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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
Let radius be ‘r’
∴ 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.