Assume that CC' = 2 ft ; B'C' = 4 ft ; A'B' = 6 ft

We see that A'C is the longest distance between any 2 vertices

\(\Delta A'B'C'\) right at B' has : 

\(A'C'^2=A'B'^2+B'C'^2=6^2+4^2=52\)

\(\Delta A'C'C\) right at C' has :

\(A'C=\sqrt{A'C'^2+CC'^2}=\sqrt{52+2^2}=\sqrt{56}\) (ft)

So, the answer is : \(\sqrt{56}\approx7.48\) (ft)

The longest distance between any two vertices is:

           2:1 x 3=6(ft)=6 x 0,3048=1,8288\(\approx\)1,83(m)

Answer:1,83 m

The longest distance between any two vertices is:

           2:1 x 3=6(ft)=6 x 0,3048=1,8288≈1,83 m

Answer:1,83 m