Question

If $$h, S, V$$ be the height, curved surface area and volume of a cone respectively, then $$(3\pi Vh^{3} - S^{2}h^{2} + 9V^{2})$$ is equal to:

Answer 1

Correct option is B. $$0$$
$$3\pi V{ h }^{ 2 }-{ S }^{ 2 }{ h }^{ 3 }+9{ V }^{ 2 }=3\pi \times \dfrac { 1 }{ 3 } \pi { r }^{ 2 }h\times { h }^{ 3 }-{ \left( \pi r\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \right) }^{ 2 }h^2+9\times { \left( \dfrac { 1 }{ 3 } \pi { r }^{ 2 }h \right) }^{ 2 }$$

$$={ \pi }^{ 2 }r^{ 2 }{ h }^{ 4 }-{ h }^{ 2 }\left( { \pi }^{ 2 }{ r }^{ 2 }\left( { r }^{ 2 }+{ h }^{ 2 } \right) \right) +9\times \dfrac { 1 }{ 9 } { \pi }^{ 2 }{ r }^{ 4 }{ h }^{ 2 }={ \pi }^{ 2 }{ r }^{ 2 }{ h }^{ 4 }-{ \pi }^{ 2 }{ r }^{ 4 }{ h }^{ 2 }-{ \pi }^{ 2 }{ r }^{ 2 }{ h }^{ 4 }+{ \pi }^{ 2 }{ r }^{ 4 }{ h }^{ 2 }$$

$$\therefore \quad 3\pi V{ h }^{ 3 }-{ S }^{ 2 }{ h }^{ 2 }+9{ V }^{ 2 }=0$$