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If a regular polygon of number of $N$ sides lies within a circle with radius $R$ where the the circle touches every vertex of the the polygon. Can I obtain the area of the circle by increasing the number of sides of the polygon?

Area of circle should equal the limit as $N$ approaches infinity?

Lim as $N$ -> infinity of $nr^2\cos{(\theta/2)}\sin{(\theta/2)}$ but that always gives infinity.

How can I increase the number of sides of a polygon inside a circle but not get an area bigger than the area of the circle.

I know how stupid that sounds.

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    $\begingroup$ $\theta$ depends on the number of vertices, and $\sin (\pi/n) \to 0$. $\endgroup$ Aug 1, 2013 at 11:37
  • $\begingroup$ You can use any other equation that doesn't deal with angles such as $1/2*perimeter*radius$ $\endgroup$ Aug 1, 2013 at 11:43
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    $\begingroup$ It doesn't sound stupid! This is how Archimedes estimated area and perimeter of a circle. You're in good company with the idea. $\endgroup$
    – Mark Ping
    Aug 1, 2013 at 16:21

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For a regular $n$-gon, the angle at the centre between two adjacent vertices is $\theta_n = \frac{2\pi}{n}$. Thus you get

$$A_n(r) = nr^2\cos \frac{\pi}{n}\sin \frac{\pi}{n}.$$

Now, $\cos \frac{\pi}{n} \to 1$ and $r^2$ is independent of $n$, but $\sin\frac{\pi}{n} \to 0$ in a way that just cancels the contribution of the factor $n$,

$$\lim_{n\to\infty} n\sin \frac{x}{n} = x$$

for every $x\in\mathbb{R}$. For the $n$-gon, we have $x = \pi$, and hence $A_n(r) \to \pi r^2$.

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