Ex 12.2, 6
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)
In a given circle,
Radius (r) = 15 cm
And, 𝜃 = 60°
Area of segment APB = Area of sector OAPB – Area of ΔOAB
Area of sector OAYB = 𝜃/360×𝜋𝑟2
= 60/360×3.14×(15)2
= 1/6×3.14×15×15
= 1/2×3.14×5×15
= 117.75 cm2
Finding area of Δ AOB
Area Δ AOB = 1/2 × Base × Height
We draw OM ⊥ AB
∴ ∠ OMB = ∠ OMA = 90°
In Δ OMA & Δ OMB
∠ OMA = ∠ OMB
OA = OB
OM = OM
∴ Δ OMA ≅ Δ OMB
⇒ ∠ AOM = ∠ BOM
∴ ∠ AOM = ∠ BOM = 1/2 ∠ BOA
⇒ ∠ AOM = ∠ BOM = 1/2 × 60° = 30°
Also, since Δ OMB ≅ Δ OMA
∴ BM = AM
⇒ BM = AM = 1/2 AB
From (1)
AM = 1/2AB
2AM = AB
AB = 2AM
Putting value of AM
AB = 2 × 15/2
AB = 15
Now,
Area of Δ AOB = 1/2 × Base × Height
= 1/2 × AB × OM
= 1/2 × 15 × √3/2 × 15
= 1/2 × 15 × 1.73/2 × 15
= 97.3125 cm2
Area of segment APB
= Area of sector OAPB – Area of ΔOAB
= 117.75 – 97.3125
= 20.4375 cm2
Thus, area of minor segment = 20.4375 cm2
Now,
Area of major segment = Area of circle – area of minor segment
= πr2−20.4375
= 3.14 ×152− 20.4375
= 3.14×15 ×15−"20.4375"
= 706.5 – 20.4375
= 686.0625 cm2

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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