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 cm2
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 cm2
Put the values of area of triangle and area of sector in equation 1
⇒ Area of minor segment = 1848 – 762.93
= 1085.07 cm2
Area of major segment = πR2 – Area of minor segment
Put the value of area of minor segment and R in above equation
= π(422) – 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 cm2
Area of major segment is 4458.93 cm2 and of minor segment is 1085.07 cm2.