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Learn9709Paper 1 · Pure Mathematics 1Circular measureSegments, chords and composite shapes
Circular measure · Subtopic 3 of 3

Segments, Chords and Composite Shapes

Cutting a sector with a chord: the segment formula, chord lengths, and the exam-style shaded regions built from sectors, triangles and tangents.

Cambridge 9709 Paper 1Edexcel YMA01 WMA11Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 45 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Find the area of a segment as sector minus triangle.
  • Find chord lengths with 2rsin⁡θ22r\sin\tfrac{\theta}{2} or the cosine rule, and angles from chords.
  • Find the perimeter of a segment (arc plus chord).
  • Break a shaded region into sectors, triangles and segments, adding and subtracting areas.
Pause and think

Cut a slice of round cake with one straight cut that does not go through the centre. The smaller piece is a segment. How could you find its area using shapes you already know?

Section 2 of 14

Core idea

Start hereThe missing bite

Cut straight across a circular cake, not through the centre: the smaller piece is a segment. It's not a sector (no point at the centre) and not a triangle. But a sector is exactly a triangle plus a segment. So segment = sector − triangle, and both of those we can calculate.

The big idea

Segment area =12r2θ−12r2sin⁡θ=12r2(θ−sin⁡θ)=\tfrac12r^2\theta-\tfrac12r^2\sin\theta=\tfrac12r^2(\theta-\sin\theta) (θ\theta in radians). The chord length is 2rsin⁡θ22r\sin\frac\theta2.

1. Segment area

Segment = sector − triangle
triangleθsegmentr
The shaded segment is what is left of the sector after removing the triangle. The triangle's area is 12absin⁡C=12r2sin⁡θ\tfrac12ab\sin C=\tfrac12r^2\sin\theta.
Segment area formula
  1. sector=12r2θ\text{sector}=\tfrac12r^2\theta
    From the previous lesson.
  2. triangle=12absin⁡C=12r⋅rsin⁡θ=12r2sin⁡θ\text{triangle}=\tfrac12ab\sin C=\tfrac12r\cdot r\sin\theta=\tfrac12r^2\sin\theta
    Two sides rr with angle θ\theta between them.
  3. segment=12r2θ−12r2sin⁡θ=12r2(θ−sin⁡θ)\text{segment}=\tfrac12r^2\theta-\tfrac12r^2\sin\theta=\tfrac12r^2(\theta-\sin\theta)
    Subtract; factor.
Chord length
  1. Drop a perpendicular from the centre to the chord.\text{Drop a perpendicular from the centre to the chord.}
    It bisects the chord and the angle (isosceles triangle).
  2. 12chord=rsin⁡θ2\tfrac12\text{chord}=r\sin\tfrac\theta2
    Right-angled triangle: hypotenuse rr, angle θ2\frac\theta2, opposite side = half chord.
  3. chord=2rsin⁡θ2\text{chord}=2r\sin\tfrac\theta2
    Double.
Worked exampler=8r=8 cm, θ=1.2\theta=1.2 rad

Find the area and perimeter of the minor segment.

Goal
Area and perimeter.
Strategy
Area by the formula; perimeter = arc + chord.
  1. 12(64)(1.2−sin⁡1.2)=32(1.2−0.9320)=8.58 cm2\tfrac12(64)(1.2-\sin1.2)=32(1.2-0.9320)=8.58\text{ cm}^2
    Radians mode for sin⁡1.2\sin1.2.
  2. arc=8×1.2=9.6\text{arc}=8\times1.2=9.6
    rθr\theta.
  3. chord=16sin⁡0.6=16(0.5646)=9.03\text{chord}=16\sin0.6=16(0.5646)=9.03
    2rsin⁡θ22r\sin\frac\theta2.
  4. P=9.6+9.03=18.6 cmP=9.6+9.03=18.6\text{ cm}
    Add.
AnswerArea 8.588.58 cm², perimeter 18.618.6 cm.
✓ CheckThe chord (9.039.03) is a little shorter than the arc (9.69.6), as a straight line must be ✓.
Worked exampleChord given, find the segment

A chord of length 66 cm is in a circle of radius 55 cm. Find the area of the minor segment.

Goal
Area.
Strategy
Find θ\theta from the chord, then use the formula.
  1. 2(5)sin⁡θ2=6⇒sin⁡θ2=0.62(5)\sin\tfrac\theta2=6\Rightarrow\sin\tfrac\theta2=0.6
    Chord formula.
  2. θ2=0.6435⇒θ=1.287\tfrac\theta2=0.6435\Rightarrow\theta=1.287
    Inverse sine (radians).
  3. 12(25)(1.287−sin⁡1.287)=12.5(1.287−0.960)=4.09\tfrac12(25)(1.287-\sin1.287)=12.5(1.287-0.960)=4.09
    sin⁡1.287=0.96\sin1.287=0.96.
Answer4.094.09 cm².
✓ CheckSmall compared with the circle (78.578.5 cm²), as expected for a short chord ✓.
Edge caseMajor segment The larger piece: circle minus minor segment, or use θ′=2π−θ\theta'=2\pi-\theta in the formula (then sin⁡θ′\sin\theta' is negative, which correctly adds the triangle).
Pause and think

Why must θ\theta be in radians in 12r2(θ−sin⁡θ)\frac12r^2(\theta-\sin\theta)?

Summary

  • Segment = sector − triangle.
  • Chord =2rsin⁡θ2=2r\sin\frac\theta2.
  • Split complicated shapes into these pieces.
Key words
Segment
Region between a chord and an arc.
Chord
A straight line joining two points on a circle.
HistoryThe first curved area found exactly. Around 440 BC Hippocrates of Chios (not the doctor) showed that a crescent-shaped region called a lune, bounded by two circular arcs, has exactly the same area as a certain triangle. It was the first curved shape whose area was found exactly, and it was done by adding and subtracting segments.
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