CBSE Class 10 Sample Paper for 2019 Boards

Class 10
Solutions of Sample Papers for Class 10 Boards

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. Letβs first draw the figure We first have to find the angle We know that Length of arc = π/360 Γ 2Οr 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 ΞOAB Area of sector OAPB Area of sector OAPB = π/360Γππ2 = 60/360Γ22/7Γ42Γ42 = 1/6Γ22/7Γ42Γ42 = 1/6Γ22Γ6Γ42 = 22Γ42 = 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 = (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 30Β° = ππ/π΄π β3/2=ππ/21 OM = β3/2 Γ 42 = 21β3 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 21β3 = 441β3 cm2 Now, Area of segment APB = Area of sector OAPB β Area of ΞOAB = (924 β 441β3) cm2