## Deriving the Area of a Circle

November 4, 2013 by richbeveridge

During the 1998-99 school year I was teaching math at Ellsworth High School in Maine. I got to talking with one of the other math teachers there about deriving the area of a circle. We started approaching the idea from a similar perspective as Archimedes had 2300 years ago in his work with circles and . We inscribed a polygon in a circle and then took the limit of the area of the polygon as the number of sides () approached infinity. For the area of the polygon, we divided it into triangles and used the area formula .

In this diagram, both and are radii of the circle and are labeled as . The central angle which will be used to find the area of the triangle is actually equal to or (and this will be VERY important later) .

So the area of the polygon is times the area of the triangle, or:

For this to be the area of the circle we want to take the limit of the previous expression as .

Since the don’t involve , we can pull them out from the limit.

Now comes the calculus – we were just talking about the derivative of the sine function in Calculus I last week, which is what made me think about this. Remember that:

So, we need to establish this limit somewhere in our derivation and then replace it with . The most likely place for this would be in here:

but we need to have:

OK – so, let’s multiply our origninal expression by .

or

which is equivalent to

Let’s bring back the whole expression:

as ,

or

When we were working through this the first time, we kept the in the formula and ended up with , which didn’t make any sense until we realized that is radians!

Archimedes’ conclusion was that the area of the circle was equal to one-half the radius times the circumference:

or, since the mathematicians in the Greek tradition like Archimedes thought in terms of geometry – the area of a circle is the same as the area of triangle with height equal to the radius of the circle and base equal to the circumference of the circle!

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