Arc Length and Circle Area

Circumference

As we inscribe regular $n$-gons in a circle with increasing $n$, their perimeters approach the circle’s circumference. Two circles with radii $R$ and $R'$ satisfy $\dfrac{C}{C'} = \dfrac{2R}{2R'}$, so the ratio $\dfrac{C}{2R}$ is the same for all circles.

This common ratio is denoted $\pi$ (read: “pi”):

$\boxed{C = 2\pi R}$

The number $\pi$ is irrational: $\pi = 3.14159265\ldots$ For calculations, $\pi \approx 3.14$.

Archimedes (3rd century BCE) proved $3\tfrac{10}{71} < \pi < 3\tfrac{1}{7}$, confirming $\pi > 3.14$.

Arc Length

A circle of radius $R$ has circumference $2\pi R$, corresponding to $360°$. An arc of $n°$ has length:

$\boxed{l = \frac{\pi R n}{180}}$

Circle Area

Using the same limiting argument with inscribed polygons, the area of a circle of radius $R$ is:

$\boxed{S = \pi R^2}$

Adjust the central angle to see the sector (blue), the triangle OAB (red), and the segment (their difference) update live:

Circular Sector

A full circle has area $\pi R^2$ corresponding to $360°$. A sector with arc measure $n°$ has area:

$\boxed{S_{\text{sector}} = \frac{\pi R^2 n}{360}}$

Circular Segment

To find the area of a segment:

• Minor segment (arc less than semicircle): $S_{\text{seg}} = S_{\text{sector}} - S_{\triangle AOB}$, where $O$ is the center.
• Major segment: $S_{\text{seg}} = S_{\text{sector}} + S_{\triangle AOB}$ (for reflex sector).

The area of triangle $AOB$ with $OA = OB = R$ and central angle $\angle AOB = \varphi$:

$S_{\triangle AOB} = \frac{1}{2}R^2\sin\varphi$

Semicircle

A semicircle is a segment whose chord is a diameter:

$S_{\text{semicircle}} = \frac{\pi R^2}{2}$

Summary

Exercises