(No Greek letters were harmed in the writing of this blogpost.)

πr^{2} is the area of a circle, but for a lot of people, this is another one of these ‘faith’ formulae. So here, I offer a proof of this fact:

It is known that x^{2} + y^{2} = r^{2} is the equation of a circle. (For an explanation of this, see this site.)

The circumference, by definition, is 2π times the radius. So, circumference = 2πr.

(I can justify this by saying that π = circumference/diameter = c/2r. So c = 2r*π = 2πr.)

Now, imagine a small part of this circle, like this (but with a far smaller angle):

The white line to the left of the circumference is *y*, and the angle is θ.

Circumference of wedge = 2 πr * (θ/360)

It follows that if is very small, c = *y*. That means that the wedge is nothing but a simple triangle.

*y* = 2 πr * (θ/360)

So, the area of the triangle is ½ *y*x.

Area^{2} = x^{2}y^{2}/4

= (x^{2}/4) * (2πr * (θ/360))^{2}

= x^{2 }π^{2}r^{2} (θ/360)^{2}

Therefore, area = πrx (θ/360)

Since this triangle is right-angled (although, thanks to my poor MS Paint drawing skills, it doesn’t look like it) at the vertex common to x and *y*, you can say that cos = x/r, and so, it follows that x = r cos θ. So, if you substitute r cos θ in the formula, you get:

πr^{2} (θ/360) cos θ

This is the formula for the area of a circle wedge, of angle θ. So, if you work it out for θ = 360 degrees, you get:

πr^{2} (360/360) cos θ = πr^{2} cos 360.

cos 360 = 1, so the area of a circle of radius r is πr^{2}.

QED.

With this formula, it’s possible to express the area of a circle in other ways, too! I’ll show you two of them.

c = 2πr = πd.

A = πr^{2} = π(2r)^{2}/4 = πd^{2}/4 = cd/4.

This formula completely avoids π!

The second one is one which I derived myself, and I could not find any mention of it until recently, when I Googled it and found that I was not the first one (as it had happened before).

I derived this at the age of 10, when there were lots of maths questions of the form “Here is the circumference. Find the area.” or vice-versa. The ‘proper method’ (one thing which I have never been very good at sticking to) was to divide by , find the radius, then square it and multiply by to get the area. To me, it seemed inefficient, so I worked out a formula that used the circumference to directly find the area:

c = 2πr

c^{2} = 4π^{2}r^{2}

c^{2}/4π = πr^{2}

This is a formula that I haven’t found in any books or similar resources – it’s one of those secrets of maths, really.