Arc lengths and sectors - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 36 exam-style questions on Arc lengths and sectors for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

36 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A sector of a circle has radius 12 cm and angle 118°.

Work out the area of the sector.
Give your answer correct to 3 significant figures.

Model answer

Area \(= \dfrac{118}{360} \times \pi \times 12^2 = 148.28...\)
148 cm² (3 s.f.)

Explanation

The area of a sector is \(\frac{\theta}{360} \times \pi r^2\). Square the radius before multiplying, and use the angle of this sector (118°) as the fraction of 360°.

Mark scheme

1 mark
M1 for \(\dfrac{118}{360} \times \pi \times 12^2\)
1 mark
A1 for 148 cm²
awrt 148 (148.2...); correct answer with no working scores full marks

Question 2

3 marks

A sector of a circle of radius 8 cm has an area of 50 cm².

Work out the angle of the sector.
Give your answer correct to 1 decimal place.

Model answer

\(\dfrac{\theta}{360} \times \pi \times 8^2 = 50\)
\(\dfrac{\theta}{360} \times 64\pi = 50\)
\(\theta = \dfrac{50 \times 360}{64\pi} = 89.524...\)
\(\theta = 89.5^\circ\) (1 d.p.)

Explanation

Substitute into the sector area formula and solve for the angle: multiply by 360 and divide by \(64\pi\). Check: the area of the whole circle is \(64\pi \approx 201\) cm², and 50 is about a quarter of that, so the angle should be about 90°.

Mark scheme

1 mark
M1 for \(\dfrac{\theta}{360} \times \pi \times 8^2 = 50\)
1 mark
M1 for \(\theta = \dfrac{50 \times 360}{64\pi}\)
oe
1 mark
A1 for 89.5°
awrt 89.5 (89.52...); correct answer with no working scores full marks

Question 3

4 marks

The diagram shows a sector OAB of a circle with centre O.
The radius is 10 cm and angle AOB = 70°.
The shaded region is the segment between the chord AB and the arc AB.

Work out the area of the shaded segment.
Give your answer correct to 3 significant figures.

Geometry diagram, marked "Diagram NOT accurately drawn". A minor arc from A to B with centre O (the region between the arc and AB is shaded). Line segments OA, OB and AB are drawn. Angle AOB is marked 70°. OA is labelled 10 cm.

Model answer

Sector OAB \(= \dfrac{70}{360} \times \pi \times 10^2 = 61.086...\) cm²
Triangle OAB \(= \frac{1}{2} \times 10 \times 10 \times \sin 70^\circ = 46.984...\) cm²
Segment \(= 61.086... - 46.984... = 14.101...\)
14.1 cm² (3 s.f.)

Explanation

A segment is the sector with the triangle OAB removed. The triangle has two sides of 10 cm with the 70° angle between them, so its area is \(\frac{1}{2}ab \sin C\). Treating the triangle as right-angled is a common error.

Mark scheme

1 mark
P1 for the area of the sector, \(\dfrac{70}{360} \times \pi \times 10^2\) (= 61.0...)
1 mark
P1 for the area of triangle OAB, \(\frac{1}{2} \times 10 \times 10 \times \sin 70^\circ\) (= 46.9...)
1 mark
P1 for sector − triangle
1 mark
A1 for 14.1 cm²
awrt 14.1 (14.10...); correct answer with no working scores full marks

Question 4

5 marks

The diagram shows a sector OAB of a circle with centre O.
The radius is 9 cm and angle AOB = 80°.
The shaded region is the segment between the chord AB and the arc AB.

Work out the perimeter of the shaded segment.
Give your answer correct to 3 significant figures.

Model answer

Arc AB \(= \dfrac{80}{360} \times 2 \times \pi \times 9 = 4\pi = 12.566...\) cm
Triangle OAB is isosceles. The perpendicular from O bisects angle AOB and the chord AB:
\(AB = 2 \times 9 \sin 40^\circ = 11.570...\) cm
Perimeter of segment \(= 12.566... + 11.570... = 24.136...\)
24.1 cm (3 s.f.)

Explanation

The segment's perimeter is the arc plus the straight chord AB - the radii are not part of it. Find the chord with the cosine rule in triangle OAB, or by splitting the isosceles triangle into two right-angled triangles with angle 40° and hypotenuse 9 cm.

Mark scheme

1 mark
P1 for the arc length, \(\dfrac{80}{360} \times 2 \times \pi \times 9\) (= 12.5...)
1 mark
P1 for a correct method for the chord, e.g. \(AB^2 = 9^2 + 9^2 - 2 \times 9 \times 9 \times \cos 80^\circ\) or \(AB = 2 \times 9 \sin 40^\circ\)
1 mark
A1 for AB = 11.5...
accept 11.5 to 11.6
1 mark
P1 for their arc + their chord
1 mark
A1 for 24.1 cm
awrt 24.1 (24.13...); correct answer with no working scores full marks

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