Home · IGCSE Add Maths 0606 · Circular measure · Radians, arc length and sector area · Practice
IGCSE · Practice questions · Radians, arc length and sector area

Radians make it short.

Six original Cambridge-style questions on circular measure: converting angles, the formulas \(s = r\theta\) and \(A = \frac{1}{2}r^{2}\theta\), exact answers in terms of \(\pi\), and finding a radius or angle from an area or a perimeter.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Radians first, then \(s = r\theta\) and \(A = \frac{1}{2}r^{2}\theta\).

01
Recall
[2 marks]

(a) Write \(210^{\circ}\) in radians as an exact multiple of \(\pi\).

(b) Convert \(0.9\) radians to degrees, giving your answer to 1 decimal place.

  • (a) \(210 \times \dfrac{\pi}{180} = \dfrac{7\pi}{6}\). ✓
  • (b) \(0.9 \times \dfrac{180}{\pi} = 51.56\ldots\), so \(51.6^{\circ}\). ✓
02
Calculation
[3 marks]

A sector of a circle has radius 7.5 cm and angle 1.2 radians. Find the arc length, the area and the perimeter of the sector.

  • Arc: \(s = 7.5 \times 1.2 = 9\) cm. ✓
  • Area: \(A = \dfrac{1}{2} \times 7.5^{2} \times 1.2 = 0.6 \times 56.25 = 33.75\) cm2. ✓
  • Perimeter: \(7.5 + 7.5 + 9 = 24\) cm. ✓
03
Calculation
[3 marks]

Do not use a calculator in this question.

A sector has radius 12 cm and angle \(\dfrac{5\pi}{6}\) radians. Find, in terms of \(\pi\), its arc length, its area and its perimeter.

  • Arc: \(s = 12 \times \dfrac{5\pi}{6} = 10\pi\) cm. ✓
  • Area: \(A = \dfrac{1}{2} \times 144 \times \dfrac{5\pi}{6} = 60\pi\) cm2. ✓
  • Perimeter: \((24 + 10\pi)\) cm. ✓
04
Calculation
[3 marks]

A sector of a circle has area 54 cm2 and angle 0.75 radians. Find the radius and the perimeter of the sector.

  • \(\dfrac{1}{2}r^{2}(0.75) = 54\), so \(r^{2} = \dfrac{108}{0.75} = 144\). ✓
  • \(r = 12\) cm (a length, so the positive root). ✓
  • Arc \(= 12 \times 0.75 = 9\) cm, so the perimeter is \(12 + 12 + 9 = 33\) cm. ✓
05
Analysis
[5 marks]

A sector of a circle has radius \(r\) cm and angle \(\theta\) radians. Its perimeter is 26 cm and its area is 40 cm2.

(a) Show that \(r^{2} - 13r + 40 = 0\).

(b) Hence find the two possible pairs of values of \(r\) and \(\theta\).

  • (a) Perimeter: \(2r + r\theta = 26\), so \(r\theta = 26 - 2r\). ✓
  • Area: \[ \tfrac{1}{2}r(r\theta) = \tfrac{1}{2}r(26 - 2r) = 13r - r^{2} \] This equals 40, so \(r^{2} - 13r + 40 = 0\). ✓
  • (b) \((r - 5)(r - 8) = 0\), so \(r = 5\) or \(r = 8\). ✓
  • \(r = 5\): \(\theta = \dfrac{26 - 10}{5} = 3.2\) rad. \(r = 8\): \(\theta = \dfrac{26 - 16}{8} = 1.25\) rad. ✓
  • Both angles are less than \(2\pi\), so both pairs are valid. ✓
06
Analysis
[5 marks]

A piece of wire 40 cm long is bent to form the boundary of a sector of a circle of radius \(r\) cm and angle \(\theta\) radians.

(a) Show that \(\theta = \dfrac{40 - 2r}{r}\).

(b) Show that the area of the sector is \(A = 20r - r^{2}\).

(c) Find the two values of \(r\) for which the area is 96 cm2, and the angle of the sector in each case.

  • (a) The boundary is two radii and the arc: \(2r + r\theta = 40\), so \(r\theta = 40 - 2r\) and \(\theta = \dfrac{40 - 2r}{r}\). ✓
  • (b) \[ A = \tfrac{1}{2}r^{2} \times \frac{40 - 2r}{r} = \tfrac{1}{2}r(40 - 2r) \] so \(A = 20r - r^{2}\). ✓
  • (c) \(20r - r^{2} = 96\), so \(r^{2} - 20r + 96 = 0\) and \((r - 8)(r - 12) = 0\). ✓
  • \(r = 8\): \(\theta = \dfrac{24}{8} = 3\) rad. \(r = 12\): \(\theta = \dfrac{16}{12} = \dfrac{4}{3}\) rad. ✓
  • Both angles lie between 0 and \(2\pi\), so both sectors can be made. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.