## Elegant proofs 2 – The area of a circle

We are so familiar with the formula for the area enclosed by a circle that we tend not to think much about how it was derived, at least I don’t.

The proofs of the formula are in fact many and varied; the first one found by Google is at:
http://www.artofproblemsolving.com/LaTeX/Examples/AreaOfACircle.pdf

Don’t worry, that’s not the elegant one.

There are many proofs that don’t (directly) involve the use of calculus, and Wikipedia gives a good sample of them:

http://en.wikipedia.org/wiki/Area_of_a_disk

of which the rearrangement proof is perhaps the most elegant.  Another presentation of this proof is given here (along with Archimedes’ equally elegant derivation of the volume of a sphere):

http://www.mathreference.com/geo,circle.html

Yesterday I came across an approach that to me seems even simpler, based on a post at:

http://foxmath.wordpress.com/2008/06/24/perimeter-area/ This shows that for any regular polygon with an area equal to its circumference, the length of the apothem (the red line in the diagram to the left) is 2.  This is immediately obvious from the fact that the area of each individual triangle is equal to the base length, when the height equals 2.

In the limit as the number of sides of a regular polygon tends to infinity the polygon approaches a circle, and the length of the apothem approaches the radius of the enclosing circle.  It therefore follows that the area of a circle of radius 2 is equal to its circumference; i.e. 4.pi.

A circle of radius R may be scaled to radius 2 by multiplying the radius by 2/R.  The radius of this circle is then 4pi x (R/2)^2 = pi.R^2.

Finally a “wordless” proof provided by the people at SSSF:

http://www.maa.org/pubs/Calc_articles/ma018.pdf

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### 4 Responses to Elegant proofs 2 – The area of a circle

1. mrt says:

Hi –

Troubled by the thought of an area equal to a circumference. Implicit in that is that units-squared equals units. Physicists would disagree. 😉

…mrt

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2. dougaj4 says:

mrt – you are quite right that an area can’t be physically equal to a length, but they can have the same numerical value, which is what the proof is concerned with.

Watch out for an upcoming post on the history of dimensional analysis! 🙂

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3. Piyush Gupta says:

Thanks a lot dude!! 😀

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4. SasQ says:

@mrt: You’re right, the area is not the same as length, so they cannot be equated *directly*. But if `A` is area, and `L` is length, then you can make an equation like this: `A = 1*L`. Multiplying by `1` doesn’t change anything. But now you can interpret `1*L` as a *rectangle* of `1` by `L`, which has the area of `L`. So you are now comparing areas with areas, and this is OK.
To better understand this concept, you can see that any length is a multiple of a unit length, called `1`. So `L` is really `L*1` (read it: “`L` units of length”). Similarly, an area is a multiple of a unit area, which is a `1` by `1` square. So when you say that something has an area `A`, what you really say is this: `A * 1*1` (read it: “`A` units of area, that is, those little 1×1 squares”). Can you see now how can you jump between lenghts and areas and volumes etc. and why? It’s all about units and the fact that you can multiply by `1` as many times as you need, and use as many dimensions as you need. You can construct a unit of any dimension by multiplying all these `1`’s together. Any other number just tells you how many times you repeat this unit.

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