Question

The height, curved surface area and volume of a cone are h,c, and V respectively. Prove that $3\pi {Vh}^3-c^2{h}^2+9V^2=0$

Answer 1

Curved surface area of cone $=\pi rl=c\left(\text{Given}\right)$

Volume of cone $=\frac{1}{3}\pi r^{2}h=V\left(\text{Given}\right)$

Height of cone $=h$

Now,

$3\pi V{h}^3-c^2{h}^2+9V^2=0$

Taking L.H.S.-

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

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

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

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

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

$=0$

$=$ R.H.S.

Hence proved.