# Homework Help: Spatial Geometry Help?

1. Apr 11, 2007

### .hacker//Kazu

My math teacher is currently teaching us spatial geometry and recently handed out a package of worksheets. And I...don't...understand...a word of it (My math is rather terrible.). He asked us to cut out the nets and build the following shapes: a cube, an octahedron, a dodecahedron, an icosahedron, a truncated cube, and cuboctahedron.

We are required to find the volume, dimensions of the edges, areas of the faces, and the # of vertices/edges.

I have managed to solve the cube and octahedron and can build the shapes fine. I even have the formulas and dimensions. I just can't seem to solve it.

He said we did not have to solve the dodecahedron, but the icosahedron, truncated cube and cuboctahedron are driving me nuts.

First the icosahedron. The dimensions for one of the edges are 3.8cm. The formula to get the volume is (15+5*square root of 5)/12*s cubed. I don't know if that makes any sense...

Let's see if anyone can help me on the first one, before I post the second...

2. Apr 11, 2007

### .hacker//Kazu

If I don't figure it out, my math teacher will kill me...

3. Apr 11, 2007

### neurocomp2003

do the shapes have symmetry/similarity that you can notice.

and one algorithm you should think of Divide & Conquer. So its up to you to think how you can "divide".

Also use past/builtup knowledge

Whats teh area of a triangle, volume of tetrahedron, area of a square, volume of a cube/box. whats the area of a N-polygon, whats the volume of a N-polyhedra. etc... DIVIIDE & CONQUER

4. Apr 11, 2007

### .hacker//Kazu

The area of one of the faces of the icosahedron is 6.46cm square.

5. Apr 11, 2007

### neurocomp2003

so ...how will that help you find the volume...i guess you can say picture yourself in the volume...and then DIVIDE & CONQUER

6. Apr 11, 2007

### Dick

First, your math teacher is unlikely to kill you. Second, are you being given formulas for each of these shapes? If not, then deriving them is actually a pretty difficult exercise in solid geometry. If so, then it's simply a matter of solving the given equations for a variable and anyone can help you if you state the problems clearly and one by one instead of all at once in a blur of panic.

7. Apr 12, 2007

### HallsofIvy

Don't just memorize formulas (or copy them from your book) without being sure what the letters MEAN. I say that because the volume of a figure can be calculated based on a number of things (length of an edge, distance between opposite vertices, etc.) Here, I'm pretty sure s is "length of an edge" but you had better check that. If it is then you have the formula
$$\frac{15+ 5\sqrt{5}}{12} s^3$$
and you know that s= 3.8 cm. Replace s in the formula with 3.8 and do the arithmetic. It's easy with a calculator, a little harder without.

8. Apr 12, 2007

### .hacker//Kazu

Umm, the problem is my answer is definetely not right when I do it. I don't quite get how you can square root a 5 (do I leave it as a decimal?).

9. Apr 12, 2007

### Integral

Staff Emeritus
This is not a helpful or even meaningful post. Numeric answers are not what you should give us. Where did that number come from? Show us how you arrived at it. Did you do a calculation involving the side length? If so show us that.

Specific numbers are more obscuring then helpful, avoid them until you are done with the problem.

10. Apr 12, 2007

### .hacker//Kazu

I measured the side with a ruler. How would I show that?

11. Apr 12, 2007

### Dick

In fact, you really have not been at all clear on what you are doing. Are you cutting out shapes with paper and scissors and being asked to verify equations?

12. Apr 12, 2007

### neurocomp2003

yes that is what he stated in his first post

"He asked us to cut out the nets and build the following shapes: a cube, an octahedron, a dodecahedron, an icosahedron, a truncated cube, and cuboctahedron."

once he gets those shapes up he stated this
"He asked us to cut out the nets and build the following shapes: a cube, an octahedron, a dodecahedron, an icosahedron, a truncated cube, and cuboctahedron."

Now he needs the formulas and how to solve them...since he does not know how to build the equations by breaking the volumes into simpler polyhedrons(simple volume equations like the tetra and cube) ...which is the divide & conquer algo.

its best to go look at www.mathworld.com

13. Apr 12, 2007

### Dick

My point is that I don't think the poster is trying to derive any equations. Just try to verify them experimentally.

14. Apr 12, 2007

### neurocomp2003

I think the policy of the forum is to not give out answers to homework but to guide them along the way towards getting the answer.

The only way for him to understand how to calculate those volumes is to attempt to derive the equations.

15. Apr 12, 2007

### Dick

Another, perhaps unstated and subtle policy, is to try to figure out what the poster is really asking and at what level they are capable of following the advice proferred. I don't think somebody cutting out paper polyhedra should be expected to have a detailed grasp of solid geometry.

16. Apr 13, 2007

### HallsofIvy

You use a calculator of course! I'm not clear on what you mean by "leave it as a decimal". What else could you do with it?

You said
or, in LaTex:
$$\frac{15+ 5\sqrt{5}}{12}s^3$$

Okay, that is
[tex]\frac{15+ 5\sqrt{5}}{12} (3.8)^3[/itex]

On my calculator I would enter
"15+ 5*sqrt(5)" and press enter
then "/5" and press enter (my calculator, a TI83, automatically enters the previous line before an operation like "/") then "*3.8^3". Again, my calculator would enter the previous line before the operation "*".

Last edited by a moderator: Apr 13, 2007
17. Apr 13, 2007

### .hacker//Kazu

Oh thank you, all of you.

I tried doing the formula for a while. I am not sure, but is the answer 119.7139675 cm cubed?

By the way, I am a girl.

18. Apr 13, 2007

### Dick

Good job. That's the same thing as I get.

19. Apr 13, 2007

### .hacker//Kazu

Alright then. What would be the formula to find the volume of a cuboctahedron?

20. Apr 13, 2007

### Dick

Where did you get the formula for the icosahedron?

21. Apr 13, 2007

### .hacker//Kazu

On a worksheet. But it only has the inradius, circumradius, dihedral angle, surface area and volume for the cube, tetrahedron, octahedron, dodecahedron, and icosahedron. My teacher said we don't need to learn the inradius, circumradius and dihedral angle until next year though.

22. Apr 13, 2007

### Dick

Sorry. I would probably be hard put to derive the equation for volume of a cuboctohedron even if I knew exactly what one is. And I don't. I doubt you are expected to derive one either. So if you don't have one, are you SURE that computing its volume is part of the assignment?

23. Apr 13, 2007

### Data

The general strategy of "divide & conquer" suggested earlier suffices to find the volume expressions for all these polyhedra; However, I doubt that your teacher expects you to do so for the cuboctahedron (it'd be a rather tedious task in any case, and the trigonometry needed to simplify the problem even a little probably hasn't been taught to you)!

The formula's available at Wikipedia: http://en.wikipedia.org/wiki/Cuboctahedron.

Last edited: Apr 13, 2007
24. Apr 13, 2007

### .hacker//Kazu

I'm only in Gr. 8, I don't know if I'm supposed to know this stuff...If so, I'm pretty behind then. But, yes, I'm sure it is part of the assignment. The only one my teacher told me not to compute is the dodecahedron. But he didn't say anything about the cuboctahedron.

The formula on Wiki doesn't make sense to me. Area=(6 + 2 x square root of 3) x a squared. Volume=5/3 x square root of 2 x a cubed.

25. Apr 13, 2007

### Data

a is the edge length.