A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle ? (Use π = 3.14 and √ 3 = 1.73)

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Solution:
⇨ Given
    r = 12 cm
    θ = 120°
⇨ Area Of Sector
⇨ 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
⇨ Now Area of △AOB

⇨ Area of Minor segment 
Area Of Sector - Area of △AOB
= (150.72 - 62.28) 
= 88.44 


  Exercise 12.2  
















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