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The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease? |
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Answer» `d_2=d_1-1/4d_1` `d_2=3/4d_1` =`(d_2-d_1)/d_1*100` =`(4pi(r_2^2-r_1^2))/(4pir_1^2)*100` =`(((3d_1)/4)-d_1^2)/d_1^2 *100` =43.75. |
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