in a circle of radius 12cm, a chord subtens an angle 120° at the centre.find the area of the corresponding minor segment of the circle
Answers
EXPLANATION.
Circle of the radius = 12cm.
Chord subtends at an angle of = 120° at the center.
As we know that,
Area of minor segment = Area of sector - Area of triangle
Area of segment AMP = Area of sector OAMB - Area of ΔOAB.
Area of sector OAMB = θ/360 x πr².
⇒ 120/360 x (3.14) x (12)².
⇒ 1/3 x (3.14) x 12 x 12.
⇒ 3.14 x 4 x 12 == 150.72cm².
Area of ΔOAB.
Draw OP ⊥ AB.
⇒ ΔOPA = ΔOPB.
⇒ OA = OB [Same radius].
⇒ OP = OP. [Common].
⇒ ∠OPA = ∠OPB. [Each at 90°].
⇒ ∠AOP = ∠BOP = ∠BOA/2.
⇒ ∠AOP = 60°.
⇒ ∠BOP = 60°.
In right angled triangle ΔOPA.
sinθ = Perpendicular/Hypotenuse.
⇒ sin60° = AP/AO.
⇒ √3/2 = AP/12.
⇒ AP = 6√3cm.
cosθ = Base/Hypotenuse.
⇒ cos60° = OP/AO.
⇒ 1/2 = OP/12.
⇒ OP = 6cm.
⇒ AB = AP + PB.
⇒ AB = 2 x AP.
⇒ AB = 2 x 6√3.
⇒ AB = 12√3.
Area of triangle = 1/2 x Base x Height.
Area of ΔOAB = 1/2 x 12√3 x 6.
Area of ΔOAB = 36√3cm.
Area of ΔOAB = 62.28cm².
Area of segment AMP = Area of sector OAMB - Area of ΔOAB.
Area of segment AMP = 150.72 - 62.28.
Area of segment AMP = 88.44cm².
Given :-
In a circle of radius 12cm, a chord subtens an angle 120°
To Find :-
Area
Solution :-
We know that
Area = θ/360 × πr²
Area = 120/360 × 3.14 × (12)²
Area = 1/3 × 3.14 × 144
Area = 48 × 3.14
Area = 150.72 cm²
Now
Let say that there is a point M that is AB ⊥ OM
Now
AM = BM
AM = 1/2 AB
sin 60 = AM/12
√3/2 = AM/12
√3/2 × 12 = AM
√3 × 6 = AM
6√3 = AM
Now
cos 60 = OM/12
1/2 = OM/12
1/2 × 12 = OM
6 = OM
Now
AB × 1/2 × AM
AB = 2 × AM
AB = 2AM
AB = 2 × 6√3
AB = 12√3
Area = 1/2 × 12√3 × 6
Area = 1 × 12√3 × 3
Area = 36√2
Now
Area of minor segment = 150.72 - 36√3 = 88.4 cm²
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