Question

if h, s, v be the height,csa and volume respectively of a cone,then(3πvhcube-Ssquarehsquare+9vsquare is equal to​

Answer 1

Answer:

Given

Height of cone = h

CSA of cone = s

Volume of Cone = v

Find

${ \sf(3\pi \: v {h}^{3} - {s}^{2} {h}^{2} + 9 {v}^{2} )}$

Solution

${ \sf{ \to{area \: of \: cone \: = \pi \: r \sqrt{ {h}^{2} + {r}^{2} } }}}$

${ \to{ \sf{volume \: \: of \: \: cone = \frac{\pi}{3} {r}^{2} h}}}$

${ \sf{ \to{3\pi \: v {h}^{3} + 9 {v}^{2} - {s}^{2} {h}^{2} }}}$

${ \to{ \sf{(3\pi \: \times \frac{\pi}{3} \times {r}^{2} h \times {h}^{3}) + 9( { \frac{\pi}{3}) }^{2} {r}^{4} {h}^{2} - {\pi}^{2} {r}^{2}( {h}^{2} + {r}^{2} ) \times {h}^{2} }}}$

${ \to{ \sf{ {\pi}^{2} {r}^{2} {h}^{2}( {h}^{2} + {r}^{2} - {h}^{2} - {r}^{2} )}}}$

${ \to{ \sf{0}}}$

${ \boxed{ \therefore{ \sf{3\pi \: v {h}^{3} + 9 {v}^{3} - {s}^{2} {h}^{2} = 0 }}}}$