## Book: RS Aggarwal - Mathematics

### Chapter: 18. Area of Circle, Sector and Segment

#### Subject: Maths - Class 10th

##### Q. No. 56 of Exercise 18B

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56
##### In the given figure, PQ = 24 cm, PR = 7 cm and 0 is the centre of the circle. Find the area of the shaded region. [Take π = 3.14.]

Here we will subtract the area of right angle triangle PQR and semicircle from the area of entire circle.

Given PQ = 24 cm, PR = 7 cm

Consider ∆PQR, QPR = 90°

RQ2 = 242 + 72

RQ2 = 576 + 49

RQ2 = 625

Therefore Radius of the circle = half of RQ

r = 12.5 cm

Area of ∆PQR = 7×12

Area of ∆PQR = 84 cm2 eqn1

Area of semicircle = 245.3125 cm2 eqn2

Area of circle = πr2

Area of circle = π(12.52)

Area of circle = 3.14×156.25 (putting π = 3.14)

Area of circle = 490.625 cm2 eqn3

Area of shaded region = eqn3 – eqn2 – eqn1

Area of shaded region = 490.625 – 245.3125 – 84

Area of shaded region = 161.3125 cm2

Area of shaded region is 161.3125 cm2.

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