with a volume of a cube how can I obtain a tetraedre ? I would know how writting calcul with google calc.

I'm not a specialist so could you write a model for me as an example bellow ?

Thx by advance.

Response : Thx CPhill(+5) :-) , finally it is easy !

unzip
Jun 7, 2014

#1**+16 **

I'm assuming you want to find the volume of a tetrahedron inscribed in a cube....I don't know much about this, but there is a way it can be done.....(see below)....the volume of this is just 1/3 the volume of the cube = s^3/3 where s is the side of the cube....

There may be other answerers who could give you more help, but if not, here's a page that will probably tell you more.....

http://en.wikipedia.org/wiki/Tetrahedron

CPhill
Jun 7, 2014

#1**+16 **

Best Answer

I'm assuming you want to find the volume of a tetrahedron inscribed in a cube....I don't know much about this, but there is a way it can be done.....(see below)....the volume of this is just 1/3 the volume of the cube = s^3/3 where s is the side of the cube....

There may be other answerers who could give you more help, but if not, here's a page that will probably tell you more.....

http://en.wikipedia.org/wiki/Tetrahedron

CPhill
Jun 7, 2014

#2**0 **

I just want to play with this one for a moment. (This has been edited)

I don't think Chris's is right I think it is 1/3 of a different cube which i could find but I want to try a different way.

Let the side lenght of the cube be 1.

Inside the cube is the tranglular required trianglular pyramid but

the space surrounding this consist of 3(no 4 as alan has pointed out) congruent right triangular pyramids

The volume for each of these is

$$V=1/3 \mbox{ x area of base x height}\\

V=\frac{1}{3}\times (1/2 * 1*1)*1\\

V=\frac{1}{6}\qquad u^3\\$$

So for all three added together the volume is 1/6*4 = 2/3 u^{3}

The volume of the whole cube is 1u^{3}

Volume of the tetrahedron (triangular pyramid) = 1-2/3 = (1/3)u^{3}

Which is one third the volume of the whole cube.

Did I make a mistake? I can't find one. (yes I did there were 4 not 3)

Melody
Jun 7, 2014

#3**0 **

The space surrounding the tetrahedron consists of *four*, not three, congruent right-triangular pyramids Melody.

Alan
Jun 7, 2014

#4**+5 **

After looking at this again, let's analyze it to see if I blew a fuse (which is always a faint possibility)....I know that a tetrahedron can be put in a cube (as the picture indicates) and that it can be oriented in such a way as to produce a tetrahedron that is equal to (1/3) of the cube's volume, but I'm having second thoughts as to whether or not * THIS* pariticular orientation produces such a thing.

Let's call the side length of the cube, s. Notice that every edge length appears to be just √(s^{2} + s^{2}) = s√2. And the "formula" for the volume of a tetrahedron with edge length "a" is just

(a)^{3}/(6√2) ....so we have.....

(s√2)^{3}/(6√2) = [(s^{3})2√2] / [6√2] = 2s^{3}/6 =

........s^{3}/3 ..........

Mmmmm.........interesting!!

Did I make a mistake somewhere along the line??

CPhill
Jun 7, 2014