Arc length and sector area

Covers AS91259

The parts of a circle

An arc is part of the circumference — the curved edge of a slice.
A sector is the "pizza slice" shape bounded by two radii and an arc.
The angle at the centre between the two radii is $\theta$ (theta), measured in degrees.
Both an arc and a sector are just a fraction of the whole circle — the fraction $\dfrac{\theta}{360}$ , because a full turn is $360^\circ$ .

A circle of radius r with a shaded sector, the centre angle theta between two radii, and the curved arc highlighted.

Arc length

The arc is that same fraction of the whole circumference $2\pi r$ :

$\ell = \frac{\theta}{360} \times 2\pi r$

$\ell$ — the arc length.
$\theta$ — the centre angle in degrees.
$r$ — the radius.

Sector area

The sector is that fraction of the whole area $\pi r^2$ :

$A = \frac{\theta}{360} \times \pi r^2$

$A$ — the sector area.
Same $\theta$ and $r$ as the arc — only the "whole" changes (circumference for arc, area for sector).

Using the formulas

Write the fraction $\dfrac{\theta}{360}$ first — it is common to both.
Multiply by the circumference $2\pi r$ for an arc, or the area $\pi r^2$ for a sector.
Keep units straight: arc length is in cm or m; sector area is in cm² or m².

Worked ExampleLength of a curved path

A circular garden has radius $8$ m. A curved path runs along the edge of a sector with a centre angle of $45^\circ$ . Find the length of the path.

Step 1 — Write the fraction of the circle

$\frac{\theta}{360} = \frac{45}{360} = \frac{1}{8}$

Step 2 — Multiply by the whole circumference

$\ell = \frac{1}{8} \times 2\pi(8) = \frac{1}{8} \times 16\pi = 2\pi$

Step 3 — Evaluate

$\ell = 2\pi \approx 6.28 \text{ m}$

Worked ExampleArea of a sector

A lawn sprinkler sweeps a sector of radius $5$ m through an angle of $120^\circ$ . Find the area it waters.

Step 1 — Fraction of the circle

$\frac{120}{360} = \frac{1}{3}$

Step 2 — Multiply by the whole area

$A = \frac{1}{3} \times \pi(5)^2 = \frac{1}{3} \times 25\pi$

Step 3 — Evaluate

$A = \frac{25\pi}{3} \approx 26.2 \text{ m}^2$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A sector has radius $6$ cm and centre angle $90^\circ$ . Calculate the arc length.

Merit

A sector of a circle has area $30\pi$ cm² and radius $10$ cm. Find the centre angle $\theta$ .

Excellence

A windscreen wiper has a blade $30$ cm long attached to an arm; the blade's far end is $50$ cm from the pivot. The wiper sweeps through $110^\circ$ . Find the area cleaned by the blade (the ring-shaped region between radius 20 cm and radius 50 cm).

Arc length and sector area

Covers AS91259

The parts of a circle

An arc is part of the circumference — the curved edge of a slice.
A sector is the "pizza slice" shape bounded by two radii and an arc.
The angle at the centre between the two radii is $\theta$ (theta), measured in degrees.
Both an arc and a sector are just a fraction of the whole circle — the fraction $\dfrac{\theta}{360}$ , because a full turn is $360^\circ$ .

A circle of radius r with a shaded sector, the centre angle theta between two radii, and the curved arc highlighted.

Arc length

The arc is that same fraction of the whole circumference $2\pi r$ :

$\ell = \frac{\theta}{360} \times 2\pi r$

$\ell$ — the arc length.
$\theta$ — the centre angle in degrees.
$r$ — the radius.

Sector area

The sector is that fraction of the whole area $\pi r^2$ :

$A = \frac{\theta}{360} \times \pi r^2$

$A$ — the sector area.
Same $\theta$ and $r$ as the arc — only the "whole" changes (circumference for arc, area for sector).

Using the formulas

Write the fraction $\dfrac{\theta}{360}$ first — it is common to both.
Multiply by the circumference $2\pi r$ for an arc, or the area $\pi r^2$ for a sector.
Keep units straight: arc length is in cm or m; sector area is in cm² or m².

Worked ExampleLength of a curved path

A circular garden has radius $8$ m. A curved path runs along the edge of a sector with a centre angle of $45^\circ$ . Find the length of the path.

Step 1 — Write the fraction of the circle

$\frac{\theta}{360} = \frac{45}{360} = \frac{1}{8}$

Step 2 — Multiply by the whole circumference

$\ell = \frac{1}{8} \times 2\pi(8) = \frac{1}{8} \times 16\pi = 2\pi$

Step 3 — Evaluate

$\ell = 2\pi \approx 6.28 \text{ m}$

Worked ExampleArea of a sector

A lawn sprinkler sweeps a sector of radius $5$ m through an angle of $120^\circ$ . Find the area it waters.

Step 1 — Fraction of the circle

$\frac{120}{360} = \frac{1}{3}$

Step 2 — Multiply by the whole area

$A = \frac{1}{3} \times \pi(5)^2 = \frac{1}{3} \times 25\pi$

Step 3 — Evaluate

$A = \frac{25\pi}{3} \approx 26.2 \text{ m}^2$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A sector has radius $6$ cm and centre angle $90^\circ$ . Calculate the arc length.

Merit

A sector of a circle has area $30\pi$ cm² and radius $10$ cm. Find the centre angle $\theta$ .

Excellence