Question

Q1 A chord of a circle of radius 5cm subtends an angle of

450 at the centre. Find the area of minor segments of circle​

Answer 1

Step-by-step explanation:

ANSWER

In a given circle, Radiusr=15cm

And θ=60

∘

Area of segmentAPB=Area of sector OAPB−Area of △OAB

Ares of sectorOAYB=

360

θ

×πr

2

360

60

×3.14×15×15=117.75cm

2

Area of △AOB=

2

1

×b×h

We draw OM perpendicular to AB

∴∠OMB=∠OMA=90

∘

In △OMA and △OMB

∠OMA=∠OMB=90

∘

OA=OB(both radius)

OM=OM(common)

∴△OMA≅△OMB(by R.H.S congruency)

⇒∠AOM=∠BOM by C.P.C.T

∴∠AOM=∠BOM=

2

1

∠BOA

⇒∠AOM=∠BOM=

2

1

×60

∘

Also, since △OMB≅△OMA

∴BM=AM by CPCT

⇒BM=AM=

2

1

AB .......(1)

In rightangled triangle OMA

sin30

∘

=

AO

AM

=

15

AM

⇒AM=

2

15

In △OMA

cos30

∘

=

AO

OM

⇒OM=

2

3

×15

From (1)

AM=

2

1

AB

⇒2AM=AB

Put the value of AM

⇒AB=2×

2

15

=15cm

Now, Area of △AOB=

2

1

bh=

2

1

×AB×OM

=

2

1

×15×

2

3

×15=97.3125cm

2

Area of segment APB=Area of sector OAPB−Area of △OAB

=117.75−97.3125=20.4375cm

2

Thus, area of minor segment=20.4375cm

2

Now, Area of major segment=Area of circle−Area of minor segment

=πr

2

−20.4375=3.24×15×15−20.4375=706.5−10.4375=686.0625cm

2