Today i am solving the ncert class 10 question of areas related to circles of exercise 12.2 question no. 7 its very hard to solve for me please help me to solve this question in a simplest form also give me the best solution of this question.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)
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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 segment of the circle. (Use π = 3.14 and √3 = 1.73) Q.7
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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°)×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
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 cm2
∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB
= 150.72 cm2– 62.28 cm2 = 88.44 cm2