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Question 7
A chord of a circle of radius 12 cm subtends an angle of 120 at the center. Find the area of the corresponding segment of the circle. (Use π=3.14 and 3=1.73)

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Solution

Radius of the circle, r = 12 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
A=180(90+60)=30
cos θ=BaseHypotenuse
cos 30=ADOA
32=AD12
AD=63 cm
AB=2×AD=123 cm
sin 30=ODOA
12=OD12
OD=6 cm
Area of ΔAOB=12×base×height
=12×123×6=363 cm
=36×1.73=62.28 cm2

Angle made by Minor sector =120
Area of the sector making angle θ
=(θ360)×πr2
Area of the sector making angle 120
=(120360)×πr2 cm2
=(13)×122πcm2=1443π cm2
=48×3.14 cm2=150.72 cm2
Area of the corresponding Minor segment = Area of the Minor sector - Area of ΔAOB
=150.72 cm262.28 cm2
=88.44 cm2

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