What is G18: Arcs and sectors?

A sector or arc occupies the same fraction of a circle as its central angle occupies of 360°. For angle θ, arc length is θ/3602π r and sector area is θ/360×π r^2.

How many marks are G18 questions worth?

In verified past-paper appearances, G18 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has G18 appeared in?

G18 has appeared in 10 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G18

Calculate arc lengths, angles and areas of sectors of circles

Arcs and sectors

Worked answers, methods and verified real exam appearances for G18 on Edexcel GCSE Maths 1MA1.

Explanation

A sector or arc occupies the same fraction of a circle as its central angle occupies of $360^\circ$ . For angle $\theta$ , arc length is $\frac{\theta}{360}\times2\pi r$ and sector area is $\frac{\theta}{360}\times\pi r^2$ .
Rearrange these relationships when the angle or radius is unknown.
An arc length contains only the curved boundary; a sector perimeter also contains the two radii.
For an annular sector, subtract the inner sector area from the outer one and include both arcs in the perimeter.
Examiners expect exact answers in terms of $\pi$ unless a decimal accuracy is specified.

A sector's arc and area are the fraction

\theta/360

of the full circle.

Worked example

A sector has radius $9$ cm and angle $120^\circ$ . Find its arc length and area in terms of $\pi$ .

1.Arc length $=\frac{120}{360}\times2\pi(9)=6\pi$ cm.
2.Sector area $=\frac{120}{360}\times\pi(9^2)=27\pi\text{ cm}^2$ .
3.Keep the exact $\pi$ forms because no decimal accuracy is requested.

Answer: Arc length $=6\pi$ cm; area $=27\pi\text{ cm}^2$ .

Common mistakes

Don't use $\theta/180$ instead of $\theta/360$ for the sector fraction.
Don't add the two radii when the question asks for arc length rather than sector perimeter.

Exam tip

Write the fraction $\theta/360$ first; it earns the method whether the question asks for an arc, area or rearranged angle.

Worked practice

Tier 1 · Easy

Work out the arc length of a quarter-circle with radius

8\text{ cm}

. Give the answer in terms of

\pi

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$4\pi\text{ cm}$	2	A quarter-circle has angle $90^\circ$ . Its arc length is $\frac{90}{360}\times2\pi(8)=\frac14\times16\pi=4\pi\text{ cm}$ .

Tier 2 · Standard

A sector has radius

9\text{ cm}

and arc length

7\pi\text{ cm}

. Work out the angle of the sector and its area in terms of

\pi

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	Angle $=140^\circ$ Area $=\frac{63\pi}{2}\text{ cm}^2$	4	Use $7\pi=\frac{\theta}{360}\times2\pi(9)$ . Cancelling $\pi$ and solving gives $\theta=140^\circ$ . The sector area is then $\frac{140}{360}\times\pi(9^2)=\frac{63\pi}{2}\text{ cm}^2$ .

Tier 3 · Hard

A shape is an annular sector with angle

144^\circ

, outer radius

12\text{ cm}

and inner radius

7\text{ cm}

. Work out its area and perimeter in terms of

\pi

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Area $=38\pi\text{ cm}^2$ Perimeter $=10+\frac{76\pi}{5}\text{ cm}$	5	The area is $\frac{144}{360}\pi(12^2-7^2)=\frac25\pi(144-49)=38\pi\text{ cm}^2$ . The two arcs have total length $\frac25\times2\pi(12+7)=\frac{76\pi}{5}\text{ cm}$ . The two straight radial edges each have length $12-7=5\text{ cm}$ , so the perimeter is $10+\frac{76\pi}{5}\text{ cm}$ .

Tier 1 · Easy

A semicircle has diameter

22\text{ cm}

. Work out its perimeter in terms of

\pi

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$22+11\pi\text{ cm}$	2	The radius is $11\text{ cm}$ , so the curved edge has length $\pi r=11\pi\text{ cm}$ . Including the diameter, the perimeter is $22+11\pi\text{ cm}$ .

Tier 2 · Standard

A semicircle has area

50\pi\text{ cm}^2

. Work out the length of its curved edge in terms of

\pi

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$10\pi\text{ cm}$	3	The area is $\frac12\pi r^2$ , so $\frac12\pi r^2=50\pi$ . Hence $r^2=100$ and $r=10\text{ cm}$ . The curved edge has length $\pi r=10\pi\text{ cm}$ .

