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प्रश्न
A chord PQ of length 12 cm subtends an angle of 120° at the centre of a circle. Find the area of the minor segment cut off by the chord PQ.
उत्तर
We know that the area of minor segment of angle θ in a circle of radius r is,
`A={(piθ)/360°-sin θ/2 cos θ/2}r^2`
It is given that the chord PQ divides the circle in two segments.
We have ∠POQ=120° and`PQ=12 cm`. So,
`PL=(PQ)/2 cm`
`=12/2 cm`
`= 6 cm`
Since `∠POQ=120°`
`∠POL=∠QOL`
`=60°`
In ΔOPQ, We have
`sin θ=(PL)/(OA)`
`sin 60°=6/(OA)`
`sqrt3/2=6/(OA)`
`OA=12/sqrt3`
Thus the radius of circle is `OA=4sqrt3 cm`
Now using the value of radius r and angle θ we will find the area of minor segment
`A={(120°pi)/(360°)-sin (120°)/2 cos (120°)/2} (4sqrt3)^2`
`= {pi/3-sqrt3/2xx1/2}xx48`
`= 4{4pi-3sqrt3}cm^2`
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