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Question

There is a circle with radius 6 cm. A chord is drawn in it. Find the angle subtended by the minor segment of the chord at the centre of the circle if the area of segment is 22.1 cm2.


  1. 120

  2. 60

  3. 90

  4. 180


Solution

The correct option is A

120


(since OS is perpendicular to PQ) 

OQ = OP (radius of the circle)

OSP = OSQ =  90 

OS is the common side to both the triangles

Thus, ΔPSO is congruent to  ΔQSO by RHS congruence.

This means that corresponding POS and QOS are equal.

So, POS = QOS = θ

POQ =POS +QOS = 2θ

Also PS = SQ (corresponding sides of 2 congruent triangles)  

Area of segment PSQR = Area of sector POQR - Area of ΔPOQ

Area of sector = 2θ360×πr2        (Where r is radius)    

                    = 2θ360×π(6)2    

                    = 0.63θ    

In ΔOSQ, OSQ = 90

 sinθ = oppositehypotenuse   (SQ is opposite side for angle θ since the opposite side has no common vertex with angle)  

      = SQOQ

 sinθ = SQ6

Hence, SQ = 6sinθ

Similarly, OS = 6cosθ  

PQ = PS + SQ  = 2SQ    (Since PS = SQ)

⇒ PQ = 12sinθ

  Area of triangle OPQ = 12×base×height   

     = 12×PQ×OS 

     = 12×12sinθ×6cosθ     

     = 36(sinθ)(cosθ)  

Area of segment = Area of sector – Area of triangle    

                    = 0.63θ - 36(sinθ)(cosθ)        

         22.1 = 0.63θ - 36(sinθ)(cosθ)                            ...........(1)  

Here we can get the value of θ by substituting values of answer options. But take care! The answer is not θ. What is asked is POQ = 2θ

By substituting 2θ = 120, θ = 60 

When θ = 60 the above equation (1) is satisfied. Hence, the angle subtended by segment at centre is 120.

Note that sin (120) = sin (90+30), which is sin (30) = 12. This is because sin (90θ ) = sin(θ)

sin (180)  = 0, when we substitute 2θ as 180

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