Area of segment of circle and length of arc

Chapter 11 Class 10 Areas related to Circles
Concept wise

### Transcript

Ex 11.1, 6 A chord of a circle of radius 15 cm subtends an angle of 60Β° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use Ο = 3.14 and β3 = 1.73) In a given circle, Radius (r) = 15 cm And, π½ = 60Β° Now, Area of segment APB = Area of sector OAPB β Area of ΞOAB Finding Area of sector OAPB Area of sector OAPB = π/360 Γ ππ2 = 60/360 Γ 3.14 Γ (15)2 = 1/6 Γ 3.14 Γ 15 Γ 15 = 1/2 Γ 3.14 Γ 5 Γ 15 = 117.75 cm2 Finding area of Ξ AOB 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/15 15/2 = 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 15 = OM OM = βπ/π Γ 15 From (1) AM = π/πAB 2AM = AB AB = 2AM Putting value of AM AB = 2 Γ 1/2 Γ 15 AB = 15 cm Now, Area of Ξ AOB = 1/2 Γ Base Γ Height = π/π Γ AB Γ OM = 1/2 Γ 15 Γ β3/2 Γ 15 = 1/2 Γ 15 Γ 1.73/2 Γ 15 = 97.3125 cm2 Therefore, Area of segment APB = Area of sector OAPB β Area of ΞOAB = 117.75 β 97.3125 = 20.4375 cm2 Thus, Area of minor segment = 20.4375 cm2 Now, Area of major segment = Area of circle β Area of minor segment = Οr2β 20.4375 = 3.14 Γ πππβ 20.4375 = 3.14 Γ 15 Γ 15β"20.4375" = 706.5 β 20.4375 = 686.0625 cm2