Area of segment of circle and length of arc

Chapter 11 Class 10 Areas related to Circles
Concept wise

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Example 2 Find the area of the segment AYB shown in figure, if radius of the circle is 21 cm and β AOB = 120Β°. (Use Ο = 22/7 ). In a given circle, Radius (r) = 21 cm And, π½ = 120Β° Now, Area of segment AYB = Area of sector OAYB β Area of ΞOAB Finding Area of sector OAYB Area of sector OAYB = π/360Γ ππ2 = 120/360 Γ 22/7Γ(21)2 = 1/3Γ22/7 Γ 21 Γ 21 = 22 Γ 21 = 462 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/21 β3/2 Γ 21 = AM AM = βπ/π Γ 21 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/21 21/2 = OM OM = ππ/π From (1) AM = π/πAB 2AM = AB AB = 2AM Putting value of AM AB = 2 Γ β3/2 Γ 21 AB = β3 Γ 21 AB = 21βπ cm Now, Area of Ξ AOB = 1/2 Γ Base Γ Height = π/π Γ AB Γ OM = 1/2 Γ 21β3 Γ 21/2 = (πππβπ)/π cm2 Therefore, Area of the segment AYB = Area of sector β Area of β π΄ππ΅ = (462 β πππ/π βπ ) cm2