# 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}

\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}

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