Question

If $h,c$ and $v$ be the height, curved surface and volume of a cone, show that $3\pi v{h}^3-c^2{h}^2+9v^2=0$.

Answer 1

Let r and l denote respectively the radius of the base and slant height of the cone.

$l=\sqrt{r^2+h^2}$

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

$C=\pi rl$

Therefore,

$3\pi V{h}^3-C^2{h}^2+9V^2$

$=3\pi \times \frac{1}{3}\pi r^{2}h\times h^3-(\pi rl{)}^2{h}^2+9\times (\frac{1}{3}\pi r^{2}h{)}^2$

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

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

$={\pi }^2{r}^2{h}^4-{\pi }^2{r}^4{h}^2-{\pi }^2{r}^2{h}^4+{\pi }^2{r}^4{h}^2=0$