Find the area of the shaded region in the figure if PQ-24cm. PR - 10cm and 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
2
+PR
2
[By pythagoras theorem]
RQ²=24²+7²
RQ²=576+49
RQ²=625
RQ=√625cm
RQ=25cm
radius of the circle (OQ)=
2
RQ
=
2
25
cm
Area of right ∆RPQ=
2
1
×Base×height
Area of right ∆RPQ=
2
1
×RP×PQ
Area of right ∆RPQ=
2
1
×7×24=7×12=84cm²
Area of right ∆RPQ=84cm²
Area of semicircle=
2
πr²
=
7
22
×
2
25
×
2
25
×
2
1
=
28
11×25×25
=
28
6875
cm
2
Area of the shaded region = Area of semicircle - Area of right ∆ RPQ
=
28
6875
−84
=
28
6875−2352
=
28
4523
=161.54 cm
2
Hence, the area of the shaded region = 161.54 cm²
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