TMUA: Radians, arcs and sectors
Radians, arcs, sectors and exact values.
- Section 1 Part 1 · MM4 Trigonometry · spec MM4.2-MM4.3
- Papers 1 and 2
- 7 practice questions
- No calculator
What the specification covers
- Arc length rθ, sector area ½r²θ, segments
- Exact values at 0, 30, 45, 60, 90 degrees
Key ideas
- \(\pi\) radians \(=180^\circ\). Arc length \(=r\theta\), sector area \(=\tfrac12r^2\theta\) (\(\theta\) in radians).
- Segment area \(=\tfrac12r^2(\theta-\sin\theta)\).
- Exact values: \(\sin30^\circ=\tfrac12\), \(\sin45^\circ=\tfrac{\sqrt2}{2}\), \(\sin60^\circ=\tfrac{\sqrt3}{2}\), \(\tan45^\circ=1\), \(\tan60^\circ=\sqrt3\).
- A sector angle must satisfy \(0<\theta\le2\pi\): reject solutions outside that range.
Common mistakes
- Perimeter of a sector includes the two radii: \(2r+r\theta\).
- Using degrees in \(r\theta\) gives nonsense.
Exam tip
Two equations (perimeter and area) usually give a quadratic in \(r\); check both roots.
Worked example
Worked example
A sector of a circle has perimeter 20 cm and area 16 cm². What is the angle of the sector, in radians?
- \(\frac{1}{4}\)
- \(4\)
- \(\frac{1}{2}\)
- \(8\)
- \(2\)
Answer: C: \(\frac{1}{2}\)
\(2r+r\theta=20\) and \(\tfrac12r^2\theta=16\). So \(r\theta=20-2r\) and \(\tfrac12r(20-2r)=16\): \(r^2-10r+16=0\), \(r=2\) or \(r=8\). \(r=2\) gives \(\theta=8>2\pi\), impossible for a sector. \(r=8\) gives \(\theta=\tfrac{4}{8}=\tfrac12\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
A sector of a circle has perimeter \(18\) cm and area \(14\) cm². What is the angle of the sector, in radians?
- \(\frac{4}{7}\)
- \(\frac{8}{7}\)
- \(\frac{2}{7}\)
- \(7\)
- \(\frac{18}{7}\)
Show the answer and solution
Answer: A: \(\frac{4}{7}\)
\(2r+r\theta=18\) and \(\tfrac12r^2\theta=14\). So \(r\theta=18-2r\) and \(\tfrac12r(18-2r)=14\): \(r^2-9r+14=0\), \(r=2\) or \(r=7\). \(r=2\) gives \(\theta=7>2\pi\), impossible for a sector. \(r=7\) gives \(\theta=\frac{4}{7}\).
Question 2
A sector of a circle has perimeter \(10\) cm and area \(4\) cm². What is the angle of the sector, in radians?
- \(1\)
- \(8\)
- \(\frac{1}{2}\)
- \(\frac{1}{4}\)
- \(\frac{5}{2}\)
Show the answer and solution
Answer: C: \(\frac{1}{2}\)
\(2r+r\theta=10\) and \(\tfrac12r^2\theta=4\). So \(r\theta=10-2r\) and \(\tfrac12r(10-2r)=4\): \(r^2-5r+4=0\), \(r=1\) or \(r=4\). \(r=1\) gives \(\theta=8>2\pi\), impossible for a sector. \(r=4\) gives \(\theta=\frac{1}{2}\).
Question 3
A sector of a circle has perimeter \(20\) cm and area \(16\) cm². What is the angle of the sector, in radians?
- \(\frac{5}{2}\)
- \(\frac{1}{2}\)
- \(8\)
- \(1\)
- \(\frac{1}{4}\)
Show the answer and solution
Answer: B: \(\frac{1}{2}\)
\(2r+r\theta=20\) and \(\tfrac12r^2\theta=16\). So \(r\theta=20-2r\) and \(\tfrac12r(20-2r)=16\): \(r^2-10r+16=0\), \(r=2\) or \(r=8\). \(r=2\) gives \(\theta=8>2\pi\), impossible for a sector. \(r=8\) gives \(\theta=\frac{1}{2}\).
Question 4
A sector of a circle has perimeter \(12\) cm and area \(5\) cm². What is the angle of the sector, in radians?
- \(\frac{1}{5}\)
- \(\frac{2}{5}\)
- \(10\)
- \(\frac{4}{5}\)
- \(\frac{12}{5}\)
Show the answer and solution
Answer: B: \(\frac{2}{5}\)
\(2r+r\theta=12\) and \(\tfrac12r^2\theta=5\). So \(r\theta=12-2r\) and \(\tfrac12r(12-2r)=5\): \(r^2-6r+5=0\), \(r=1\) or \(r=5\). \(r=1\) gives \(\theta=10>2\pi\), impossible for a sector. \(r=5\) gives \(\theta=\frac{2}{5}\).
Question 5
A sector of a circle has perimeter \(26\) cm and area \(30\) cm². What is the angle of the sector, in radians?
- \(\frac{6}{5}\)
- \(\frac{20}{3}\)
- \(\frac{3}{10}\)
- \(\frac{3}{5}\)
- \(\frac{13}{5}\)
Show the answer and solution
Answer: D: \(\frac{3}{5}\)
\(2r+r\theta=26\) and \(\tfrac12r^2\theta=30\). So \(r\theta=26-2r\) and \(\tfrac12r(26-2r)=30\): \(r^2-13r+30=0\), \(r=3\) or \(r=10\). \(r=3\) gives \(\theta=\frac{20}{3}>2\pi\), impossible for a sector. \(r=10\) gives \(\theta=\frac{3}{5}\).
Question 6
A sector of a circle has perimeter \(22\) cm and area \(18\) cm². What is the angle of the sector, in radians?
- \(\frac{22}{9}\)
- \(\frac{4}{9}\)
- \(9\)
- \(\frac{2}{9}\)
- \(\frac{8}{9}\)
Show the answer and solution
Answer: B: \(\frac{4}{9}\)
\(2r+r\theta=22\) and \(\tfrac12r^2\theta=18\). So \(r\theta=22-2r\) and \(\tfrac12r(22-2r)=18\): \(r^2-11r+18=0\), \(r=2\) or \(r=9\). \(r=2\) gives \(\theta=9>2\pi\), impossible for a sector. \(r=9\) gives \(\theta=\frac{4}{9}\).
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