Question

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.]
​

Answer 1

Answer:

please mention your figure without figure question cannot be solved

Answer 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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