Ex 11.1, 5 - Chapter 12 Class 10 Areas related to Circles (Important Question)

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Ex 11.1, 5 In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: the length of the arc Length of Arc APB = 𝜽/𝟑𝟔𝟎 × (𝟐𝝅𝒓) = (60°)/(360°) × 2 × 22/7 × 21 = 1/6 × 2 × 22 × 3 = 22 cm Ex 11.1, 5 In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (ii) area of the sector formed by the arc Area of sector OAPB = 𝜃/360×𝜋𝑟2 = 𝟔𝟎/𝟑𝟔𝟎 × 𝟐𝟐/𝟕 × 𝟐𝟏 × 𝟐𝟏 = 1/6 × 22/7 × 21 × 21 = 1/6 × 22 × 3 × 21 = 231 cm2 Ex 11.1, 5 In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (iii) area of segment formed by the corresponding chord Area of segment APB = Area of sector OAPB – Area of ΔOAB From last part, Area of sector OAPB = 231 cm2 Finding area of Δ AOB Area Δ AOB = 1/2 × Base × Height 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 𝟑𝟎° = 𝐀𝑴/𝑨𝑶 1/2=𝐴𝑀/21 21/2 = AM AM = 𝟐𝟏/𝟐 In right triangle Δ OMA cos O = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos 𝟑𝟎° = 𝑶𝑴/𝑨𝑶 √3/2=𝑂𝑀/21 √3/2 × 21 = OM OM = √𝟑/𝟐 × 21 From (1) AM = 𝟏/𝟐AB 2AM = AB AB = 2AM Putting value of AM AB = 2 × 1/2 × 21 AB = 21cm Now, Area of Δ AOB = 1/2 × Base × Height = 𝟏/𝟐 × AB × OM = 1/2 × 21 × √3/2 × 21 = (𝟒𝟒𝟏√𝟑)/𝟒 cm2 Therefore, Area of segment APB = Area of sector OAPB – Area of ΔOAB = (231 – 𝟒𝟒𝟏/𝟒 √𝟑) cm2