Question

plzz solve this i didn't need any irrelevant answer​

Answer 1

Answer:

nxisis hsusis hdusud hzushd hsudb hdid

Answer 2

Answer:

Area of sector OACBO=

360

θ

×πr

2

cm

2

=

360

120

×

7

22

×42×42=1848 cm

2

Area of ΔOAB=

2

1

r

2

sinθ

= (

2

1

×42×42×sin120

0

)=(21×42×

2

3

) cm

2

=(21×21×1.73) cm

2

=762.93 cm

2

Now,

Area of minor segment ACBA = (area of sector OACBO) - (area of triangle OAB) = 1848−762.93=1085.07 cm

2

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

=

7

22

×42×42−1085.07=4458.93 cm

2

solution