Question

22
Unless stated otherwise, use t =
7
O
1. Find the area of the shaded region in Fig. 5.19, ifPQ
= 24 cm, PR=7 cm and O is the centre of the circle.
R
P
Fig. 5.19​

Answer 1

Answer:

Given :  

PQ= 24 cm ,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= 25 cm

radius of the circle (OQ)= 25 / 2 cm

Area of right ∆ RPQ= ½ × Base × height

Area of right ∆ RPQ= ½ × RP × PQ

Area of right ∆ RPQ = ½ × 7 × 24 = 7 × 12 = 84 cm²

Area of right ∆ RPQ = 84 cm²

Area of semicircle= πr²/2

= (22/7) × (25/2)² / 2

= (22 × 25 × 25)/ (7× 2 × 2 × 2)

= 11 × 625 /28 = 6875/28 cm²

Area of semicircle = 6875/28 cm²

Area of the shaded region = Area of semicircle - Area of right ∆ RPQ

=  (6875/28  - 84 )cm²

= (6875 - 2532)/ 28

Area of the shaded region = 4523 / 28= 161.54 cm²

Hence, the area of the shaded region = 161.54 cm²

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