1. Cone

A right circular cone is inscribed in a regular tetrahedron whose edge is 10cm.Find the volume of the cone. Help please.

2. 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 $ABC,$
. . with: . $AB = BC = CA = 10.$
$AD$ is an altitude of the triangle: . $BD = DC = 5$

The base of the cone is the inscribed circle with center $O.$
. . Its radius is $OD.$

We find that $\Delta ODC$ is a 30°-60° right triangle with $DC = 5.$
. . Hence: . $OD \,=\,\frac{5}{\sqrt{3}},\;OC \,=\,\frac{10}{\sqrt{3}}$

The radius of the cone is: . $\boxed{r \,=\,\frac{5}{\sqrt{3}}}$

Consider a vertical cross-section of the tetrahedron,
. . cut through the top vertex $V$ and $OC.$
Code:
    V *
|\
| \
|  \
|   \
h |    \ 10
|     \
|      \
|       \
|        \
O *---------* C
_
10/√3

We have: . $h^2 + \left(\frac{10}{\sqrt{3}}\right)^2 \:=\:10^2 \quad\Rightarrow\quad h^2 + \frac{100}{3} \:=\:100 \quad\Rightarrow\quad h^2 \:=\:\frac{200}{3}$

The height of the cone is: . $\boxed{h \:=\:\frac{10\sqrt{6}}{3}}$

Therefore, the volume of the cone is:

. . . $V \;=\;\frac{\pi}{3}r^2h \;=\;\frac{\pi}{3}\left(\frac{5}{\sqrt{3}}\right)^ 2\left(\frac{10\sqrt{6}}{3}\right) \;=\;\frac{250\pi\sqrt{6}}{27} \text{ cm}^3$