Find the area of the major segment (in terms of 𝜋) of a circle of

Question 24 (B) Find the area of the major segment (in terms of 𝜋) of a circle of radius 5 cm, formed by a chord subtending an angle of 90˚ at the centre.Given that OA = OB = radius = 5 cm θ=90° First lets’ find Area of minor Segment Area of minor segment = Area of sector OAPB – Area of ΔAOB Area of sector OAPB Area of sector OAPB = θ/(360°)× πr2 = 𝟗𝟎/𝟑𝟔𝟎 × 𝝅 × (𝟓)𝟐 = 1/4 × 𝜋 × 25 = 𝟐𝟓𝝅/𝟒 cm2 Area of ΔAOB Now, ΔAOB is a right triangle, where ∠ O = 90° having Base = OA & Height = OB Area of Δ AOB = 1/2 × Base × Height = 𝟏/𝟐 × OA × OB = 1/2 × 5 × 5 = 𝟐𝟓/𝟐 cm2 Now, Area of minor segment = Area of sector OAPB – Area of ΔAOB = (𝟐𝟓𝝅/𝟒−𝟐𝟓/𝟐) Area of major segment Area of major segment = Area of circle – Area of minor segment = 𝝅𝒓^𝟐− (𝟐𝟓𝝅/𝟒−𝟐𝟓/𝟐) = 𝜋 × 5^2− (𝟐𝟓𝝅/𝟒−𝟐𝟓/𝟐) = 25𝜋− (𝟐𝟓𝝅/𝟒−𝟐𝟓/𝟐) = 𝟐𝟓𝝅−𝟐𝟓𝝅/𝟒+𝟐𝟓/𝟐 = 25𝜋(1−𝟏/𝟒)+𝟐𝟓/𝟐 = 25𝜋 ×3/4+𝟐𝟓/𝟐 = 𝟕𝟓𝝅/𝟒+𝟐𝟓/𝟐

Question 24 (B) - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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