1. ## Two problems

1)http://img.photobucket.com/albums/v2...kid/wrench.gif

ABCDEFGH is a regular octagon with each side of length 2. Find AF and AE.

2)http://img.photobucket.com/albums/v2...id/hexagon.png

Is AF 3 and AE 4 (it's probably not, because the answer seems too simple)?

Thanks!

2. Is there any more information? Like are any of these line segments medians? Maybe bisectors?

3. Below is a the picture. There are 8 triangles. And each one has 45 degrees. Now use trigonometry (or whatever) to find the length of the "red" line. Double it to get x.

4. Originally Posted by Jonboy
Is there any more information? Like are any of these line segments medians? Maybe bisectors?
Sorry, that's all. My teacher has a tendency to provide us with vague problems... that's why I went here--I was wondering if I was missing something.

ThePerfectHacker: Just to clarify: that equation is for which problem? The octagon? If it is, what is x equal to AE or AF?

5. Originally Posted by icebreaker09

ABCDEFGH is a regular octagon with each side of length 2. Find AF and AE.

Is AF 3 and AE 4 (it's probably not, because the answer seems too simple)?

Thanks!
See regular hexagon & octagon properties

6. Originally Posted by icebreaker09
...
2)http://img.photobucket.com/albums/v2...id/hexagon.png

Is AF 3 and AE 4 (it's probably not, because the answer seems too simple)?

Thanks!
Hello,

to #2.:

I'll pick up TPH's hint how to do this problem.
I've modified your drawing of an octogon (which you saved as a hexagon ).

1. $AE = 2r$
Take one of the 8 isosceles triangles and split it into 2 right triangles. Then you get:
$\frac{1~cm}{r}=\sin(22.5^\circ)~~\Longrightarrow ~~r=\frac{1~cm}{\sin(22.5^\circ)}\approx 2.613... ~cm$

Therefore: $AE = 2r \approx 5.226...~cm$

2. $AF = 2x$
$\frac{1~cm}{x}=\tan(22.5^\circ)~~\Longrightarrow~~ x=\frac{1~cm}{\tan(22.5^\circ)}\approx 2.414...~cm$

Therefore: $AE = 2r \approx 4.828...~cm$

7. Originally Posted by icebreaker09
1)http://img.photobucket.com/albums/v2...kid/wrench.gif

ABCDEFGH is a regular octagon with each side of length 2. Find AF and AE.

2)http://img.photobucket.com/albums/v2...id/hexagon.png

Is AF 3 and AE 4 (it's probably not, because the answer seems too simple)?

Thanks!
Hello,

I've attached a modified drawing of the nut.
You deal with a regular hexagon which consists of 6 equilateral triangles. Split one of these triangles into 2 right triangles and use Pythagorean theorem.

For confirmation only: $x = \sqrt{3}~cm \approx 1.732...~cm$

8. Hello, icebreaker09!

2) Find $AF$ and $AE$.
Code:
              F   2   E
* * * * *
2  * | _       *
*   |√2         *
*     |             *
G * - - - +               * D
*       |               *
2 *       | 2             *
*       |               *
H * - - - +               * C
*     | _           *
2  *   |√2         *
* |         *
* * * * *
A   2   B

In the upper-left and lower-left, we have isosceles right triangles.
. . Their hypotenuse is $2$, hence their sides are $\sqrt{2}$

Therefore: . $AF \;=\;2+2\sqrt{2}$

Draw $AF$.

In right triangle $AFE$, we have: . $AE^2\:=\:AF^2 + FE^2$

That is: . $AE^2 \;=\;(2 + 2\sqrt{2})^2 + 2^2\;=\;16 + 8\sqrt{2}\;=\;4(4+2\sqrt{2})$

Therefore: . $AE \;=\;2\sqrt{4+2\sqrt{2}}$