## What is the Area of a Circle?

June 2007

Suppose that we are asked to find the area enclosed by a circle of given radius. A simple way to go about this is to draw such a circle on graph paper and count the number of small squares within it. Then

area contained ≈ number of small squares within circle × area of a small square.

 A circle drawn on graph paper - the area inside is approximately the number of small squares times the area of each small square If we doubled all the lengths involved then the new circle would have 4 times the area contained in the old circle

We notice that if we doubled all the lengths involved then the new circle would have twice the radius and each of the small squares would have four times the area. Thus

area contained in new circle of twice the radius

≈ number of small squares × area of a new small square
= number of small squares × 4 × area of a old small square
≈ 4 × area contained in old circle.

By imagining what would happen if we used finer and finer graph paper, we conclude that doubling the radius of a circle increases the area by a factor of 4.

The same argument shows that, if we multiply the the radius of a circle by , its area is multiplied by . Thus If, following tradition, we write we obtain the memorable formula

We can now play another trick. Consider our circle of radius r as a cake and divide it into a large number of equal slices. By reassembling the slices so that their pointed ends are alternately up and down we obtain something which looks like a rectangle of height r and width half the length of the circle.

The area covered by a cake is unchanged by cutting it up and moving the pieces about.

But the area of a rectangle is

width × height

and the area of of the cake is unchanged by cutting it up and moving the pieces about. So

Dividing by and multiplying by , we get By increasing the number of slices, we can make our approximations better and better, obtaining another memorable result The formulae we have obtained for the area enclosed and the length of a circle are very nice but we can not actually use them unless we know a good approximation to .

### Approximating Pi

One approximation goes back to the ancient Greeks who looked at the length of a regular polygon inscribed in a circle of unit radius. As we increase the number of sides of the polygon the length of the polygon will get closer and closer to the length of the circle that is to .

Can you compute the total length of an inscribed square? Of an inscribed equilateral triangle?

Francois Vieta

Many of the ideas in this article go back to Archimedes, but developments in mathematical notation and computation enabled the great 16th century mathematician Vieta to put them in a more accessible form. (Among other things, Vieta broke codes for the King of France. The King of Spain, who believed his codes to be unbreakable, complained to the Pope that black magic was being employed against his country.)

We can approximate the circle with n-sided polygon,
in this case an hexagon with n=6.

We use trigonometry to find a general formula for the length of the perimeter of an inscribed -sided regular polygon. Observe that the polygon is made up of triangles of exactly the same shape. If we look at one of these triangles, say, we see that (as we have drawn things) If is the mid point of then is a right angled triangle with hypotenuse of length one and angle By trigonometry and so We have shown that Since we are interested in which is half the length of the perimeter of the circle of radius , we consider When is large, we expect that will be close to . Trying out our formula on a hand calculator, we get

If you calculated the length of the perimeter for an inscribed square or triangle, does our general formula for an n-sided polygon agree for n = 3 and 4? If you try to use your calculator to calculate , , you'll observe that the results aren't a very good approximation for .

There are two problems with our formula for . The first is that we need to take large to get a good approximation to . The second is that we cheated when we used our calculator to evaluate since the calculator uses hidden mathematics substantially more complicated than occurs in our discussion.

### Doubling sides

How can we calculate ? The answer is that we cannot with the tools presented here. However, if instead of trying to calculate for all , we concentrate on , , , ... (in other words we double the number of sides each time), we begin to see a way forward.

Ideally, we would like to know how to calculate from . We can not quite do that, but we do know the formulae (from the standard trigonometric identities) and Thus and We can use these forumulae to find , ... . Try it out on your caculator (but don't use the trigonometric function keys) and hence find and .

We can make the pattern of the calculation clear by writing things algebraically. Let us take and write and Then and, continuing as far as we like, we see that Thus for very large Since , and , and we can write .

Therefore

### Vieta's formula and beyond

Although Vieta lived long before computers, this approach is admirably suited to writing a short computer program. The definitions of and lead to the rules and we can use these to easily compute successive values of . If you have a computer or programmable calculator, see if you can compute .

Nowadays we leave computation of square roots to electronic calculators, but, already in Vieta's time, there were methods for calculating square roots by hand which were little more complicated than long division. Vieta used his method to calculate to 10 decimal places. We have shown that and the larger is, the better the approximation. It is natural to write where, in some sense, we multiply together an infinite number of terms. Using the rule above to write the successive values for (starting with ), we obtain Vieta's formula:

From an elementary point of view this formula is nonsense, but it is beautiful nonsense and the theory of the calculus shows that, from a more advanced standpoint, we can make complete sense of such formulae.

The Rhind Papyrus, from around 1650 BC,
is thought to contain the earliest
approximation of pi
; as 3.16.

The accuracy of this approximation increases fairly steadily, as you will have seen if you used your calculator to compute the successive values of sn. Roughly speaking the number of correct decimal places after the nth step is proportional to n. Other methods were developed using calculus, but it remained true for these methods that number of correct decimal places after the nth step was proportional to n.

In the 1970's the mathematical world was stunned by the discovery by Brent and Salamin of method which roughly doubled the number of correct decimal places at each step. To show that it works requires hard work and first year university calculus. However, the method is simple enough to be given here.

Take and . Let Then is a good approximation to . Using this algorithm, it is easy to compute the first few approximations, say P1, P2, P3 and P4, on your calculator.

Since then even faster methods have been discovered. Although has now been calculated to a trillion (that is to say 1,000,000,000,000) places, the hunt for better ways to do the computation will continue.

Tom Körner is a lecturer in the Department of Pure Mathematics and Mathematical Statistics at the University of Cambridge, and Director of Mathematical Studies at Trinity Hall, Cambridge. His popular mathematics book, The Pleasures of Counting, was reviewed in Issue 13 of Plus.

He works in the field of classical analysis and has a particular interest in Fourier analysis. Much of his research deals with the construction of fairly intricate counter-examples to plausible conjectures.