Arcs and Sectors – GCSE Maths

An arc is part of a circumference and a sector is part of an area. Both are found the same way: work out what fraction of the whole circle you have, then take that fraction of the whole.

The key fraction

$\dfrac{\theta}{360}$ — where $\theta$ is the angle at the centre, in degrees

The two formulae

Arc length $= \dfrac{\theta}{360} \times 2\pi r$

Sector area $= \dfrac{\theta}{360} \times \pi r^2$

Angle $\theta$	Fraction	Name
$360^\circ$	$1$	the whole circle
$180^\circ$	$\tfrac{1}{2}$	semicircle
$120^\circ$	$\tfrac{1}{3}$	a third
$90^\circ$	$\tfrac{1}{4}$	quarter circle
$60^\circ$	$\tfrac{1}{6}$	a sixth
$45^\circ$	$\tfrac{1}{8}$	an eighth

Write down $r$ and $\theta$.
Find the whole circumference, $2\pi r$.
Multiply by $\dfrac{\theta}{360}$.
Give the answer in the units of length (cm, m…), not squared.

Worked Example 1 — Arc length

A sector has radius $12$ cm and angle $50^\circ$. Find the arc length, to 2 d.p.

①Whole circumference $= 2\pi \times 12 = 24\pi$

②Fraction $= \dfrac{50}{360}$

③Arc $= \dfrac{50}{360} \times 24\pi = \dfrac{1200\pi}{360} = \dfrac{10\pi}{3}$

④$= 10.47$ cm

Worked Example 2 — Finding the angle

An arc of length $15$ cm is drawn on a circle of radius $10$ cm. Find the angle at the centre, to the nearest degree.

①$\dfrac{\theta}{360} \times 2\pi \times 10 = 15$

②$\dfrac{\theta}{360} \times 20\pi = 15$

③$\theta = \dfrac{15 \times 360}{20\pi} = \dfrac{5400}{62.832} = 85.94$

④$\theta \approx 86^\circ$

Worked Example 3 — Sector area

A sector has radius $8$ cm and angle $135^\circ$. Find its area, to 1 d.p.

①Whole circle area $= \pi r^2 = 64\pi$

②Fraction $= \dfrac{135}{360} = \dfrac{3}{8}$

③Area $= \dfrac{3}{8} \times 64\pi = 24\pi$

④$= 75.4\text{ cm}^2$

Worked Example 4 — Finding the radius

A sector of angle $72^\circ$ has area $50\text{ cm}^2$. Find its radius, to 2 d.p.

①$\dfrac{72}{360} \times \pi r^2 = 50$

②$\dfrac{1}{5} \pi r^2 = 50$

③$\pi r^2 = 250$, so $r^2 = \dfrac{250}{\pi} = 79.577$

④$r = \sqrt{79.577} = 8.92$ cm

Perimeter of a sector

$P = \text{arc length} + 2r$

Don't forget the two radii. The perimeter is the whole way round the outside of the slice: the curved arc plus the two straight edges. Leaving out the $2r$ is the most common error on this topic.

Worked Example 5 — Perimeter of a sector

A sector has radius $9$ cm and angle $80^\circ$. Find its perimeter, to 1 d.p.

①Arc $= \dfrac{80}{360} \times 2\pi \times 9 = \dfrac{80}{360} \times 18\pi = 4\pi = 12.566$ cm

②Two radii $= 2 \times 9 = 18$ cm

③Perimeter $= 12.566 + 18 = 30.6$ cm

A segment is what is left when you cut the triangle off a sector.

Segment area

segment $=$ sector $-$ triangle

$= \dfrac{\theta}{360}\pi r^2 - \dfrac{1}{2}r^2 \sin\theta$

The triangle formed by the two radii and the chord has two sides of length $r$ with the angle $\theta$ between them, so its area is $\tfrac{1}{2}ab\sin C = \tfrac{1}{2}r^2\sin\theta$.

Worked Example 6 — Segment area

A circle has radius $10$ cm. A chord subtends an angle of $70^\circ$ at the centre. Find the area of the minor segment, to 2 d.p.

①Sector area $= \dfrac{70}{360} \times \pi \times 100 = 61.087\text{ cm}^2$

②Triangle area $= \tfrac{1}{2} \times 10^2 \times \sin 70^\circ = 50 \times 0.93969 = 46.985\text{ cm}^2$

③Segment $= 61.087 - 46.985 = 14.10\text{ cm}^2$

Make sure your calculator is in degrees mode before using $\sin$.

Minor means the smaller piece (angle less than $180^\circ$).
Major means the larger piece (angle more than $180^\circ$).
The two angles always add to $360^\circ$.

Worked Example 7 — Major sector

A minor sector has angle $110^\circ$ and radius $6$ cm. Find the area of the corresponding major sector, to 1 d.p.

①Major angle $= 360 - 110 = 250^\circ$

②Area $= \dfrac{250}{360} \times \pi \times 36 = \dfrac{250}{360} \times 113.097$

③$= 78.5\text{ cm}^2$

Check: the minor sector is $\dfrac{110}{360} \times 113.097 = 34.6\text{ cm}^2$, and $78.5 + 34.6 = 113.1$ — the whole circle ✓

The fraction

$\dfrac{\theta}{360}$ of the whole circle.

