proof of finding the diagonals of n sided polygon??? Share with your friends Share 0 Ambilimon answered this Dear Student, Please find below the solution to the asked query: Consider a polygon of n sides.Name the vertices as 1, 2, 3, ..., n.Consider the vertex 1. Draw lines starting from this point.Refer the figure.Each line joins a pair of vertices.The possible pairs starting at 1 are,1,1, 1,2, 1,3,...1,n-2, 1,n-1, 1,nThere are n number of such pairs.Of these,1,1 represents a point, not a line. Hence it does not qualify for a diagonal.1,2 and 1,n represent two sides of the polygon. They also do not qualify for diagonals.Number of pairs that are not diagonals=3Of n possible pairs, the number of diagonals=n-3That is, for each vertex, there are n-3 diagonal pairs, and there are n vertices.Hence the total number of diagonal pairs=n×n-3Till now we have seen the number of different pairs.We have counted, for example, 1,3 and 3,1 as different pairs.But geometrically, 1,3 and 3,1 represent a single diagonal.We have counted all diagonals twice.Hence the number of diagonals actually is half of n×n-3The number of diagonals in an n-sided plygon=nn-32This completes the proof. If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible. 0 View Full Answer
