# I think this should be simple, but i need help!

1. Dec 7, 2007

### ethanmuncher

[SOLVED] i think this should be simple, but i need help!

1. The problem statement, all variables and given/known data
this isn't really homework, but i'm having trouble figuring this problem out...
the problem says:

In this diagram, the red part has an area of 800 square centimetres.

What is the area of the green part, in square centimetres? Please round up to the nearest whole number.

2. Relevant equations

i have no idea how to solve this, since the only information they give is that the pentagon is 800 square centimeters...

3. The attempt at a solution

Well, I figured out that you could probably fold the green triangle into the pentagon, and make it 3 triangles, the 2 on the sides being congruent... but how do i figure out an area???
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 7, 2007

### rocomath

i like this problem, what book did you get it from?

i sorta have an idea but i don't think i'm the best person to help.

3. Dec 7, 2007

### Office_Shredder

Staff Emeritus
I think you're missing a piece of information on the triangles... as is, you don't have any information on how the edges not part of the pentagon act, except that the triangle is isosceles, so you can't actually get an answer

4. Dec 7, 2007

### Diffy

I really enjoyed this problem

The nicest I got the area to reduce to was:
$$\sqrt{\phi^2 - \frac{1}{4}} * \frac{1600}{\sqrt{25 + 10\sqrt{5}}}$$

Where $$\phi$$ is the golden ratio.

Can anyone confirm? Is this right?

Last edited: Dec 7, 2007
5. Dec 7, 2007

### Diffy

I don't agree.

Assuming it is a regular pentagon, the side of a pentagon is in a golden relation with the side of the triangle.

6. Dec 7, 2007

### Office_Shredder

Staff Emeritus
I completely discounted that the diagram shows the triangles edges simply as extensions of the edges of the pentagon. Whoops

7. Dec 8, 2007

### ethanmuncher

i really don't know what that means. i'm taking a 10th grade geometry class right now. if there was like a simple whole number you could use to solve, what would you, well estimate it to be?

8. Dec 8, 2007

### Avodyne

http://mathcentral.uregina.ca/QQ/database/QQ.09.04/susan1.html
I got that a pentagon with side length "2a" has area

Area_pentagon = 5 tan(54) a^2

where 54 means 54 degrees. From that page we can also get the angles in the triangle, and then I get

Area_triangle = tan(72) a^2

And so

Area_triangle / Area_pentagon = tan(72)/(5 tan(54))

And, amazingly, this works out to something very simple, namely $1/\sqrt{5}=0.447$. The triangle looks to me to have an area between 1/3 and 1/2 the pentagon, so numerically this fits as well.

9. Dec 8, 2007

### Dick

That agrees with Diffy's result as well, which could have been simplified using phi=(sqrt(5)+1)/2.

10. Dec 8, 2007

### Diffy

Both answers work out to be around 357.7 This is good enough for me to confirm mine and Avodyne's answers are good.

11. Dec 12, 2007

### ethanmuncher

thanks everyone! i rounded 357.7 to 358, and apparently it was the right answer. :)