Arc length

$\dfrac{\theta}{360} \times 2\pi r$ — a length, so units are cm not cm².

Sector area

$\dfrac{\theta}{360} \times \pi r^2$.

Sector perimeter

arc $+ 2r$ — never forget the two radii.

Segment

sector $-$ triangle, with triangle $= \tfrac{1}{2}r^2\sin\theta$.

Major / minor

The two angles add to $360^\circ$.

Finding $\theta$ or $r$

Substitute into the formula and rearrange.

Accuracy

Keep $\pi$ exact until the final line.

Question 1

Find the arc length of a sector with radius $15$ cm and angle $60^\circ$, in terms of $\pi$.

▶ Show solution

$\dfrac{60}{360} = \dfrac{1}{6}$

Arc $= \dfrac{1}{6} \times 2\pi \times 15 = \dfrac{30\pi}{6} = 5\pi$ cm

Question 2

Find the area of a sector with radius $10$ cm and angle $45^\circ$, to 2 d.p.

▶ Show solution

$\dfrac{45}{360} = \dfrac{1}{8}$

Area $= \dfrac{1}{8} \times \pi \times 100 = 12.5\pi = 39.27\text{ cm}^2$

Question 3

Find the perimeter of a quarter circle of radius $14$ cm, to 1 d.p.

▶ Show solution

Arc $= \dfrac{90}{360} \times 2\pi \times 14 = \dfrac{1}{4} \times 28\pi = 7\pi = 21.99$ cm

Two radii $= 2 \times 14 = 28$ cm

Perimeter $= 21.99 + 28 = 50.0$ cm

Question 4

A sector has radius $8$ cm and arc length $10$ cm. Find the angle at the centre, to the nearest degree.

▶ Show solution

$\dfrac{\theta}{360} \times 2\pi \times 8 = 10$

$\dfrac{\theta}{360} \times 16\pi = 10$

$\theta = \dfrac{10 \times 360}{16\pi} = \dfrac{3600}{50.265} = 71.6$

$\theta \approx 72^\circ$

Question 5

A sector of angle $120^\circ$ has area $30\text{ cm}^2$. Find its radius, to 2 d.p.

▶ Show solution

$\dfrac{120}{360} \times \pi r^2 = 30$

$\dfrac{1}{3}\pi r^2 = 30$, so $\pi r^2 = 90$

$r^2 = \dfrac{90}{\pi} = 28.648$

$r = 5.35$ cm

Question 6

A minor sector has angle $95^\circ$ and radius $12$ cm. Find the arc length of the major arc, to 1 d.p.

▶ Show solution

Major angle $= 360 - 95 = 265^\circ$

Arc $= \dfrac{265}{360} \times 2\pi \times 12 = \dfrac{265}{360} \times 75.398$

$= 55.5$ cm

Question 7

A circle of radius $9$ cm has a chord subtending $100^\circ$ at the centre. Find the area of the minor segment, to 2 d.p.

▶ Show solution

Sector $= \dfrac{100}{360} \times \pi \times 81 = 70.686\text{ cm}^2$

Triangle $= \tfrac{1}{2} \times 81 \times \sin 100^\circ = 40.5 \times 0.98481 = 39.885\text{ cm}^2$

Segment $= 70.686 - 39.885 = 30.80\text{ cm}^2$

Question 8

A goat is tethered by a $7$ m rope to a post at the corner of a large rectangular field, so it can graze a quarter circle. Find the grazing area, to 1 d.p.

▶ Show solution

The corner gives a $90^\circ$ sector of radius $7$ m.

Area $= \dfrac{90}{360} \times \pi \times 49 = \dfrac{1}{4} \times 153.938$

$= 38.5\text{ m}^2$

Question 9

A sector of a circle of radius $10$ cm has a perimeter of $35$ cm. Find the angle at the centre, to the nearest degree.

▶ Show solution

Perimeter $=$ arc $+ 2r$, so arc $= 35 - 20 = 15$ cm.

$\dfrac{\theta}{360} \times 2\pi \times 10 = 15$

$\theta = \dfrac{15 \times 360}{20\pi} = \dfrac{5400}{62.832} = 85.94$

$\theta \approx 86^\circ$

Question 10

A running track is made of a rectangle $84.4$ m long with a semicircular end of radius $36.5$ m at each end.

(a) Find the total perimeter of the track, to 1 d.p. (b) Find the total area enclosed, to the nearest m². (c) A runner completes $4$ laps. How far has she run, to the nearest metre?

▶ Show solution

(a) The two semicircular ends together form one full circle of radius $36.5$ m.

Curved part $= 2\pi \times 36.5 = 229.336$ m

Straight parts $= 2 \times 84.4 = 168.8$ m

Perimeter $= 229.336 + 168.8 = \mathbf{398.1}$ m (1 d.p.)

(b) Rectangle $= 84.4 \times (2 \times 36.5) = 84.4 \times 73 = 6161.2\text{ m}^2$

Full circle $= \pi \times 36.5^2 = \pi \times 1332.25 = 4185.4\text{ m}^2$

Total $= 6161.2 + 4185.4 = \mathbf{10\,347}\text{ m}^2$ (nearest m²)

(c) $4 \times 398.136 = 1592.5$, so about $\mathbf{1593}$ m.