Question

hey guys if anybody knows to find the no. of polynomials then it's ok & if anyone don't knows then have a look:)

Answer 1

No. of diagonals in a polygon if it has 'n' vertices:-
\[no. \space of \space diagonals=\frac{n(n-3)}{2}\]

Answer 2

Example:
Polygon with 22 vertices.
No. of diagonal=\[\frac{n(n-3)}{2}\]\[22(22-3)\over2\]\[22 \times 19\over2\]\[11 \times 19=209\]

Answer 3

and derivation of that?

Answer 4

Haha! I just went through this formula in my class :)

Answer 5

Also, please correct your question to "no. of diagonals."
You've written - 'no. of polynomials'. :P

Answer 6

sorry ! @ParthKohli

Answer 7

Nah. This was very helpful for curious people like me =)

Answer 8

thanx @ParthKohli

Answer 9

the derivation is done through example @shubhamsrg

Answer 10

lol..
i meant how do you get that formulla for a n sided figure ?

Answer 11

the derivation, the proof!! ??

Answer 12

i study about this formula through a site there they don't taught me the proof

Answer 13

didn't*

Answer 14

@shubhamsrg

Answer 15

hmm..

Answer 16

It can be derived if you know a bit of Combinatorics.


Number of ways to select any two corners = \(^{n}C_{2}\)
Now among all selections of any two corners, it wont form a diagonal if the two points are adjacent to each other.

No. of ways of selecting two adjacent points = \(n\) (Why?)

So number of diagonals = \(^{n}C_{2} - n\)\[= \frac{n(n-1)}{2}-n= \frac{n(n-3)}{2}\]