Answer to Question #93180 in Geometry for daryl_toledo@yahoo.com

Question #93180
the composite area shown consist of semicircle and a region bounded by line BC and two congruent arcs AB and AC with radii 12 inches each and central angle of 60 degrees each. find the area of the figure?
1
Expert's answer
2019-08-26T11:57:22-0400



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°\degree - area of \triangle BOC)   

Area of semicircle== 0.50.5 ×3.14×r×r\times 3.14 \times r\times r =0.5=0.5 ×3.14×12×12\times 3.14 \times 12 \times 12 =226.08=226.08 square inch

Area traversed at centre by arc ==(3.14×r×r\times r\times r)÷3\div3 =(3.14×12×12)÷3=150.72=(3.14\times 12 \times 12)\div3=150.72 square inch

Using trigonometry in triangle OBD,

sin(OBD)=sin(30°)=OD÷OB=1÷2\sin(\angle OBD)=\sin(30\degree)=OD\div OB=1\div2

OD=OB÷2\div 2 = 6

Using Pythagorean theorem

BD2=OB^2=OB 2^2 OD2-OD^2 =144-36=108

BD=10.4

BC=2×BD=20.82\times BD=20.8 inch

Area of triangle=(OD×BC)÷2=(OD\times BC)\div 2 =62.4 square inch.

Hence, 

Area of figure == 226.08226.08 ++ (150.7262.4)=314.4(150.72 −62.4)=314.4 square inch


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