A sector and the triangle inside it share two radii and the angle at the centre. Their difference is the segment, the region between the chord and the arc. This lesson in radians, arcs and sectors joins the sector formula from the previous lesson with the triangle area you know from earlier years.
How do the two areas compare?
For a sector OAB with radius r and angle θ in radians:
| Shape | Area |
|---|---|
| Sector OAB | ½r²θ |
| Triangle OAB | ½r² sin θ |
| Segment (between chord and arc) | ½r²θ − ½r² sin θ = ½r²(θ − sin θ) |
The sector is always the bigger of the two when θ is between 0 and π. The difference is the segment, which is why it is worth drawing both shapes on the diagram.
How do you solve it, step by step?
- Sketch the sector and draw the chord, labelling O, A, B, r and θ.
- Find the sector area ½r²θ with θ in radians.
- Find the triangle area ½r² sin θ, with the calculator in radian mode.
- Subtract triangle from sector to get the segment.
- Keep full working values until the final line, then round once.
Worked example
A sector has radius 10 cm and angle 1.2 rad. Find the area of the segment, and the length of the chord AB. Give answers to 3 significant figures.
Step 1, sector: ½ × 10² × 1.2 = ½ × 100 × 1.2 = 60 cm².
Step 2, triangle: ½ × 10² × sin 1.2 = 50 × 0.93204 = 46.602 cm².
Step 3, segment: 60 − 46.602 = 13.398, so 13.4 cm².
Step 4, chord: 2r sin(θ/2) = 20 × sin 0.6 = 20 × 0.56464 = 11.293, so 11.3 cm.
Check: the chord (11.3 cm) is shorter than the arc rθ = 12 cm, as it must be, since a straight line is the shortest path between two points. ✓
The mistake to watch for
A common slip is to evaluate sin θ in degree mode, or to find the triangle with the wrong angle.
Mistaken working: triangle = 50 × sin 1.2 = 50 × 0.0209 = 1.05 cm², so segment = 58.95 cm²
The calculator was in degree mode, so it used 1.2°. The segment cannot be almost as big as the whole sector.
The correction is to set radian mode, and to sense-check: the segment should be a small slice of the sector, here 13.4 out of 60. The test sin 1 = 0.8415 in radian mode confirms the setting.
Check yourself
1. A sector has radius 8 cm and angle π/2. Find the exact area of the segment, then give it to 3 significant figures.
Show answer
Sector: ½ × 64 × π/2 = 16π. Triangle: ½ × 64 × sin(π/2) = 32 × 1 = 32. Segment: 16π − 32 = 18.265…
16π − 32 cm², which is 18.3 cm²
2. A sector has radius 5 cm and angle 2 rad. Find the area of the segment to 3 significant figures.
Show answer
Sector: ½ × 25 × 2 = 25. Triangle: ½ × 25 × sin 2 = 12.5 × 0.90930 = 11.366. Segment: 25 − 11.366 = 13.634.
13.6 cm²
3. A sector has radius 6 cm and angle π/6. Find the exact area of the segment.
Show answer
Sector: ½ × 36 × π/6 = 3π. Triangle: ½ × 36 × sin(π/6) = 18 × ½ = 9. Segment: 3π − 9 = 0.4248…
3π − 9 cm² (about 0.425 cm²). The segment is thin because the angle is small.
Where this leads next
Now combine arcs, sectors and perimeters in solving a perimeter condition involving an arc. The non-calculator working trainer is useful for exact answers such as 16π − 32.
If you can do each area alone but the shaded-region diagrams still confuse you, a teacher can go through them with you in online one-to-one Additional Mathematics tuition.