Area of segment of circle and length of arc

Chapter 11 Class 10 Areas related to Circles
Concept wise

### Transcript

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 β3 = 1.73) In a given circle, Radius (r) = 12 cm And, π½ = 120Β° Now, Area of segment APB = Area of sector OAPB β Area of ΞOAB Finding Area of sector OAPB Area of sector OAPB = π/360Γ ππ2 = 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 β3 = 1.73) In a given circle, Radius (r) = 12 cm And, π½ = 120Β° Now, Area of segment APB = Area of sector OAPB β Area of ΞOAB Finding Area of sector OAPB Area of sector OAPB = π/360Γ ππ2 = 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 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 β3/2=π΄π/12 β3/2 Γ 12 = AM AM = 6β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 1/2=ππ/12 12/2 = OM OM = 6 From (1) AM = π/πAB 2AM = AB AB = 2AM Putting value of AM AB = 2 Γ 6 β3 AB = 12βπ cm Now, Area of Ξ AOB = 1/2 Γ Base Γ Height = π/π Γ AB Γ OM = 1/2 Γ 12β3 Γ 6 = 36β3 = 36 Γ 1.73 = 62.28 cm2 Therefore, Area of segment APB = Area of sector OAPB β Area of ΞOAB = 150.72 β 62.28 = 88.44 cm2