maths

Let the base is △ABC (AB = BC = CA = 3) , the vertex be D , O be the centre of △ABC and the height = h :

△DAO is an right-angle △ :

DA^2 = h^2 + OA^2........*

find OA :

(3/2) / OA = sin [ (1/2)(360/3) ]

(3/2)/OA = sin60 = √3 / 2

OA = √3 , sub to * :

DA^2 = h^2 + 3

3^2 = h^2 + 3

h^2 = 6

h = √6


The height of the tetrahedron = √6 cm