Question

How to find this class 10

Answer 1

Answer:

Here, we have

r=15 cm

θ=60

o

Let OACBO be the given sector and OAB is a triangle.

then,

(i) Area of the sector (OACBO) =πr

2

×

360

θ

=3.14×15×15×

360

60

=117.75 cm

2

(ii) Area of triangle (AOB) =

2

1

r

2

sinθ

=

2

1

×15×15×

2

3

=

4

15×15×1.73

=97.313 cm

2

Now,

Area of the Minor segment (ACBA)

= Area of sector (OACBO) − Area of triangle (AOB)

=117.75−97.313

=20.437 cm

2

and

Area of major segment (ABDA)

= Area of circle − Area of Minor segment

=πr

2

−20.437

=3.14×15×15−20.437

=706.5−20.437

=686.063 cm