hint: recall that 1+2+...+k = k(k+1)/2
Now start counting your diagonals for an n-gon.
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
3 answers
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
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
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
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