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

Question

I have a square with sides of 10cm and I have a circle with radius of 6cm. Now I've to find the area of the circle that is inside of the square.Here is the graph

I had an idea of finding the area of the arc(90 degrees) and subtracting it from 25(100/4), but then I noticed that the area of arc would still include the areas which are outside of the square.

Answer 1

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.

Answer 2

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: $$A=4(2\cdot A_{\triangle}+A_{\text{sector}})$$

Answer 3

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:

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

Answer 4

Consider 1st quadrant then consider the overlapped region and find area of two rectangles and one triangle, add them up and then multiply by 4. one rectangle will be having points (0,0), (0,5), (3.2,5), (3.2,0). other rectangle will have points (3.2,3.5), (5,3.5), (5,0), (3.2, 0). the triangle will have points around (3.2, 3.5), (3.2,5), (5, 3.5). use distance formula to measure the side lengths and then find areas.