First draw a picture (see attached).
You want to find the diagonal of the cube, length AB.
The right triangle formed is trianagle ABD.
AD = 10m
DB will be the hypotenuse of triangle BCD.
[tex](BC)^2 + (CD)^2 = (DB)^2 \\
(10)^2 + (10)^2 = (DB)^2 \\
DB = \sqrt{200}= 10 \sqrt{2} [/tex].
If we know the length of AD and DB, we can find AB.
[tex](AD)^2+(BC)^2 = (AB)^2 \\
(10)^2 + (10 \sqrt{2})^2 = (AB)^2 \\
AB = \sqrt{300} = 10 \sqrt{3} [/tex]
In fact, the diagonal of any cube is √3 times the side length of the cube.
Let s be the side length (as in AD or CD in the attached) and
h be the hypotenuse of the base, (DB in the attached) and
d be the diagonal of the cube (AB in the attached).
The hypotenuse of the base will be:[tex]s^2+s^2 = h^2 \\
[/tex].
The cube's diagonal will be:
[tex]h^2+s^2 = d^2[/tex].
Substituting [tex]s^2+s^2[/tex] as [tex]h^2[/tex], you have
[tex]s^2+s^2+s^2 = d^2 \\
3s^2 = d^2 \\
d = \sqrt{3s^2} = s\sqrt{3}[/tex]
