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

Practise 30 exam-style questions on Arc lengths and sectors for Edexcel GCSE Maths (Non-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.

30 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 9 cm and angle 40°.

Work out the arc length of the sector.
Give your answer in terms of \(\pi\).

Model answer

Arc length \(= \dfrac{40}{360} \times 2 \times \pi \times 9\)
\(= \dfrac{1}{9} \times 18\pi\)
\(= 2\pi\) cm

Explanation

The arc is a fraction of the whole circumference \(2\pi r\), and the fraction is \(\dfrac{\text{angle}}{360}\). Since \(\dfrac{40}{360} = \dfrac{1}{9}\) and the circumference is \(18\pi\), the arc is \(2\pi\) cm. A common mistake is to use \(\pi r^2\), which gives an area, not a length.

Mark scheme

1 mark
M1 for \(\dfrac{40}{360} \times 2 \times \pi \times 9\)
oe, e.g. \(\dfrac{1}{9} \times 18\pi\) or \(\dfrac{40}{360} \times \pi \times 18\); correct answer with no working scores full marks
1 mark
A1 for \(2\pi\)
accept \(2\pi\) cm

Question 2

3 marks

The diagram shows a sector of a circle with centre O and radius 6 cm.
The area of the sector is \(15\pi\) cm².
The angle of the sector is \(x^\circ\).

Work out the value of \(x\).

Geometry diagram, marked "Diagram NOT accurately drawn". A shaded sector with centre O, bounded by the radii OA and OB and the minor arc from A to B. Angle AOB is marked x°. OA is labelled 6 cm.

Model answer

\(\dfrac{x}{360} \times \pi \times 6^2 = 15\pi\)
\(\dfrac{x}{360} \times 36 = 15\)
\(\dfrac{x}{10} = 15\)
\(x = 150\)

Explanation

Write the sector area formula with the angle as the unknown and set it equal to the given area. The \(\pi\) cancels from both sides, leaving a simple equation. Check: 150° is \(\dfrac{5}{12}\) of the circle, and \(\dfrac{5}{12} \times 36\pi = 15\pi\).

Mark scheme

1 mark
M1 for \(\dfrac{x}{360} \times \pi \times 6^2 = 15\pi\)
oe, e.g. \(\dfrac{15\pi}{36\pi}\) as the fraction of the circle; correct answer with no working scores full marks
1 mark
M1 for \(x = \dfrac{15 \times 360}{36}\)
oe, e.g. \(\dfrac{x}{10} = 15\) or \(\dfrac{15}{36} \times 360\)
1 mark
A1 for 150
cao

Question 3

3 marks

The diagram shows a square ABCD with sides of length 10 cm.
BD is an arc of a circle with centre A and radius 10 cm.

Work out the perimeter of the shaded region.
Give your answer in terms of \(\pi\).

Geometry diagram, marked "Diagram NOT accurately drawn". Square ABCD (shaded). A white sector with centre A, bounded by the radii AB and AD and the minor arc from B to D. Side AB is labelled 10 cm. Angle BAD is marked as a right angle (90°).

Model answer

Arc BD \(= \dfrac{1}{4} \times 2 \times \pi \times 10 = 5\pi\)
The shaded region is bounded by BC, CD and arc BD.
Perimeter \(= 10 + 10 + 5\pi = 20 + 5\pi\) cm

Explanation

Trace the edge of the shaded region: side BC, side CD and the curved arc DB. The arc is a quarter of the circumference of a circle with radius 10 cm. Sides AB and AD are not on the edge of the shaded region, so they are not included.

Mark scheme

1 mark
M1 for arc length \(\dfrac{1}{4} \times 2 \times \pi \times 10\) \((= 5\pi)\)
oe; correct answer with no working scores full marks
1 mark
M1 for adding BC and CD to their arc, e.g. \(5\pi + 10 + 10\)
ft their arc length; do not award if AB or AD is included
1 mark
A1 for \(20 + 5\pi\)
oe, e.g. \(5(4 + \pi)\); accept with units cm

Question 4

4 marks

The diagram shows a square ABCD with sides of length 6 cm.
Two arcs join B and D: one is an arc of a circle with centre A, the other is an arc of a circle with centre C. Both circles have radius 6 cm.
The shaded region lies between the two arcs.

Show that the area of the shaded region is \(18(\pi - 2)\) cm².

Geometry diagram, marked "Diagram NOT accurately drawn". Square ABCD. A minor arc from B to D with centre A (the region between the arc and BD is shaded). A minor arc from D to B with centre C (the region between the arc and DB is shaded). Side AB is labelled 6 cm.

Model answer

Area of quarter circle with centre A \(= \dfrac{1}{4} \times \pi \times 6^2 = 9\pi\)
Area of triangle ABD \(= \dfrac{1}{2} \times 6 \times 6 = 18\)
The diagonal BD splits the shaded region into two equal segments.
Area of one segment \(= 9\pi - 18\)
Shaded area \(= 2(9\pi - 18) = 18\pi - 36 = 18(\pi - 2)\) cm²

Explanation

The diagonal BD cuts the shaded shape into two identical segments, each a quarter circle minus a right-angled triangle. Alternatively, the two quarter circles together cover the square once plus the shaded region a second time, so shaded area = \(9\pi + 9\pi - 36\).

Mark scheme

1 mark
M1 for area of a quarter circle \(\dfrac{1}{4} \times \pi \times 6^2\) \((= 9\pi)\)
oe
1 mark
M1 for area of triangle ABD (or BCD) \(\dfrac{1}{2} \times 6 \times 6\) \((= 18)\), or area of square 36
1 mark
M1 for a complete method, e.g. \(2 \times (9\pi - 18)\) or \(9\pi + 9\pi - 36\)
oe
1 mark
A1 for \(18\pi - 36 = 18(\pi - 2)\) with full working
answer only scores 0 (show that)

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