# Thread: volume of a cube using space diagonals

1. ## volume of a cube using space diagonals

so i believe the volume of a cube using space diagonals is

V = d^3 / 3sqrt(3) where d is the space diagonal

can anyone help me prove this? the question this relates to is actually the volume of a rectangular prism in terms of space diagonal, but to understand the cube scenario would be a good starter.

apologies for not using latex, i havent used it in a while and need to re-educate myself

2. Yes. Using a 1by1by1 cube: space diagonal = sqrt(3)
From a right triangle 1-sqrt(2)-sqrt(3)

3. Hello, walleye!

$\text{I believe the volume of a cube using space diagonals is:}$

. . $V \:=\:\dfrac{d^3}{3\sqrt{3}}\;\text{ where }d\text{ is the space diagonal.}$ . Right!

We have a cube with side $\,x.$

Code:
             x
*--------C
/       /|
/       / | x
*-------*  |
|       |  *B
x |       | /
|       |/ x
A*-------*
x

Consider the "floor" of the cube.

Code:
      *-------* B
|     * |
|   *   | x
| *     |
A *-------*
x

We see that: . $AB^2 \:=\:x^2+x^2 \quad\Rightarrow\quad AB = x\sqrt{2}$

Now "slice" the cube through vertices $A, B,$ and $C.$

We have this cross-section:

Code:
                  * C
d   *  |
*     | x
*        |
A *-----------* B
_
x√2

And we have: . $(x\sqrt{2})^2 + x^2 \:=\:d^2 \quad\Rightarrow\quad 2x^2 + x^2 \:=\:d^2$

. . $3x^2 \:=\:d^2 \quad\Rightarrow\quad x^2 \:=\:\dfrac{d^2}{3} \quad\Rightarrow\quad x \:=\:\dfrac{d}{\sqrt{3}}$

The volume of the cube is $\,x^3$

$\text{Therefore: }\;V \;=\;\left(\dfrac{d}{\sqrt{3}}\right)^3 \;=\;\dfrac{d^3}{3\sqrt{3}}$