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**A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle. (Use π = 3.14 and √3 = 1.73)**

Answer:

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°)×πr^{2}

= (120/360)×(22/7)×12^{2}

= 150.72 cm^{2}

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

Or, ½ = OD/12

∴ OD = 6 cm

So, the area of ΔAOB = ½ × base × height

Here, base = AB = 12√3 and

Height = OD = 6

So, area of ΔAOB = ½×12√3×6 = 36√3 cm = 62.28 cm^{2}

∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB

= 150.72 cm^{2}– 62.28 cm^{2} = 88.44 cm^{2}

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