A chord of a circle of radius 12 cm subtends an angle of 120° at the centre.Find the area of the corresponding segment of the circle.
[Take π = 3.14 and √3 = 1.732]
(Class 10 Maths Sample Question Paper)
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FIGURE IS IN THE ATTACHMENT
SOLUTION:
Let PQ is a chord of a circle with Centre O and
∠POQ= 120°.
OP = OQ = 12 cm (radii of the circle) (GIVEN)
Draw OR ⟂PQ
Since, ∆POQ is an isosceles triangle and OR ⟂PQ.
OR is the angle bisector as well as median.
In right angled ∆ PRO,
∠R = 90° and ∠POR = 60°
OR /PO = cos 60°
OR = PO × cos 60°
OR = 12 × (½)
[ Cos 60° = ½]
OR = 6 cm
PR /PO = sin 60°
PR = PO × sin 60°
OR = 12 × (√3/2)
[sin 60° =√3/2]
OR = 6√3 cm
PQ = 2PR = 2 × 6√3= 12√3
PQ= 12√3 cm.
Area of minor segment = area of minor sector OPMQO - area of ∆POQ
=( ϴ/360°)(πr²) - ½ × PQ × OR
= (120°/360°) (3.14 × 12 ×12) - ½ × 12√3 × 6
=( ⅓)(3.14× 144) - 36 √3
= 48 × 3.14 - 36 × 1.73
= 150.72 - 62.35
= 88.37 cm².
HOPE THIS WILL HELP YOU....
SOLUTION:
Let PQ is a chord of a circle with Centre O and
∠POQ= 120°.
OP = OQ = 12 cm (radii of the circle) (GIVEN)
Draw OR ⟂PQ
Since, ∆POQ is an isosceles triangle and OR ⟂PQ.
OR is the angle bisector as well as median.
In right angled ∆ PRO,
∠R = 90° and ∠POR = 60°
OR /PO = cos 60°
OR = PO × cos 60°
OR = 12 × (½)
[ Cos 60° = ½]
OR = 6 cm
PR /PO = sin 60°
PR = PO × sin 60°
OR = 12 × (√3/2)
[sin 60° =√3/2]
OR = 6√3 cm
PQ = 2PR = 2 × 6√3= 12√3
PQ= 12√3 cm.
Area of minor segment = area of minor sector OPMQO - area of ∆POQ
=( ϴ/360°)(πr²) - ½ × PQ × OR
= (120°/360°) (3.14 × 12 ×12) - ½ × 12√3 × 6
=( ⅓)(3.14× 144) - 36 √3
= 48 × 3.14 - 36 × 1.73
= 150.72 - 62.35
= 88.37 cm².
HOPE THIS WILL HELP YOU....
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