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Ex 6.4, 5 A chord of a circle of radius 15 cm subtends an angle of 60^∘ at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use πœ‹β‰ˆ3.14 and √3β‰ˆ1.73.) Let the minor segment be APB Given that Radius = r = 15 cm πœƒ = 60Β° Now, Area of segment APB = Area of sector OAPB – Area of Ξ”OAB Let’s find separately Area of sector OAPB Area of sector OAPB = πœƒ/360Γ—πœ‹π‘Ÿ2 = πŸ”πŸŽ/πŸ‘πŸ”πŸŽ Γ— πŸ‘.πŸπŸ’ Γ— πŸπŸ“ Γ— πŸπŸ“ = 1/6 Γ— 3.14 Γ— 15 Γ— 15 = 1/2 Γ— 3.14 Γ— 5 Γ— 15 = 1.57 Γ— 5 Γ— 15 = 117.75 cm2 Area of Ξ” AOB In βˆ† AOB, OA = OB = 15 cm ∠ AOB = 60Β° We know that Angles opposite equal side are equal ∴ ∠ OAB = ∠ OBA By Angle sum property ∠ OAB + ∠ AOB + ∠ OBA = 180Β° ∠ OAB + 60Β° + ∠ OBA = 180Β° Putting ∠ OAB = ∠ OBA ∠ OAB + 60Β° + ∠ OAB = 180Β° 2 Γ— ∠ OAB + 60Β° = 180Β° 2 Γ— ∠ OAB = 180Β° – 60Β° 2 Γ— ∠ OAB = 120Β° ∠ OAB = (120Β°)/2 ∠ OAB = 60Β° Thus, ∠ OBA = ∠ OAB = 60Β° So, all 3 angles are 60Β° Which means βˆ† OAB is an equilateral triangle We need to find area of equilateral triangle Area of equilateral triangle = βˆšπŸ‘/πŸ’ Γ— π‘Ίπ’Šπ’…π’†^𝟐 Putting = = βˆšπŸ‘/πŸ’ Γ— π‘Ίπ’Šπ’…π’†^𝟐 Area Ξ” AOB = 1/2 Γ— Base Γ— Height We draw OM βŠ₯ AB ∴ ∠ OMB = ∠ OMA = 90Β° And, by symmetry M is the mid-point of AB ∴ BM = AM = 1/2 AB 2 Γ— ∠ OAB + 60Β° = 180Β° 2 Γ— ∠ OAB = 180Β° – 60Β° 2 Γ— ∠ OAB = 120Β° ∠ OAB = (120Β°)/2 ∠ OAB = 60Β° Thus, ∠ OBA = ∠ OAB = 60Β° So, all 3 angles are 60Β° Which means βˆ† OAB is an equilateral triangle We need to find area of equilateral triangle Area of equilateral triangle = βˆšπŸ‘/πŸ’ Γ— π‘Ίπ’Šπ’…π’†^𝟐 Putting = = βˆšπŸ‘/πŸ’ Γ— π‘Ίπ’Šπ’…π’†^𝟐 Area Ξ” AOB = 1/2 Γ— Base Γ— Height We draw OM βŠ₯ AB ∴ ∠ OMB = ∠ OMA = 90Β° And, by symmetry M is the mid-point of AB ∴ BM = AM = 1/2 AB Area of equilateral triangle = βˆšπŸ‘/πŸ’ Γ— π‘Ίπ’Šπ’…π’†^𝟐 Putting √3 = 1.73, Side = 15 = (𝟏.πŸ•πŸ‘)/πŸ’ Γ— πŸπŸ“^𝟐 = 1.73/4 Γ— 225 = 1.73/4 Γ— 15^2 = 389.25/4 = 389.25/4 = 97.3125 cm2 Thus, Area of segment APB = Area of sector OAPB – Area of Ξ”OAB = 117.75 – 97.3125 = 20.4375 cm2 Thus, Area of minor segment = 20.4375 cm2 And, Area of major segment = Area of circle – Area of minor segment = πœ‹π‘Ÿ^2 – 20.4375 = 3.14 Γ— 15^2 – 20.4375 = πŸ‘.πŸπŸ’ Γ— πŸπŸπŸ“ – 20.4375 = 706.5 – 20.4375 = πŸ”πŸ–πŸ”.πŸŽπŸ”πŸπŸ“ cm2

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