A segment is the region of a circle cut off by a chord. The chord splits the circle into a minor segment and a major segment, and unless the question says otherwise you want the minor one. To find its area, work out the sector area (the "pizza slice" bounded by the two radii and the arc), then subtract the triangle area formed by the two radii and the chord. In one formula: segment area = ½r²θ − ½r²sin θ, which factorises to ½r²(θ − sin θ) when θ is in radians.
The segment is often confused with the sector. A sector is bounded by two straight radii and the arc, it's the whole "slice". A segment is bounded by the straight chord and the arc, it's what's left of the slice once you cut off the triangular part with a straight line joining the two ends of the arc. If you calculate the sector area and hand that in when the question asks for the segment, you lose the marks even though every number in your working is correct.
This topic sits in two places on the syllabus, Chapter 9 (Circle Properties) and Chapter 11 (Mensuration), because it uses circle theorem setup and mensuration formulas together. The method is identical in both, only the surrounding question changes.
Step 1
Find sector area: ½r²θ (radians) or (θ/360) × πr² (degrees)
Step 2
Find triangle area: ½r²sin θ
Step 3
Segment area = sector area − triangle area
Shortcut
Segment area = ½r²(θ − sin θ), θ in radians
A circle has radius r = 8 cm. The chord AB subtends an angle θ = 1.2 rad at the centre. Find the area of the minor segment cut off by AB.
Solution
½r²θ = ½ × 8² × 1.2 = ½ × 64 × 1.2 = 38.4 cm².½r²sin θ = ½ × 64 × sin(1.2 rad) = 32 × 0.932 = 29.83 cm² (4 s.f.).38.4 − 29.83 = 8.57 cm² (3 s.f.).Calculator check: on the Casio fx-97SG (both the X and the CW ClassWiz), switch the Angle Unit to Radian before finding sin(1.2), then switch straight back to Degree once the question is done. The top-left of the display shows R in radian mode and D in degree mode. Full steps are in the degree vs radian mode check. Working entirely in radians is what lets you use ½r²θ directly for the sector, no degree conversion needed.
A circle has radius r = 6 cm. The chord PQ subtends an angle of θ = 100° at the centre. Find the area of the minor segment.
Solution
(θ/360) × πr² = (100/360) × π × 6² = 0.2778 × π × 36 = 31.42 cm² (4 s.f.).½r²sin θ = ½ × 36 × sin(100°) = 18 × 0.9848 = 17.73 cm² (4 s.f.).31.42 − 17.73 = 13.7 cm² (3 s.f.).If the question asks for the major segment (the larger piece), find the minor segment first using the method above, then subtract it from the full circle area πr². Do not try to use a reflex angle directly in the sector formula unless your calculator and working are set up carefully for it, subtracting from the whole circle is safer and less error-prone.
Example: major segment, using worked example 2's circle
πr² = π × 6² = 113.1 cm² (4 s.f.).113.1 − 13.7 = 99.4 cm² (3 s.f.).The step students get wrong
Subtracting the sector area from the triangle area instead of the other way round, giving a negative answer that gets written down anyway. The sector is always bigger than the triangle it contains (for θ between 0° and 180°), so segment area must be positive. If you get a negative number, you subtracted in the wrong order or used the wrong angle. Also common: using degrees in the ½r²θ formula, which only works when θ is in radians. If your angle is in degrees, use (θ/360) × πr² for the sector instead.
What's the difference between a sector and a segment?
A sector is bounded by two radii and an arc, like a slice of pizza with a straight point at the centre. A segment is bounded by a chord and an arc, it's the region cut off when you draw a straight line between the two ends of the arc. Every segment sits inside a sector that shares the same arc.
Can I use ½r²(θ − sin θ) with θ in degrees?
No. The formula ½r²(θ − sin θ) only gives the correct area when θ is in radians. If your angle is in degrees, either convert it to radians first, or find sector area and triangle area separately using the degree-based sector formula (θ/360) × πr² and ½r²sin θ, then subtract.
How is this related to arc length and sector area?
Segment area builds directly on arc length and sector area: you need the sector area formula as step 1 of this method. It also reuses ½r²sin θ from area of a triangle with ½ab sin C, since the triangle inside the sector has two sides of length r with included angle θ.
Why does this topic appear in two chapters?
Area of a segment appears under both Circle Properties and Mensuration on the syllabus because it uses circle theorem reasoning (angle at the centre, isosceles radii triangle) together with mensuration formulas (sector and triangle area). The method taught here is exactly the same regardless of which chapter the question is filed under.
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