a right circular cylinder and a …
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Sia ? 6 years, 4 months ago
A cylinder and a cone have equal heights and equal radii of their bases.
So, As per given condition
{tex}\frac{Curved \ surface \ area \ of \ cylinder}{Curved \ surface \ area \ of \ cone}{/tex} = {tex}\frac{2 \pi rh}{\pi rl}{/tex} = {tex}\frac{2 \pi rh}{\pi r \sqrt{r^{2}+h^{2}}}{/tex}
{tex}\frac{8}{5}{/tex} = {tex}\frac{2 h}{\sqrt{r^{2}+h^{2}}}{/tex}
{tex}\Rightarrow{/tex}{tex}\frac{64}{25}{/tex} = {tex}\frac{4h^2}{r^2 + h^2}{/tex}
{tex}\Rightarrow{/tex}64r2 + 64h2 = 100h2
{tex}\Rightarrow{/tex} 64r2 = 100h2 - 64h2
{tex}\Rightarrow{/tex}64r2 = 36h2
{tex}\Rightarrow{/tex}{tex}\frac{r^2}{h^2}{/tex} = {tex}\frac{36}{64}{/tex} = {tex}\frac{9}{16}{/tex}
{tex}\Rightarrow{/tex}{tex}\frac{r}{h}{/tex} = {tex}\frac{3}{4}{/tex}
{tex}\therefore{/tex}r : h = 3 : 4
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