cone/ surface area/volume..

**snigdha** · February 20th 2010, 08:07 AM

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

**Archie Meade** · February 20th 2010, 08:38 AM

Originally Posted by snigdha

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

$C={\pi}r\sqrt{r^2+h^2}$ for cone curved surface area.

Therefore $V^2=\frac{1}{9}{\pi^2}r^4h^2$

$C^2={\pi^2}r^2\left(r^2+h^2\right)$

Then

$3{\pi}Vh^2={\pi}^2r^2h^3$

$-C^2h^2=-{\pi}^2r^2\left(r^2+h^2\right)h^2$

$9V^2={\pi}^2r^4h^2$

If these are added

${\pi^2}r^2h^2\left(h-\left(h^2+r^2\right)+r^2\right)$

we get zero only if h=1.

Therefore should your equation have $3{\pi}Vh^3$ instead of $3{\pi}Vh^2$ ?

**snigdha** · February 21st 2010, 07:19 AM

umm..i guess you are right..... the correct equation will be 3 pi V h^3 – C² h²+ 9 V² = 0

thanks...!

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