Given, ∠POQ = 120o and PQ = 12 cm
Draw OV ⊥ PQ,
PV = PQ × (0.5) = 12 × 0.5 = 6 cm
Since, ∠POV = 120o
∠POV = ∠QOV = 60o
In triangle OPQ, we have
sin θ = PV/ OA
sin 60o = 6/ OA
√3/2 = 6/ OA
OA = 12/ √3 = 4√3 = r
Now, using the above we shall find the area of the minor segment
We know that,
Area of the segment = area of sector OPUQO – area of △OPQ
= θ/360 x πr2 – 1/2 x PQ x OV
= 120/360 x π(4√3)2 – 1/2 x 12 x 2√3
= 16π – 12√3 = 4(4π – 3√3)
Therefore, the area of the minor segment = 4(4π – 3√3) cm2