A pyramid has altitude h and a square base of side m. The four edges that meet at V, the vertex of the pyramid, each have length e, If e = m, what is the value of h in terms of m?

The answer I got was (m*sqrt(6))/3, but that's wrong

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- Nov 1st 2012, 06:33 AMrasen58Pyramid height problem
A pyramid has altitude h and a square base of side m. The four edges that meet at V, the vertex of the pyramid, each have length e, If e = m, what is the value of h in terms of m?

The answer I got was (m*sqrt(6))/3, but that's wrong - Nov 1st 2012, 10:21 AMMarkFLRe: Pyramid height problem
Consider a right triangle whose horizontal leg runs from one of the vertices of the square base to the center of the base (it's length is half the diagonal of the square base), the vertical leg runs from the center of the base to the vertex of the pyramid (the altitude of the pyramid) and the hypotenuse is the edge of length e.

So, you know the horizontal leg and the hypotenuse in terms of m, you may find the altitude using the Pythagorean theorem. - Nov 1st 2012, 10:25 AMskeeterRe: Pyramid height problem
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- Nov 1st 2012, 10:57 AMrasen58Re: Pyramid height problem
Thank you, but do you need the slant height? Also how did you get the slant height?

i got up to the [m*sqrt(2)]/2 for the base diagonal, but could you explain how you got the same for the height? - Nov 1st 2012, 11:04 AMMarkFLRe: Pyramid height problem
You don't need the slant height, that was thrown in for good measure I am assuming.

To find $\displaystyle h$, use the Pythagorean theorem:

$\displaystyle \left(\frac{m}{\sqrt{2}} \right)^2+h^2=m^2$

Now, solve for $\displaystyle h$. - Nov 1st 2012, 04:31 PMrasen58Re: Pyramid height problem
OOOOH thank you! I didn't even realize

- Nov 8th 2012, 07:16 PMChalama123Re: Pyramid height problem
- Nov 9th 2012, 06:58 PMrasen58Re: Pyramid height problem
[ root(2)*s ]/2