Question

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?

Answer 1

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$=$ $0.5$ $\times 3.14 \times r\times r$ $=0.5$ $\times 3.14 \times 12 \times 12$ $=226.08$ square inch

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

Using trigonometry in triangle OBD,

$\sin(\angle OBD)=\sin(30\degree)=OD\div OB=1\div2$

OD=OB$\div 2$ = 6

Using Pythagorean theorem

BD$^2=OB$ $^2$ $-OD^2$ =144-36=108

BD=10.4

BC=$2\times BD=20.8$ inch

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

Hence, 

Area of figure $=$ $226.08$ $+$ $(150.72 −62.4)=314.4$ square inch