# Is it possible to turn this geometric demonstration of the area of a circle into a rigorous proof?

In this New York Times article, Steven Strogatz offers the following argument for why the area of a circle is $$\pi r^2$$. Suppose you divide the circle into an even number of pizza slices of equal arc length, and wedge them together in such a way that half of the slices have an arc at the bottom, and half of the slices have an arc at the top:

Then, the base of the shape created has length $$\pi r$$, and its height is $$r$$. As the number of slices tends to infinity, the limiting case is that of a rectangle:

Hence, the area of the circle is $$\pi r^2$$. Although this argument is very geometrically appealing, it also seems fairly difficult to make rigorous. I suppose the most challenging part is showing that the base of the shape really does become arbitrarily flat, and its height becomes arbitrarily vertical, if that makes sense. How might we convert this intuitive argument into a rigorous proof?

• Arbitrarily vertical can be explained by $\sin\theta\sim\theta$ for small $\theta$. Commented May 29, 2021 at 21:13
• More slices= less at length per piece= less deviation from a line segment ... Commented May 29, 2021 at 21:15
• You can approximate the area of the shape from above and from below using parallelograms, whose base and height you can calculate with a bit of trigonometry. You can then show that as the number of slices increases, the difference between the approximation from above and below goes to $0$, so the approximations have to converge to the actual area. Commented May 29, 2021 at 21:33
• Archimedes used inscribed and circumscribed polygons to approximate $\pi$ since he could calculate the size of polygonal wedges exactly. You could use the same idea and the squeeze theorem to prove this. Commented May 29, 2021 at 22:09
• There's a subtlety here that a lot of answers are glossing over: the fact that the circumference of a circle is less than the perimeter of a circumscribed polygon is not trivial. Commented May 30, 2021 at 16:47

For simplicity I will assume the number of slices $$n$$ is even.

If you connect the four "corners" of the wedge figure, you obtain an "inner" parallelogram with one pair of sides having length exactly $$r$$, and another pair of sides having length approximately $$\pi r$$. The height of the parallelogram is $$r \cos \frac{\pi}{n}$$ (which tends to $$r$$); the length is $$n r \sin \frac{\pi}{n}$$ (which tends to $$\pi r$$). So the area of the "inner" parallelogram tends to $$\pi r^2$$.

You can also consider inscribing the wedge figure inside a slightly larger parallelogram. The height is again $$r \cos \frac{\pi}{n}$$ but with two additional "crusts" each having thickness $$r(1 - \cos \frac{\pi}{n})$$, so the final height is $$r(2-\cos \frac{\pi}{n})$$ (which also tends to $$r$$). I think the length is the same as the that of the inner parallelogram: $$nr \sin \frac{\pi}{n}$$ (which tends to $$\pi r$$).

So the area of the wedge figure is between the areas of the two parallelograms $$\pi r^2 cos \frac{\pi}{n} \frac{\sin(\pi/n)}{\pi/n}$$ and $$\pi r^2 (2-\cos \frac{\pi}{n}) \frac{\sin(\pi/n)}{\pi/n}$$. Since both of these areas converge to $$\pi r^2$$, so does the area of the circle.

Picture of the inner and outer parallelograms:

• The length of each side is only $nr\sin(\pi/n)$, because it's made up of only half the triangle bases (the other half are on the other side of the parallelogram). Your answer is off by a factor of 2. Commented May 30, 2021 at 14:08
• @eyeballfrog Thanks, I've now corrected this. Commented May 30, 2021 at 16:59
• I don't get the larger parallelogram part. Can you make a drawing of it, or perhaps a more rigorous definition of how you make a "slightly larger parallelogram"? Commented May 30, 2021 at 18:57
• @justhalf Look at each pizza slice. You can "cut off the crust" to get a smaller triangle with height $r \cos (\pi/n)$. (This builds the inner parallelogram.) You can also "add more crust" to get a larger triangle with height $r$. (This builds the larger parallelogram.) More generally you can think of my answer as sandwiching a circle between an inscribed polygon and a circumscribed polygon. Commented May 30, 2021 at 19:05
• @justhalf Ah, I see, sorry for the unclear wording. I've changed it to hopefully be clearer. Commented May 31, 2021 at 4:46

The circle and the rectangle must have the same area if it is the case that no matter how small you make a positive number $$\varepsilon,$$ the difference between their areas is less than $$\varepsilon.$$ As you make $$\varepsilon$$ smaller, the areas of the circle and the rectangle are not changing, yet they always continue to differ in area by less than $$\varepsilon.$$

So the idea is this: First observe that the sum of the areas of the wedges is equal to the area of the circle. Then, given the small positive number $$\varepsilon,$$ you can make the number of wedges so large that the difference between the area of the rectangle and the sum of the areas of the wedges is less than $$\varepsilon.$$ That establishes the inequality discussed in my first paragraph above.

• Hi Michael. I don't fully understand your answer: the 'limiting' rectangle shown in the diagram has $\pi r^2$, whereas the sum of the areas of the wedges is also $\pi r^2$. So it doesn't seem necessary to 'make the numbers of wedges so large that the difference between the area of the rectangle and the sum of the areas of the wedges is less than $\epsilon$'—if the two shapes have the same area, then any number of wedges will suffice. What am I missing?
– Joe
Commented Jun 3, 2021 at 21:35
• @Joe : What you're missing is just about the whole point. You DO NOT KNOW that the sum of the areas of the wedges is $\pi r^2$ because that is just the thing that you are trying to prove. Commented Jun 4, 2021 at 3:04

Whether this is rigorous will depend on your definitions of area and length, and whether this is a proof will depend on what we are allowed to presuppose, but I will try to show how control the error.

We only need to show that for small $$x$$, one of those segments with arc length $$rx$$ (making $$x$$ the angle) will be close enough in area to a rectangle with area $$\frac{rx\cdot r}2$$. Now from your usual diagram used to introduce trigonometric functions, we see that the area is between $$\frac{r\cos x\cdot r\sin x}2$$ and $$\frac{r\cdot r\tan x}2$$. Since the circumference of the circle is $$2\pi r$$, we will need $$\frac{2\pi}x$$ of these segments. (You can choose $$x$$ to make this an integer if you like, i.e. set $$x=\frac{2\pi}n$$ and let $$n\to\infty$$.) This sandwiches the area of the circle between $$\lim_{x\to0} \frac{2\pi}x \cdot \frac{r\cos x\cdot r\sin x}2 = \lim_{x\to0} \pi r^2\cdot \cos x\cdot\frac{\sin x}x = \pi r^2$$ and $$\lim_{x\to0} \frac{2\pi}x \cdot \frac{r\cdot r\tan x}2 = \lim_{x\to0} \pi r^2\cdot \frac1{\cos x}\cdot\frac{\sin x}x = \pi r^2.$$ So the area must be $$\pi r^2$$. Of course, this assumes that we already know that $$\lim_{x\to0}\frac{\sin x}x=1$$.