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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 $$v =\frac{\sqrt3a^2h}{12}\;?$$

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

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Why is this tagged differential equations? – Jebruho Dec 17 '12 at 1:48
Are you presupposing knowledge of integral calculus? – JohnD Dec 17 '12 at 1:50
Fixed to just calculus. – Kyle H Dec 17 '12 at 2:53
up vote 4 down vote accepted

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.

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 $\dfrac{s^2\sqrt{3}}{4}$.

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The volume is given by

$V=\frac{1}{3}B\cdot h$

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, $B=\frac{\sqrt{3}a^2}{4}$, which is pretty easy to show. The result follows.

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