IF h, C , V are height , curved surface and volume of a cone respectively,
then prove that 3 π V h² – C² h²+ 9 V² = 0
Hi snighda,
this is just a practice question for cones.
Use $\displaystyle V=\frac{1}{3}{\pi}r^2h$ for cone volume
$\displaystyle C={\pi}r\sqrt{r^2+h^2}$ for cone curved surface area.
Therefore $\displaystyle V^2=\frac{1}{9}{\pi^2}r^4h^2$
$\displaystyle C^2={\pi^2}r^2\left(r^2+h^2\right)$
Then
$\displaystyle 3{\pi}Vh^2={\pi}^2r^2h^3$
$\displaystyle -C^2h^2=-{\pi}^2r^2\left(r^2+h^2\right)h^2$
$\displaystyle 9V^2={\pi}^2r^4h^2$
If these are added
$\displaystyle {\pi^2}r^2h^2\left(h-\left(h^2+r^2\right)+r^2\right)$
we get zero only if h=1.
Therefore should your equation have $\displaystyle 3{\pi}Vh^3$ instead of $\displaystyle 3{\pi}Vh^2$ ?