The diameter of a sphere is decreased by 25percent.By what,t per cent

1.	The diameter of a sphere is decreased by 25percent.By what,t per cent does its curved surface area decrease?
Answer» $\large{\underbrace{\sf{\red{Answer:}}}}$ $\huge{\boxed{\sf{\orange{ 43 \dfrac{3}{4} \% \:or\: 43.75 \%}}}}$ $\large{\underbrace{\sf{\purple{Explanation:}}}}$ Let the radius of the sphere be $\displaystyle{\sf{ \dfrac{ \pi}{2}cm.}}$ Then its DIAMETER = $\implies{\sf{ \big( \dfrac{ \pi}{2} \big) cm.}}$ Curved SURFACE area of the ORIGINAL sphere ⠀⠀: $\implies{\sf{ 4 \pi \dfrac{r}{2}^{2} = \pi r^{2}cm^{2}}}$ New diameter(decreased) of the sphere ⠀: $\implies{\sf{ r-r \times \dfrac{25}{100}}}$ ⠀⠀: $\implies{\sf{ r- \dfrac{r}{4}= \dfrac{3r}{4}}}$ ${ \therefore}$ Radius of the new sphere : $\implies{\sf{ \dfrac{1}{2} \big( \dfrac{3r}{4} \big) = \frac{3r}{8}cm }}$ ${ \therefore}$ New curved surface area of the sphere : $\implies{\sf{ 4\pi \big( \frac{3r}{8}cm \big) = \dfrac{9\pi r^{2}}{16}cm^{2}}}$ ${ \therefore}$ Decrease in the original curved surface area : $\implies{\sf{ \pi r^{2} - \dfrac{9 \pi r^{2}}{16}}}$ : $\implies{\sf{ \pi r^{2} - \dfrac{7 \pi r^{2}}{16}}}$ ${ \therefore}$ Percentage of decrease in the original curved surface area : $\implies{\sf{ \dfrac{\dfrac{7 \pi r^{2}}{16}}{ \pi r^{2}} \times 100 \%}}$ $\large{:}\implies{\boxed{\sf{\pink{ 43 \dfrac{3}{4} \%\:or\:43.75 \%}}}}$ Hence, the original curved surface area decreases by 43.75%.