Ex 11.1, 7 - Chapter 11 Class 10 Areas related to Circles

Last updated at April 16, 2024 by Teachoo

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 β3 = 1.73)
In a given circle,
Radius (r) = 12 cm
And, π½ = 120Β°
Now,
Area of segment APB = Area of sector OAPB β Area of ΞOAB
Finding Area of sector OAPB
Area of sector OAPB = π/360Γ ππ2
= 120/360Γ3.14Γ(12)2
= 1/3Γ3.14Γ12Γ12
= 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 β3 = 1.73)
In a given circle,
Radius (r) = 12 cm
And, π½ = 120Β°
Now,
Area of segment APB = Area of sector OAPB β Area of ΞOAB
Finding Area of sector OAPB
Area of sector OAPB = π/360Γ ππ2
= 120/360Γ3.14Γ(12)2
= 1/3Γ3.14Γ12Γ12
= 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 ππΒ° = ππ΄/π¨πΆ
β3/2=π΄π/12
β3/2 Γ 12 = AM
AM = 6βπ
In right triangle Ξ OMA
cos O = (π πππ ππππππππ‘ π‘π πππππ π)/π»π¦πππ‘πππ’π π
cos ππΒ° = πΆπ΄/π¨πΆ
1/2=ππ/12
12/2 = OM
OM = 6
From (1)
AM = π/πAB
2AM = AB
AB = 2AM
Putting value of AM
AB = 2 Γ 6 β3
AB = 12βπ cm
Now,
Area of Ξ AOB = 1/2 Γ Base Γ Height
= π/π Γ AB Γ OM
= 1/2 Γ 12β3 Γ 6
= 36β3
= 36 Γ 1.73
= 62.28 cm2
Therefore,
Area of segment APB
= Area of sector OAPB β Area of ΞOAB
= 150.72 β 62.28
= 88.44 cm2

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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