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

## 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 \(= 43.75\%\)