Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
$\theta$ depends on the number of vertices, and $\sin (\pi/n) \to 0$. –  Daniel Fischer Aug 1 '13 at 11:37
    
You can use any other equation that doesn't deal with angles such as $1/2*perimeter*radius$ –  Mohamed Ayman Aug 1 '13 at 11:43
1  
It doesn't sound stupid! This is how Archimedes estimated area and perimeter of a circle. You're in good company with the idea. –  Mark Ping Aug 1 '13 at 16:21
add comment

1 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 $A_n(r) \to \pi r^2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.