If 'h' be the height of a pyramid standing on a base which is an equilateral triangle of side 'a' units, then what is the slant height?
What I did:-
We know in equilateral triangle height= $\sqrt3/2*(side)^2$= $\sqrt3/2*a^2$ Total height= h+$\sqrt3/2*a^2$ base=a/2 Applying pythagoras
$(h+\sqrt3/2*a^2)^2+a^2/4$ It should give me the answer $\sqrt (h^2+a^2/3)$
Where I am wrong
$\endgroup$ 31 Answer
$\begingroup$The altitude $AB$ of pyramid will intersect base at the centroid $B$ of base-equilateral triangle. and the distance of centroid $BC$ from a vertex of base is $\frac{a}{\sqrt{3}}$. You can verify this using pythagorean theorem or simple trigonometry.
Thus we have to apply pythagoras theorem on the $\triangle ABC$ whose legs are $AB = h$ and $BC = \frac{a}{\sqrt{3}}$ So the hypotenuse, ie, slant height $AC$ is : $$s=\sqrt{h^2 + \frac{a^2}{3}}$$
$\endgroup$ 2
