Area of a segment (central angles 60°, 90° and 120°)
A segment is the region between a chord and its arc. Segment = sector − triangle. For 60°, 90° and 120°, the triangle's area has a simple formula.
Minor and major segment
A chord AB divides a circle into two segments. The smaller one, between the chord and the minor arc, is the minor segment. The larger one is the major segment. To find the minor segment, take the sector OAB and remove the triangle OAB. So minor segment = sector − triangle, and major segment = circle − minor segment. In this chapter the central angle is 60°, 90° or 120°, so the triangle is easy to find.
Triangle OAB for the three angles
| Central angle | Shape of ΔOAB | Area of ΔOAB | Chord AB |
|---|---|---|---|
| 60° | Equilateral (OA = OB, so all angles are 60°) | (√3/4) r² | r |
| 90° | Right isosceles (right angle at O) | (1/2) × r × r = r²/2 | √2 r |
| 120° | Isosceles; OM ⟂ AB makes two 30°-60°-90° triangles with OM = r/2 | (1/2) × √3 r × r/2 = (√3/4) r² | √3 r |
How to draw the segment figure
- Draw a circle with centre O.
- Draw a chord AB and join the radii OA and OB.
- Mark ∠AOB with its value.
- Shade the region between the chord AB and the minor arc: this is the minor segment.
- For 120°, also draw OM ⟂ AB. Mark M as the midpoint of AB and ∠AOM = 60°.
Segment with a 60° angle
A chord of a circle of radius 42 cm subtends 60° at the centre. Find the area of the minor segment. (Use π = 22/7 and √3 = 1.73)
- Sector = (60/360) × (22/7) × 42 × 42 = (1/6) × 5544 = 924 cm².
- ΔOAB is equilateral, so its area = (√3/4) × 42² = (√3/4) × 1764 = 441√3.
- = 441 × 1.73 = 762.93 cm².
- Minor segment = 924 − 762.93 = 161.07 cm².
161.07 cm²
Segment with a 120° angle
A chord of a circle of radius 35 cm subtends 120° at the centre. Find the areas of the minor and the major segments. (Use π = 22/7 and √3 = 1.73)
- Sector = (120/360) × (22/7) × 35 × 35 = (1/3) × 3850 = 1283.33 cm² (to 2 decimal places).
- Draw OM ⟂ AB. ∠AOM = 60°, so OM = 35 × cos 60° = 17.5 cm and AM = 35 × sin 60° = 17.5√3 cm.
- AB = 2 × AM = 35√3 cm. Area of ΔOAB = (1/2) × 35√3 × 17.5 = 306.25√3.
- = 306.25 × 1.73 = 529.81 cm² (to 2 decimal places).
- Minor segment = 1283.33 − 529.81 = 753.52 cm².
- Major segment = 3850 − 753.52 = 3096.48 cm².
Minor segment = 753.52 cm², major segment = 3096.48 cm²
When you cut straight across a round papad or dosa, the small piece is a segment. Carpenters work out such areas to order glass for arched windows and temple doorways.
1 marks: “In a circle of radius 6 cm, ∠AOB = 90°. Find the area of ΔOAB.” Area = (1/2) × 6 × 6 = 18 cm².
2 marks: “A chord of a circle of radius 14 cm makes a right angle at the centre. Find the area of the minor segment. (Use π = 22/7)” Quadrant = (1/4) × 616 = 154 cm². Triangle = (1/2) × 14 × 14 = 98 cm². Segment = 154 − 98 = 56 cm².
3 marks: “A chord of a circle of radius 28 cm subtends 60° at the centre. Find the areas of the minor and major segments. (Use π = 22/7 and √3 = 1.73)” Sector = (1/6) × 2464 = 410.67 cm². Triangle = (√3/4) × 784 = 196 × 1.73 = 339.08 cm². Minor = 71.59 cm². Major = 2464 − 71.59 = 2392.41 cm².
Segment = Slice − Triangle. For 60° and 120° the triangle is (√3/4)r²; for 90° it is r²/2.