What is the volume of this triangular pyramid? Given a pyramid, the base is a isosceles triangle $ABC$ where $AB=AC=10$ and $BC=12$. All lateral faces of the pyramid ( triangles) have altitudes equal to $20$. Find the volume of this pyramid.
 A: 
$AB < 20$, therefore an altitude with length $20$ in $\triangle ABD$ can be drawn only from the vertex $D$. Denote the foot of this altitude by $E$, and the foot of the altitude of the pyramid by $H$. $E$ and $H$ are different because the other two lateral altitudes also have length $20$. Now we have two lines in the plane $DEH$ orthogonal to $AB$ (the line $DE$ and the line through $E$ parallel to $DH$), therefore the plane $DEH$ is orthogonal to $AB$, and $EH \perp AB$. 
Similarly, the altitudes in $\triangle ACD$ and $\triangle BCD$ are the altitudes to $AC$ and $BC$, and if $DF \perp AC$ and $DG \perp BC$, then $FH \perp AC$ and $GH \perp BC$.
The altitudes of the faces are equal, therefore $EH = FH = GH$, and $H$ is the incenter of $\triangle ABC$. From this,
$$V = \frac 1 3 S_{\triangle ABC} \sqrt {DE^2 - r_{\triangle ABC}^2} =
16 \sqrt {391}.$$
A: Here is the short version of Maxim's answer:
The three sides of the base triangle $\triangle$ are tangents to the sphere $S$ of radius $20$ centered at the apex of the pyramid.  Therefore $S$ intersects $\triangle$ in its incircle of radius $\rho={A\over s}$, whereby $A=48$ is the area of $\triangle$ and $s=16$ its semiperimeter. The height $h$ of the pyramid is then given by $h=\sqrt{400-\rho^2}=\sqrt{391}$, so that we obtain the volume $V={1\over3} A\,h=16\sqrt{391}$.
