Question

If $h$, $s$, $V$ be the height, curved surface area and volume of a cone respectively, then $\left(3\pi V{h}^3+9V^2-s^2{h}^2\right)=?$

Answer 1

$3\pi v{h}^3+9v^2-5^2{h}^2$

$3\pi \left(\frac{1}{3}\pi r^{2}h\right)x^3+9{\left(\frac{1}{3}\pi r^{2}h\right)}^2-{\left(\pi rl\right)}^2{h}^2$

${\pi }^2{r}^2{h}^4+{\pi }^2{r}^4{h}^2-{\pi }^2{r}^2{l}^2{h}^2$

${\pi }^2{r}^2{h}^4+{\pi }^2{r}^4{h}^2-{\pi }^2\left(h^2+r^2\right)h^2$

${\pi }^2{r}^2{h}^4+{\pi }^2{h}^2{r}^4-{\pi }^2{r}^2{h}^4-{\pi }^2{r}^4{h}^2$\

$=0$