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 !
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
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
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 x area of base x heightV=13×(1/2∗1∗1)∗1V=16u3
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)
The space surrounding the tetrahedron consists of four, not three, congruent right-triangular pyramids Melody.
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??