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

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?

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

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

Can you proceed from here?

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Thanks I can for sure, we have the semicircle areaPerhaps 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 areaAin 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?

the quarter area of the circle minus the area of the red we will get the semi area of below sector.