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प्रश्न
The radius of a circle with centre O is 5 cm (Figure). Two radii OA and OB are drawn at right angles to each other. Find the areas of the segments made by the chord AB (Take π = 3.14).

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उत्तर
Given:
Radius OA = OB = 5 cm, OA ⟂ OB ⇒ central angle ∠AOB = 90°.
Take π = 3.14. Method: minor segment area = area of sector AOB – area of ΔAOB; major segment area = area of circle – minor segment.
Step-wise calculation:
1. Area of circle = πr2
= 3.14 × 25
= 78.50 cm2
2. Area of sector AOB (θ = 90°)
= `(θ/360) xx πr^2`
= `(90/360) xx 78.50`
= `1/4 xx 78.50`
= 19.625 cm2
3. Area of ΔAOB (two radii with included angle 90°)
= `(1/2)·OA·OB·sin90^circ`
= `(1/2)·5·5·1`
= 12.50 cm2
4. Area of minor segment (closer to the chord AB)
= Sector – Triangle
= 19.625 – 12.50
= 7.125 cm2
5. Area of major segment
= Area of circle – Minor segment
= 78.50 – 7.125
= 71.375 cm2
Minor segment area = 7.125 cm2 (≈ 7.13 cm2).
Major segment area = 71.375 cm2 (≈ 71.38 cm2).
