Let O be the centre of circle
Let OD be the perpendicular to BC
The area of the figure = Area of semicircle + (Area traversed by arc of radius 12 inch and central angle 120° − area of △ BOC)
Area of semicircle= 0.5 ×3.14×r×r =0.5 ×3.14×12×12 =226.08 square inch
Area traversed at centre by arc =(3.14×r×r)÷3 =(3.14×12×12)÷3=150.72 square inch
Using trigonometry in triangle OBD,
sin(∠OBD)=sin(30°)=OD÷OB=1÷2
OD=OB÷2 = 6
Using Pythagorean theorem
BD2=OB 2 −OD2 =144-36=108
BD=10.4
BC=2×BD=20.8 inch
Area of triangle=(OD×BC)÷2 =62.4 square inch.
Hence,
Area of figure = 226.08 + (150.72−62.4)=314.4 square inch
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