If PQ = 24 cm, PR = 7 cm and O is the center of the circle.

NCERT class 10 Chapter 12 AREAS RELATED TO CIRCLES

Page No. 234

Exercise 12.3

Question No. 1

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QR is diameter of circle.

Therefore, ∠RPQ = 90° [Angle in semicircle is right angle]

In APQR, by Pythagoras theorem,

RP² + PQ² = RQ²

(7)² + (24)² = RQ²

⇒ RQ² = 576 + 49 = 625

⇒ RQ = √625 = 25

Therefore, the radius of circle = RQ/2 = 25/2 cm

Area of shaded region = Area of semicircle – Area of APQR

= 1/2 × πr² – 1/2 PR × PQ = 1/2 × π(25/2)² – 1/2 × 7 × 24 = 1/2 × 22/7 × 25/2 × 25/2 – 7 × 12

= 6875/28 – 84 = (6875 – 2352)/28 = 4523/28 cm²

Here is the video explanation ✌😇