Question 64953
There's an error in Stan's solution above. He has

D^2=10.89 + 1.7sqrt2   <--That should be D^2=10.89 + (1.7sqrt2)^2
D^2=13.29              <--That should be 16.67
D=sqrt13.29            <--That should be sqrt16.67
D=3.646..              <--That should be 4.08289...

Here's my explanation:

<pre><font size = 4><b>My son is trying to find the length from the top left
corner to the bottom right corner of a cube. 
This is something I have not been taught, and have 
tried to find the formula on the internet, without success.
For example, how would this be done for a cube with the 
dimensions 3.3m x 1.7m x 1.7m 

First of all, the dimensions 3.3m x 1.7m x 1.7m, cannot 
apply to a cube, because to be correctly called a "cube", 
all 6 of its faces must be congruent squares.  

Now either of the dimensions 1.7m x 1.7m x 1.7m or the dimensions
3.3m x 3.3m x 3.3m could apply to a cube, but not the dimensions 
3.3m x 1.7m x 1.7m. 

The dimensions 3.3m x 1.7m x 1.7m could only apply to a 
rectangular solid, two of the faces of which are 1.7m x 1.7m 
squares and the other four faces 3.3m x 1.7m rectangles.

Anyway, with the semantics of the word "cube" out of the way, I 
will answer your question about the longest possible rigid rod 
that could fit inside a 3.3m x 1.7m x 1.7m rectangular solid 
without bending the rod, that is, the distance from A to G
in the drawing below:

   A____B  
   |\D__\C     
  E||   |    <--(H is back behind, below B, right of E, and behind G.     
   \|___|        You can't see H in the picture.   
   F    G 
 
Suppose AB = BC = CD = DA = FG = GH = HE = EF = 1.7m

Suppose AE = BH = CG = DF = 3.3m

You want to find AG. Plan:

1. Draw (or think of drawing) AC. Then ABC is a right triangle
2. Find AC² using AC² = AB²+BC²   
3. Draw (or think of drawing) AG. Then ACG is a right triangle
4. Find AG² using AG² = AC²+CG²
5. Find AG by taking the square root of AG² 


AC² = AB² + BC²
AC² = 1.7² + 1.7² = 5.78

AG² = AC² + CG²
AG² = 5.78 + 3.3² = 16.67
      _____
AG = <font face = "symbol">Ö</font>16.67 = 4.082891133 approximately  

Edwin</pre>