What is the longest pole that will fit into a 4' x 3' x 2' box?

Question

can't grasp it HELP please

Answer 1

The question is - What is the length of the diagonal of the box ?

Answer = sqaure root of (4x4 + 3x3 + 2x2)  = 5,38 feet

Answer 2

The square of the length of the long diagonal d is

d^2 = 4^2 + 3^2 + 2^2

d = 5.38'

Answer 3

sqr( 4^2 + 3^2 + 2^2)

pole would go from one corner to the opposite.

using right triangles

say the floor has sides A and B

a diagonal across the floor would be sqr(a^2 + b^2) = diag

another triangle would be the floor diagonal plus side C (height)

pole = sqr( diag^2 + c^2)

::: diag^2 = a^2 + b^2

pole = sqr ( a^2 + b^2 + c^2)