IGCSE Additional Maths · Cambridge 0606 · Edexcel 4PM1

Circular measure: notes and questions

Revision for radians, arc length, sector area, chords and segments. Both boards. You must learn the formulas: neither puts them on the formula sheet.

Cambridge 0606: section 9
Edexcel 4PM1: section 10A
14 questions

Practise circular measure →

Key points

π radians = 180°. To convert degrees to radians multiply by π/180; radians to degrees multiply by 180/π.
With θ in radians: arc length s = rθ and sector area A = ½r²θ.
Segment area = sector area − triangle area = ½r²θ − ½r² sin θ.
A chord subtending θ at the centre has length 2r sin(θ/2).
Perimeter of a sector = 2r + rθ. Many questions give the perimeter or area and ask for r or θ: form an equation.

Watch out: Set your calculator to radians before using sin θ in a segment area. Using degrees mode with a radian angle is the most common lost mark here.

Worked example · 3 marks

(a) Convert \(135^\circ\) to radians, giving your answer as a multiple of \(\pi\).

(b) Convert 2.4 radians to degrees, correct to 1 decimal place.

B1 (a) 3π/4
M1 (b) 2.4 × 180/π
A1 (b) 137.5°

π radians = 180°, so 135° = 135/180 × π = 3π/4.
2.4 rad = 2.4 × 180/π = 137.509…° = 137.5°.

Answer: (a) \(\frac{3\pi}{4}\) (b) \(137.5^\circ\)

Circular measure questions

Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).

Question 1 · 4 marks06064PM1no calculator

A sector of a circle has radius 8 cm and angle 0.9 radians.

Find (a) the arc length (b) the area of the sector.

Show the mark scheme and worked solution

M1 (a) rθ
A1 (a) 7.2 cm
M1 (b) ½r²θ
A1 (b) 28.8 cm²

Arc length s = rθ = 8 × 0.9 = 7.2 cm.
Area A = ½r²θ = ½ × 64 × 0.9 = 28.8 cm².

Answer: (a) \(7.2\) cm (b) \(28.8\) cm²

Question 2 · 3 marks06064PM1

A chord \(AB\) of a circle with centre \(O\) and radius 10 cm subtends an angle of \(\frac{\pi}{3}\) radians at \(O\).

Find the area of the minor segment cut off by \(AB\), correct to 3 significant figures.

Show the mark scheme and worked solution

M1 sector area ½ × 100 × π/3
M1 triangle area ½ × 100 × sin(π/3)
A1 9.06 cm²

Sector area = ½ × 10² × π/3 = 50π/3 = 52.359… cm².
Triangle OAB = ½ × 10 × 10 × sin(π/3) = 25√3 = 43.301… cm².
Segment = 52.359… − 43.301… = 9.058… = 9.06 cm².

Answer: \(9.06\) cm²

Question 3 · 2 marks06064PM1no calculator

A sector has radius 6 cm and angle \(\frac{2\pi}{3}\) radians. Find, in terms of \(\pi\), its arc length and its area.

Show the mark scheme and worked solution

B1 arc 4π cm
B1 area 12π cm²

Arc = rθ = 6 × 2π/3 = 4π cm.
Area = ½r²θ = ½ × 36 × 2π/3 = 12π cm².

Answer: arc \(4\pi\) cm, area \(12\pi\) cm²

Question 4 · 3 marks06064PM1

An arc of length 15 cm is part of a circle of radius 9 cm. Find the angle subtended at the centre, in radians and in degrees (to 1 decimal place).

Show the mark scheme and worked solution

M1 θ = 15/9
A1 5/3 rad
A1 95.5°

θ = s/r = 15/9 = 5/3 rad.
In degrees: 5/3 × 180/π = 95.49…° = 95.5°.

Answer: \(\frac{5}{3}\) radians, \(95.5^\circ\)

9 more circular measure questions

Each with a mark scheme and a worked solution. Included with IGCSE plans, free trials and school licences.

Next steps

Cambridge 0606: papers, syllabus map and official past papers
Edexcel 4PM1: papers, syllabus map and official past papers
Going on to A Level? Circular measure continues in A Level trigonometry (A Level Math Revision). See also from Additional Maths to A Level.