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

Answers (1)
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}

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

\thereforeDecrease 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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