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If, inside a big circle, exactly $n (n ≥ 3)$ small circles, each of radius $r$, can be drawn in such a way that each small circle touches the big circle and also touches both its adjacent circles (as shown in the picture), then the radius of the big circle is? enter image description here

My approach: Let the circumference of the circle formed by joining the centres of the small circles be $2\pi R$.

Circumference of the circle formed by joining the centres of the small circles=$2rn$

Equating, $2rn=2\pi R$ $=>R=\frac {rn}{\pi}$

Now, the radius of the bigger circle is $r$ more than the $R$ found out previously. Therefore,

Radius of bigger circle$=r+\frac{nr}{\pi}$ = $r(1+\frac{n}{\pi})$

The answer given in my book is, $r(1+cosec\frac{\pi}{n})$ Where exactly am I going wrong?

Edit: I think writing the circumference as $2 rn$ is incorrect because the circumference is actually curved and I'm not sure that I can just keep adding the diameter n times to get the circumference. How should I apply trigonometry here?

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The mistake is in the claim "Circumference of the circle formed by joining the centres of the small circles =$2rn$". It is true that the line segment joining the centers of two touching circles of radius $r$ is $2r$. However, what you really want to calculate is the length of a circular arc joining those two centers, which will be slightly longer (and trigonometry will help you calculate how much longer).

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    $\begingroup$ How exactly should I apply trigonometry? I can just think of $l=r \theta$. Would that help? $\endgroup$ – Tapi Jun 14 at 18:47

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