Arc length and sector area

By the end of this lesson

By the end you can find the arc length and area of any sector with $s = r\theta$ and $A = \tfrac12 r^2\theta$ , find its perimeter, and run the formulas backwards to find $r$ or $\theta$ .

Two formulas, one condition

Think of a slice of pizza. The sector is the whole slice, the arc is the crust, and the two straight edges are radii.

For a sector with radius $r$ and angle $\theta$ at the centre:

$s = r\theta \qquad\qquad A = \tfrac12 r^2 \theta$

$s$ is the arc length (the length of the crust)
$A$ is the area of the sector
$r$ is the radius
$\theta$ (the Greek letter "theta") is the angle at the centre

The condition that matters: $\theta$ must be in radians. Both formulas are wrong in degrees.

(Where the pizza breaks down: real slices have thickness. Ours are perfectly flat.)

Where the formulas come from

A full turn is $2\pi$ radians. So a sector with angle $\theta$ is the fraction $\dfrac{\theta}{2\pi}$ of the whole circle.

(Check: a half circle has $\theta = \pi$ , and $\dfrac{\pi}{2\pi} = \dfrac12$ ✓)

So take that fraction of the whole circle's measurements.

Arc: the whole circumference is $2\pi r$ .

$s = \frac{\theta}{2\pi} \times 2\pi r = r\theta$

(The $2\pi$ cancels.)

Area: the whole circle's area is $\pi r^2$ .

$A = \frac{\theta}{2\pi} \times \pi r^2 = \tfrac12 r^2 \theta$

(The $\pi$ cancels, leaving $\frac{1}{2}$ .)

Even quicker for the arc: by definition, 1 radian cuts off an arc of length $r$ . So $\theta$ radians cuts off $\theta$ lots of $r$ .

Worked example: Arc, area and perimeter

A sector has radius $8$ cm and angle $0.9$ radians. Find its arc length, its area and its perimeter.

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