Centroid of a Tetrahedron

[redflag64]

Find the location of the centroid of a tetrahedron. I want the tetrahedron to have one face in the x,y plane and one vertex at (a,0,0). Two other vertices are located at (-a/2, a * sqrt(3)/2, 0) and (-a/2, -a*sqrt(3)/2, 0). You have to find the 3rd vertex yourself (as well as the location of the centroid).
During the course of this problem, you will have to compute the volume of the tetrahedron. Explicitly compute the volume of the tetrahedron. Finally, what is the radius of the smallest sphere which contains this tetrahedron.

 

[Deleted member 4993]

Find the location of the centroid of a tetrahedron. I want the tetrahedron to have one face in the x,y plane and one vertex at (a,0,0). Two other vertices are located at (-a/2, a * sqrt(3)/2, 0) and (-a/2, -a*sqrt(3)/2, 0). You have to find the 3rd vertex yourself (as well as the location of the centroid).
During the course of this problem, you will have to compute the volume of the tetrahedron. Explicitly compute the volume of the tetrahedron. Finally, what is the radius of the smallest sphere which contains this tetrahedron.

Straight forward problem.

First draw an approximate sketch.

Slice it and dice it and add those up!

What have you been taught about calculation of volume of a solid?

What have you been taught about calculation of centroid of a solid?

Please share your work with us .

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

 

[HallsofIvy]

(Edited thanks to lookagain) There is a general rule that the centroid of a "minimal figure" (a figure, in n-dimensional space, made with the fewest n-1 dimensional "hyper-planes"- a triangle in 2 space, tetrahedron in 3 space, etc.) is the point whose coordinates are the means of the corresponding coordinates of the vertices.

That is, if the vertices of a triangle are given by \(\displaystyle (x_0, y_0)\), \(\displaystyle (x_1, y_1)\), and \(\displaystyle (x_2, y_2)\) then its centroid is \(\displaystyle ((x_0+ x_1+ x_2)/3, (y_0+ y_1+ y_2)/3)\) while if the vertices of a tetrahedron are given by \(\displaystyle (x_0, y_0, z_0)\), \(\displaystyle (x_1, y_1, z_1)\), \(\displaystyle x_2, y_2, z_2)\), and \(\displaystyle (x_3, y_3, z_3)\) then its centroid is
\(\displaystyle ((x_0+ x_1+ x_2+ x_3)/4, (y_0+ y_1+ y_2+ y_3)/4, (z_0+ z_1+ z_2+ z_3)/4)\)

 

[Deleted member 4993]

The post above by HoI is an excellent suggestion for checking your answer.

However, your problem defines explicit method of calculating the co-ordinates of CG (you must calculate the volume on the way)

 

[HallsofIvy]

Yes, IF the problem explicetly says that integration must be used, it is a harder but still not impossible problem.

 

[lookagain]

That is, if the vertices of a triangle are given by \(\displaystyle (x_0, y_0)\), \(\displaystyle (x_1, y_1)\), and \(\displaystyle (x_2, y_2, z_2)\) then its centroid is \(\displaystyle ((x_0+ x_1+ x_2)/3, (y_0+ y_1+ y_2)/3, (z_0+ z_1+ z_2)/3)\)

HallsofIvy,

don't you mean that if the vertices of a triangle are given by \(\displaystyle (x_0, y_0)\), \(\displaystyle (x_1, y_1), \ \) and \(\displaystyle \ (x_2, y_2), \ \)


then its centroid is \(\displaystyle \ \bigg((x_0 + x_1 + x_2)/3, \ (y_0 + y_1 + y_2)/3\bigg)\ ?\)

 

[HallsofIvy]

Yes, I started to give the case for the tetrahedron, then went back and tried to change to triangle and messed it up! Thanks.

 

[redflag64]

My Work So far...

The centroid of the triangle is:
x = (a - a/2 - a/2)/3 = 0
y = (0 + a*√(3)/2 - a*√(3)/2)/3 = 0
z = (0 + 0 + 0)/3 = 0

all side lengths should be the same. In this case, it's easy to determine the height along the x-axis (a - (-a/2)) = 3a/2, so the final vertex should be at (0,0,3a/2).

The centroid of a tetrahedron is 1/4 the distance from base to apex, so (0,0,3a/8).

You can use the centroid to find the radius of the sphere by using the distance formula between any of the vertices:
r = √[a² + 9a²/64] = √[73a²/64] = a*√(73)/8

is this reasoning correct?