Inner diagonals?

**evilgummybear** · February 16th 2013, 04:29 PM

How do you find inner diagonals of a cube? I was wondering that for a very long time.

**veileen** · February 16th 2013, 11:42 PM

Let a be the length of every side. D'B is an inner diagonal of the cube ABCDA'B'C'D'. B'D, A'C, C'A are also inner diagonals.

DD' $\perp$ (ABCD), so DD' is perpendicular to DB (since DB is in (ABCD)), that means that BDD' is a right triangle:

$D'D^2+DB^2=D'B^2 \Rightarrow D'B=\sqrt{D'D^2+DB^2}$

ABD is a right triangle, so: $AD^2+AB^2=DB^2 \Rightarrow DB^2=2a^2$

$D'B=\sqrt{a^2+2a^2}=\sqrt{3a^2}=a\sqrt3$

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