Number of diagonals in a polygon formula derivation
Answers
Step-by-step explanation:
The number of diagonals in a polygon = n(n-3)/2, where n is the number of polygon sides. For a convex n-sided polygon, there are n vertices, and from each vertex you can draw n-3 diagonals, so the total number of diagonals that can be drawn is n(n-3).
Answer: If we try to count the diagonals of a polygon U will find that the 1st vertex is making (n-3) diagonals same as 2nd
But the 3rd will make (n-4) diagonals
Step-by-step explanation:
We can say that we are adding the diagonals made by every vertex so as to get total number of diagonals.
Example Consider a hexagon
Here n=6
So number of diagonals are
2(6-3)+(6-4)+(6-5) STOP here because 6-5 is 1
It sums up to 9
Total number of diagonals in hexagon are 9
Basically we have
2(n-3)+Sum of (n-4) terms
Where (n-4) terms are in AP
Where d=-1 and a=n-4
So it become
2(n-3)+(n-4)/2[2(n-4)+(n-4-1)d]
=2(n-3)+(n-4)/2[n-3]
=4(n-3)/2+[(n-4)(n-2)]/2
=(n-3)/2[4+n-4]
=n(n-3)/2