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)

The perpendicular OV drown form O to chord ST bisects ST.
Therefore, SV = VT
In △OVS,
OV/OS = cos 60° ⇒ OV/12 = 1/2 ⇒ OV = 6 cm
SV/OS = sin 60° ⇒ SV/12 = √3/2 ⇒ OV = 6√3 cm
ST = 2 x SV = 2 x 6√3 = 12√3
Area of △OST
= 1/2 × ST × OV = 1/2 × 12√3 × 6 = 36√3 = 36 x 1.73 = 62.28 cm²
Area of sector OSUT
= 120°/360° x πr² = 1/3 x π(12)²
= 1/3 x 3.14 x 12 x 12 = 150.72 cm²
Area of minor segment
= Area of sector OSUT – Area of △OST
= (150.72 – 62.28) cm²
= 88.44 cm²

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