Area of segment of circle and length of arc

Chapter 11 Class 10 Areas related to Circles
Concept wise

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Ex 11.1, 5 In a circle of radius 21 cm, an arc subtends an angle of 60Β° at the centre. Find: the length of the arc Length of Arc APB = π½/πππ Γ (πππ) = (60Β°)/(360Β°) Γ 2 Γ 22/7 Γ 21 = 1/6 Γ 2 Γ 22 Γ 3 = 22 cm Ex 11.1, 5 In a circle of radius 21 cm, an arc subtends an angle of 60Β° at the centre. Find: (ii) area of the sector formed by the arc Area of sector OAPB = π/360Γππ2 = ππ/πππ Γ ππ/π Γ ππ Γ ππ = 1/6 Γ 22/7 Γ 21 Γ 21 = 1/6 Γ 22 Γ 3 Γ 21 = 231 cm2 Ex 11.1, 5 In a circle of radius 21 cm, an arc subtends an angle of 60Β° at the centre. Find: (iii) area of segment formed by the corresponding chord Area of segment APB = Area of sector OAPB β Area of ΞOAB From last part, Area of sector OAPB = 231 cm2 Finding area of Ξ AOB Area Ξ AOB = 1/2 Γ Base Γ Height We draw OM β₯ AB β΄ β  OMB = β  OMA = 90Β° And, by symmetry M is the mid-point of AB β΄ BM = AM = 1/2 AB In right triangle Ξ OMA sin O = (side opposite to angle O)/Hypotenuse sin ππΒ° = ππ΄/π¨πΆ 1/2=π΄π/21 21/2 = AM AM = ππ/π In right triangle Ξ OMA cos O = (π πππ ππππππππ‘ π‘π πππππ π)/π»π¦πππ‘πππ’π π cos ππΒ° = πΆπ΄/π¨πΆ β3/2=ππ/21 β3/2 Γ 21 = OM OM = βπ/π Γ 21 From (1) AM = π/πAB 2AM = AB AB = 2AM Putting value of AM AB = 2 Γ 1/2 Γ 21 AB = 21cm Now, Area of Ξ AOB = 1/2 Γ Base Γ Height = π/π Γ AB Γ OM = 1/2 Γ 21 Γ β3/2 Γ 21 = (πππβπ)/π cm2 Therefore, Area of segment APB = Area of sector OAPB β Area of ΞOAB = (231 β πππ/π βπ) cm2