Ex 11.1, 7
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
(Use π = 3.14 and √3 = 1.73)
In a given circle,
Radius (r) = 12 cm
And, 𝜽 = 120°
Now,
Area of segment APB = Area of sector OAPB – Area of ΔOAB
Finding Area of sector OAPB
Area of sector OAPB = 𝜃/360× 𝜋𝑟2
= 120/360×3.14×(12)2
= 1/3×3.14×12×12
= 1 × 3.14 × 4 × 12
= 150.72 cm2
Ex 11.1, 7
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
(Use π = 3.14 and √3 = 1.73)
In a given circle,
Radius (r) = 12 cm
And, 𝜽 = 120°
Now,
Area of segment APB = Area of sector OAPB – Area of ΔOAB
Finding Area of sector OAPB
Area of sector OAPB = 𝜃/360× 𝜋𝑟2
= 120/360×3.14×(12)2
= 1/3×3.14×12×12
= 1 × 3.14 × 4 × 12
= 150.72 cm2
Finding area of Δ AOB
We draw OM ⊥ AB
∴ ∠ OMB = ∠ OMA = 90°
And, by symmetry
M is the mid-point of AB
∴ BM = AM = 1/2 AB
In right triangle Δ OMA
sin O = (side opposite to angle O)/Hypotenuse
sin 𝟔𝟎° = 𝐀𝑴/𝑨𝑶
√3/2=𝐴𝑀/12
√3/2 × 12 = AM
AM = 6√𝟑
In right triangle Δ OMA
cos O = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝟔𝟎° = 𝑶𝑴/𝑨𝑶
1/2=𝑂𝑀/12
12/2 = OM
OM = 6
From (1)
AM = 𝟏/𝟐AB
2AM = AB
AB = 2AM
Putting value of AM
AB = 2 × 6 √3
AB = 12√𝟑 cm
Now,
Area of Δ AOB = 1/2 × Base × Height
= 𝟏/𝟐 × AB × OM
= 1/2 × 12√3 × 6
= 36√3
= 36 × 1.73
= 62.28 cm2
Therefore,
Area of segment APB
= Area of sector OAPB – Area of ΔOAB
= 150.72 – 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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