IGCSE Maths 0580 Circles, Arcs and Sectors Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Mensuration · 5.3

Every circles, arcs and sectors question from 16 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 5.3. 16 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/12/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Mensuration past paper questions.

0580 Circles, Arcs and Sectors past paper questions: Core (Papers 1 and 3)

12 marksCore

May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q23

The circumference of a circle is 18π cm. Work out the radius of the circle.

Show solution
  1. The circumference of a circle is C = 2πr.
  2. 2πr = 18π. Divide both sides by π: 2r = 18.
  3. Divide by 2: r = 9 cm
Answer 9 cm
22 marksCore

May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q16

The area of a circle is 36π cm². Find the radius of the circle.

Show solution
  1. The area of a circle is A = πr².
  2. πr² = 36π. Divide both sides by π: r² = 36.
  3. Take the square root: r = 6 cm
Answer 6 cm
33 marksCore

Feb/Mar 2026 · Paper 1 · 0580/12/F/M/26 · Q17

The diagram shows two circles. The small circle has diameter 8 cm. The large circle has diameter 20 cm. Work out the shaded area (the large circle outside the small circle). Give your answer in terms of π.

0580/12 February/March 2026 Q17 – small circle of diameter 8 cm inside a large circle of diameter 20 cm20 cm8 cmNOT TOSCALE
Show solution
  1. Radius = diameter ÷ 2. Large circle: 20 ÷ 2 = 10 cm. Small circle: 8 ÷ 2 = 4 cm.
  2. Area of large circle = πr² = π × 10² = 100π cm².
  3. Area of small circle = π × 4² = 16π cm².
  4. Shaded area = 100π − 16π = 84π cm²
Answer 84π cm²
43 marksCore

Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q19

An ant walks at a constant speed of 4.5 cm/s. The ant takes 13 seconds to walk once around the circumference of a circular plate. Calculate the diameter of the plate.

Show solution
  1. Distance walked once round = speed × time = 4.5 × 13 = 58.5 cm. This is the circumference.
  2. Circumference = πd, so πd = 58.5.
  3. d = 58.5 ÷ π = 18.62… = 18.6 cm
Answer 18.6 cm
53 marksCore

Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q27

The diagram shows a semicircle. The radius of the semicircle is 8 cm. Find the perimeter of the semicircle. Give your answer in terms of π in its simplest form.

0580/11 October/November 2025 Q27 – semicircle with radius 8 cm8 cmNOT TOSCALE
Show solution
  1. The diameter is 2 × 8 = 16 cm.
  2. The full circumference would be πd = 16π cm, so the curved edge is half of this: 8π cm.
  3. The perimeter also includes the straight edge, which is the diameter, 16 cm.
  4. Perimeter = 8π + 16 cm
Answer 8π + 16 cm
62 marksCore

Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q18

The diameter of a circle is 16 cm. Find the area of the circle. Leave your answer in terms of π.

Show solution
  1. Radius = diameter ÷ 2 = 16 ÷ 2 = 8 cm.
  2. Area = πr² = π × 8² = 64π cm²
Answer 64π cm²
73 marksCore

May/June 2025 · Paper 1 · 0580/12/M/J/25 · Q25

The diagram shows a shape made from two different semicircles, with the same centre. The radius of the large semicircle is 7 cm. The radius of the small semicircle is 4 cm. Work out the perimeter of the shape. Give your answer in terms of π.

0580/12 May/June 2025 Q25 – shape made from semicircles of radius 7 cm and 4 cm with the same centre4 cm7 cmNOT TOSCALE
Show solution
  1. The perimeter is all the way round the outside: two curved arcs and two short straight edges.
  2. Large arc = half the circumference = 1⁄2 × 2π × 7 = 7π cm.
  3. Small arc = 1⁄2 × 2π × 4 = 4π cm.
  4. Each straight edge = 7 − 4 = 3 cm, and there are two of them: 6 cm.
  5. Perimeter = 7π + 4π + 6 = 11π + 6 cm
Answer 11π + 6 cm
82 marksCore

May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q19

The diagram shows a sector of a circle with radius 3 cm and sector angle 60°. Calculate the area of the sector. Give your answer in terms of π in its simplest form.

