Question

A chord 10 cm long is drawn in a circle of radius V50 cm. Find the area of minor segment​

Answer 1

Answer:

In △OAB,

OA

2

+OB

2

=50+50=100

AB

2

=10

2

=100

i.e., OA

2

+OB

2

=AB

2

which satisfies the Pythagoras theorem. Therefor,e θ=90°

Area of minor segment=area of sector-Area of △AOB

=

360°

θ

×πr

2

−

2

1

×OA×OB

=

360°

90°

×

7

22

×5

2

×5

2

−

2

1

×5

2

×5

2

=39.29−25=14.29cm

2

Area of major segment=area of circle-area of minor segment

=πr

2

−14.29=

7

22

×5

2

×5

2

−14.29

=142.85cm

2