Cone

Hello, reiward!

This one takes a lot of preliminary work . . .

Quote:

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:

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