Minor sector is (60/360) x 3.14 x 15² = 117.75 cm². Equilateral triangle area is (√3/4) x 15² = 97.31 cm². Minor segment is 117.75 – 97.31 = 20.44 cm². Major segment is 706.5 – 20.44 = 686.06 cm².
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π ≈ 3.14 and √3 ≈ 1.73.)
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Given r = 15 cm, θ = 60°, π = 3.14 and √3 = 1.73.
Total circle area = π x r² = 3.14 x 15² = 3.14 x 225 = 706.5 cm².
Area of minor sector = (60 / 360) x 3.14 x 225 = (1/6) x 706.5 = 117.75 cm².
The triangle formed is equilateral (angle 60°):
Area of triangle = (√3 / 4) x r² = (1.73 / 4) x 225 = 97.3125 cm².
Minor segment = Minor sector – Triangle = 117.75 – 97.3125 = 20.4375 cm² (or 20.44 cm²).
Major segment = Circle area – Minor segment = 706.5 – 20.4375 = 686.0625 cm² (or 686.06 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/