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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 11.1, 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° Now, Area of segment APB = Area of sector OAPB – Area of ΔOAB Finding Area of sector OAPB Area of sector OAPB = 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 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/15 15/2 = AM AM = 𝟏𝟓/𝟐 In right triangle Δ OMA cos O = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos 𝟑𝟎° = 𝑶𝑴/𝑨𝑶 √3/2=𝑂𝑀/21 √3/2 × 15 = OM OM = √𝟑/𝟐 × 15 From (1) AM = 𝟏/𝟐AB 2AM = AB AB = 2AM Putting value of AM AB = 2 × 1/2 × 15 AB = 15 cm 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 Therefore, 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.