## Frequently Asked Questions

#### 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 4`d`; so we deduce that the circumference of a circle, `C` < 4`d`. A regular hexagon is made up of six equilateral triangles, so in the case of an inscribed hexagon, its perimeter will be 6`r` = 3`d`, because `d` = 2`r`. Clearly the circumference of the circle is greater than the perimeter of the hexagon; so we deduce further that `C` > 3`d`.

Hence 3`d` < `C` < 4`d`.

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 `P _{c}`

Let the perimeter of the inscribed

`n`-gon be

`P`

_{i}For all values of `n`, `P _{i} < C < P_{c}`.

By using trigonometry, `x = r` `sin`θ.

As `x` is half a side length, `P _{i} = n` × 2

`x = n`× 2

`r`

`sin`θ =

`n`

`sin`θ ×

`d`.

Similarly, `y = r` `tan`θ, so `P _{c} = n` × 2

`y = n`× 2

`r`

`tan`θ =

`n`

`tan`θ ×

`d`.

Therefore, `n` `sin`θ × `d < C < n` `tan`θ × `d`

As `n` increases, the value of `P _{i}` continues to increase towards the circumference of the circle and in the same way the value of

`P`continues to decrease. In addition we can see that both

_{c}`P`and

_{i}`P`are multiples of

_{c}`d`.

Hence `C` becomes the limit of the upper bound of `P _{i}` and the limit of the lower bound of

`P`, such that

_{c}`C = k × d`, where

`k`is the perfect proportion which we define as Pi, π.