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  1. Chapter 13 Class 9 Surface Areas and Volumes
  2. Serial order wise

Transcript

Ex 13.9, 3 (Optional) 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 = ๐‘‘_0/2 Now, Original C.S.A of sphere (Ao) = 4๐œ‹r02 = 4๐œ‹ (๐‘‘_0/2)^2 = 4๐œ‹ ร— (๐‘‘_0 )^2/4 = ๐œ‹ do2 Now, Diameter of sphere is decreased by 25% Diameter of sphere after decrease= do โˆ’ 25/100 do d = (1 โˆ’25/100) do = ((100 โˆ’ 25)/100) do = 75/100 do = 3/4 do Radius of sphere after decrease = ๐‘‘/2 = 1/2 (3/4 ๐‘‘_0 ) = 3/8 do CSA of sphere after decrease (A) = 4๐œ‹r2 A = 4 ร— ๐œ‹ ร— (3/8 ๐‘‘_0 )^2 A = 4 ร— ๐œ‹ ร— 9/64 do2 A = 9/16 ๐œ‹ do2 Now, Percentage of Area decreased = (๐ท๐‘’๐‘๐‘Ÿ๐‘’๐‘Ž๐‘ ๐‘’ ๐‘–๐‘› ๐ด๐‘Ÿ๐‘’๐‘Ž)/(๐‘‚๐‘Ÿ๐‘–๐‘”๐‘–๐‘›๐‘Ž๐‘™ ๐ด๐‘Ÿ๐‘’๐‘Ž) ร— 100 = (๐ด_0 โˆ’ ๐ด)/๐ด_0 ร— 100% = (๐œ‹ใ€–๐‘‘0ใ€—^2 โˆ’ 9/16 ๐œ‹ใ€–๐‘‘0ใ€—^2)/(๐œ‹ใ€–๐‘‘0ใ€—^2 ) ร— 100% = (๐œ‹ใ€–๐‘‘0ใ€—^2 (1 โˆ’ 9/16))/(๐œ‹ใ€–๐‘‘0ใ€—^2 ) ร— 100% = (1โˆ’9/16)ร—100 = (16 โˆ’ 9)/16 ร— 100% = 7/16 ร— 100 % = 7/4 ร— 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.