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 
. . with: . 
is an altitude of the triangle: . 
The base of the cone is the inscribed circle with center 
. . Its radius is 
We find that 
is a 30°-60° right triangle with 
. . Hence: . 
The radius of the cone is: . 
Consider a vertical cross-section of the tetrahedron,
. . cut through the top vertex 
and 
Code:
V *
|\
| \
| \
| \
h | \ 10
| \
| \
| \
| \
O *---------* C
_
10/√3
We have: . ^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: . 
Therefore, the volume of the cone is:
. . . ^ 2\left(\frac{10\sqrt{6}}{3}\right) \;=\;\frac{250\pi\sqrt{6}}{27} \text{ cm}^3)
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