A right circular cylinder and a …

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Sia ? 6 years, 4 months ago
Let radius of cylinder = Radius of cone = r cm
And, height of cylinder = Height of cone = h cm
Given, {tex}\frac { \text { Curved surface area of cylinder } } { \text { Curved surface area of cone } } = \frac { 8 } { 5 }{/tex}
{tex}\Rightarrow \quad \frac { 2 \pi r h } { \pi rl } = \frac { 8 } { 5 }{/tex}
{tex}\Rightarrow \quad \frac { 2 h } { l} = \frac { 8 } { 5 }{/tex}
{tex}\Rightarrow \quad \frac {h } {l } = \frac { 4 } { 5 }{/tex}
{tex}\Rightarrow \quad \frac { h ^ { 2 } } { l ^ { 2 } } = \frac { 16 } { 25 }{/tex}
{tex}\Rightarrow \quad \frac { h ^ { 2 } } { r ^ { 2 } + h ^ { 2 } } = \frac { 16 } { 25 }{/tex} [for a cone l2 = r2 + h2]
{tex}\Rightarrow{/tex} 25h2 = 16r2 + 16h2
{tex}\Rightarrow{/tex} 9h2 = 16r2
{tex}\Rightarrow \quad \frac { 9 } { 16 } = \frac { r ^ { 2 } } { h ^ { 2 } }{/tex}
{tex}\Rightarrow \quad \frac { r } { h } = \frac { 3 } { 4 }{/tex}
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