Given Octagon Diameter, Find Side LengthDate: 04/09/2003 at 18:30:25 From: James Subject: Octagonal Diameter vs Sides If you have the diameter of an octagon, what formula gives you the length of the sides? I've found a formula where if you have the sides, you can find the diameter, but I don't know how to reverse the equation. I have the short diameter or the side to side, but it would also be nice to know the vertex to vertex as well. I'm currently designing a house with several octagonal rooms. I appreciate all input you can give me. Thanks! James Fields Date: 04/11/2003 at 09:24:51 From: Doctor Ian Subject: Re: Octagonal Diameter vs Sides Hi James, Draw a regular octagon, and make a mark at the center. From the center, draw a line to each vertex. You end up with 8 identical isosceles triangles. The angle at each vertex is (8 - 2)*180 ----------- = 135 degrees 8 How do we know that? Pick any vertex, and draw diagonals to all but the two adjacent vertices. You get 6 triangles. The sum of the interior angles of each triangle is 180 degrees, and the sum of the interior angles of all the triangles is the sum of the interior angles of the octagon, i.e., 6*180 degrees. In a regular octagon, the interior angles are all equal, i.e., each one is (6*180)/8 degrees. You can use this strategy to find the interior angles of any regular polygon. So the base angle of each isosceles triangle is half that, or 67.5 degrees. Now pick one of the sides of the octagon, and draw a line from the center of the octagon to the center of the side. You will have two right triangles. Let's look at one of them: b ________ /\t | \ t = 67.5 deg / \ | \ c \ |a a = (1/2)(diameter of octagon) \| From trigonometry, we know that tan(t) = a/b b = a/tan(t) = a/2.414 (approximately) If you know the longer diameter, it's essentially the same story, except that now you know c instead of a, and cos(t) = b/c b = c cos(t) = c * 0.383 (approximately) Does this help? By the way, that sounds like it's going to be an interesting house! - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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