The radius of a sphere is decreased by 20 % . | vedclass

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(C) Let the initial radius of the sphere be $r_1 = r$.The initial surface area is $A_1 = 4 \pi r^2$.The radius is decreased by $20 \%$,so the new radi

Vedclass · Accepted Answer

$36$

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(C) Let the initial radius of the sphere be $r_1 = r$.The initial surface area is $A_1 = 4 \pi r^2$.The radius is decreased by $20 \%$,so the new radius $r_2 = r - 0.20r = 0.8r$.The new surface area is $A_2 = 4 \pi (0.8r)^2 = 4 \pi (0.64r^2) = 0.64(4 \pi r^2) = 0.64 A_1$.The decrease in surface area is $A_1 - A_2 = A_1 - 0.64 A_1 = 0.36 A_1$.The percentage decrease is $\frac{0.36 A_1}{A_1} 	imes 100 \% = 36 \%$.

The radius of a sphere is decreased by $20 \% .$ By what per cent does its curved surface area decrease (in $\%$)?

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