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Ex 12.2, 6 - A chord of a circle of radius 15 cm subtends 60 - Ex 12.2

Ex 12.2, 6 - Chapter 12 Class 10 Areas related to Circles - Part 2
Ex 12.2, 6 - Chapter 12 Class 10 Areas related to Circles - Part 3
Ex 12.2, 6 - Chapter 12 Class 10 Areas related to Circles - Part 4
Ex 12.2, 6 - Chapter 12 Class 10 Areas related to Circles - Part 5

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Ex 12.2, 6 A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73) In a given circle, Radius (r) = 15 cm And, 𝜃 = 60° Area of segment APB = Area of sector OAPB – Area of ΔOAB Area of sector OAYB = 𝜃/360×𝜋𝑟2 = 60/360×3.14×(15)2 = 1/6×3.14×15×15 = 1/2×3.14×5×15 = 117.75 cm2 Finding area of Δ AOB Area Δ AOB = 1/2 × Base × Height We draw OM ⊥ AB ∴ ∠ OMB = ∠ OMA = 90° In Δ OMA & Δ OMB ∠ OMA = ∠ OMB OA = OB OM = OM ∴ Δ OMA ≅ Δ OMB ⇒ ∠ AOM = ∠ BOM ∴ ∠ AOM = ∠ BOM = 1/2 ∠ BOA ⇒ ∠ AOM = ∠ BOM = 1/2 × 60° = 30° Also, since Δ OMB ≅ Δ OMA ∴ BM = AM ⇒ BM = AM = 1/2 AB From (1) AM = 1/2AB 2AM = AB AB = 2AM Putting value of AM AB = 2 × 15/2 AB = 15 Now, Area of Δ AOB = 1/2 × Base × Height = 1/2 × AB × OM = 1/2 × 15 × √3/2 × 15 = 1/2 × 15 × 1.73/2 × 15 = 97.3125 cm2 Area of segment APB = Area of sector OAPB – Area of ΔOAB = 117.75 – 97.3125 = 20.4375 cm2 Thus, area of minor segment = 20.4375 cm2 Now, Area of major segment = Area of circle – area of minor segment = πr2−20.4375 = 3.14 ×152− 20.4375 = 3.14×15 ×15−"20.4375" = 706.5 – 20.4375 = 686.0625 cm2

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.