Find volume of a cube as a function of diagonal length.

**choy777** · September 29th 2010, 10:15 AM

Find the relationship describing the volume (V) of a cube as a function of the length of the diagonal going through the cube (d). and evaluate it for a diagonal length of d = 1.2.

**Archie Meade** · September 29th 2010, 01:11 PM

Originally Posted by choy777

Find the relationship describing the volume (V) of a cube as a function of the length of the diagonal going through the cube (d). and evaluate it for a diagonal length of d = 1.2.

Pythagoras' theorem will give you this.

If you have a regular cube of side length "x", the volume is $x^3$

The diagonal across the base square has length $\sqrt{x^2+x^2}=\sqrt{2x^2}=\sqrt{2}x$

The diagonal going through the cube has length $\sqrt{2x^2+x^2}=\sqrt{3x^2}=\sqrt{3}x$

$\displaystyle\ V=x^3=\frac{\left(\sqrt{3}x\right)^3}{\left(\sqrt{ 3}\right)^3}$

$=\displaystyle\frac{d^3}{\left(\sqrt{3}\right)^3}= \left(\frac{d}{\sqrt{3}}\right)^3$

You can then perform your calculation.

**deadman** · September 30th 2010, 11:03 AM

I got lost how did it end up over root3 cubed

**Archie Meade** · September 30th 2010, 11:34 AM

Originally Posted by deadman

I got lost how did it end up over root3 cubed

After applying Pythagoras' theorem twice, we get... "internal diagonal"= $x\sqrt{3}$

where "x" is the length of a side of the cube.

The volume of the cube is $x^3$

Since $\displaystyle\frac{\sqrt{3}}{\sqrt{3}}=1$

then $\displaystyle\frac{(x\sqrt{3})^3}{(\sqrt{3})^3}=x^ 3$

which is the cube volume.
This allows us to express the cube volume as a function of the internal diagonal length.
The internal diagonal goes from the bottom right-corner of the base of the cube to the
top-left corner of the facing side,
or from the bottom left-corner to the facing side's top right-corner.

I'll be the first to admit that my sketch is not drawn well!
The sides should appear to be the same lengths.

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