a polygon with n number of sides will have how many digonals
a)n
b)n(n-3)
c)n/2
d) n-3/2
e)n(n-3)/2
or, think of it this way:
The number of possible lines between n points = C(n,2)
but n of these are sides of the polygon, the rest are diagonals
So, the number of diagonals
= C(n,2) - n
= n!((n-2)!2!) - n
= n(n-1)/2 - 2n/2
= (n^2 - n - 2n)/2
= n(n-3)/2
hint: recall that 1+2+...+k = k(k+1)/2
Now start counting your diagonals for an n-gon.
c does not work if n is odd.
Also, sorry about my hint. That does not apply here.
There are n vertices.
From each vertex, you can draw n-3 diagonals. (don't count your vertex or its two neighbors)
So, there are n*(n-3) diagonals to draw, using each vertex as a base.
But, now you have drawn each one twice, from both ends.
So the answer is E: n(n-3)/2
To find out how many diagonals a polygon with n sides has, we need to understand the concept of a diagonal in a polygon.
A diagonal is a line segment that connects two non-adjacent vertices of a polygon. In other words, it is a line segment that does not intersect any of the sides.
To count the number of diagonals, we can start with one vertex of the polygon and count the number of line segments we can draw to connect it with the other vertices. However, we need to keep in mind that we cannot connect it to adjacent vertices, as that would not be a diagonal.
Let's consider a polygon with n sides:
- From one vertex, we can draw diagonals to n-3 other vertices. (We subtract 3 because we cannot connect to the adjacent vertex and the two vertices adjacent to it.)
Now, we have accounted for one diagonal from the starting vertex. However, we need to repeat this process for all vertices of the polygon, excluding the starting vertex.
- Since there are n vertices in total, we have to repeat the process n-1 times.
Finally, to calculate the total number of diagonals, we multiply the number of diagonals from one vertex by the number of times we repeat the process for all vertices:
Total number of diagonals = (n-3) * (n-1)
Simplifying this expression, we get:
Total number of diagonals = n(n-3)
Therefore, the correct answer is e) n(n-3)/2.