Teacher2Teacher Q&A #3829

Diagonals of polygons

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From: Jeanne (for Teacher2Teacher Service)
Date: May 03, 2000 at 21:18:30
Subject: Re: Diagonals of polygons

Here are a couple of approaches to the solution of your question. Both begin with the idea of counting all of the segments in the polygon and then subtracting the number of sides. 1. A counting technique that will help you do this is typically taught in beginning probability and statistics courses, called combinations. The teaching of combinations is usually paired with the teaching of permutations. Anyway, to count the number of line segments in an n-gon (both diagonals and sides), one would compute a "combination of n vertices taken 2 at a time." A shorter way of saying this is "n choose 2." Two at a time since each line segment is determined by 2 vertices. The formula for this is n!/(2!*(n-2)!). "n!" is read "n factorial." It equals n*(n-1)*(n-2)* ...3* 2* 1. Therefore, the number of line segments in an n-gon is [n*(n-1)*(n-2)* ...3* 2* 1]/[(2*1)*(n-2)* ...3* 2* 1] which reduces to n(n-1)/2. The number of diagonals, d, in an n-gon is the total number of segment minus the number of sides, n. d = n(n-1)/2 - n A little algebraic manipulation gives us d = n(n-3)/2. Note: Every scientific calculator I've seen has a factorial button (x!) and a "choose" button (nCr). 2. A different approach has you counting how many new segments can be drawn at each vertex and then adding up the numbers from each vertex. Let's look at hexagon ABCDEF first. From vertex A, we can draw 5 new segments. (AB, AC, AD, AE, AF) From vertex B, we can draw 4 new segments. (BC, BD, BE, BF) From vertex C, we can draw 3 new segments. (CD, CE, CF) From vertex D, we can draw 2 new segments. (DE, DF) From vertex E, we can draw 1 new segment. (EF) From vertex F, we can draw 0 new segments. The total number of segments (including diagonals) is 1 + 2 + 3 + 4 + 5 or 15. Notice that this is the sum of the first 5 counting numbers and that the highest number of segments is one less than the number of vertices, 6. This observation will be important in a bit. The number of diagonals is 15 minus 6, the number of sides, or 9. You may want to try to try this method with other polygons on your own. Now let's move on to counting the diagonals of an n-gon. The highest number of new segments that can be draw from a vertex is (n-1). The total number of segments is 1 + 2 + 3 + ... + (n-3) + (n-2) + (n-1), the sum of the first (n-1) counting numbers. A formula for the sum of the first x counting numbers is x(x+1)/2. (Dr. Math can answer where this comes from.) So, the expression for the total number of segments can be rewritten as n(n-1)/2. The number of diagonal, d, is n(n-1)/2 - n. Remember, we had to subtract the number of sides in an n-gon. Again, algebraic manipulations gives us, d = n(n-3)/3. Hope this helps. I can see doing this question with several different levels of classes. Thanks for asking it. -Jeanne, for the Teacher2Teacher service

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