Consider a pyramid with an equilateral triangle as its base. Suppose each side of the base is $a$ and each slant edge is $s$. How to find the height, $h$.

if the base is a square, we can use the following approach.

Suppose the base is a square with each side $a$ and slant edge $s$. Then we can easily find the diagonal($d$) of the base from sides. then there is a right angled triangle where $(d/2)^2+h^2=s^2$

So, there might be a similar approach but i am not able to find how this concept can be applied when the base is an equilateral triangle.

We may be able to use the same concept.

first find height of the base, $h_b$ using the formula $\dfrac{\sqrt{3}a}{2}$.

then use the formula for Centroid, $c$ = $\dfrac{2h_1}{3}$.

now use $c^2+h^2=s^2$. is this a correct solution?

Please help.

  • $\begingroup$ That is the right idea, but you need the distance from the base to the centroid and not from the angle to the centriod. $\endgroup$ – Doug M Mar 15 '16 at 19:25
  • $\begingroup$ @DougM, we are trying to find relation from slant edge, not slant height. so is my approach right? $\endgroup$ – Kiran Mar 15 '16 at 19:27
  • $\begingroup$ the slant edge, yes.. What you have is correct. $\endgroup$ – Doug M Mar 15 '16 at 19:28

You want to drop a line from the vertex of this pyramid down to the base. Where does it hit the base? If the pyramid is regular, it is going to come smack down in the center of the equilateral triangle. Draw the equilateral triangle. And draw the medians of the triangle. Where do they intersect? That is your center.

As it turns out it is 2/3 of the from the vertex to the base along any of the medians.

the slant edge ... now use the Pythagorean theorem.

$s^2 = (\frac{2}{3} a \frac{\sqrt 3}{2})^2 +h^2$

  • $\begingroup$ please see my edit also, is that also a correct option? $\endgroup$ – Kiran Mar 15 '16 at 19:24
  • $\begingroup$ we are trying to find relation from slant edge, not slant height . could you please update the answer $\endgroup$ – Kiran Mar 15 '16 at 19:27

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