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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 enter image description here

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.

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  • $\begingroup$ You can always use integration $\endgroup$
    – John Glenn
    Mar 24, 2018 at 12:24
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    $\begingroup$ @JohnGlenn i'm still in high school,haven't studied calculus $\endgroup$
    – Murad
    Mar 24, 2018 at 12:25
  • $\begingroup$ Compute the intersection points. And draw some figures. $\endgroup$ Mar 24, 2018 at 12:25
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    $\begingroup$ for each quadrant, you can divide the target figure in two triangles and one circle sector, being the two triangles of same area. $\endgroup$ Mar 24, 2018 at 12:26
  • $\begingroup$ You mean the area of the intersection of circle and square? What formulae did you do for circles etc. ? $\endgroup$ Mar 24, 2018 at 12:26

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

enter image description here

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

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  • $\begingroup$ It looks better now, though it's easier with acos in radians. See my answer. $\endgroup$ Mar 24, 2018 at 17:59
  • $\begingroup$ Say that you don't have this graph and can't draw.How could you find out that intersection point is at sqrt(11)? $\endgroup$
    – Murad
    Mar 25, 2018 at 5:59
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    $\begingroup$ The graph is just for illustration, just for intuition. As I pointed out, you know your circle is defined by: $$x^2+y^2=6^2$$ Which you can rewrite as: $$y=\sqrt{6^2-x^2}$$ Now you must find the intersection points with $x=5$ and $y=5$ because these are lines that make up the sides of your square. $\endgroup$
    – John Glenn
    Mar 25, 2018 at 6:03
  • $\begingroup$ @Murad: You'd use [Pythagorean Theorem](Pythagorean theorem). Solving a geometry problem without drawing is kinda hard, though. $\endgroup$ Mar 25, 2018 at 9:00
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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.

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  • $\begingroup$ Thanks for your help! $\endgroup$
    – Murad
    Mar 24, 2018 at 12:38
  • $\begingroup$ Circle area should be 36pi $\endgroup$
    – Murad
    Mar 24, 2018 at 12:39
  • $\begingroup$ @pasabaporaqui quite right. I edited. $\endgroup$ Mar 24, 2018 at 12:41
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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:

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

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    $\begingroup$ Inclusion-exclusion doesn't seem to help much here: It's not appreciately easier to compute the area of the union than it is to compute the area of the intersection directly. $\endgroup$ Mar 25, 2018 at 0:56
  • $\begingroup$ @HenningMakholm you're right, I'll edit. $\endgroup$ Mar 25, 2018 at 2:28
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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.

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