Area and Perimeter in PolygonsDate: 06/24/99 at 15:15:39 From: Mike Hansen Subject: Geometry Dear Dr. Math, I teach Algebra I to some very precocious 13-year-olds, and one in particular who came up with a formula for area that I would like you to look over for me. According to James, he saw a pattern in the areas of regular polygons and decided to dedicate some time to trying to find a formula that would calculate the areas of regular polygons when the lengths of the sides and the number of sides is known. He sent me the following: This formula finds the area of any regular polygon where a is the length of each side, and b is the number of sides. A = [((1/2)a) / (tan(360/2b))] * (1/2)a * b With James' permission I rewrote the formula as the following: A = (a^2b) / ((4tan(360/2b)) The formula works perfectly. Because of my general belief that seldom will any of us have an original thought I would like to pose the following question to you: Do you recognize either form of this formula? Have you seen a general formula for areas of regular polygons that is similar? Is there any historic precedent for this formula? I appreciate any time you devote to this query. I respectfully await your response. Sincerely, Mike Hansen Date: 06/24/99 at 23:37:25 From: Doctor Peterson Subject: Re: Geometry Hi, Mike. Yes, this is a good formula, and not too unusual (except coming from a 13-year-old). If you look at our FAQ on regular polygons http://mathforum.org/dr.math/faq/formulas/faq.regpoly.html you will find a formula K = na^2 cot(alpha/2)/4 where n is the number of sides and a is the length of a side. The angle alpha is 360/n. If we rewrite cot as 1/tan, this becomes n a^2 K = ------------ 4 tan(180/n) which is exactly the formula you give. (I wrote 180/n rather than 360/2n to avoid ambiguity and the need to write 360/(2n) for clarity.) You may be interested in the following answer I gave a few days ago, in the course of which I derived essentially this formula, except that I used the perimeter P = na rather than a. James probably did it more or less the same way. The result is pretty neat; if you graph it you see that it approaches 1 as N increases. James sounds like me at that age, and my son too, who is now 14. Encourage him! >Dear Dr. Math, >A circle is the largest area you can have with a given circumference. >Any regular polygon takes up X% of the area of a circle with the same >circumference/perimeter. I really need a formula where I could plug >in the number of sides of a regular polygon and see what % of a >circle it covers if they have the same perimeter. Please help me. Here's a sample polygon: s +-------+-------+ / \ | / \ / \ | / \ / \ |a / \ / \ | / \ / \ | / \ / \ | / \ / \|/2pi/N \ +---------------+---------------+ \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / +---------------+ The perimeter of an N-gon with apothem a is N times the side length s: P = N * s = N * 2 * a * tan(pi/N) The area will be N times the area of one triangle: A = N * 1/2 * s * a = N * a * a * tan(pi/N) Solving the first for a in terms of P, we get P a = ------------ 2N tan(pi/N) Substituting this into the area formula, P^2 A_p = ------------- 4 N tan(pi/N) A circle with the same perimeter will have radius r = P/(2 pi), so the area is pi P^2 P^2 A_c = pi r^2 = ------ = ---- 4 pi^2 4 pi Divide these and you're done: P^2 4 pi pi A_p/A_c = ------------- * ---- = ----------- 4 N tan(pi/N) P^2 N tan(pi/N) - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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