Home / A-Level Maths 9709 / Circular measure / Arc length and sector area
Cambridge International AS & A Level · Topic 1.4
A-Level Maths 9709 / Topic 1.4 / AS & A Level

Two formulae, endless shapes.

With the angle in radians, the arc of a sector is just \(r\theta\) and its area is \(\tfrac{1}{2}r^2\theta\). Add a triangle or two and you can find the perimeter and area of segments, tangents and every other shape Paper 1 builds from circles.

Topic 1.4All circular measure lessons →
The key idea

For a sector of radius \(r\) and angle \(\theta\) radians: arc length \(s = r\theta\) and area \(A = \tfrac{1}{2}r^2\theta\). A segment is a sector minus a triangle, so its area is \(\tfrac{1}{2}r^2\theta - \tfrac{1}{2}r^2\sin\theta\), and its chord is \(2r\sin\dfrac{\theta}{2}\).

O θ r r arc s segment A B arc s = rθ sector ½r²θ triangle ½r² sin θ segment ½r²(θ − sin θ)
Fig. 1: sector OAB with angle θ radians. The shaded segment is the sector minus triangle OAB; the dashed chord AB has length 2r sin(θ/2)
Section 01

Arc length, s = rθ.

A sector with angle \(\theta\) is the fraction \(\dfrac{\theta}{2\pi}\) of the whole circle, so its arc is \(\dfrac{\theta}{2\pi} \times 2\pi r = r\theta\). That tidy formula is the reason we measure in radians: in degrees you would carry \(\dfrac{\theta}{360} \times 2\pi r\) everywhere.

Worked example Arc and perimeter
A sector has radius 9 cm and angle 0.8 radians. Find the arc length and the perimeter of the sector.
  1. Arc length \(s = r\theta = 9 \times 0.8 = 7.2\) cm.
  2. The perimeter is the arc plus two radii: \(7.2 + 9 + 9\).
Answer: Arc \(= 7.2\) cm, perimeter \(= 25.2\) cm
Worked example Finding the angle
A sector of radius 8 cm has a perimeter of 30 cm. Find the angle of the sector in radians.
  1. Perimeter \(= 2r + r\theta\), so \(16 + 8\theta = 30\).
  2. \(8\theta = 14\), giving \(\theta = \dfrac{14}{8}\).
Answer: \(\theta = 1.75\) radians
Examiner trap

\(s = r\theta\) only works with \(\theta\) in radians. If the angle is given as \(60^\circ\), convert it to \(\dfrac{\pi}{3}\) first; \(9 \times 60 = 540\) cm is nonsense.

Section 02

Sector area.

The same fraction gives the area: \(\dfrac{\theta}{2\pi} \times \pi r^2 = \tfrac{1}{2}r^2\theta\). When a question gives two facts about a sector (arc, area, perimeter), write each as an equation in \(r\) and \(\theta\) and solve them together.

Worked example Exact sector area
A sector has radius 6 cm and angle \(\dfrac{5\pi}{6}\). Find the exact area and the exact arc length.
  1. Area \(= \tfrac{1}{2} \times 6^2 \times \dfrac{5\pi}{6} = 18 \times \dfrac{5\pi}{6}\).
  2. \(18 \times \dfrac{5\pi}{6} = 15\pi\).
  3. Arc \(= 6 \times \dfrac{5\pi}{6} = 5\pi\).
Answer: Area \(= 15\pi\ \text{cm}^2\), arc \(= 5\pi\) cm
Worked example Two facts, two unknowns
A sector has area \(48\ \text{cm}^2\) and arc length 12 cm. Find the radius and the angle.
  1. Write both facts: \(\tfrac{1}{2}r^2\theta = 48\) and \(r\theta = 12\).
  2. Notice \(\tfrac{1}{2}r^2\theta = \tfrac{1}{2}r \times (r\theta)\), so \(\tfrac{1}{2}r \times 12 = 48\).
  3. \(6r = 48\), so \(r = 8\).
  4. Then \(\theta = \dfrac{12}{8} = 1.5\).
Answer: \(r = 8\) cm, \(\theta = 1.5\) radians
Your turn
A sector has radius 10 cm and angle \(\dfrac{\pi}{5}\). Find the exact arc length and area.
Examiner trap

Forgetting the \(\tfrac{1}{2}\) doubles the answer. A useful check: the sector must be smaller than the whole circle, \(\pi r^2\).

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Find the arc length of a sector with radius 5 cm and angle 1.4 radians.

Quick check+10 XP

Find the exact area of a sector with radius 4 cm and angle \(\dfrac{\pi}{4}\).

