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Rs aggarwal solutions

A chord of a circle of radius 30 cm makes an angle of 60^0 at the centre of the circle. Find the areas of the minor-major segments. (take pi = 3.14 and √3 = 1.7)

Radius of circle = r = 30 cm

Area of minor segment = Area of the sector – Area of the triangle ...(1)

Area of major segment = Area of the circle – the area of the minor segment ...(2)

Area of sector = θ/360 x πr^2

= 60/360 x 3.14 x 30 x 30

= 471 cm^2

Area of triangle = √3/4 (side)^2 (Since it forms an equilateral triangle)

= √3/4 x 30 x 30

= 389.7 cm^2

(1) =>

Area of minor segment = 471 – 389.7 = 81.3 cm^2

(2) =>

Area of major segment = π(30^2) – 81.3 = 2744.7 cm^2

**Area of the major segment is 2744.7 cm^2 and of the minor segment is 81.3 cm^2.**

Radius of circle = r = 30 cm

Area of minor segment = Area of the sector – Area of the triangle ...(1)

Area of major segment = Area of the circle – the area of the minor segment ...(2)

Area of sector = θ/360 x πr^2

= 60/360 x 3.14 x 30 x 30

= 471 cm^2

Area of triangle = √3/4 (side)^2 (Since it forms an equilateral triangle)

= √3/4 x 30 x 30

= 389.7 cm^2

(1) =>

Area of minor segment = 471 – 389.7 = 81.3 cm^2

(2) =>

Area of major segment = π(30^2) – 81.3 = 2744.7 cm^2

**Area of the major segment is 2744.7 cm^2 and of the minor segment is 81.3 cm^2.**

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