at $y=5$. Compute these. This gives two triangles, that you can compute the ara of, hopefully. The circle part inbetween them is just a fraction of the circle area depending on the angle $\alpha$ (in radians) between these two points (inner product can help to compute the cosine e.g.) namely $\frac{\alpha }{2\pi}A_c$ where $A_c = 36\pi$, the area of the circle.

Then times 4 as we have 4 quadrants.

The desired area is the area of the disk ($\pi\cdot6^2$) minus the area of the 4 segments outside the square. You can calculate the area of one of those segments as the difference between a circular sector and a triangle.

Putting all this together, you get:

$$A = \pi\cdot6^2 - 4\cdot(6^2\cdot acos(\frac{5}{6}) - 5 \cdot\sqrt{11})\\ \approx 95.091113 cm^2$$