Date: 03/14/98 at 14:16:52 From: Molly Thompson Subject: Polygons How many diagonals can be drawn for a polygon with 15 sides? How many diagonals can be drawn for a polygon with n sides? I tried to find a pattern in how many diagonals there are for a polygon with 4 sides, 5 sides, etc., but I couldn't extract a pattern. Please show me a method for this question.
Date: 03/16/98 at 13:59:36 From: Doctor Wilkinson Subject: Re: polygons Looking for a pattern with polygons of 4 sides, 5 sides, etc. was an excellent idea. Sometimes, though, it's pretty hard to see a pattern in a sequence of numbers. It may be better to look for the pattern in the whole process of counting the diagonals. That is, look for a systematic way of counting diagonals as you look at specific cases like 4 and 5. So let's take another look at polygons with 4 and 5 sides. For 4 sides, we have a quadrilateral ABCD. The diagonals are: AC BD There's one diagonal containing A, one containing B, one containing C, and one containing D. Let's leave it at that for a minute, and look at case 5. Here we have a pentagon ABCDE. The diagonals are: AC AD BD BE CE Here we have two diagonals containing A, two containing B, two containing C, two containing D, and two containing E. So for 4 sides there is one diagonal for each vertex and for 5 there are two. Why? Well, there is a diagonal from each vertex to each other vertex except the two adjacent vertices. For 4 vertices, there are three other vertices for each vertex, and two of them are adjacent, which leaves 1; for 5 vertices, there are four other vertices for each vertex, and two of them are adjacent, which leaves 2. Okay now, we're starting to get somewhere. For four vertices there is one diagonal containing each vertex, but we don't end up with four diagonals but just two. For five vertices there are two diagonals containing each vertex, but we don't end up with ten diagonals but just five. What's going on? Aha! If you count one diagonal for every vertex of the quadrilateral, you're counting every diagonal twice, because every diagonal contains two vertices. And similarly for the pentagon. So you have to divide by two. I hope this helps you with the general problem. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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