Question

How many diagonals can be constructed from one vertex of an n-gon? State your answer in terms of n and, of course, justify your answer.
IM REPOSTING THIS QUESTION, BUT THIS TIME FOR 15 POINTS!!!!!!!!!!!!!!
PLEASE HELP :)

Answer 1

Answer:

$The \ total \ number \ of \ distinct \ diagonals \ in \ a \ polygon \ with \ n \ sides= \dfrac{n \times (n - 3)}{2}$

Step-by-step explanation:

A diagonal is defined in geometry as a line connecting to two non adjacent  vertices.

Therefore, the minimum number of sides a polygon must have in order to have a diagonal n - 3 sides as the 3 comes from the originating vertex and the other two adjacent vertices

Given that the polygon has n sides, the number of diagonals that can be drawn from each of those n sides gives the total number of diagonals as follows;

Total possible diagonals = n × (n - 3)

However, half of the diagonals drawn within the polygon are the same diagonals drawn in reverse. Therefore, the total number of distinct diagonals that can be drawn in a polygon is given as follows;

$The \ total \ number \ of \ distinct \ diagonals \ in \ a \ polygon \ with \ n \ sides= \dfrac{n \times (n - 3)}{2}$