Question 20
Find the area of the minor segment of a circle of radius 42 cm, if length of the corresponding arc is 44 cm.
Last updated at Oct. 1, 2019 by Teachoo
Question 20
Find the area of the minor segment of a circle of radius 42 cm, if length of the corresponding arc is 44 cm.
Transcript
Question 20 Find the area of the minor segment of a circle of radius 42cm, if length of the corresponding arc is 44cm. Letβs first draw the figure We first have to find the angle We know that Length of arc = π/360 Γ 2Οr Putting values 44 = π/360 Γ 2 Γ 22/7 Γ 42 44 = π/180 Γ 22/7 Γ 42 44 = π/180 Γ 22 Γ 6 (44 Γ 180)/(22 Γ 6) = ΞΈ 2 Γ 30 = ΞΈ 60 = ΞΈ ΞΈ = 60Β° Now, we have to find Area of segment Area of segment APB = Area of sector OAPB β Area of ΞOAB Area of sector OAPB Area of sector OAPB = π/360Γππ2 = 60/360Γ22/7Γ42Γ42 = 1/6Γ22/7Γ42Γ42 = 1/6Γ22Γ6Γ42 = 22Γ42 = 924 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° In Ξ OMA & Ξ OMB β OMA = β OMB (Both 90Β°) OA = OB (Both radius) OM = OM (Common) β΄ Ξ OMA β Ξ OMB (By R.H.S congruency) β β AOM = β BOM (CPCT) β΄ β AOM = β BOM = 1/2 β BOA β β AOM = β BOM = 1/2 Γ 60Β° = 30Β° Also, since Ξ OMB β Ξ OMA (CPCT) β¦(1) β΄ BM = AM β BM = AM = 1/2 AB In right triangle Ξ OMA sin O = (Side opposite to angle O)/Hypotenuse sin 30Β° = AM/AO 1/2=π΄π/42 AM = 42/2 = 21 In right triangle Ξ OMA cos O = (ππππ ππππππππ‘ π‘π πππππ π)/π»π¦πππ‘πππ’π π cos 30Β° = ππ/π΄π β3/2=ππ/21 OM = β3/2 Γ 42 = 21β3 From (1) AB = 2AM Putting value of AM AB = 2 Γ 21 AB = 42 cm Now, Area of Ξ AOB = 1/2 Γ Base Γ Height = 1/2 Γ AB Γ OM = 1/2 Γ 42 Γ 21β3 = 441β3 cm2 Now, Area of segment APB = Area of sector OAPB β Area of ΞOAB = (924 β 441β3) cm2
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