Why a circle's area is π·r²

2026-05-05

A rectangle's area is the easiest thing in geometry. Width times height. The eye does it before the formula has a chance. A circle's area, on the other hand, has a π in it - and π is supposed to be the awkward number that lives on the rim, not inside.

Cut the disc into wedges. Lay them out alternately, tip up, tip down. The bumpy outline straightens as you go.

Drag n up. The bumps shrink. By the time you have sixty wedges, the figure is a rectangle within the width of a hair. And it has settled on a rectangle of π·r across by r tall.

Why those numbers

The wedges' tips alternate along the top and bottom. So the top edge of the rectangle is built from half of the wedges' arcs - and the bottom edge is built from the other half. Together, the arcs along the top and the arcs along the bottom add up to the whole rim, which is the circumference. Half of the circumference goes on top; half on the bottom. The top edge therefore has length (2·π·r) ⁄ 2 = π·r.

The height is just the radius - that's how far each wedge reaches from its tip to its arc. As n grows, the wedges get thinner and their arcs flatten, so the top and bottom edges become true straight lines.

So the rectangle is π·r wide and r tall. Its area is the product:

area = π·r · r = π·r²

Why we can rearrange at all

Cutting a shape and rearranging the pieces never changes the area. The pieces are still the same pieces; they cover the same amount of paper. So whatever the area of the disc is, the area of the rectangle has to be the same. We have not derived π·r² from nothing - we have noticed that the same area, looked at sideways, is a rectangle whose area we already knew how to compute.

This is the oldest move in mathematics. The rearrangement proof of Pythagoras is the same trick: cut, slide, recompute. Whatever you can re-arrange, you understand.

A small honesty

At any finite n, the figure is not exactly a rectangle. The top edge is faintly bumpy. The width is n·sin(π⁄n), which is almost π but a hair short. The height is cos(π⁄n), which is almost 1 but a hair short. Multiply them: n·sin(π⁄n)·cos(π⁄n), which is almost π but a hair short.

As n goes to infinity, all three "almosts" become equal. The wedges become slivers; the slivers stack into a true rectangle; the formula lands on π·r² exactly. The viz lets you see the convergence by hand: pull n up and watch the readout walk toward π.

π·r². A rectangle of π·r by r, drawn sideways out of a disc.