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Solution:
⇨ Given
r = 12 cm
θ = 120°
⇨ Area Of Sector
⇨ In △AOC
∠AOC = 120/2 = 60° (As θ = 120°)
∴ sin 60° = p/h
⇨ √ 3/2 = AC/AO
⇨ √ 3/2 = AC/12
⇨ AC = √ 3 x 12/2
⇨ AC = 6√ 3 cm
∵ AB = 2AC (As C is the mid point)
⇨ Again
⇨ cos 60° = b/h
⇨ 1/2 = b/h
⇨ 1/2 = OC/AO
⇨ 1/2 = OC/12
⇨ OC = 12/2
⇨ OC = 6 cm
⇨ Now Area of △AOB
⇨ Area of Minor segment
= Area Of Sector - Area of △AOB
= (150.72 - 62.28)
= 88.44
Exercise 12.2