the diameter of a sphere is decreased by 25% by what percent its curved surface ...

September 2021 1 41 Report

the diameter of a sphere is decreased by 25% by what percent its curved surface area decrease

Let r be the radius and S be the curved surface area. Then,

It is given that the diameter decreases by 25%. Therefore,

$\mathfrak{\underline{\underline{Decrease\:in\: diameter:-}}}$

$\tt = 25\% \: of \: 2r = ( \frac{25}{100} \times 2r) = \frac{r}{2} \\$

$\therefore$ $\underline\mathfrak{Decrease\:Diameter}$

$= \tt 2r - \frac{r}{2} = \frac{3r}{2} \\$

$\implies$ Decreased Radius

$\sf = \frac{3r}{4} \\$

↝Let $\sf{s}_{1}$ be the new curved surface area. Then,

$\tt \: s_1 = 4\pi \times ( \frac{3r}{4}) {}^{2} = \frac{9\pi {r}^{2} }{4} \\$

$\therefore$ Decrease in curved surface area.

$\tt \: s - s_1 = 4\pi {r}^{2} - \frac{9\pi {r}^{2} }{4} \\ \\ \tt = \frac{7\pi {r}^{2} }{4}$

• So, percentage decrease in curved surface area is equal to

$\tt \frac{s - s_1}{s} \times 100\% \\ \\ \tt = \frac { 7\pi {r}^{2} }{4 \div 4\pi {r}^{2} } \times 100\% = \frac{700}{16} \% \\ \\ \tt = 43.75\%$