Question

If h, s, V 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

As real surface area of cone

$=s=\pi rl=\pi r\sqrt{h^2+r^2}$$l$= slant height

Volume of cone $=V=\frac{1}{3}\pi r^{2}h$

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

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

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

$=0$