Trapezoid board length

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Laddcarl

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First of all, I’m not sure what category this question belongs and hope someone can help.

I have a 24’ round pool with in essence a flat metal cap with rounded edges all the way around it. This cap is made up of 16 sections which has created a 16 sided octagon.

I want to build a deck around this pool made up of 16 trapezoid sections that when bolted sided by side it would surround the pool.

What I know is that each section needs to be 5’ out from the edge of the pool and the edge running along side the pool is 55.75”. My question is: How do I determine the length of the board at the 5’ end of the trapezoid and the angle of cut needed at the end of each board to form the trapezoid. To summarize: I want to build 16 sets of trapezoids that when bolted together side by side they will encompass the pool.

Once I know how to figure out the length of the 5’ end I figure I’ll have this problem dealt with.

Thank you to whom ever may assist me with this.

Best regards,

Carl
 
First (not important but I just have to say it) there are no "16 sided octagons"! An octagon, by definition, has eight sides. What you have is a 16 sided polygon. ("Octa" is from the Latin for "eight", "poly" for "many".)

On each of those 16 sides you want a trapezoid (yes, that is the right word!) with one base along the side of the polygon and the other 5" away. Each of the 16 sides has length 55.75/16= 3.48 inches which you can round to 3 and 1/2 inches. The critical thing to calculate is what angle the side makes with that base.

Bear with me while I think this out! A side of each trapezoid is an extension of a line from the center of the polygon. Drawing every line from the center of an n side polygon to a vertex divides the polygon into n triangles. Every t18riangle has angle sum 180 degrees so those n triangles have angle sum 180n. But that includes the angles at the center of the polygon. Since the triangles go all the way around the center the angles at the center sum to 360 degrees. So the angles at the edges sum to 180n- 360 degrees. There are n equal angles so each is 180- 360/n degrees. (For example in a triangle n= 3 so each angle is 180- 360/3= 180- 120= 60 degrees. For a square n= 5 so each angle is 180- 360/4= 180- 90= 90 degrees.) Your 16 sided polygon has each angle 180- 360/16= 157.5 degrees.
 
Thank you for your reply and thank you for the correction on the description of what I have in regards to the shape. I’m not the greatest at math, especially in the area that you described here in. I would be lying if I said I completely understand what you described so I will have to try to let that sink in and hope I can break it down in my mind.

What I see from what you described is that the angle of cut at the end of the boards can be determined by this math but the length of the board at the 5’ end is yet to be determined.

Did you see the picture I included with my thread? I understand that each trapezoids shape is derived from the center point of the pool and when fanned out (like a slice of pizza) to the point of intersecting with the left and right section screws the length of the board closest to the pool should be 55.75”. Continuing out 5’ from there increases the width of the fan (crust end of the pizza) and this of course is the length of board I’m trying to figure out along with the needed angle of cut.
 
HallsofIvy said:
First (not important but I just have to say it) there are no "16 sided octagons"! An octagon, by definition, has eight sides. What you have is a 16 sided polygon. ("Octa" is from the Latin for "eight", "poly" for "many".)

On each of those 16 sides you want a trapezoid (yes, that is the right word!) with one base along the side of the polygon and the other 5" away. Each of the 16 sides has length 55.75/16= 3.48 inches which you can round to 3 and 1/2 inches. The critical thing to calculate is what angle the side makes with that base.

Bear with me while I think this out! A side of each trapezoid is an extension of a line from the center of the polygon. Drawing every line from the center of an n side polygon to a vertex divides the polygon into n triangles. Every t18riangle has angle sum 180 degrees so those n triangles have angle sum 180n. But that includes the angles at the center of the polygon. Since the triangles go all the way around the center the angles at the center sum to 360 degrees. So the angles at the edges sum to 180n- 360 degrees. There are n equal angles so each is 180- 360/n degrees. (For example in a triangle n= 3 so each angle is 180- 360/3= 180- 120= 60 degrees. For a square n= 5 so each angle is 180- 360/4= 180- 90= 90 degrees.) Your 16 sided polygon has each angle 180- 360/16= 157.5 degrees.
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I’ve been looking your reply over and somewhat understand what you are saying. Can you offer any additional explanation so I can move on to figuring out the board length at the 5 foot side of the Trapezoid? BTW: I remeasured for the shorter length, the one closet to the pool and found that that length is actually 58.5 inches long. Thank you
 
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