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)
Radius, r = 12 cm
Now, draw a perpendicular OD on chord AB and it will bisect chord AB.
AD = DB
Now, the area of the minor sector = (θ/360°) × πr2
= (120/360) × (22/7) × 122
= 150.72 cm2
Consider the 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
1/2 = OD/12
∴ OD = 6 cm
The area of ΔAOB = 1/2 × base × height
Base = AB = 12√3 and
Height = OD = 6
Area of ΔAOB = 1/2 × 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