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.
- 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)