# Diagonal in a box

What is the relation between the length of the *space diagonal* \(a\) and the lengths of the edges \(x\), \(y\) and \(z\) in a three-dimensional box?

If we draw the *face diagonal* \(b\) for the rectangle defined by the edges \(x\) and \(y\), we can see that we have a *right triangle* (blue below) with the sides \(x\), \(y\) and the hypotenuse \(b\).

Now we can see that the space diagonal \(a\) is the hypotenuse of a right triangle (red below), with the sides \(b\) and \(z\).

Thanks to Pythagoras, we have these relations between the variables

So the length of the space diagonal in a box is expressed by a simple equation between the length of the three edges.

## Diagonal in a cube

In the special case of a *cube* all edges are equal (\(x\) = \(y\) = \(z\))

Note: If you want to bring your 165 cm skis when traveling, but the maximum package dimension for a side is 100 cm, you can solve your problem by putting the skis in a 100x100x100 cm box.

## Diagonal in a square

We leave the three-dimensional geometry and examine the two-dimension instead. For a *square* with edges of length \(x\), the face diagonal \(b\) is

## Diagonal in a line

Now to the one-dimension. For a *line* of length \(x\), the "diagonal" \(c\) can be expressed as

Note: The length of the "diagonal" of a line is of course equal to the length of the line itself, but we want to follow the pattern from the previous equations.

## Diagonal in an n-dimensional hypercube

It seems to be so nice, that for the \(n\)-dimensional hypercube with edges of length \(x\), the diagonal \(d\) is calculated by