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Question
A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the area of the minor and major segments.
Solution
Let AB be the chord of a circle with centre O and radius 30 cm such that ∠AOB = 60°
Area of the sector OACBO `= (pi"r"^2theta)/360`
`=(3.14xx30xx30xx60/360)"cm"^2`
= 471 cm2
Area of ΔOAB`=1/2"r"^2"sin" theta`
`=(1/2xx30xx30xx"sin "60°)"cm"^2`
= (225 × 1.732) cm2
= 389.7 cm2
Area of the minor segment = (Area of the sector OACBO) - (Area of ΔOAB)
= (471 - 389.7) cm2
= 81.3 cm2
Area of the major segment =(Area of the circle) - (Area of minor sregment)
`=|(3.14xx30xx30) - 81.3| "cm"^2`
= 2744.7 cm2
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