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°

RQ^{2} = 24^{2} + 7^{2}

⇒ RQ^{2} = 576 + 49

⇒ RQ^{2} = 625

Therefore Radius of the circle = half of RQ

Let radius be ‘r’

∴ r = 12.5 cm

Area of ∆PQR = 7×12

∴ Area of ∆PQR = 84 cm^{2}→ eqn1

∴ Area of semicircle = 245.3125 cm^{2}→ eqn2

Area of circle = πr^{2}

Area of circle = π(12.5^{2})

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

∴ Area of circle = 490.625 cm^{2}→ eqn3

Area of shaded region = eqn3 – eqn2 – eqn1

⇒ Area of shaded region = 490.625 – 245.3125 – 84

∴ Area of shaded region = 161.3125 cm^{2}

Area of shaded region is 161.3125 cm^{2}.

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