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Question

A chord of a circle of radius 12 cm subtends an angle of1200 at the centre. Find the area of the corresponding segment of the circle.
[takeπ=3.14and3=1.73]

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Solution

We have,

In a circle,

Radius

r=12cm.

θ=120o

Area of segment APB= Area of sector OAPB -Area of ΔOAB

Area of sector OAPB

=θ360o×πr2

=120o360×3.14×(12)2

=13×3.14×144

=3.14×4×12

=3.14×48

=150.72cm2

Finding area of triangle ΔAOB

AreaofΔAOB=12×Base×Height

We draw

$\begin{align}

OMAB

OMB=OMA=90o

In

ΔOMAandΔOMB

OMA=OMB(Both90o)

OM=OM(Bothradius)

OA=OB(Commonline)

ΔOMAΔOMB(S.A.S.Congurency)

AOM=BOM

Then,

BM=AM=12AB.......(1)

Now,

AOM=BOM=12BOA

=12×1200

=60o

In right angle triangle OMA

sinO=AMAO

sin60o=AM12

32=AM12

AM=63

Again, In right angle triangle OMA

cosO=OMAO

cos60o=OMAO

12=OM12

OM=6cm.

Now, from equation (1) and we get,

AM=12AB

2AM=AB

AB=2AM

AB=2×63AM=63

AB=123cm.

Now,

AreaofΔAOB=12×Base×Height

=12×AB×OM

=12×123×6

=363

=36×1.73

62.28cm2

Hence, the Area of segment of

APB=AreaofsectorOAPBAreaofΔOAB

=150.7262.28

=88.44cm2

Hence, this is the answer.
1164720_1153482_ans_cbb436815d1344c8ad3c19a7f5bcf794.png

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