Geometry (surface area)

Hello! I am having difficulty answering this question. If someone can please tell me how to solve these geometry problems, I will be very grateful!

Find the surface area of the following regular polyhedra to the nearest cm^2 (squared).

A. Tetrahedron with edge length of 5 cm.
B. Octahedron with edge length of 4 (sqaure root) 3 cm.
C. Icosahedron with edge length of 12 cm.

I know the answers because they are behind the book.
They are:
A. SA= 43 cm^2
B. SA= 166 cm^2
C. SA= 1247 cm^2

 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #2 Feb 13, 2007, 06:53 AM
A tetrahedron is made of 4 equilateral triangles,
An octahedron is made of 8 equilateral triangles,
An Icosahedron is made of 20 equilateral triangles.

The area of a single equilateral triangle should be easily derivable from

$A = frac{1}{2}bh$

Where b is the length of the base, and h is the perpendicular height (you can work this out in terms of the base by splitting the triangle into 2 right angled triangles).

Let me know if you need any more help.
 JillValentine724 Posts: 8, Reputation: 1 New Member #3 Feb 13, 2007, 06:56 AM
Can you do me the favor of working one of them out? That way I have an idea of what exactly I have to write on the paper.
Thank you!
 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #4 Feb 13, 2007, 06:57 AM
Only if you first derive the height interms of the base for an equilateral triangle.

It's fairly easy, I just worked it in 20 seconds from first principles.

Then i'll show you how to do the tetrahedron
 JillValentine724 Posts: 8, Reputation: 1 New Member #5 Feb 13, 2007, 07:19 AM
OOOO I got it! I did this:

(sqaure root) of 3 * 5^2 = 43 cm^2

Now if only I could figure out the others hehe. =P
 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #6 Feb 13, 2007, 07:20 AM
There's no logic in what you did there.

Where did you get $\sqrt{3}$ from?

 JillValentine724 Posts: 8, Reputation: 1 New Member #7 Feb 13, 2007, 07:24 AM
I got 3 squared from the formula in the book.
And I do want you to help me. I'm sorry if I got you upset. I did not mean to. =/
 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #8 Feb 13, 2007, 07:25 AM
If you have the formula from the book then it shouldn't be a problem
 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #9 Feb 13, 2007, 07:26 AM
I'm trying to make you do a little bit of work, in order to do this question, you need to relate the height of an equilateral triangle to it's base, then you can get the area of an equilateral triangle, and then you can get the surface area of all the shapes because they are based on equilateral triangles.
 JillValentine724 Posts: 8, Reputation: 1 New Member #10 Feb 13, 2007, 07:28 AM
It says here that the Surface area of an octahedron is 2 sqare root 3 ts^2. Now, what I am doing for the next problem is this:

2 sqare root 3 * 4 square root 3 ^ 2

Am I doing it right?

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