MHB Find the area of the four sectors of the given circle

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To find the area of the four sectors in the given circle, the area of the circular sector is calculated using the formula A_S = (1/2)r^2θ, where θ is derived from θ = cos^(-1)(k/r). The area of the triangle within the sector is given by A_T = (1/2)k√(r^2-k^2). The area of the red section is then determined by A = A_S - A_T, leading to A = (1/2)(r^2cos^(-1)(k/r) - k√(r^2-k^2)). The discussion emphasizes the relationship between the semicircle and quarter areas to derive the total area of the sectors.
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if we have the circle in the picture given x,y,z
View attachment 528

the middle line pass through the circle center
find the area of the four sectors with respect to x,y,z
parallel lines
Thanks
 

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Perhaps this can get you started. Please refer to the following diagram:

View attachment 529

The area of the circular sector (the sum of the red and green areas) is:

$\displaystyle A_S = \frac{1}{2}r^2\theta$

Now, we see that:

$\displaystyle \cos(\theta)=\frac{k}{r}\,\therefore\,\theta=\cos^{-1}\left(\frac{k}{r} \right)$

and so we have:

$\displaystyle A_S = \frac{1}{2}r^2\cos^{-1}\left(\frac{k}{r} \right)$

The area of the green triangle is:

$\displaystyle A_T=\frac{1}{2}k\sqrt{r^2-k^2}$

And thus, the area A in red is:

$\displaystyle A=A_S-A_T=\frac{1}{2}\left(r^2\cos^{-1}\left(\frac{k}{r} \right)-k\sqrt{r^2-k^2} \right)$

Can you proceed from here?
 

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Last edited:
MarkFL said:
Perhaps this can get you started. Please refer to the following diagram:

View attachment 529

The area of the circular sector (the sum of the red and green areas) is:

$\displaystyle A_S = \frac{1}{2}r^2\theta$

Now, we see that:

$\displaystyle \cos(\theta)=\frac{k}{r}\,\therefore\,\theta=\cos^{-1}\left(\frac{k}{r} \right)$

and so we have:

$\displaystyle A_S = \frac{1}{2}r^2\cos^{-1}\left(\frac{k}{r} \right)$

The area of the green triangle is:

$\displaystyle A_T=\frac{1}{2}k\sqrt{r^2-k^2}$

And thus, the area A in red is:

$\displaystyle A=A_S-A_T=\frac{1}{2}\left(r^2\cos^{-1}\left(\frac{k}{r} \right)-k\sqrt{r^2-k^2} \right)$

Can you proceed from here?

Thanks I can for sure, we have the semicircle area
the quarter area of the circle minus the area of the red we will get the semi area of below sector.
 
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