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


Transcript

Ex 11.1, 4 A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : minor segment (Use π= 3.14) Given that OA = OB = radius = 10 cm θ=90° Now, Area of segment APB = Area of sector OAPB – Area of ΔAOB Area of sector OAPB Area of sector OAPB = θ/(360°)× πr2 = 90/360 × 3.14 × (10)2 = 𝟏/𝟒 × 𝟑.𝟏𝟒 × 𝟏𝟎𝟎 = 1/4× 314 = 78.5 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 × 10 × 10 = 50 cm2 Now, Area of segment APB = Area of sector OAPB – Area of ΔAOB = 78.5 – 50 = 28.5 cm2 Ex 11.1, 4 (Method 1) A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (ii) major sector (Use π= 3.14) Here, θ = 90° & radius = r = 10 cm Area of major sector = ((𝟑𝟔𝟎° − 𝜽))/(𝟑𝟔𝟎°)×𝛑𝐫𝟐 = ((360 − 90))/360 × 3.14 × (10)2 = 𝟐𝟕𝟎/𝟑𝟔𝟎 × 𝟑.𝟏𝟒 × 𝟏𝟎𝟎 = 3/4× 314 = 235.5 cm2 Ex 11.1, 4(Method 2) A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (ii) major sector (Use π= 3.14) Area of major sector = Area of circle – Area of sector OAPB Area of circle = πr2 = 3.14 × (10)2 = 3.14 × 100 = 314 cm2 Now, Area of major sector = Area of circle – Area of sector OAPB = 314 – 78.5 = 235.5 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.