0580/13 May/June 2025 Q19 – sector with radius 3 cm and angle 60°60°3 cmNOT TOSCALE
Show solution
  1. A 60° sector is 60⁄360 = 1⁄6 of a full circle.
  2. Area of the full circle = πr² = π × 3² = 9π cm².
  3. Sector area = 1⁄6 × 9π = 9⁄6π = 3⁄2π cm²
Answer 3⁄2π cm²

0580 Circles, Arcs and Sectors past paper questions: Extended (Papers 2 and 4)

94 marksExtended

May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q15

Shape A is one quarter of a circle with radius t cm. Shape B is a sector of a circle with radius 2t cm and sector angle 135°. The area of shape B is k times the area of shape A. Work out the value of k.

0580/21 May/June 2026 Q15 – quarter circle radius t and sector radius 2t with angle 135°r cmshape A135°2 cmshape BNOT TOSCALE
Show solution
  1. Area of shape A (a quarter circle) = 1⁄4 × πt² = 1⁄4πt².
  2. Area of shape B = 135⁄360 × π × (2t)².
  3. 135⁄360 = 3⁄8 and (2t)² = 4t², so area B = 3⁄8 × 4πt² = 3⁄2πt².
  4. k = area B ÷ area A = 3⁄2 ÷ 1⁄4 = 3⁄2 × 4 = 6
Answer 6
102 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q13

A sector of a circle has sector angle 90° and radius 12 cm. Calculate the arc length of the sector. Give your answer in terms of π.

Show solution
  1. Arc length = angle⁄360 × 2πr.
  2. A 90° sector is 90⁄360 = 1⁄4 of the circle.
  3. Arc length = 1⁄4 × 2 × π × 12 = 1⁄4 × 24π = 6π cm
Answer 6π cm
113 marksExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q23

A minor sector of a circle of radius 6 cm has sector angle 50°. Find the area of the major sector. Give your answer in terms of π in its simplest form.

0580/23 May/June 2026 Q23 – circle radius 6 cm with a 50° minor sector50°6 cmNOT TOSCALE
Show solution
  1. The major sector is the larger part of the circle: angle = 360 − 50 = 310°.
  2. Area of the whole circle = πr² = π × 6² = 36π.
  3. Major sector area = 310⁄360 × 36π = 31⁄36 × 36π = 31π cm²
Answer 31π cm²
123 marksExtended

May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q6

A semicircle has a diameter of 15.8 cm. Calculate the perimeter of the semicircle.

Show solution
  1. The perimeter of a semicircle is the curved edge plus the straight edge (the diameter).
  2. Curved edge = half the circumference = 1⁄2 × π × 15.8 = 24.818… cm.
  3. Perimeter = 24.818… + 15.8 = 40.618…
  4. 40.6 cm
  5. Tip: Do not forget the straight edge. Leaving it out is the most common mistake.
Answer 40.6 cm
133 marksExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q16(c)

A circle has centre O and radius 6 cm. Obtuse angle EOC = 150°. Work out the area of the major sector EOC. Give your answer in terms of π in its simplest form.

Show solution
  1. The major sector is the bigger part of the circle, so its angle is 360 − 150 = 210°.
  2. Area of the whole circle = πr² = π × 6² = 36π.
  3. Sector area = 210⁄360 × 36π. 210⁄360 simplifies to 7⁄12.
  4. 7⁄12 × 36π = 7 × 3π = 21π cm²
Answer 21π cm²
144 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q12

The diagram shows a sector of a circle with centre O and radius 9 cm. The perimeter of the sector is (18 + 2π) cm. Find the area of the sector. Give your answer in terms of π.

