Question

The radius of circle with centre O is 15 cm.
The chord AB subtends an angle of 60° at
the centre of the circle. Find the areas of two
segments made by AB.

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Answer 1

Answer:

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Step-by-step explanation:

⇒ Here, r=5cm and θ=90

o

⇒ Area of sector OABO=

360

θ

πr

2

=

360

o

90

o

×3.14×(5)

2

∴ Area of sector OABO=

4

3.14×25

=19.62cm

2

⇒ Area of △AOB=

2

1

×AO×BO=

2

1

×5×5

∴ Area of △AOB=

2

25

=12.5cm

2

∴ Area of minor segment made bye the chord AB = Area of sector OABO - Area of △AOB

∴ Area of minor segment made bye the chord AB =19.62−12.5=7.12cm

2

⇒ Area of circle =πr

2

=3.14×(5)

2

∴ Area of circle =78.5cm

2

⇒ Area of major segment made by chord AB = Area of a circle - Area of minor segment.

∴ Area of major segment made by chord AB=78.5−7.12=71.38cm

2