A right circular cylinder and a right circular cone have equal

Solution:

For right circular cylinder, let r1 = r, h1 = h.

Then, curved surface area, $s_{1}$ of cylinder $=2 \pi r_{1} h_{1}=2 \pi r h$ (i)

For right circular cone, let $r_{2}=r, h_{2}=h$

Then, curved surface area, $s_{2}$ of cone $=\pi r_{2} l$ where $l=\sqrt{r_{2}{ }^{2}+h^{2}}=\sqrt{r^{2}+h^{2}}$

$=\pi r \sqrt{r^{2}+h^{2}} \quad \ldots \ldots .(i i)$

Divide (i) and (ii),

$\frac{s_{1}}{s_{2}}=\frac{2 \pi r h}{\pi r \sqrt{r^{2}+h^{2}}}$

$\frac{8}{5}=\frac{2 h}{\sqrt{r^{2}+h^{2}}}\left[\frac{s_{1}}{s_{2}}=\frac{8}{5}\right]$

$\frac{64}{25}=\frac{4 h^{2}}{r^{2}+h^{2}} \quad$ [Squaring]

$64 r^{2}+64 h^{2}=100 h^{2}$

$64 r^{2}=36 h^{2}$

$16 r^{2}=9 h^{2}$

\frac{r^{2}}{h^{2}}=\frac{9}{16}

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

$\therefore r: h=3: 4$