Question

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. 

Answer 1

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`