SOLUTION: A metal cube of side l (this is the letter L lower case not no.1) is heated and each side increases by 1%. To the nearest percent find the increase in surface area & volume.
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Question 733250: A metal cube of side l (this is the letter L lower case not no.1) is heated and each side increases by 1%. To the nearest percent find the increase in surface area & volume.
Please show workings down page if you can. It makes more sense to me. Thank you. Answer by sachi(548) (Show Source):
You can put this solution on YOUR website! A metal cube of side l is heated and each side increases by 1%
so the original side= l & after heating side=1.01l
so the original surface area=6l^2 & after heating surface area=6x(1.01)^2xl^2
increase in surface area=6x(1.01)^2xl^2-6l^2=6l^2x(1.0201-1)=0.0201*6l^2
% increase in surface area=6l^2x(0.0201)*100/6l^2=2.01 %
now the original volume=l^3 & after heating volume=(1.01l)^3=1.030301l^3
so the increase in volume=1.030301l^3-l^3=0.030301*l^3
% increase in volume=0.030301*l^3*100/l^3=3.0301 %
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