Exercise Set 6.3
Last updated at June 3, 2026 by Teachoo
Transcript
Ex 6.4, 5 A chord of a circle of radius 15 cm subtends an angle of 60^β at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use πβ3.14 and β3β1.73.) Let the minor segment be APB Given that Radius = r = 15 cm π = 60Β° Now, Area of segment APB = Area of sector OAPB β Area of ΞOAB Letβs find separately Area of sector OAPB Area of sector OAPB = π/360Γππ2 = ππ/πππ Γ π.ππ Γ ππ Γ ππ = 1/6 Γ 3.14 Γ 15 Γ 15 = 1/2 Γ 3.14 Γ 5 Γ 15 = 1.57 Γ 5 Γ 15 = 117.75 cm2 Area of Ξ AOB In β AOB, OA = OB = 15 cm β AOB = 60Β° We know that Angles opposite equal side are equal β΄ β OAB = β OBA By Angle sum property β OAB + β AOB + β OBA = 180Β° β OAB + 60Β° + β OBA = 180Β° Putting β OAB = β OBA β OAB + 60Β° + β OAB = 180Β° 2 Γ β OAB + 60Β° = 180Β° 2 Γ β OAB = 180Β° β 60Β° 2 Γ β OAB = 120Β° β OAB = (120Β°)/2 β OAB = 60Β° Thus, β OBA = β OAB = 60Β° So, all 3 angles are 60Β° Which means β OAB is an equilateral triangle We need to find area of equilateral triangle Area of equilateral triangle = βπ/π Γ πΊππ π^π Putting = = βπ/π Γ πΊππ π^π Area Ξ AOB = 1/2 Γ Base Γ Height 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 2 Γ β OAB + 60Β° = 180Β° 2 Γ β OAB = 180Β° β 60Β° 2 Γ β OAB = 120Β° β OAB = (120Β°)/2 β OAB = 60Β° Thus, β OBA = β OAB = 60Β° So, all 3 angles are 60Β° Which means β OAB is an equilateral triangle We need to find area of equilateral triangle Area of equilateral triangle = βπ/π Γ πΊππ π^π Putting = = βπ/π Γ πΊππ π^π Area Ξ AOB = 1/2 Γ Base Γ Height 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 Area of equilateral triangle = βπ/π Γ πΊππ π^π Putting β3 = 1.73, Side = 15 = (π.ππ)/π Γ ππ^π = 1.73/4 Γ 225 = 1.73/4 Γ 15^2 = 389.25/4 = 389.25/4 = 97.3125 cm2 Thus, Area of segment APB = Area of sector OAPB β Area of ΞOAB = 117.75 β 97.3125 = 20.4375 cm2 Thus, Area of minor segment = 20.4375 cm2 And, Area of major segment = Area of circle β Area of minor segment = ππ^2 β 20.4375 = 3.14 Γ 15^2 β 20.4375 = π.ππ Γ πππ β 20.4375 = 706.5 β 20.4375 = πππ.ππππ cm2