A chord of a circle of the radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73).
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Answer:
Heya,
In a given circle,
Radius (r) = 12 cm
And, 0 = 120°
Area of segment APB Area of sector OAPB - Area of AOAB
Area of sector OAPB =
360
120 = x 3.14 x (12)² 360
x 3.14 x 12 x 12
= 1 x 3.14 x 4×12
= 150.72 cm²
Finding area of triangle AOB
Area A AOB = Base x Height
We draw OM 1 AB
LOMB = 2 OMA = 90°
In triangle OMA & OMB
OA = OB. (Both radius)
angle OMB = angle OMA (Both 90°)
··ΔΟΜΑ Ξ ΔΟΜΕ
→ ANGLE AOM = BOM
angle AOM= BOM= 1/2 BOA
→ angle AOM= BOM = 1/2 * 120° = 60°
Also, since triangle OMB is congruent OMA
BM = AM (CPCT)
⇒BM = AM = 1/2 AB
In right triangle OMA
sin O = side opposite to angle 0/Hypotenuse
sin 60° AM / AO
√3/2 AM/12
√3/2 x 12 = AM
In right triangle OMA
Cos 0 = side adjacent to angle /Hypotenuse
cos 60°= OM / AO
1/2 = OM/12
12/2 = OM
6= OM
OM = 6
From (1)
AM = 1/2 AB
2AM = AB
AB= 2AM
Putting value of AM
AB=2× 6√3
AB=12√3
Now,
Area of triangle AOB = 1/2 x Base x Height
=1/2 x AB x OM
=×1/2 * 12√3x6
= 36√3
= 36 × 1.73
= 62.28 cm²
Area of segment APB
= Area of sector OAPB Area of AOAB
= 150.72-62.28
= 88.44 cm²
Hope it helps
Step-by-step explanation:
Let us draw a perpendicular OV on chord ST. It will bisect the chord ST and the angle O.
SV = VT
In ΔOVS,
As,
= Cos 60o
=OV = 6 cm
= Sin 60o
SV = 6√3 cm
ST = 2 × SV
= 2 × 6√3
= 12√3 cm
Area of ΔOST =
× 12√3 × 6
= 36√3
= 36 × 1.73
= 62.28 cm2
Area of sector OSUT =
× π × (12)2
= × 3.14 × 144
= 150.72 cm2
Area of segment SUT = Area of sector OSUT - Area of ΔOST
= 150.72 - 62.28
= 88.44 cm2