A cone of height 24 cm …
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Neeraj Sharma 7 years, 1 month ago
{tex}h = 24\,cm{/tex}
{tex}CSA = 550\,c{m^2}{/tex}
{tex}\pi rl = 550\,c{m^2}{/tex}
{tex}{{22} \over 7} \times r \times \sqrt {{h^2} + {r^2}} = 550{/tex}
{tex}r \times \sqrt {576 + {r^2}} = 175{/tex}
{tex}squaring\,on\,both\,sides{/tex}
{tex}{r^2}\left( {576 + {r^2}} \right) = 30625{/tex}
{tex}on\,solving{/tex}
{tex}r = \,7cm{/tex}
{tex}volume = {1 \over 3}\pi {r^2}h = {1 \over 3} \times {{22} \over 7} \times {\left( 7 \right)^2} \times 24{/tex}
{tex}volume = 1232\,c{m^3}{/tex}
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