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Ex 12.2, 4 - A chord of a circle of radius 10 cm subtends - Ex 12.2

 

  1. Chapter 12 Class 10 Areas related to Circles
  2. Serial order wise
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Ex 12.2, 4 A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : minor segment (Use π= 3.14) Given that OA = OB = radius = 10 cm θ=90° Area of segment APB = Area of sector OAPB – Area of ΔAOB Area of sector OAPB = θ/(360°)×πr2 = 90/360×3.14×(10)2 = 1/4×3.14×100 = 1/4× 314 = 78.5 cm2 Area of ΔAOB Now, ΔAOB is a right triangle, where ∠ O = 90° having Base = OA & Height = OB Area of Δ AOB = 1/2 × Base × Height = 1/2 × OA × OB = 1/2 × 10 × 10 = 50 cm2 Now, Area of segment APB = Area of sector OAPB – Area of ΔAOB = 78.5 – 50 = 28.5 cm2 Ex 12.2, 4(Method 1) A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (ii) major sector (Use π= 3.14) Here, θ = 90° & radius = r = 10 cm Area of major sector = ((360° − θ))/(360°)×πr2 = ((360 − 90))/360×3.14×(10)2 = 270/360×3.14×100 = 3/4× 314 = 235.5 cm2 Ex 12.2, 4(Method 2) A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (ii) major sector (Use π= 3.14) Area of major sector = Area of circle – Area of sector OAPB Area of circle = πr2 = 3.14 × (10)2 = 3.14 × 100 = 314 cm2 Now, Area of major sector = Area of circle – Area of sector OAPB = 314 – 78.5 = 235.5

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