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Ex 13.9, 3 (Optional) - Chapter 13 Class 9 Surface Areas and Volumes - Part 2
Ex 13.9, 3 (Optional) - Chapter 13 Class 9 Surface Areas and Volumes - Part 3
Ex 13.9, 3 (Optional) - Chapter 13 Class 9 Surface Areas and Volumes - Part 4
Ex 13.9, 3 (Optional) - Chapter 13 Class 9 Surface Areas and Volumes - Part 5

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Transcript

Question 3 The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease? Let Original diameter of sphere = do Original radius of sphere = ro = 𝑑_0/2 Now, Original C.S.A of sphere (Ao) = 4𝜋r02 = 4𝜋 (𝑑_0/2)^2 = 4𝜋 × (𝑑_0 )^2/4 = 𝜋 do2 Now, Diameter of sphere is decreased by 25% Diameter of sphere after decrease= do − 25/100 do d = (1 −25/100) do = ((100 − 25)/100) do = 75/100 do = 3/4 do Radius of sphere after decrease = 𝑑/2 = 1/2 (3/4 𝑑_0 ) = 3/8 do CSA of sphere after decrease (A) = 4𝜋r2 A = 4 × 𝜋 × (3/8 𝑑_0 )^2 A = 4 × 𝜋 × 9/64 do2 A = 9/16 𝜋 do2 Now, Percentage of Area decreased = (𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝐴𝑟𝑒𝑎)/(𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐴𝑟𝑒𝑎) × 100 = (𝐴_0 − 𝐴)/𝐴_0 × 100% = (𝜋〖𝑑0〗^2 − 9/16 𝜋〖𝑑0〗^2)/(𝜋〖𝑑0〗^2 ) × 100% = (𝜋〖𝑑0〗^2 (1 − 9/16))/(𝜋〖𝑑0〗^2 ) × 100% = (1−9/16)×100 = (16 − 9)/16 × 100% = 7/16 × 100 % = 7/4 × 25% = 175/4 % = 43.75% 43.75 4 175 16 15 12 30 28 20 20 0

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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.