How do you find the circumference of a circle?

Consider the two diagrams.

The circumscribed square clearly has a perimeter greater than that of the circle, and in fact it can be seen to be 4d; so we deduce that the circumference of a circle, C < 4d. A regular hexagon is made up of six equilateral triangles, so in the case of an inscribed hexagon, its perimeter will be 6r = 3d, because d = 2r. Clearly the circumference of the circle is greater than the perimeter of the hexagon; so we deduce further that C > 3d.

Hence 3d < C < 4d.

By increasing the number of sides on the inscribed and circumscribed polygons it is possible to improve the lower and upper limits that estimate the circumference. Archimedes painstakingly developed this method, such that he trapped a circle between a pair of 96-gons.

In fact, for a given diameter, d, the circumference of a circle, C = π d, where π = 3.14 (3 s.f.). The value of π is not only irrational, but it is also transcendental.

Theorem
The circumference of a circle, C is given by C = π d, where d is the diameter and π is that perfect proportion.

Proof
Consider a circle with a fixed diameter, d. For any circle there must exist an inscribed n-gon and a circumscribed n-gon.

Let the circumfernce of the circle be C
Let the perimeter of the circumscribed n-gon be Pc
Let the perimeter of the inscribed n-gon be Pi

For all values of n, Pi < C < Pc.

By using trigonometry, x = r sinθ.
As x is half a side length, Pi = n × 2x = n × 2r sinθ = n sinθ × d.

Similarly, y = r tanθ, so Pc = n × 2y = n × 2r tan θ = n tanθ × d.

Therefore, n sinθ × d < C < n tanθ × d

As n increases, the value of Pi continues to increase towards the circumference of the circle and in the same way the value of Pc continues to decrease. In addition we can see that both Pi and Pc are multiples of d.

Hence C becomes the limit of the upper bound of Pi and the limit of the lower bound of Pc, such that C = k × d, where k is the perfect proportion which we define as Pi, π.