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Ex 11.1, 7 A chord of a circle of radius 12 cm subtends an angle of 120Β° at the centre. Find the area of the corresponding segment of the circle. (Use Ο€ = 3.14 and √3 = 1.73) In a given circle, Radius (r) = 12 cm And, 𝜽 = 120Β° Now, Area of segment APB = Area of sector OAPB – Area of Ξ”OAB Finding Area of sector OAPB Area of sector OAPB = πœƒ/360Γ— πœ‹π‘Ÿ2 = 120/360Γ—3.14Γ—(12)2 = 1/3Γ—3.14Γ—12Γ—12 = 1 Γ— 3.14 Γ— 4 Γ— 12 = 150.72 cm2 Ex 11.1, 7 A chord of a circle of radius 12 cm subtends an angle of 120Β° at the centre. Find the area of the corresponding segment of the circle. (Use Ο€ = 3.14 and √3 = 1.73) In a given circle, Radius (r) = 12 cm And, 𝜽 = 120Β° Now, Area of segment APB = Area of sector OAPB – Area of Ξ”OAB Finding Area of sector OAPB Area of sector OAPB = πœƒ/360Γ— πœ‹π‘Ÿ2 = 120/360Γ—3.14Γ—(12)2 = 1/3Γ—3.14Γ—12Γ—12 = 1 Γ— 3.14 Γ— 4 Γ— 12 = 150.72 cm2 Finding area of Ξ” AOB We draw OM βŠ₯ AB ∴ ∠ OMB = ∠ OMA = 90Β° And, by symmetry M is the mid-point of AB ∴ BM = AM = 1/2 AB In right triangle Ξ” OMA sin O = (side opposite to angle O)/Hypotenuse sin πŸ”πŸŽΒ° = 𝐀𝑴/𝑨𝑢 √3/2=𝐴𝑀/12 √3/2 Γ— 12 = AM AM = 6βˆšπŸ‘ In right triangle Ξ” OMA cos O = (𝑠𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’ 𝑂)/π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ cos πŸ”πŸŽΒ° = 𝑢𝑴/𝑨𝑢 1/2=𝑂𝑀/12 12/2 = OM OM = 6 From (1) AM = 𝟏/𝟐AB 2AM = AB AB = 2AM Putting value of AM AB = 2 Γ— 6 √3 AB = 12βˆšπŸ‘ cm Now, Area of Ξ” AOB = 1/2 Γ— Base Γ— Height = 𝟏/𝟐 Γ— AB Γ— OM = 1/2 Γ— 12√3 Γ— 6 = 36√3 = 36 Γ— 1.73 = 62.28 cm2 Therefore, Area of segment APB = Area of sector OAPB – Area of Ξ”OAB = 150.72 – 62.28 = 88.44 cm2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo