Formulas: Width, Side Length of Octagon
Date: 03/24/97 at 22:45:29 From: cash perkins Subject: Width, Side Length of Octagon Hi, I trying to find some simple geometric formulas: 1. Given the width of a regular octagon, find the side lengths. 2. Given the side length, find the width of the octagon. Thanks in advance, Cash Perkins
Date: 03/25/97 at 07:29:54 From: Doctor Anthony Subject: Re: Width, Side Length of Octagon If you draw a circle of radius r and construct the octagon to lie inside the circle with the vertices on the circumference of the circle, then drop a perpendicular from the centre of the circle to the mid-point of one of the sides, it is easy to see that the length of this perpendicular is r.cos(22.5), so width of octagon is 2r.cos(22.5) Also, half the length of the side of the octagon is r.sin(22.5), so the length of a side of the octagon is 2r.sin(22.5) ratio side/width = 2rsin(22.5)/2rcos(22.5) = tan(22.5) = 0.414213 You could stop here if you like, but if you would prefer the answer in surd form, we can use the double angle formula for tan(x) tan(45) = 2tan(22.5)/[1-tan^2(22.5)] If tan(22.5) = x we get 1 = 2x/(1-x^2) so 1-x^2 = 2x x^2 + 2x - 1 = 0 x = [-2 +or- sqrt(4+4)]/2 x = [-2 +or- sqrt(8)]/2 we can ignore the negative option since x must be positive, so x = [-2 + 2sqrt(2)]/2 = -1 + sqrt(2) So ratio side/width = sqrt(2)-1 = 0.414213 as before. Knowing the ratio, then if given the width the side =(sqrt(2)-1)width. Given the side, then width = side/[sqrt(2)-1] -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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