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

AcademicMathematicsNCERTClass 9

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Given:

The diameter of a sphere is decreased by $25 \%$.

To do:

We have to find the per cent by which the curved surface area decreases.

Solution:

Let $d$ be the diameter of the sphere initially.

This implies,

Radius of the sphere initially $r=\frac{d}{2}$

Surface area of the sphere initially $=4 \pi(\frac{d}{2})^{2}$

$=4\times \pi \times \frac{d^{2}}{4}$

$=\pi d^2$

On decreasing the diameter by $25 \%$, the new diameter formed $d_{1}=d-\frac{25 d}{100}$

$=\frac{75}{100} \times d$

$=\frac{3 d}{4}$

The surface area of the new sphere $=4 \pi(\frac{d_{1}}{2})^{2}$

$=4 \pi(\frac{1}{2} \times \frac{3 d}{4})^{2}$

$=4 \pi \frac{9 d^{2}}{64}$

$=\frac{\pi d^{2} 9}{16}$

Decrease in the surface area of the sphere $=\pi d^{2}(1-\frac{9}{16})$

$=\pi d^{2}(\frac{7}{16})$

Therefore,

Percentage decrease in curved surface area $=\frac{\pi d^{2}(\frac{7}{16})}{\pi d^{2}} \times 100 \%$

$=\frac{700}{16} \%$

$=43.75 \%$

The curved surface area decreases by $43.75 \%$.

Updated on 10-Oct-2022 13:46:41

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