Connect sectors, chords, triangles and circular segments
Use the sector's isosceles triangle to find a chord, separate a segment from its sector and calculate mixed boundaries.
Section 1.4 · Circular measure
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use radians in arc and sector formulae. Written practice is self-checked.
Before you begin
- Use s = rθ and A = r²θ/2.
- Use elementary sine and cosine with the correct angle mode.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which straight boundary belongs to a circular segment?
What you will learn
- Calculate chord lengths and triangle areas using a central angle.
- Find minor segment areas, major complements and boundaries containing arcs and chords.
Halve the isosceles triangle to find the chord
Join both arc endpoints to the centre. The two radii form an isosceles triangle with included angle θ. For 0 < θ < π, bisecting the angle produces a right triangle with hypotenuse r and opposite side half the chord. Therefore chord length c = 2r sin(θ/2), and the perpendicular centre-to-chord distance is r cos(θ/2). The arc length rθ and chord length are different.
The triangle area is (1/2)r² sin θ, using the two-side included-angle formula. For r = 6 cm and θ = π/3, the chord is 12 sin(π/6) = 6 cm and the triangle area is 18 sin(π/3) = 9√3 cm². The curved arc is 2π cm, slightly longer than the chord. Evaluate the sine in radian mode when the supplied angle is in radians.
A segment is the region between a chord and its arc
For a minor central angle 0 < θ < π, the minor segment area equals sector minus triangle: Asegment = r²(θ − sin θ)/2. Its perimeter is its arc plus its chord, not the two radii. With r = 6 and θ = π/3, the area is 6π − 9√3 cm² and the perimeter is 2π + 6 cm. Check that the segment area is positive and smaller than the sector area.
For a major segment bounded by the same chord, subtract the minor segment area from πr² and use the major arc angle 2π − θ for its boundary. Do not subtract an unsigned triangle area blindly from a sector with a reflex angle; identify the region first. Compound regions can be split into sectors and triangles, then added or subtracted without double-counting shared internal edges.
Pause and explain
For r = 6 and θ = π/3, what is the minor segment area?
Worked example
For a circle of radius 6 cm, a minor segment has central angle π/3. Find its exact area and perimeter.
Show the worked solution
- The sector area is 6²(π/3)/2 = 6π cm².
- The triangle area is 6² sin(π/3)/2 = 9√3 cm², so subtract it to get 6π − 9√3 cm².
- The chord is 2(6)sin(π/6) = 6 cm and the arc is 2π cm, so their sum is 6 + 2π cm.
Answer Area 6π − 9√3 cm²; perimeter 6 + 2π cm.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A minor segment has radius 4 cm and central angle π/2. Find its area and chord length.
Give me a hint
Calculate the sector and triangle separately, and halve the angle for the chord.
Compare my reasoning
- The sector area is 4²(π/2)/2 = 4π cm².
- The triangle area is 4² sin(π/2)/2 = 8 cm², giving segment area 4π − 8 cm².
- The chord is 2(4)sin(π/4) = 4√2 cm.
Area 4π − 8 cm²; chord 4√2 cm.
Look for these in your work
- Subtract the correct triangle from the sector.
- Use the half-angle when calculating a chord.
A circle has radius 10 cm and a minor chord of length 10 cm. Find the minor central angle and segment area exactly.
Give me a hint
Use half the chord and radius in a right triangle, then determine the minor angle.
Compare my reasoning
- The half-chord relation gives sin(θ/2) = 10/(2 × 10) = 1/2.
- Because the minor angle is below π, θ/2 = π/6, hence θ = π/3.
- The segment area is 100(π/3 − sin(π/3))/2 = 50π/3 − 25√3 cm².
Angle π/3 radians; area 50π/3 − 25√3 cm².
Look for these in your work
- Use the minor-angle restriction to choose the correct inverse value.
- Retain exact sector and triangle terms.
A chord cuts a circle of radius 6 cm into minor and major segments. The minor angle is π/3. Find the major segment area and its perimeter exactly.
Give me a hint
Use the whole circle as an area complement, and the major arc with the same chord.
Compare my reasoning
- The minor segment area is 6π − 9√3, so the major area is 36π − (6π − 9√3) = 30π + 9√3 cm².
- The major arc angle is 2π − π/3 = 5π/3, so its length is 6(5π/3) = 10π cm.
- The unchanged chord has length 6 cm, giving major segment perimeter 10π + 6 cm.
Major area 30π + 9√3 cm²; perimeter 10π + 6 cm.
Look for these in your work
- Subtract the minor area from the full circle.
- Use the major arc and count the chord once.
Common mistakes
- A segment is not a sector: its straight edge is the chord.
- Use θ/2 for a half-chord triangle, but θ in the full triangle area.
- Identify minor and major regions before choosing a subtraction.
What can you explain now?
A minor segment in a circle of radius 4 has angle π/2. Find its area.
Compare with the explanation
4π − 8 square units.
The sector area is 8(π/2) = 4π and the triangle area is 8 sin(π/2) = 8.
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Make it stick
Draw a sector, its isosceles triangle and its minor segment. Label which angle each sine uses and explain why the sector and segment have different perimeters.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.