Perimeter of OctagonDate: 04/02/97 at 21:59:09 From: Anonymous Subject: Perimeter of Octagon Dr.Math, The person wants to make a vegetable garden in an octagonal shape with a six foot diameter. She plans to use special laminated coated logs for the perimeter of the garden. She wanted to know how long each side of the octagon would have to be to give a diamter of 6 feet. It is a regular polygon. We decided to use the formula of circumference of a circle formula and figured out that the circumference of a circle with a six foot diamter would be 18.84 and divided by 8 would be 2.355 feet for each part. We know that the octagon would be smaller than this but wondered if there was a formula for finding the perimeter of an octagon. Also, please note that the diamter used was not from one vertex to another. It was from the midpoint of one side to the midpoint of the opposite side. I figured that maybe the Pythagorean formula was necessary here and came out with an answer of approximately 2.1213 feet for the lengths of each section of the octagon. Was this correct or is there another way to find the perimeter? Thanks for your help. Anonymous Sewell, New Jersey Date: 04/03/97 at 11:20:38 From: Doctor Wilkinson Subject: Re: Perimeter of Octagon Your idea of using the circle as an approximation was a good one. However, if the diameter of the octagon is measured from one side to the other and not from vertex to vertex, you are making a mistake in saying that the octagon will be smaller. The circle you get will be inside the octagon, not outside, so its circumference will actually be smaller. Again you are right in thinking of using the Pythagorean Theorem, but something seems to have gone wrong, because you are not getting the right answer. This problem is somewhat tricky, however, so don't feel too bad. If you take two consecutive vertices P and Q of the octagon and look at the angle POQ, where O is the center of the octagon, you can see that POQ measures 45 degrees, because it has to be 1/8 of 360 degrees. Now if you let X be the midpoint of PQ, you have a right triangle POX, where OX is 3 (half the diameter of the octagon), and angle POX is half of 45 degrees. So the problem is to find PX, which is half the side of the octagon. (I suggest you draw a picture of this.) I don't quite know what to do with the angle of 22 1/2 degrees, so let's extend the line segment XP to a point Y so that YOX is 45 degrees. So now we have a new right triangle YOX, and it's an isosceles right triangle because of that 45-degree angle. Let's let x be PX (the length we're trying to find), and let y be YP. We know that x + y = OX = 3. Can we get another equation? We need to add one more point to the picture. Draw a perpendicular from P to OY, meeting OY at Z. Now we have something really useful. We have congruent triangles OPZ and OPX, because they share a side, they're both right triangles, and the angles ZOP and POX are both 22 1/2 degrees. So PZ is the same as PX, which is x. We also have similar triangles YZP and YOX, because they're both right triangles and they share the angle OYX. So the ratio of y to x is the same as the ratio of OY to OX. But that ratio is just the square root of 2, by the Pythagorean Theorem. So we get y/x = the square root of 2, or 1.414 approximately. y = sqrt(2) x x + y = 3 and we can now solve for x, getting approximately x = 1.242 or 2.484 for the side of the octagon. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 04/03/97 at 21:40:02 From: Anonymous Subject: Re: Perimeter of Polygon Doctor Wilkinson, Thanks a lot. I understand your explanation completely. You did a nice job with your explanation. :) Anonymous Date: 10/02/2002 at 09:58:00 From: David J. Dainelis Subject: Perimeter of an Octagon Dr. Math, While following along the explanation I could not visualize "extend a the line segment xp to a point Y so that YOX is 45 degrees" and "we need to add one more point to the picture. Draw a perpendicular from P to OY, meeting at Z." Could you create a small graphic with the points on it? Thank you in advance. David J. Dainelis Adjunct Electronics Faculty Grand Rapids Community College. Date: 10/02/2002 at 11:08:02 From: Doctor Jerry Subject: Re: Perimeter of an Octagon Hi David, Here is a simpler solution, involving trig. We don't need any extra points. We just notice that angle XOP is 45/2 = 22.5 degrees. So tan(22.5) = XP/OX = XP/3. So, XP = 3*tan(22.5) = 1.2426... I'll describe the construction in a slightly different way. Draw a set of axes, and draw any circle with center at the origin. Draw the 45-degree line through the origin, and draw the 135-degree line through the origin. Let O be the center of the circle, P the point on the positive x-axis where the circle crosses the x-axis. Let Q be the point in the first quadrant where the 45 deg line intersects the circle. Connect P and Q. The line PQ is a side of the octagon. Let X be the midpoint of PQ. Extend QP downward and to the right until you find point Y on this extension. Choose Y so that angle YOX is 45 degrees. The point Z is found by drawing from P a line that meets YO at a right angle. The meeting point is Z. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 10/02/2002 at 11:22:20 From: David J. Dainelis Subject: Thank you (Perimeter of an Octagon) Thanks that helps a lot. I knew there would be a simpler description. djd Date: 10/02/2002 at 11:37:13 From: Doctor Greenie Subject: Re: Perimeter of an Octagon Hello, David - I read with interest the referenced page in the Dr. Math archives and Dr. Jerry's response to your message. The method described for finding the perimeter of the octagon is far more complex than it needs to be. Below is a derivation which uses nothing more complex than the ratios of the lengths of the sides of an isosceles right triangle. We will think of the octagon as having been cut from a square 6 feet on a side. So to get started, we will sketch a square with side 6 feet and draw lines at 45-degree angles in each corner indicating where we make the cuts to form the hexagon. If the length of a side of the octagon is x, then the hypotenuse of each of the triangles is x, and so the length of each leg of the triangles is x/sqrt(2). And the length of each side of the original square is composed of one side of the octagon and two legs of the triangles. So we have 6 = x + 2(x/sqrt(2)) from which it follows that x = 6/(1+sqrt(2)) = 2.485281374239 - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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