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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.


A

60

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B

90

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C

120

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D

180

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Solution

The correct option is C

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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