Question

In the given figure, find the shaded area if angle in sector OPAQ is 60˚ and PQB is a semicircle.​

Answer 1

Answer:

For a Segment PMQ,

radius r=10cm

Measure of are θ=60

∘

Area of Segment PMQ=r

2

[

360

πθ

−

2

sinθ

]

=10

2

[

360

3.14×60

−

2

sin60

]

=100[

6

3.14

−

2

3

×

2

1

]

=100[

6

3.14

−

4

3

]

=100[

12

6.28−3(1.73)

]

=100[

12

6.28−5.19

]

=

12

100×1.09

=

12

109

∴areaofsegmentPMQ=9.08cm

2

△OPQsegOP≅segOQ

∴∠OPQ≅∠OQP

Let , m∠OPQ=m∠OQP=x

∴m∠OPQ+m∠OQP+m∠POQ=180

∘

∴x+x+60=180

∴2x=180−60

∴2x=120

∴x=

2

120

∴x=60

∘

∴m∠OPQ=m∠OQP=m∠POQ=60

∘

∴△OPQisanequilateraltriangle

∴OP=OQ=PQ=10cm

DiameterPQ=10cm

Radiusr=

2

10

=5cm

Area of Semicircle =

2

1

πr

2

=

2

1

×3.14×5×5=39.28cm

2

Area of the shaded portion = Area of Semi Circle - Area of segment PMQ

=39.25−9.08=30.17cm

2

Answer 2

Answer:

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