Question

If $h,S,V$ be the height, curved surface area and volume of a cone respectively, then $(3\pi V{h}^3-S^2{h}^2+9V^2)$ is equal to:

• A. $8$
• B. $0$
• C. $4\pi$
• D. $32{\pi }^2$

Answer 1

The correct option is B $0$

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

$={\pi }^2{r}^2{h}^4-h^2\left({\pi }^2{r}^2(r^2+h^2)\right)+9\times \frac{1}{9}{\pi }^2{r}^4{h}^2={\pi }^2{r}^2{h}^4-{\pi }^2{r}^4{h}^2-{\pi }^2{r}^2{h}^4+{\pi }^2{r}^4{h}^2$

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