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Ex 12.3, 9 - AB and CD are two diameters of a circle - Ex 12.3

  1. Chapter 12 Class 10 Areas related to Circles
  2. Serial order wise
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Ex 12.3, 9 In figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region. Area of shaded region = Area of circle with diameter OD + Area of semicircle ACB – Area of triangle ABC Area of circle with diameter OD Since OD & OA are radius of circle with centre O OD = OA = 7 cm Diameter = OD = 7 cm radius = r = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 = 7/2 cm Area of circle with diameter OD = 𝜋r2 = 22/7×(7/2)^2 = 22/7×7/2×7/2 = 77/2 cm2 ∴ Area of circle with diameter OD = 77/2 cm2 Area of semicircle ACB Diameter = AB So, radius = OA = 7 cm Area of semicircle ACB = 1/2 × Area of circle = 1/2 × πr2 = 1/2 × 22/7 × 7 × 7 = 11 × 7 = 77 cm2 So, Area of semicircle with diameter AB = 77 cm2 Area Δ ABC Given that diameters AB & CD are perpendicular So, AB ⊥ DC ⇒ ∠ BOC = ∠ AOC = 90° Area Δ ABC = 1/2 × Base × Height Here, Base = AB = 2 × radius = 2OA = 2 × 7 = 14 & Height = OC = 7 cm Putting values Area Δ ABC = 1/2 × Base × Height = 1/2 × AB × OC = 1/2×14×7 = 49 cm2 Area of shaded region = Area of circle with diameter OD + Area of semicircle with diameter AB – Area of triangle ABC = 77/2 + 77 – 49 = 77/2 + 28 = (77 + 2 × 28)/2 = 133/2 = 63.5

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