56. In the given figure, PQ = 24 cm, PR = 7 cm
and O is the centre of the circle. Find the
area of the shaded region. [Take t = 3.14.]
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0
Answer:
please mention your figure without figure question cannot be solved
Answered by
2
Answer:
161.54 cm2
Step-by-step explanation:
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
=722×225×225×2
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