Hello, reiward!
This one takes a lot of preliminary work . . .
A right circular cone is inscribed in a regular tetrahedron with edge 10 cm.
Find the volume of the cone. Consider the base of the tetrahedron (and the cone).
Code:
A
*
/|\
/ | \
/ | \
/ | \
/ | \
/ * * * \
/* | *\
10 * | * 10
* | *
/ | \
/* | *\
/ * O * * \
/ * | * * \
/ | 60° * \
/ * | * \
/ * | * * \
/ * | * 30° * \
B *---------------*-*-*---------------* C
: - - - 5 - - - D - - - 5 - - - :
The base of the tetrahedron is equilateral triangle $\displaystyle ABC,$
. . with: .$\displaystyle AB = BC = CA = 10.$
$\displaystyle AD$ is an altitude of the triangle: .$\displaystyle BD = DC = 5$
The base of the cone is the inscribed circle with center $\displaystyle O.$
. . Its radius is $\displaystyle OD.$
We find that $\displaystyle \Delta ODC$ is a 30°-60° right triangle with $\displaystyle DC = 5.$
. . Hence: .$\displaystyle OD \,=\,\frac{5}{\sqrt{3}},\;OC \,=\,\frac{10}{\sqrt{3}}$
The radius of the cone is: .$\displaystyle \boxed{r \,=\,\frac{5}{\sqrt{3}}}$
Consider a vertical cross-section of the tetrahedron,
. . cut through the top vertex $\displaystyle V$ and $\displaystyle OC.$ Code:
V *
|\
| \
| \
| \
h | \ 10
| \
| \
| \
| \
O *---------* C
_
10/√3
We have: .$\displaystyle h^2 + \left(\frac{10}{\sqrt{3}}\right)^2 \:=\:10^2 \quad\Rightarrow\quad h^2 + \frac{100}{3} \:=\:100 \quad\Rightarrow\quad h^2 \:=\:\frac{200}{3}$
The height of the cone is: .$\displaystyle \boxed{h \:=\:\frac{10\sqrt{6}}{3}}$
Therefore, the volume of the cone is:
. . . $\displaystyle V \;=\;\frac{\pi}{3}r^2h \;=\;\frac{\pi}{3}\left(\frac{5}{\sqrt{3}}\right)^ 2\left(\frac{10\sqrt{6}}{3}\right) \;=\;\frac{250\pi\sqrt{6}}{27} \text{ cm}^3$