A chord of a circle of radius 14 cm subtends 120° at the centre. Find the areas of the minor and major segments. (Use π = 22/7 and √3 = 1.73)

1. 1 · Write what is given and what is to be found Given: r = 14 cm, ∠AOB = 120°. To find: the areas of the minor and major segments.
2. 2 · Draw the figure Circle with centre O; chord AB; radii OA and OB with ∠AOB = 120°; OM ⟂ AB; shade the minor segment.
3. 3 · Write the formula or theorem Minor segment = sector − ΔOAB, with sector = (θ/360) × πr².
4. 4 · Find the sector Sector = (120/360) × (22/7) × 14 × 14 = (1/3) × 616 = 205.33 cm².
5. 5 · Find the triangle ∠AOM = 60°, so OM = 14 cos 60° = 7 cm and AM = 14 sin 60° = 7√3 cm. AB = 14√3 cm. Area = (1/2) × 14√3 × 7 = 49√3 = 84.77 cm².
6. 6 · Find the minor segment Minor segment = 205.33 − 84.77 = 120.56 cm².
7. 7 · Find the major segment Major segment = 616 − 120.56 = 495.44 cm².
8. 8 · Check and box the answer Minor 120.56 cm² and major 495.44 cm² add up to 616 cm², the whole circle.

Scoring tip: Show how you got the triangle's base and height (OM ⟂ AB). Writing only (√3/4)r² for 120° may lose the method mark.