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Question

In the given diagram, find the area of segment PRQS. Sides OS and SQ have lengths a and b respectively. Let the area of circle be A.


A

- ab

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B

- 2ab

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C

-(ab)

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D

- (ab)

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Solution

The correct option is A

- ab


OQ2 = OS2 + SQ2 (Using Pythagoras Theorem)

OQ2 = a2 + b2

OQ is the radius of the circle.

=> OQ=(a2+b2

Area of ΔOPQ = (12)×Base×Height

= (12)×PQ×OS

= 12×2b×a= (ab) PQ = 2SQ = 2b cm (OS is perpendicular bisector of PQ))

Area of sector POQR =POQ360×πr2 (POQ=2(SOQ)=2θ)

= (2θ360)×π(a2+b2) (Since r2=a2+b2 )

Area of segment PRQS= Area of sector POQR - Area of triangle OPQ = (θ180)×π(a2+b2) -(ab)

Also we know area of circle = πr2

= π(a2+b2)

= A(given)

Therefore;Area of segment PRQS = (θ180)×A - (ab)


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