in figure PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Find the area of shaded region.
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PQ=24cm ,PR =7 cm
We know that any angle made by the diameter QR in the semicircle is 90°.
∴∠RPQ=90°
In right angled ∆RPQ
RQ2=PQ2+PR2 [By pythagoras theorem]
RQ²=24²+7²
RQ²=576+49
RQ²=625
RQ=√625cm
RQ=25cm
radius of the circle (OQ)=2RQ=225cm
Area of right ∆RPQ=21×Base×height
Area of right ∆RPQ=21×RP×PQ
Area of right ∆RPQ=21×7×24=7×12=84cm²
Area of right ∆RPQ=84cm²
Area of semicircle=2πr²
=722×225×225×
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