22. Find the area of the shaded in figure, PO = 24cm, OR = 7cm,
O is the centre of the circle
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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
RQ^2 =PQ^+ +PR ^2
[By pythagoras theorem]
RQ²=24²+7²
RQ²=576+49
RQ²=625
RQ=√625cm
RQ=25cm
radius of the circle (OQ)= RQ /2= 25/2cm
Area of right ∆RPQ= 1/2 ×Base×height
Area of right ∆RPQ= 1/2 ×RP×PQ
Area of right ∆RPQ= 1/2×7×24=7×12=84cm²
Area of right ∆RPQ=84cm²
Area of semicircle=πr²/2
=22/7×25/2×25/2×1/2= 11×25×25/28= 6875/28cm ^2
Area of the shaded region = Area of semicircle - Area of right ∆ RPQ 6875−2352/28
= 4523/28
=161.54 cm ^2
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