SOLUTION: 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 o

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Question 64953: 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
Regards
Gordon
Found 3 solutions by 303795, stanbon, Edwin McCravy:
Answer by 303795(602)   (Show Source): You can put this solution on YOUR website!
A cube has all 3 dimensions the same length so ( 3.3m x 1.7m x 1.7m) can't be a cube.

This problem involves 3D but you need to consider only a 2D slice of it at a time. There are two steps to finding the solution.
First part is find the length diagonally across the base. Consider it as a square and use the Pythagorean Theorem to find the length.
The second part is to consider a slice taken from one top corner diagonally across to the opposite corner and slice through to the bottom of the cube. The diagonal on this rectangular slice is what you are trying to find. You calculated the length across the bottom in part one above and the vertical height is the side length of the cube. Again use the Pythagorean Theorem to find the desired length.
The general idea can be used on any rectangular prism. (including your square prism above). An answer of about 4.1 metres for your example would be correct.
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
For example, how would this be done for a cube with the dimensions 3.3m x 1.7m x 1.7m
-------------
Sketch the cube so you can follow along with the following instructions.
Find the diagonal of the base.
The base is a 1.7m by 1.7m square.
Use Pythagoras to find the diagonal of that base.
Let "d" be the base diagonal:
d^2=1.7^2 + 1.7^2
d=1.7sqrt2
---------
Now find the hypotenuse of the triangle made
with the following sides:
one side is the height=3.3 m
one side is the diagonal of the base = 1.7sqrt2
the third side is the diagonal of the cube which you have been asked to find.
Let the diagonal of the cube be "D"
Use Pythagoras again to find "D":
D^2=(3.3)^2 + (1.7sqrt2)^2
D^2=10.89 + 1.7sqrt2
D^2=13.29
D=sqrt13.29
D=3.646..
Cheers,
Stan H.
Answer by Edwin McCravy(20063)   (Show Source): You can put this solution on YOUR website!
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: 
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 = Ö16.67 = 4.082891133 approximately  

Edwin
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