Quick check+10 XP

A sector has radius 6 cm and angle 2 radians. What is its perimeter?

Quick check+10 XP

Which expression gives the area of a segment with radius \(r\) and angle \(\theta\) at the centre?

Section 03

Segments and compound shapes.

A segment is the region between a chord and its arc. Paper 1 questions build regions from sectors, triangles, tangents and segments; the method is always to name the simple pieces, find each one, then add or subtract.

Worked example Segment area and perimeter
A chord \(AB\) of a circle with centre \(O\) and radius 10 cm subtends an angle of 1.2 radians at \(O\). Find the area and the perimeter of the minor segment.
  1. Sector: \(\tfrac{1}{2} \times 100 \times 1.2 = 60\). Triangle: \(\tfrac{1}{2} \times 100 \times \sin 1.2 = 46.60\).
  2. Segment area \(= 60 - 46.60 = 13.40\).
  3. Arc \(= 10 \times 1.2 = 12\); chord \(= 2 \times 10\sin 0.6 = 11.29\).
  4. Perimeter \(= 12 + 11.29 = 23.29\).
Answer: Area \(= 13.4\ \text{cm}^2\), perimeter \(= 23.3\) cm
Worked example Exact segment
Find the exact area and exact perimeter of the segment cut off by a chord that subtends \(\dfrac{\pi}{3}\) at the centre of a circle of radius 6 cm.
  1. Sector \(= \tfrac{1}{2} \times 36 \times \dfrac{\pi}{3} = 6\pi\).
  2. Triangle \(= \tfrac{1}{2} \times 36 \times \sin\dfrac{\pi}{3} = 18 \times \dfrac{\sqrt{3}}{2} = 9\sqrt{3}\).
  3. The triangle has two sides of 6 and an angle of \(\dfrac{\pi}{3}\), so it is equilateral and the chord is 6.
  4. Arc \(= 6 \times \dfrac{\pi}{3} = 2\pi\).
Answer: Area \(= 6\pi - 9\sqrt{3}\ \text{cm}^2\), perimeter \(= 2\pi + 6\) cm
Worked example Tangent and sector
Sector \(OAB\) has radius 5 cm and angle \(\dfrac{\pi}{3}\). The tangent to the arc at \(A\) meets \(OB\) extended at \(C\). Find, in exact form, the area and the perimeter of the region bounded by the arc \(AB\), \(BC\) and \(CA\).
  1. Angle \(OAC = \dfrac{\pi}{2}\) (tangent meets radius), so \(AC = 5\tan\dfrac{\pi}{3} = 5\sqrt{3}\) and \(OC = \dfrac{5}{\cos(\pi/3)} = 10\).
  2. Triangle \(OAC = \tfrac{1}{2} \times 5 \times 5\sqrt{3} = \dfrac{25\sqrt{3}}{2}\); sector \(= \tfrac{1}{2} \times 25 \times \dfrac{\pi}{3} = \dfrac{25\pi}{6}\).
  3. Region area \(= \dfrac{25\sqrt{3}}{2} - \dfrac{25\pi}{6}\).
  4. Perimeter: arc \(\dfrac{5\pi}{3}\), plus \(BC = OC - OB = 10 - 5 = 5\), plus \(AC = 5\sqrt{3}\).
Answer: Area \(= \dfrac{25\sqrt{3}}{2} - \dfrac{25\pi}{6}\ \text{cm}^2\), perimeter \(= \dfrac{5\pi}{3} + 5 + 5\sqrt{3}\) cm
Your turn
Show that the sector of radius \(r\) whose perimeter is \(4r\) has area \(r^2\).
Examiner trap

The triangle in a segment has area \(\tfrac{1}{2}r^2\sin\theta\), using the angle at the centre, not \(\tfrac{1}{2} \times\) chord \(\times r\). And in a perimeter, use the straight chord, not another arc.

Stage 2 · Exam

Exam-style questions

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Exam style+20 XP

A sector of a circle of radius 7 cm has a perimeter of 26 cm. Find the area of the sector.

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A chord subtends an angle of \(\dfrac{2\pi}{3}\) at the centre of a circle of radius 8 cm. Find the exact area of the minor segment.

Exam style+20 XP

A chord \(AB\) subtends 1.6 radians at the centre of a circle of radius 12 cm. Find the perimeter of the minor segment.

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A sector has arc length 15 cm and area \(60\ \text{cm}^2\). Find the angle of the sector.

Skill unlocked

Arcs and sectors, done.

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Go deeper · practice
Six original Cambridge-style questions
Arcs, sectors and segments, exact and decimal, finishing with tangents and a windscreen wiper.