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# from Cube to TETRAEDRE

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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

### Best Answer

#1
+80935
+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
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### 6+0 Answers

#1
+80935
+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
+91435
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  u3

The volume of the whole cube is 1u3

Volume of the tetrahedron (triangular pyramid) = 1-2/3 = (1/3)u3

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
+26399
0

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

Alan  Jun 7, 2014
#4
+80935
+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 √(s2 + s2) = 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)  = [(s3)2√2] / [6√2] = 2s3/6 =

........s3/3 ..........

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

Did I make a mistake somewhere along the line??

CPhill  Jun 7, 2014
#5
+26399
+5

No mistake Chris, you are perfectly correct!

Alan  Jun 7, 2014
#6
+91435
0

Sorry Chris. Thank you Alan.

I wasn't at all sure that I was right which is why I asked for it to be checked.

Yes I can see I missed one of the spaces!  Silly me.

Melody  Jun 8, 2014

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