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Last updated at Feb. 4, 2019 by Teachoo
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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) = 4๐r02 = 4๐ (๐_0/2)^2 = 4๐ ร (๐_0 )^2/4 = ๐ do2 Now, Diameter of sphere is decreased by 25% Diameter of sphere after decrease= do โ 25/100 do d = (1 โ25/100) do = ((100 โ 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) = 4๐r2 A = 4 ร ๐ ร (3/8 ๐_0 )^2 A = 4 ร ๐ ร 9/64 do2 A = 9/16 ๐ do2 Now, Percentage of Area decreased = (๐ท๐๐๐๐๐๐ ๐ ๐๐ ๐ด๐๐๐)/(๐๐๐๐๐๐๐๐ ๐ด๐๐๐) ร 100 = (๐ด_0 โ ๐ด)/๐ด_0 ร 100% = (๐ใ๐0ใ^2 โ 9/16 ๐ใ๐0ใ^2)/(๐ใ๐0ใ^2 ) ร 100% = (๐ใ๐0ใ^2 (1 โ 9/16))/(๐ใ๐0ใ^2 ) ร 100% = (1โ9/16)ร100 = (16 โ 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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