Question

If h, C, V are respectively the height, the curved surface and volume of a cone, prove that $3\pi V{h}^3-C^2{h}^2+9V^2=0$. [4 MARKS]

Answer 1

Concept: 1 Mark
Formula: 0.5 Mark each
Application: 1.5 Marks

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

$l^2=r^2+h^2,V=\frac{1}{3}\pi r^{2}h,C=\pi rl$

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

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

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

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

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

$=0$