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 cm2.120 60 90 180

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

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