The diameter of a sphere is decreased by 25%. | vedclass

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(A) Let the initial diameter of the sphere be $D_1$. Then the initial radius $r_1 = D_1 / 2$.The initial curved surface area $A_1 = 4 \pi r_1^2$.When

Vedclass · Accepted Answer

$43.75$

Vedclass · Answer

(A) Let the initial diameter of the sphere be $D_1$. Then the initial radius $r_1 = D_1 / 2$.The initial curved surface area $A_1 = 4 \pi r_1^2$.When the diameter is decreased by $25\%$,the new diameter $D_2 = D_1 - 0.25 D_1 = 0.75 D_1$.The new radius $r_2 = D_2 / 2 = 0.75 (D_1 / 2) = 0.75 r_1$.The new curved surface area $A_2 = 4 \pi r_2^2 = 4 \pi (0.75 r_1)^2 = 4 \pi (0.5625 r_1^2) = 0.5625 A_1$.The decrease in surface area is $A_1 - A_2 = A_1 - 0.5625 A_1 = 0.4375 A_1$.The percentage decrease is $((A_1 - A_2) / A_1) 	imes 100\% = 0.4375 	imes 100\% = 43.75\%$.

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

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