A chord of a circle of radius 12 , cm subtend | vedclass

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(C) Let us draw a perpendicular $OV$ on chord $ST$. It will bisect the chord $ST$.$SV = VT$In $	riangle OVS$,$\frac{OV}{OS} = \cos 60^{\circ}$$\frac{

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$88.44$

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(C) Let us draw a perpendicular $OV$ on chord $ST$. It will bisect the chord $ST$.$SV = VT$In $	riangle OVS$,$\frac{OV}{OS} = \cos 60^{\circ}$$\frac{OV}{12} = \frac{1}{2}$$OV = 6 \, cm$$\frac{SV}{SO} = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$$\frac{SV}{12} = \frac{\sqrt{3}}{2}$$SV = 6\sqrt{3} \, cm$$ST = 2SV = 2 	imes 6\sqrt{3} = 12\sqrt{3} \, cm$Area of $	riangle OST = \frac{1}{2} 	imes ST 	imes OV$$= \frac{1}{2} 	imes 12\sqrt{3} 	imes 6$$= 36\sqrt{3} = 36 	imes 1.73 = 62.28 \, cm^2$Area of sector $OSUT = \frac{120^{\circ}}{360^{\circ}} 	imes \pi(12)^2$$= \frac{1}{3} 	imes 3.14 	imes 144 = 150.72 \, cm^2$Area of segment $SUT = 	ext{Area of sector } OSUT - 	ext{Area of } 	riangle OST$$= 150.72 - 62.28$$= 88.44 \, cm^2$

$A$ chord of a circle of radius $12 \, cm$ subtends an angle of $120^{\circ}$ at the centre. Find the area of the corresponding segment of the circle. (in $cm^2$)
(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

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$A$ chord of a circle of radius $12 \, cm$ subtends an angle of $120^{\circ}$ at the centre. Find the area of the corresponding segment of the circle. (in $cm^2$)(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)