Ex.13.9 Q3 Surface Areas and Volumes Solution - NCERT Maths Class 9

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Question

 

Q3. The diameter of a sphere is decreased by \(25\%\). By what percent does its curved surface area decrease?

Text Solution

Reasoning:

Curved surface area of a sphere is \( 4^2 \). So, if diameter is decreased by \(25\%\), then radius also decreases by \(25\%\) so, the percentage change in curved surface area will be the ratio of difference between old area and new area by old area multiplied by \(100.\)

What is  known?

Decreased percentage of diameter.

What is  unknown?

Decreased percentage of curved surface area.

Steps:

Let the radius of the sphere be\(\begin{align}\,\frac{r}{2} \end{align}\) \(cm.\)

Then it’s diameter\(\begin{align} = 2(\frac{r}{2}) \end{align}\)\(= r \rm\,cm\)

Curved surface area of the original sphere:

\[\begin{align} = 4\pi {\left( {\frac{r}{2}} \right)^2} = \pi {r^2}c{m^2} \end{align}\]

New diameter of the sphere:

\[\begin{align}r &= r - r \times \frac{{25}}{{100}}[\text{diameter of a sphere is decreased by 25}\% ]\\&= \frac{{3r}}{4}cm \end{align}\]

Radius of the new sphere:

\[\begin{align} = \frac{1}{2}(\frac{{3r}}{4})= \frac{3}{8r} cm\end{align}\]

New curved surface area of the sphere.

\[\begin{align} =4 \pi\left(\frac{3 r}{8}\right)^{2}=\frac{9 \pi r^{2}}{16} \mathrm{cm}^{2}\end{align}\]

Decrease in the original curved Surface area \(\begin{align} = \pi {r^2} - \frac{{9\pi {r^2}}}{{16}} \end{align}\)

\[\begin{align} &= \frac{{16\pi {r^2} - 9\pi {r^2}}}{{16}}\\ &= \frac{{7\pi {r^2}}}{{16}} \end{align}\]

Percentage of decrease in the original curved surface area

\[\begin{align} &= \frac{{7\pi {r^2}}}{{16}} \times 100 \\ &= 43.75 \% \end{align}\]

Answer:

Hence the original curved surface area decrease by \(= 43.75\%\)

  
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