Solution:
Radius, r = 12 cm
Now, draw a perpendicular OD on chord AB and it will bisect chord AB.
So, AD =DB
Now, the area of the minor sector = (θ/360°) × π r2
= (120/360) × (22/7) × 122
= 150.72 cm2
Consider the ΔAOB
Area of ΔAOB =
∠OAB = 180° – (90° + 60°) = 30°
Now, cos 30° = AD/OA
=> √3/2 = AD/12
Or, AD = 6√3 cm
We know OD bisects AB. So,
AB = 2 × AD = 12√3 cm
Now, sin 30° = OD/OA
Or, ½ = OD/12
∴ OD = 6 cm
So, the area of ΔAOB = ½ × base × height
Here, base = AB = 12√3 and
Height = OD = 6
area of ΔAOB = ½ × 12√3 × 6 = 36√3 cm = 62.28 cm2
∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB
= 150.72 cm2 – 62.28 cm2 = 88.44 cm2