Question

A chord PQ of length 12 cm subtends an angle 120^o at the centre of a circle. Find the area of the minor segment cut off by the chord PQ. 

Answer 1

Given, ∠POQ = 120^o and PQ = 12 cm

Draw OV ⊥ PQ,

PV = PQ × (0.5) = 12 × 0.5 = 6 cm

Since, ∠POV = 120^o

∠POV = ∠QOV = 60^o

In triangle OPQ, we have

sin θ = PV/ OA

sin 60^o = 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