Find the area of both the segments of a circle of radius 42 cm with central angle 120°. [Given, sin 120° = √3/2 and √3 = 1.73.]

Given R = 42 cm and central angle of sector = 120°

Area of minor segment = Area of sector – Area of triangle → eqn1

Where R = radius of the circle and θ = central angle of the sector

∴ Area of sector = 1848 cm^{2}

Area of right angle triangle = 1/2×base×height×sin θ

Where θ = central angle of the sector

Area of triangle = 1/2×42×42×√3/2

Area of triangle = (42×42×√3)/4

(put √3 = 1.73)

= 762.93 cm^{2}

Put the values of area of triangle and area of sector in equation 1

⇒ Area of minor segment = 1848 – 762.93

= 1085.07 cm^{2}

Area of major segment = πR^{2} – Area of minor segment

Put the value of area of minor segment and R in above equation

= π(42^{2}) – 1085.07

⇒Area of major segment = 22/7×42×42-1085.07

(put π = 22/7)

⇒ Area of major segment = 5544 – 1085.07

∴ Area of major segment = 4458.93 cm^{2}

Area of major segment is 4458.93 cm^{2} and of minor segment is 1085.07 cm^{2}.

17