This question is similar to Chapter 11 Class 10 Areas related to Circles - Ex 11.1

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https://www.teachoo.com/1854/547/Ex-12.2--5---In-a-circle-of-radius-21-cm--an-arc-subtends-60/category/Ex-12.2/

 

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Question 30 (B) AB is a chord of a circle centred at O such that ∠AOB=60˚. If OA = 14 cm then find the area of the minor segment. (take √3 = 1.73)Area of segment APB = Area of sector OAPB – Area of ΔOAB Now, Area of sector OAPB = 𝜃/360×𝜋𝑟2 = 𝟔𝟎/𝟑𝟔𝟎 × 𝟐𝟐/𝟕 × 𝟏𝟒 × 𝟏𝟒 = 1/6 × 22/7 × 14 × 14 = 1/6 × 22 × 2 × 14 = 102.67 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/14 14/2 = AM AM = 7 In right triangle Δ OMA cos O = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos 𝟑𝟎° = 𝑶𝑴/𝑨𝑶 √3/2=𝑂𝑀/14 √3/2 × 14 = OM OM = 𝟕√𝟑 From (1) AM = 𝟏/𝟐AB 2AM = AB AB = 2AM Putting value of AM AB = 2 × 1/2 × 14 AB = 14 cm Now, Area of Δ AOB = 1/2 × Base × Height = 𝟏/𝟐 × AB × OM = 1/2 × 14 × 7√3 = 𝟒𝟗√𝟑 cm2 = 49 × 1.73 = 84.77 cm2 Therefore, Area of segment APB = Area of sector OAPB – Area of ΔOAB = 102.67 − 84.77 = 17.9 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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.