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` ]

Answer 1

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