Question

a chord pq of length 12 cm subtends an angle of 120 degree at the center of a circle find the area of minor segment cut off by the chord pq

Answer 1

Answer:

In △AOB,∠AOB=120°

OM⊥AB,∴∠AOM=∠MOB=60° & MA=MB=6

Now, sin60°=

r

6

⇒

2

3

=

r

6

⇒r=

3

12

=4

3

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

Area of sector=

360°

θ

×πr

2

=

360°

120°

×π×4

3

×4

3

=16πcm

2

Area △AOB=

s(s−a)(s−b)(s−c)

,s=

2

a+b+c

=

2

4

3

+4

3

+12

=4

3

+6

Area of △AOB=

(4

3

+6)(4

3

+6−4

3

)(4

3

+6−4

3

)(4

3

+6−12)

=

6×6(4

3

+6)(4

3

−6)

=6

(4

3

)

2

−(6)

2

=6

48−36

=6

12

=12

3

cm

2

∴Area of minor segment=16π−12

3

=4(4π−3

3

)cm

2

Answer 2

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