Tier 3 · Hard

A sector has angle

128^\circ

and area

80\pi\text{ cm}^2

. Work out the full perimeter of the sector, including the two radii. Give your answer in terms of

\pi

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$30+\frac{32\pi}{3}\text{ cm}$	5	Let the radius be $r\text{ cm}$ . From the area, $\frac{128}{360}\pi r^2=80\pi$ , so $\frac{16}{45}r^2=80$ . Hence $r^2=225$ and $r=15\text{ cm}$ . The arc length is $\frac{128}{360}\times2\pi(15)=\frac{32\pi}{3}\text{ cm}$ . Including the two radii, the full perimeter is $30+\frac{32\pi}{3}\text{ cm}$ .

Tier 2 · Standard

The minute hand of a clock is

9\text{ cm}

long. Work out the distance travelled by the tip of the minute hand from

2{:}00

2{:}25

. Give your answer in terms of

\pi

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$\frac{15\pi}{2}\text{ cm}$ (or $7.5\pi\text{ cm}$ )	3	In $25$ minutes the hand turns through $\frac{25}{60}$ of a full circle. The tip travels $\frac{25}{60}\times2\pi(9)=\frac{15\pi}{2}\text{ cm}$ .

Tier 3 · Hard

Sector A has radius

9\text{ cm}

and angle

160^\circ

. Sector B has radius

12\text{ cm}

and the same area as sector A. Work out the angle of sector B and the difference between the arc lengths of the two sectors. Give the difference in terms of

\pi

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	Angle of sector B $=90^\circ$ Difference $=2\pi\text{ cm}$	4	Sector A has area $\frac{160}{360}\pi(9^2)=36\pi\text{ cm}^2$ . If sector B has angle $x$ , then $\frac{x}{360}\pi(12^2)=36\pi$ , giving $x=90^\circ$ . The arc lengths are $\frac{160}{360}\times2\pi(9)=8\pi\text{ cm}$ and $\frac{90}{360}\times2\pi(12)=6\pi\text{ cm}$ , so the difference is $2\pi\text{ cm}$ .

Tier 3 · Hard

Higher only: A sector has area

54\pi\text{ cm}^2

and arc length

6\pi\text{ cm}

. Work out the radius and the angle of the sector.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Radius $=18\text{ cm}$ Angle $=60^\circ$	5	For the same sector fraction, dividing area by arc length gives $\frac{\pi r^2}{2\pi r}=\frac r2$ . Hence $54\pi\div6\pi=9=\frac r2$ , so $r=18\text{ cm}$ . Now $6\pi=\frac{\theta}{360}\times2\pi(18)$ . Therefore $6=\frac{\theta}{10}$ and $\theta=60^\circ$ .

Q10

Tier 3 · Hard

Higher only: A clock face has radius

18\text{ cm}

. At

4{:}20

, the hour hand has moved continuously from the

4

towards the

5

. Work out the exact area of the smaller sector between the hour hand and the minute hand.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$9\pi\text{ cm}^2$	4	At $4{:}20$ , the minute hand is $20\times6^\circ=120^\circ$ clockwise from $12$ . The hour hand is $4\times30^\circ+\frac{20}{60}\times30^\circ=130^\circ$ clockwise from $12$ . The smaller angle is $10^\circ$ , so the sector area is $\frac{10}{360}\times\pi(18^2)=9\pi\text{ cm}^2$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2022-06	1H	Q21	5	Non-calculator	Higher	QP MS
2019-06	2H	Q12	4	Allowed	Higher	QP MS
2021-11	3H	Q7	4	Allowed	Higher	QP MS
2023-11	3H	Q12	2	Allowed	Higher	QP MS
2019-11	3H	Q23	5	Allowed	Higher	QP MS
2021-11	3F	Q27	4	Allowed	Foundation	QP MS
2022-11	1H	Q15	3	Non-calculator	Higher	QP MS
2024-06	1H	Q15	4	Non-calculator	Higher	QP MS
2023-06	3H	Q14	4	Allowed	Higher	QP MS
2024-11	3H	Q11	3	Allowed	Higher	QP MS

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