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Probability

Question
CBSEENMA9003767
Wired Faculty App

The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?

Solution
Let the radius of the sphere be straight r over 2 cm.
This its diameter = 2open parentheses straight r over 2 close parentheses equals space straight r space cm
Curved surface area of the original sphere = 4 straight pi open parentheses straight r over 2 close parentheses squared equals πr squared space space cm squared
New diameter (decreased) of the sphere 
equals space straight r minus straight r cross times 25 over 100 equals straight r minus straight r over 4 equals fraction numerator 3 straight r over denominator 4 end fraction space cm
∴ Radius of the new sphere
equals 1 half open parentheses fraction numerator 3 straight r over denominator 4 end fraction close parentheses equals fraction numerator 3 straight r over denominator 8 end fraction space cm
∴ New curved surface area of the sphere
equals 4 straight pi open parentheses fraction numerator 3 straight r over denominator 8 end fraction close parentheses squared equals fraction numerator 9 πr squared over denominator 16 end fraction space cm squared
∴ Decrease in the original curved surface area
equals πr squared minus fraction numerator 9 πr squared over denominator 16 end fraction equals space fraction numerator 16 πr squared minus 9 πr squared over denominator 16 end fraction equals fraction numerator 7 πr squared over denominator 16 end fraction
∴ Percentage of decrease in the original curved surface area
equals fraction numerator begin display style fraction numerator 7 πr squared over denominator 16 end fraction end style over denominator πr squared end fraction cross times 100 percent sign equals 700 over 16 percent sign equals 175 over 4 percent sign equals 43 3 over 4 percent sign space or space 43.75 percent sign
Hence, the original curved surface area decreases by 43.75%.

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