I need to make an octagon shape. I want the back and the front to be 6 ft. wide. What do the side need to be? I'm not good in math and at my old oh well... Would someone please help me?

Results 1 to 3 of 3

- Feb 9th 2006, 09:14 PM #1dinkywGuest

- Feb 10th 2006, 02:46 AM #2

- Joined
- Nov 2005
- From
- Wethersfield, CT
- Posts
- 92

Greetings:

I assumed, in my calculations, that by 6 ft wide, you refer the distance between the centers of two opposite sides (not corner to corner). Thus your octagon will have the orientation of a stop sign (U.S.), i.e., horizontal top and bottom. (We often refer to half that distance as the apothem, vs radius, i.e., center to corner) This being the case, The length of each side must measure

6[sqrt2 - 1] ft or, in practical terms, 2 ft. 5 13/16 in (to the nearest 32nd inch).

If you require greater accuracy (nrst 64th, etc.), or additional information, do not hesitate to ask.

Regards:

Rich B.

PS: In the event that your 6 ft width is corner to corner (diameter), then each side should measure 2.2961...ft or, 2 ft., 3 9/16 in. with like accuracy. Enjoy.

- Feb 10th 2006, 02:55 AM #3

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5