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
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Answered by
11
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
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
Answered by
20
Answer:- Area of minor segment is 81. 3 square cm.
Area of major segment is 2744. 7 square cm .
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