Question

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

Answer 1

Step 1: Determine the new diameter.

The diameter of the sphere is reduced by $25\%$.

Find the new diameter:

$\begin{array}{rcl}d\text{'}& =& d-25\%d\\ & \Rightarrow & d\text{'}=d\left(1-\frac{25}{100}\right)\\ & \Rightarrow & d\text{'}=d\left(\frac{75}{100}\right)\\ & \Rightarrow & d\text{'}=\frac{3}{4}d\end{array}$

So, the new diameter is $\begin{array}{rcl}& & \frac{3}{4}d\end{array}$.

Step 2 : Find the new curved surface area.

The curved surface area of a sphere is $\pi d^2$.

Find the new curved surface area:

$\begin{array}{rcl}C\text{'}& =& \pi {\left(\frac{3}{4}d\right)}^2\\ & \Rightarrow & C\text{'}=\frac{9}{16}\pi d^2\end{array}$

So, the new curved surface area is $\begin{array}{rcl}& & \frac{9}{16}\pi d^2\end{array}$.

Step 3 : Determine the percentage of decrease.

Subtract new curved surface area from the initial area and then find the percentage of decrease:

$$\begin{gathered} \frac{\pi d^2-{\displaystyle \frac{9}{16}}\pi d^2}{\pi d^2}\times 100\%=\frac{{\displaystyle \frac{7}{16}}\pi d^2}{\pi d^2}\times 100\% \\ \Rightarrow \frac{\pi d^2-{\displaystyle \frac{9}{16}}\pi d^2}{\pi d^2}\times 100\%=\frac{7}{16}\times 100\% \\ \Rightarrow \frac{\pi d^2-{\displaystyle \frac{9}{16}}\pi d^2}{\pi d^2}\times 100\%=43.75\% \end{gathered}$$

Hence, the required percentage is $43.75\%$.