# Cone and Hemisphere

• July 8th 2007, 03:56 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
• July 8th 2007, 04:10 AM
janvdl
For b)

Use Pyhtagoras.

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

$h = \sqrt{l^2 + r^2}$
• July 8th 2007, 09:25 AM
Soroban

Quote:

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

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

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

Quote:

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

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

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

• July 8th 2007, 09:44 AM
So if $l = 2r$ then just plug that in like Soroban did.