Layman's proof that the area of a circle of radius $r$ equals $\pi r^2$.

Question

There are many, many clear and simple proofs of basic but nontrivial facts in high-school mathematics, such as Pythagoras' theorem or the identities $$\sum_{k=1}^nk=\binom{n}{2}\qquad\text{ and }\qquad\sum_{k=1}^n(2k-1)=n^2,$$ that are accessible to the lay person, having no mathematical knowledge beyond perhaps the most basic high-school curriculum, or perhaps even elementary school. See the answers to this question for many such proofs. I'm surprised I can't find such a proof for the fact that the area of a circle of radius $r$ equals $\pi r^2$ that is thoroughly convincing. Let me explain what I find unconvcing about two popular visual proofs.

Parallellogram proof

The proof I see most often is the parallellogram proof, cutting the circle radially into ever smaller wedges and joining them together to approximate a parallellogram:

It leaves the question of convergence open; do these approximations of parallellograms converge to a rectangle with side lengths $r$ and $\pi r$ as we take ever smaller wedges? It turns out that they do and the argument can be made rigorous, but to the critical lay person this need not be clear from the picture.

Triangle proof

Another proof I have seen quite often is the triangle proof, cutting the circle into ever thinner concentric rings and unwrapping them to approximate a right triangle:

Again this leaves the question of convergence; do these approximations of right triangles converge to a right triangle of height $r$ and base $2\pi r$? Again it turns out that they do and the argument can be made rigorous, but again to the critical lay person this need not be clear from the picture.

Other proofs either require techniques beyond the most basic high-school curriculum, such as calculus for the onion proof, or are computationally involved such as Archimedes' proof or variations thereof.

I do assume that the layman is familiar with the fact that the circumference of a circle of radius $r$ equals $2\pi r$.

TL;DR

Given that the circumference of a circle of radius $r$ equals $2\pi r$, is there a simple proof of the fact that its area equals $\pi r^2$, using nothing but the most elementary mathematical techniques (certainly no calculus)?

Answer 1

This is a calculus-based proof, but it is very elementary. As a reference, see Keith Ball - An Elementary Introduction to Modern Convex Geometry.

For any $r>0$, let $L(r)$ be the length of a circle with radius $r$ and let $A(r)$ be the enclosed area.
By convexity (which also ensures that $A$ and $L$ are well-defined) $$ L(r)=\lim_{\varepsilon\to 0^+}\frac{A(r+\varepsilon)-A(r)}{\varepsilon}.\tag{$\delta$}$$ Since concentric circles with any radius are homothetic, $A(r) = r^2 A(1)$ and $L(r)=r L(1)$.
On the other hand $(\delta)$ gives $L(r)=\frac{d}{dr}A(r)$, hence $L(1)=\color{red}{2}\,A(1)$.

So we may equivalently define $\pi$ as $A(1)$ or as $\frac{1}{2}L(1)$.
And we have that in $\mathbb{R}^n$ the surface area of the Euclidean unit ball is just $n$ times its volume.

$(\delta)$ also follows from this pictorial argument, where the outer "polygon" is obtained by summing (in the set-sense) a small disk to the inner polygon:

It is visually clear what happens if we let the number of sides go to $+\infty$ while preserving the circum-radii of the inner and outer polygon.

Answer 2

I dont know how to draw pictures in the site, so I hope my explanation is clear, but you may have to draw your own. Inscribe a regular $n$-gon into the circle. At the same time exscribe a regular $n$-gon. This is in such a way that the sides of the triangles of the inscribed $n$-gon (which are radii, as in your first picture) extend to the sides of the exscribed $n$-gon. The height $h$ of a triangle of the inscribed $n$-gon is $r\cos \frac{2\pi}{n}$. Thus the area of a triangle of the inscribed $n$-gon is $$\frac{1}{2}br\cos\frac{2\pi}{n},$$ where $b$ is the base. Adding these together we have the area of the inscribed $n$-gon is $$\frac{1}{2}C_i r\cos\frac{2\pi}{n}$$ where $C_i$ is the circumference of the inscribed $n$-gon.

For the exscribed $n$-gon, the height of a triangle is $r$, and so the area of the exscribed $n$-gon is $$\frac{1}{2}C_e r$$ where $C_e$ is the circumference of the exscribes $n$-gon. Thus we have

$$\frac{1}{2}C_i r\cos\frac{2\pi}{n}\leq A\leq \frac{1}{2}C_e r$$

What is the relation between $C_i$ and $C_e$ ? It follows from similar triangles that

$$\frac{C_e}{C_i}=\frac{r}{h}=\frac{1}{\cos\frac{2\pi}{n}}$$

Thus we have

$$\frac{1}{2}C_i r\cos\frac{2\pi}{n}\leq A\leq \frac{1}{2}C_i r\frac{1}{\cos\frac{2\pi}{n}}$$ Now as $n\to\infty$, $$\cos\frac{2\pi}{n}\to 1$$ and $$C_i\to C$$ where $C$ is the circumference of the circle. This is in fact the definition of the circumference. Further, $$C=2\pi r$$ is the definition of $\pi$. So we have

$$A=\frac{1}{2}Cr=\pi r^2.$$

Answer 3

The onion proof can be done without calculus, provided one assumes the following: the area of a circular ring is greater than width $\times$ inner circumference, and less than width $\times$ outer circumference. For a positive integer $n$, divide the circle into concentric rings of equal width $r/n$. If the above assumption is accepted, then the area of the $k$th ring ($k = 1, 2, \ldots, n$) lies between $(r/n) \times 2\pi r(k - 1)/n = 2\pi r^2(k - 1)/n^2$ and $2\pi r^2k/n^2$. From the formulae $$0 + 1 + 2 + \cdots + (n - 1) = (n^2 - n)/2,\quad 1 + 2 + 3 + \cdots + n = (n^2 + n)/2$$ it follows that the area of the whole circle lies between $\pi r^2(1 - 1/n)$ and $\pi r^2(1 + 1/n)$. Since this is true for any $n$, the area is $\pi r^2$.

Answer 4

We can explain to the layman that thousands of years ago Euclid knew that there was a fixed number $K$ such that the area of a circle of radius $r$ is given by $K r^2$, and that this result was obtained without calculus.

Now, we tell the layman to look at $y = \sqrt{1-x^2}$ in the first quadrant and the region $G$ bounded by the graph and $x = 0$ and $y = 0$. Without calculus, in the obvious way (uh, with the help of a computer) they can approximate the area of $G$ and then multiply by $4$ to get the area of the unit circle = $\pi$.

We explain that this brute force method certainly seems convincing that we are talking about the same number $\pi$ in both the formula for the circumference and the formula for the area of a circle, but it is not a proof. Without exploring Archimedes proof or calculus, they will need to accept some amount of 'hand waving' in any purported proof. But this quick video seems accessible,

Archimedes' Proof that the Area of a Circle is $ \pi r^2$

If the layman needs to see a Cartesian brute force method, they can inspect the google spreadsheet Calculate $\pi$. At the bottom of the spreadsheet ($\text{row } = 1027$) they will find that

$\quad 3.139603642 \lt \pi \lt 3.143509892$

with the average of the 'squeeze' = $3.141556767$.

If the layman is with us so far, they might become so interested that they begin to study calculus in their spare time.