# Cone and Hemisphere

• Jul 8th 2007, 02:56 AM
Cone and Hemisphere
(a) pi * r * l = 1/2 (4 * pi * r^2)

l = 2r

(b) I can't do

(c) I can't do

Thanks
• Jul 8th 2007, 03:10 AM
janvdl
For b)

Use Pyhtagoras.

$\displaystyle h^2 = l^2 + r^2$

$\displaystyle h = \sqrt{l^2 + r^2}$
• Jul 8th 2007, 08:25 AM
Soroban

Quote:

(b) Find the perpendicular height, $\displaystyle h$, of the cone in terms of $\displaystyle r$.
From the diagram in (a), we can use Pythagorus (as janvdl suggested):

. . $\displaystyle h^2 + r^2\:=\:L^2\quad\Rightarrow\quad h \:=\:\sqrt{L^2 - r^2}$

Since $\displaystyle L = 2r$, we have: .$\displaystyle h \:=\:\sqrt{(2r)^2 - r^2} \:=\:\sqrt{3r^2}\quad\Rightarrow\quad\boxed{h \:=\:\sqrt{3}r}$

Quote:

(c) Find the ratio of the volumes of the cone and the hemisphere.
The volume of a cone is: .$\displaystyle V \:=\:\frac{1}{3}\pi r^2h \:=\:\frac{1}{3}\pi r^2(\sqrt{3}r) \:=\:\frac{\sqrt{3}}{3}\pi r^3$

The volume of a hemisphere is: .$\displaystyle \frac{1}{2} \times \frac{4}{3}\pi r^3\:=\:\frac{2}{3}\pi r^3$

The ratio is: .$\displaystyle \frac{V_c}{V_h} \;=\;\frac{\frac{\sqrt{3}}{3}\pi r^3}{\frac{2}{3}\pi r^3} \:=\:\boxed{\frac{\sqrt{3}}{2}}$

• Jul 8th 2007, 08:44 AM
The volume of a cone is confusing me:

I can't see how you got from 1/3 * pi * sq.rt3r to sq.rt3/3 * pi * r^3
• Jul 8th 2007, 09:15 AM
janvdl
I tried googling for the formulas in a), but i struggled to find them. So i couldn't prove that formula. I think i struggled to find the hemisphere one.

So if $\displaystyle l = 2r$ then just plug that in like Soroban did.