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=3.14and√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∘ cosθ∘=BaseHypotenuse cos30∘=ADOA ⇒√32=AD12 ⇒AD=6√3cm ⇒AB=2×AD=12√3cm sin30∘=ODOA ⇒12=OD12 ⇒OD=6cm AreaofΔAOB=12×base×height =12×12√3×6=36√3cm =36×1.73=62.28cm2
Angle made by Minor sector =120∘
Area of the sector making angle θ =(θ360∘)×πr2
Area of the sector making angle 120∘ =(120∘360∘)×πr2cm2 =(13)×122πcm2=1443πcm2 =48×3.14cm2=150.72cm2 ∴ Area of the corresponding Minor segment = Area of the Minor sector - Area of ΔAOB =150.72cm2−62.28cm2 =88.44cm2