Volume of Sphere Inscribed within a Tetrahedron

So we have a tetrahedral pyramid with side length **a**. What's the volume of the inscribed sphere that is tangent to the sides of the pyramid?

I just need some feedback on a solution, as I have no idea if it is correct. And I should mention that all the faces are equilateral triangles.

Start by solving for the height of each triangular face, with side length **a**.

$\displaystyle a^2+(\frac{a}{2})^2=k$

where **k** is the height of one of our face triangles.

Solving, $\displaystyle k=\frac{a\sqrt{3}}{2}$

Now, we can use **k** to solve for the height of the pyramid, **h**; imagine a new triangle cutting the pyramid in half.

$\displaystyle k^2=(\frac{k}{2})^2+h^2$

Solving, $\displaystyle h=\frac{k\sqrt{3}}{2}=\frac{3a}{4}$

Using the relationship between a 2-d triangle's height and an inscribed circle's diameter, which is

$\displaystyle height=\frac{3}{2}*diameter$,

$\displaystyle diameter=\frac{2}{3}*height$,

we can find the diameter, **d**, and radius, **r**, of the inscribed sphere.

$\displaystyle r=\frac{d}{2}=\frac{h}{3}=\frac{3a}{4}*\frac{1}{3} =\frac{a}{4}$

Plug the radius, **r**, into the volume of a sphere, **V**, to get the solution.

$\displaystyle V=\frac{4}{3} \pi r^3=\frac{4}{3}\pi (\frac{a}{4})^3=\frac{4\pi}{3}*\frac{a^3}{64}=\fra c{\pi a^3}{48}$

...and that's the solution I've come up with. And I have no idea how to even check to see if it's correct. I'll try to get a picture up so the solution is more visible.