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.
Best Answer
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 $A_n(r) \to \pi r^2$.