A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the areas of the minor and major segments. [Take π = 3.14 and √3 = 1.732.]

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 = 471 cm^{2}

Where a = side of the triangle

Area of triangle = √3/4×30×30

Area of triangle = √3/4×900

Area of triangle = (900×√(3 ))/4

(put √3 = 1.732)

Area of triangle = (1.732×900)/4

∴ Area of triangle = 389.7 cm^{2}

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

Area of minor segment = 471 – 389.7

⇒ Area of minor segment = 81.3 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

⇒ Area of major segment = π×(30^{2}) – 81.3 (put π = 3.14)

⇒ Area of major segment = 3.14×30×30 – 81.3

⇒ Area of major segment = 2826 – 81.3

= 2744.7 cm^{2}

Area of major segment is 2744.7cm^{2} and of minor segment is 81.3 cm^{2}.

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