I have been poking at this question for the past few days and have not much to show for it.

**Hint:**It is convenient to put the pyramid on its side, with the apex at the origin. By using similar triangles you can find an expression for the area of cross-section of the pyramid "at" $x$. Then you can use the usual formula for the volume of a solid when areas of cross-section are known.

You are watching: A pyramid with height h and base an equilateral triangle with side a (a tetrahedron)

**Added:** At some stage you will want to know the area of an equilateral triangle with side $s$. By using "special angles" (and in other ways) it can be shown that this area is $dfracs^2sqrt34$.

The volume is given by

$V=frac13Bcdot h$

where $B$ is the area of the base and $h$ is the height of the pyramid.

Now, the base is an isosceles triangle of side length $a$. So, $B=fracsqrt3a^24$, which is pretty easy to show. The result follows.

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