How can I show that the formula for the volume of a pyramid with height \$h\$ and an equilateral triangle as a base with side length \$a\$ is given by \$\$v =fracsqrt3a^2h12;?\$\$

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.

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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. Thanks for contributing an answer to jonathanlewisforcongress.comematics Stack Exchange!

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