Question

A chord AB of a circle, of radius 14cm makes an angle of 60° at the centre of the circle.

Find the area of the minor segment of the circle.​

Answer 1

Answer:

Q1 In △AOB,OA=OB=14㎝

Therefore, ∠OBA=∠OAB [angles opposite to equal side]

∠AOB=60°

∴∠AOB+∠OAB+∠OBA=180°

60°+2∠OAB=180°⇒∠OAB=

2

120°

=60°=∠OBA

All angles are 60°, therefore △OAB is an equilateral △

Now, area of minor segment=area of sector−ar△OAB

=

360°

θ

×πr

2

−

4

3

(OA)

2

=

360°

60°

×

7

22

×14×14−

4

3

×14×14

=

3

308

−49

3

=

3

308−147

3

=17.80cm°

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