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Question
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. [Use π = 3.14 and `sqrt3 = 1.73` ]
Solution
Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT
In ΔOVS,
`(OV)/(OS) = cos 60º`
`(OV)/12 = 1/2`
OV = 6 cm
`(SV)/(SO) = sin 60^@ = sqrt3/2`
`(SV)/12 = sqrt3/2`
`SV = 6sqrt3 cm`
ST = 2SV
= `2xx6sqrt3`
= `12sqrt3 "cm"`
Area of ΔOST =` 1/2 xx ST xx OV`
`1/2xx12sqrt3xx6`
= `36sqrt3`
= 36 × 1.73
= 62.28 cm2
Area of sector OSUT = `120^@/360^@ xx pi(12)^2`
`=1/3xx3.14 xx 144 = 150.72 cm^2`
Area of segment SUT = Area of sector ΔSUT − Area of ΔOST
= 150.72 − 62.28
= 88.44 cm2
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