A chord of a circle of radius 5cm subtends an angle of
450 at the centre. Find the area of minor segments of circle
Answers
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 X 60 ×3.14×15×15=117.75cm
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 xAM
= 15 xAM
⇒AM= 2 x15
In △OMA
cos30 = AO xOM
⇒OM= 2 x3 ×15
From (1)
AM= 2x AB
⇒2AM=AB
Put the value of AM
⇒AB=15
Now, Area of △AOB= 2x 1
bh=2x1 ×AB×OM
=2x1 x×15× 2 x3 ×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
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
2In 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