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