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I study maths as a hobby and have come across this problem. Two circles with centres A and B intersect at points P and Q, such that $\angle APB$ is a right angle. If AB = xcm and $\angle PAQ = \frac{1}{3}\pi$ radians, find in terms of x the length of the perimeter and the area of the region common to the two circles.

I calculate the area of the segment APQ to be $\frac{\pi}{6}r^2$, where r = radius. The area of $\triangle APQ$ I calculate as $\frac{1}{2}r^2\sin \frac{\pi}{3} = \frac{1}{2}r^2\sqrt3$

I also know that to find the right hand side of the central segment I need to subtract the area of the triangle APQ from the area of the segment APQ.

But I cannot proceed any further and certainly not in terms of the length x.

This is the diagram as I visualise it:

enter image description here

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HINT

We know that $\angle PAB=\frac{\pi}{6}$, hence the radius of the larger circle is $x\cos{\frac{\pi}{6}}=\frac{x\sqrt3}{2}$. Similarly, the radius of the smaller circle is $x\sin{\frac{\pi}{6}}=\frac{x}{2}$.

Does that help? If you need any more help, please don't hesitate to ask. I have a full solution ready to post if you need it.

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  • $\begingroup$ Thank you so much. $\endgroup$ – Steblo Dec 26 '20 at 11:34
  • $\begingroup$ @Steblo you are most welcome, I'm very glad to have helped you! :) Feel free to ask any more questions. $\endgroup$ – A-Level Student Dec 26 '20 at 19:49
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Hint. For the common area, $$\text{area of sector PBQ}+\text{area of sector PAQ}=\text{area of }\square\text{PBQA}+\text{common area of circles}$$

where $$\begin{align*}&\text{area of sector PAQ}=\frac16\pi r_1^2\\ &\text{area of sector PBQ}=\frac13\pi r_2^2\\ &\text{area of }\square\text{PBQA}=AB\times PD=xr_1\sin\angle PAD\end{align*}$$

Can you find $r_1,r_2$ in terms of $x$? Use the fact that $\triangle APB$ is right-angled.

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  • $\begingroup$ What do you mean by $\text{area of }\square PBQ$? What square is there? $\endgroup$ – A-Level Student Dec 24 '20 at 16:28
  • $\begingroup$ @A-LevelStudent I meant quadrilateral PBQA, fixed now. $\endgroup$ – Shubham Johri Dec 24 '20 at 17:07
  • $\begingroup$ Ah, thank you :) $\endgroup$ – A-Level Student Dec 24 '20 at 17:09

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