# Diagonals in an Octagon

• Apr 1st 2007, 09:50 PM
DivideBy0
Diagonals in an Octagon
In a regular octagon, what is the ratio of the length of its longest diagonal to the length of its shortest diagonal?

I would easily be able to do this if it were talking about a hexagon, because its so easy to construct equilateral triangles in one, but in an octagon, all the angles and lengths are messed up and :confused: .
• Apr 1st 2007, 10:07 PM
earboth
Quote:

Originally Posted by DivideBy0
In a regular octagon, what is the ratio of the length of its longest diagonal to the length of its shortest diagonal?

I would easily be able to do this if it were talking about a hexagon, because its so easy to construct equilateral triangles in one, but in an octagon, all the angles and lengths are messed up and :confused: .

Hello,

I've attached an image of an octagon and its diagonals.

let the longest diagonal be D = 2r

then the shortest diagonal is d = √(2r²) = r√(2)

the ratio is: q = 2r/r√(2) = √(2)

EB
• Apr 1st 2007, 10:19 PM
DivideBy0
Doh! Thanks, I tried starting by giving the octagon a side length of 1 and working my way inwards :P

Say, if you tried with my approach, how would you get the length of the longer diagonal?
• Apr 2nd 2007, 01:45 AM
earboth
Quote:

Originally Posted by DivideBy0
Doh! Thanks, I tried starting by giving the octagon a side length of 1 and working my way inwards :P

Say, if you tried with my approach, how would you get the length of the longer diagonal?

Hello,

in this case you have to use a little bit trigonometrie:

let s be the length of one side of the octagon.

Then you deal with 8 isosceles triangle with the base s, the two legs r and the angle at the vertex (that is the center of the circle) is 45°.
Now divide one isoscele triangle into 2 right triangles:
The hypotenuse is r and one leg is ½s, the angle opposite of this leg is 22.5°.

Now you can calculate r:
Code:

``` ½s                        ½s ----- = sin(22.5°) ==> r = ----------   r                        sin(22.5°)```
The longest diagonal D = 2r

Let d be the shortest diagonal. Then ½d is the height in one of the isoscele triangles:
Now you can calculate ½d:
Code:

``` ½d                        ½d ----- = sin(45°) ==> r = ----------   r                        sin(45°)```
By the way: This last equation is exactly the same as the one I've posted in my previous post.

EB