Question

A chord of a circle of radius 30 cm makes an angle of 60 degree at the centre of the circle find areas of minor and major segment

Answer 1

radius of circle = 30 cm

length of chord = 2r sinA/ 2

= 2×30 × sin60/ 2


= 60 × sin30°= 60 × 1/ 2

= 30cm

each side of the formed equilateral

triangle = 30cm


area of equi.triangle AOB


= √3/ 4* (side)^2 =√3 × (30)^2/ 4


= 900√3/ 4 = 389.7cm^2


area of sector AOB

= 60°×( πr^2)/ 360°


= 22× (30)^2/ 6 ×7 = 471.4 cm^2


area of minor segment

= 471.4 - 389.7 = 81.7 cm


area of major segment

= πr^2 - 81.7

= (22 ×900/ 7) - 81.7


= 2746.8 cm^2

Answer:

area of minor segment = 81.7cm^2

area of major segment = 2746.8cm^2

Answer 2

Answer:- Area of minor segment is 81. 3 square cm.

Area of major segment is 2744. 7 square cm .