A circular segment is the region between a chord and the arc it cuts off. To find its area, calculate the area of the sector containing the arc, then subtract the area of the triangle formed by the two radii and the chord. Both steps use the angle at the centre.
What is the difference between a sector and a segment?
These two words cause a lot of confusion, so fix them now.
- A sector is like a slice of pie — the region bounded by two radii and an arc. It looks like a wedge with a curved outer edge.
- A segment is the region between a chord (a straight line joining two points on the circle) and the arc. Remove the triangular part from a sector and you have a segment.
The minor segment is the smaller region (the region cut off by the chord on the shorter arc side). The major segment is the larger region. Unless a question says otherwise, "segment" means the minor segment.
What formula do you use for segment area?
Segment area = Sector area − Triangle area
Written out:
$$\text{Segment area} = \frac{\theta}{360} \times \pi r^2 ;-; \frac{1}{2}r^2 \sin\theta$$
where θ is the angle at the centre in degrees and r is the radius.
The triangle formed by the two radii and the chord has two sides of length r (the radii) and an included angle of θ, so you use the formula Area = ½ab sin C = ½ × r × r × sin θ = ½r² sin θ.
How do you calculate segment area step by step?
Worked example: A circle has radius 9 cm. A chord subtends an angle of 80° at the centre. Find the area of the minor segment. Give your answer to 3 significant figures.
Step 1: Find the sector area.
$$\text{Sector area} = \frac{80}{360} \times \pi \times 9^2 = \frac{2}{9} \times \pi \times 81 = 18\pi \approx 56.549 \text{ cm}^2$$
Step 2: Find the triangle area.
$$\text{Triangle area} = \frac{1}{2} \times 9^2 \times \sin 80° = \frac{1}{2} \times 81 \times 0.9848 = 39.87 \text{ cm}^2$$
Step 3: Subtract.
$$\text{Segment area} = 56.549 - 39.87 = 16.679 \approx \mathbf{16.7 \text{ cm}^2}$$
What if the angle is given in radians?
At GCSE, angles are almost always given in degrees. If you ever encounter radians (more common at A-level), the combined formula becomes:
$$\text{Segment area} = \frac{1}{2}r^2(\theta - \sin\theta) \quad (\theta \text{ in radians})$$
For the GCSE exam, use the degree versions of both formulas separately.
How do you find the major segment area?
The major segment and the minor segment together make the full circle:
$$\text{Major segment area} = \pi r^2 - \text{Minor segment area}$$
Alternatively, use the major sector angle (360° − θ) in the segment formula. Both approaches give the same answer.
Example: Using the same circle (r = 9 cm, minor segment = 16.7 cm²):
Major segment area = π × 9² − 16.7 = 254.469 − 16.7 = 237.8 cm² (3 s.f.)
Check: minor + major = 16.7 + 237.8 = 254.5 ≈ π × 81 ✓
What mistakes do students commonly make?
| Mistake | What goes wrong | How to avoid it |
|---|---|---|
| Using the triangle height formula | Area = ½ × base × height requires the perpendicular height, not r | Always use ½r² sin θ when you know two sides and the included angle |
| Forgetting to subtract | Giving sector area as the final answer | Re-read the question: "segment" always means sector minus triangle |
| Using the wrong angle | Using 360° − θ for the minor segment | The minor segment corresponds to the minor (smaller) sector angle |
| Rounding too early | Rounding sin θ to 2 d.p. early inflates the error | Keep full calculator precision until the final step |
Frequently asked questions
Does the segment formula appear on the GCSE formula sheet?
No. The sector area formula (θ/360 × πr²) is given on the sheet. The triangle area formula (½ab sin C) is also given. But you must know to combine them: segment = sector − triangle. Practise this combination until it is automatic.
What is the arc length of a segment?
The arc length of a segment is the same as the arc length of the sector with the same angle: arc length = (θ/360) × 2πr. The chord length (straight line across) is a separate calculation: chord = 2r sin(θ/2).
Can a segment area ever equal the sector area?
Only if the triangle area is zero, which would require sin θ = 0 — meaning θ = 0° or θ = 180°. At θ = 0 there is no sector and no segment. At θ = 180° the two radii are collinear and the "triangle" is flat with zero area, so the segment and sector coincide: the segment is a semicircle.
How is this different from the area of a sector?
A sector is a wedge shape — two straight radii plus an arc. A segment is the region between the arc and the chord (the straight line joining the two points). The segment is always smaller than the sector (unless θ = 180°, when they are equal).
Tackle circular segment problems with Professor Pi's step-by-step hints at aitutors.me.