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.
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★Given :-
- A chord of a circle of radius = 15 cm
- Angle subtended at centre = 60°.
★To find :-
- The areas of the corresponding minor and major segments of the circle.
★Solution :-
Area of ΔAOB :
→1/2 r²sinθ
→1/2 ×(15)²×sin60°
[Sin 60° = √3/2 ]
→1/2×15×15×√3/2
→97.3 cm²
Area of sector OACBO :
→πr²×θ/360°
→ πr²×(60°/360°)
→3.14×15²×(60/360)
→117.75 cm²
Now,Area of the minor segment :
→Area of sector OACBO- Area of ΔAOB
→117.75 cm² - 97.3 cm²
→20.4 cm²
Angle made by Major sector = 360° - 60° = 300°
Area of the sector making angle 300° :
→(300°/360°) × π r² cm²
→(5/6) × 152 π cm²
→1125/6 π cm²
→(1125/6) × 3.14 cm²
→588.75 cm²
Area of major segment :
→588.75 cm ²+ 97.3 cm ²
→686.05 cm²
______________
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