# Q : 3    The diameter of a sphere is decreased by $\small 25\%$. By what per cent does its curved surface area decrease?

H Harsh Kankaria

Let the radius of the sphere be $r$

Diameter of the sphere = $2r$

According to question,

Diameter is decreased by $\small 25\%$

So, the new diameter = $\frac{3}{4}\times2r = \frac{3r}{2}$

So, the new radius = $r' = \frac{3r}{4}$

$\therefore$New surface area = $4\pi r'^2 = 4\pi (\frac{3r}{4})^2$

$\therefore$Decrease in surface area = $4\pi r^2 - 4\pi (\frac{3r}{4})^2$

$= 4\pi r^2[1 -\frac{9}{16} ]$

$= 4\pi r^2[\frac{7}{16} ]$

$\therefore$ Percentage decrease in the surface area = $\frac{Difference\ in\ areas}{Original\ surface\ area}$

$= \frac{4\pi r^2[\frac{7}{16}]}{4\pi r^2}\times100 \%$

$= \frac{7}{16}\times100 \%$

$= 43.75 \%$

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