[JillValentine724]

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!
Thank you in advance! =)

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]

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



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]

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]

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]

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]

there's no logic in what you did there.

where did you get from?

If you don't want me to help you then just say.

[JillValentine724]

I got 3 squared from the formula in the book.
And I do want you to help me. Im sorry if I got you upset. I did not mean to. =/

[Capuchin]

If you have the formula from the book then it shouldn't be a problem :)

[Capuchin]

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]

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?

[JillValentine724]

No, I'm doing it wrong.

[Capuchin]

let me tell you how I would do it.

if you split an equilateral triangle down the middle we get a right angled triangle with short side 1/2b and hypotenuse b

using pythagoras the height is given by:



putting this into
we get



Now the areas for your shapes are

where n is the number of equilateral triangles (given in my first post), and b is the length of a side.

I tried to help you do it this way, but you ignored it and refused to follow my logic, congratulations on doing no work. I can only hope that you understand what I've said, that will give me some comfort.

[Capuchin]

Apologies for being rude, it's frustrating when someone asks for help and then doesn't listen.

[JillValentine724]

I appreciate your help and patience with me during this time. The way you showed me above is completely different on how he has taught it so don't be upset at me for not understanding you from the get go. I understand that you must think I'm lazy but actually, I am at work right now and trying to do my homework at the same time. I will write down your explanation and ask my Professor tomorrow during class. Sorry for bothering you and thank you for your help.

[JillValentine724]

It's not that I don't listen, I just don't understand geometry. =/ I don't understand math in general. Its my weakness. I really do understand your frustration and I know you didn't mean to sound rude. On the contrary, I'm grateful for your assistance.

[Capuchin]

Okay thank you for clarifying the situation.

I hope you can see that the formula I derive gives the correct answers.