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12.3, 12 i.jpg

  1. Chapter 12 Class 10 Areas related to Circles
  2. Serial order wise
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Ex 12.3, 12 In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the quadrant OACB, It is given that, OACB is a quadrant Given radius = 3.5 cm We know that Area of Quadrant OACB = 1/4 × Area of a circle Area of quadrant OACB = 1/4 𝜋𝑟2 = 1/4×22/7×(3.5)2 = 1/4×22/7×3.5×3.5 = 3.625 cm2 Ex 12.3, 12 In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (ii) Shaded region Area of shaded region = Area of quadrant OACB – Area of Δ BOD We calculated Area of quadrant in part (i) Area of Δ BOD Area of ∆ 𝐵𝑂𝐷=1/2× Base × Height Since OACB is a quadrant ∠ BOA = 90° Now, Area Δ BOD = 1/2 × Base × Height = 1/2×OB×OD = 1/2 × 3.5 × 2 = 3.5 cm2 Area of shaded region = Area of quadrant OACB – Area of ∆ 𝐵𝑂𝐷 = 9.625 – 3.5 = 6.125 cm2 Hence, Area of shaded region = 6.125 cm2

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