0580/21 October/November 2025 Q12 – sector of a circle with centre O and radius 9 cm9 cm9 cmONOT TOSCALE
Show solution
  1. The perimeter of a sector = two radii + the arc. The two radii are 9 + 9 = 18 cm.
  2. So the arc length = (18 + 2π) − 18 = 2π cm.
  3. The full circumference = 2πr = 2 × π × 9 = 18π cm.
  4. The arc is 2π⁄18π = 1⁄9 of the circumference, so the sector is 1⁄9 of the circle.
  5. Area of the whole circle = πr² = π × 9² = 81π.
  6. Sector area = 1⁄9 × 81π = 9π cm²
Answer 9π cm²
155 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q19

A circle has centre O and radius 12 cm. Angle AOB = 60°.

0580/23 October/November 2025 Q19 – circle centre O with sector AOB of angle 60°60°12 cmOABNOT TOSCALE

(a) Work out the area of the minor sector AOB. Give your answer in terms of π in its simplest form.

(b) Calculate the length of the major arc AB. Give your answer in terms of π in its simplest form.

Show solution
Part (a)
  1. A 60° sector is 60⁄360 = 1⁄6 of the circle.
  2. Area of the whole circle = πr² = π × 12² = 144π.
  3. Minor sector area = 1⁄6 × 144π = 24π cm²
Answer 24π cm²
Part (b)
  1. The major arc goes the long way round, so its angle is 360 − 60 = 300°.
  2. Full circumference = 2πr = 2 × π × 12 = 24π.
  3. Major arc = 300⁄360 × 24π = 5⁄6 × 24π = 20π cm
Answer 20π cm
162 marksExtended

May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q10

The diagram shows a sector of a circle, centre O, with radius 18 cm and angle 45°. The length of the arc is nπ cm. Find the value of n.

0580/22 May/June 2025 Q10 – sector with radius 18 cm and angle 45°45°18 cmONOT TOSCALE
Show solution
  1. Arc length = angle⁄360 × 2πr.
  2. Arc length = 45⁄360 × 2 × π × 18. 45⁄360 = 1⁄8 and 2 × 18 = 36.
  3. 1⁄8 × 36π = 4.5π.
  4. n = 9⁄2 (4.5)
Answer 9⁄2

Question papers used on this page

All circles, arcs and sectors questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/12/M/J/260580_s26_qp_12May/June 2026 · Paper 1 (Core)Q23
0580/13/M/J/260580_s26_qp_13May/June 2026 · Paper 1 (Core)Q16
0580/21/M/J/260580_s26_qp_21May/June 2026 · Paper 2 (Extended)Q15
0580/22/M/J/260580_s26_qp_22May/June 2026 · Paper 2 (Extended)Q13
0580/23/M/J/260580_s26_qp_23May/June 2026 · Paper 2 (Extended)Q23
0580/42/M/J/260580_s26_qp_42May/June 2026 · Paper 4 (Extended)Q6
0580/12/F/M/260580_m26_qp_12Feb/Mar 2026 · Paper 1 (Core)Q17
0580/22/F/M/260580_m26_qp_22Feb/Mar 2026 · Paper 2 (Extended)Q16(c)
0580/32/F/M/260580_m26_qp_32Feb/Mar 2026 · Paper 3 (Core)Q19
0580/11/O/N/250580_w25_qp_11Oct/Nov 2025 · Paper 1 (Core)Q27
0580/13/O/N/250580_w25_qp_13Oct/Nov 2025 · Paper 1 (Core)Q18
0580/21/O/N/250580_w25_qp_21Oct/Nov 2025 · Paper 2 (Extended)Q12
0580/23/O/N/250580_w25_qp_23Oct/Nov 2025 · Paper 2 (Extended)Q19
0580/12/M/J/250580_s25_qp_12May/June 2025 · Paper 1 (Core)Q25
0580/13/M/J/250580_s25_qp_13May/June 2025 · Paper 1 (Core)Q19
0580/22/M/J/250580_s25_qp_22May/June 2025 · Paper 2 (Extended)Q10

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