Convergence of the perimeter of an inscribed regular polygon

Consider a nn-sided regular polygon inscribed in a circle of radius rr. The summits of the polygon lie on the circle, and all sides of the polygon are of equal length. The angle at the center of a regular polygon is the angle formed by two consecutive rays of the circle passing through the vertices of the polygon. Since the polygon is regular, all these angles at the center are equal.

Therefore, the centrale angle subtended by a side of the polygon is :

Central angle=2πn radians\text{Central angle} = \frac{2\pi}{n} \text{ radians}

Example :

For n=6n = 6 (hexagon), each side subtends an angle at the center of : 2π6=π3 radians\frac{2\pi}{6} = \frac{\pi}{3} \text{ radians}

Thanks to the inscribed angle theorem , we know that :

Base angle=Central angle2=πn\text{Base angle} = \frac{\text{Central angle}}{2} = \frac{\pi}{n}

Let’s take the isosceles triangle OABOAB where :

We have proved that :

To find the length of side ss, we can use trigonometry in the OABOAB triangle ; the definition of sine is :

sin(πn)=opposedhypotenuse\sin(\frac{\pi}{n}) = \frac{\text{opposed}}{\text{hypotenuse}}

And as :

Then we have :

sin(πn)=s/2r\sin\left(\frac{\pi}{n}\right) = \frac{s/2}{r}s2=rsin(πn)\frac{s}{2} = r \sin\left(\frac{\pi}{n}\right)

Thus, the length of a side ss of the regular polygon inscribed in a circle of radius rr is given by :

s=2rsin(πn)s = 2r \sin\left(\frac{\pi}{n}\right)

We now have all we need to demonstrate the value of the perimeter of a polygon.

The perimeter PP of the polygon is the sum of the lengths of all its sides. Since the polygon has nn sides, each of length ss, the perimeter is :

P=ns=n2rsin(πn)P = n \cdot s = n \cdot 2r \sin\left(\frac{\pi}{n}\right)

The basic question was to find out the value of the perimeter PP of the inscribed regular polygon when the number of its sides nn tends towards infinity.

limnP=2πr\lim\limits_{n \to \infty} P = 2\pi r

We know that for small angles sinx=x\sin{x}=x and when nn is very large, πn\frac{\pi}{n} is small. We can therefore use this approximation :

sin(πn)πn\sin\left(\frac{\pi}{n}\right) \approx \frac{\pi}{n}
Substituting this approximation into the perimeter expression, we obtain
Pn2rπn=2πrP \approx n \cdot 2r \cdot \frac{\pi}{n} = 2\pi r

Thus, when nn tends to infinity, the perimeter PP of the regular polygon inscribed in the circle tends to 2πr2\pi r, which is the circumference of the circle:

limnP=2πr\lim\limits_{n \to \infty} P = 2\pi r