Question

find the area of the shaded region in Fig. 12.19, if PQ=24cm, PR=7cm and O is the centre of the circle
​

Answer 1

Answer:

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

hp u understand my answer right

Answer 2

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