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Question
A chord AB of circle, of radius 14cm makes an angle of 60° at the centre. Find the area of minor segment of circle.
Solution
Given radius (r) = 14cm = OA = OB
𝜃 = angle at centre = 60°
In ΔAOB, ∠A = ∠B [angles opposite to equal sides OA and OB] = x
By angle sum property ∠A + ∠B + ∠O = 180°
x + x + 60° = 180° ⇒ 2x = 120° ⇒ x = 60°
All angles are 60°, OAB is equilateral OA = OB = AB
Area of segment = area of sector – area Δle OAB
=`theta/360^@× pir^2 − sqrt(3)/4× (−AB)^2`
=`60/360×22/7× 14 × 14 − sqrt(3)/4× 14 × 14`
=`308/3− 49sqrt(3)`
`= (308−147sqrt(3))/3cm^2`
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