How we will find out the perpendicular distance of top most point of a tetrahedron from it's triangular base??

Question

Please answer this question in 3d method

 

Answer 1

A neat way of working this problem is to note that a tetrahedron can be inscribed into a cube.   Drawings of this construction can be found in some advanced textbooks.    Analytically,if the eight corners of a cube are 

(c,c,c), (-c,c,c),(-c-c,c),(c, -c,c),(c,c,-c),(-c,c,-c),(-c,-c,-c),(c,-c,-c)  then the corners of the inscribed tetrahedron can be

(c,c,c), (-c,-c,c),(-c,c,-c), (c,-c,-c) .  With the choice of (c,c,c) as the top point , the vertices of the equilateral triangle base

are:  (-c,c, -c), (-c,-c,c) , (c, -c, -c).  The point at the center of this triangle is (-c/3 , -c/3, -c/3)  . The distance formula can be use to get the distance between this center point and the tip corner = 4 c sqrt(3)/3.   The edge length of the tetrahedron , s, is s = 2 c sqrt(2).  so c = s / (2 sqrt(2))  .   Substituting for c  gives the desired distance as

s sqrt(6) /3