How Archimedes found the area of a circle

"Squaring the Circle"

The problem of determining the area of a circle was once considered a great mathematical challenge. On this page, we will discuss a method that Archimedes (287-212 BCE) used for solving this problem. (This problem was called "squaring the circle": i.e. trying to find the square that has the same enclosed area as a circle of a given radius.)
Ofcourse, we now know that

$A = \pi r^2$

The whole point will be to actually show that this is true, by somehow computing the area of the circle from known areas of simpler shapes.

A reminder about Pi

The irrational number $\pi$ is defined as the ratio of the circumference of a circle to its diameter, that is:

$\pi = \frac{\rm circumference}{\rm diameter} = \frac{2 \pi r}{2r}$
We will use this result in the final steps of the method below.

Step 1: Approximating the circle with a square

The diagram here shows a square inscribed in a circle. (Inscribed means that it exactly fits inside, with its vertices just touching the edge of the circle). Let us look at the geometry more closely.

We notice that the diagonals of the square are the same as the diameter of the circle, and so have length . Thus the sides of the square, AB, BC which are equal ofcourse, must have length $\sqrt{2} r$ . (Since ABC is a right triangle, whose hypotenuse is AC). Thus the area of the inscribed square is thus

$A = \sqrt{2}r ~\sqrt{2}r = 2 r^2$

This is still not a very good approximation to the true area of the circle, but it is a start.

Step 2: Approximating the circle with a hexagon

If we use a polygon with more sides to try to approximate the area of the circle, we would hope to get a better result. So consider an approximation which uses the hexagon, as shown in the diagram.

To find the area of the hexagon, we might subdivide it into six triangles, whose area is easily computed if we know the height and the base of any one of the (all equal) triangles.
One such triangle is shown in the diagram at left: Observe that its sides AC and AB are of length , since they are radii of the circle. Further, notice that the angle ABC should be 60 degrees (i.e. one sixth of a revolution, 360/6 =60 degrees). This implies that the sum of the other angles, CAB plus CBA is 120 degrees (since the sum of all the angles in a triangle is 180 degrees). Further, since they are equal (opposite equal sides), they, too, must be 60 degrees. So we have, hopefully, convinced you that the triangles in this hexagon are all equilateral: all three sides are of length . We leave it up to you to further establish that the height of each of these triangles is $\sqrt{3}r/2$ , so that the area of the hexagon, it follows, is:

$A= 6 ~(\frac 12~ r~ h) = 6 (\frac 12 ~r ~\sqrt{3} ~\frac r2) = \frac{3 \sqrt{3}r^2}{2} = 2.59~r^2$
This is closer to the true area of the circle, but still not very close.

Step n: Using an n-sided polygon

Suppose we increase the number of sides in the polygon and look at the general case, in which there are n sides. Then the area of the n sided polygon will be n times the area of one of the triangles, i.e.

$A= n~ ({\rm area~ triangle})= n~ (\frac 12~ h~ b)$
where are, respectively the base and height of one of the triangles shown in this picture. Now note what happens as the number of sides, n increases:

$A= n~ (\frac 12~ h~ b) = \frac 12 ~h ~(n~b)$

In this expression, the term is the perimeter of the polygon, which, as n increases, becomes closer and closer to the circumference of the circle. Further, the height of the triangle, approaches the radius of the circle, so that, as we approximate the circle by a polygon with more sides, i.e. a greater number of (thinner) triangles, we find that the area approaches:

$A= \frac 12 ~h ~(n~b) \approx \frac 12 ~r ~(2 \pi r) = \pi r^2$

We have shown above, that the area of the circle does indeed involve that special constant $\pi$ which arose as the ratio of the circumference to the diameter of the circle.

By dragging the red ball in the diagram below, you can experiment with the number of sides in the polygon used to approximate the area of the circle. You should notice that as n increases, the area of the polygon becomes a closer and closer fit to the area of the circle.

Drag the red ball. As the number of sides, n, in the polygon increases, its area becomes a closer and closer approximation to the area of the circle.

For your consideration:

(1) Show that the length of each side of the square in the first step we did is $\sqrt{2}$ .

(2) Show that the height of the triangles that form the hexagon, in the second step we discussed is $\sqrt{3}r/2$ .

(3) Suppose you were to consider a 36-sided polygon. In that case the angle of the tip of each of the triangular subdivisions would be 360/36 = 10 degrees. This is not a "standard" angle, but we could certainly use trigonometry to establish the length of the base and the height of such a triangle. What would you get for the area of this 36-sided polygon? How does this area compare with the true area of the circle?

This is the last time we will be using triangles to compute areas for a little while. From now on, you will see that rectangular subdivisions will form a much more important unit of subdivision in computing areas of irregular regions in the plane.
However, the concept of increasing the number of units to better approximate some complicated shape will be very basic. It will appear and reappear in many examples, as it forms one of the key ideas in integral calculus.