Polygon diagonals formula

[Naruto]

Well guys I searched if this has been posted before, and didnt find it. Btw whats the best way to search if a question has been posted before or not. I simply put the whole question in the search, is there a smarter way I am missing?

Anway, here's the question

A Polygon has 20 diagonals, how many sides does it have.

A 12
B 11
C 10
D 9
E 8


Now by taking general cases I was trying to solve and I came up with this equation where No. of diagonals of a polygon with n sides = (n-3)n/2
Its applicable here, but i want to know if its general, and if there is any other shorter way to solve the problem

[tohellandback]

well this is how I solved:
0 diagonals-3 sides
2 diagonals-4 sides
5 diagonals-5sides
9 diagonals-6sides

you see 0+2=2,2+3=5,5+4=9. so next will be 9+5=14, 14+6=20
20 diagonals will have 8 sides

[Naruto]

well this is how I solved:
0 diagonals-3 sides
2 diagonals-4 sides
5 diagonals-5sides
9 diagonals-6sides

you see 0+2=2,2+3=5,5+4=9. so next will be 9+5=14, 14+6=20
20 diagonals will have 8 sides

I didnt understand please elaborate what series are u continuing?

[tohellandback]

0-2 (add 0 to 2)
2-5(added 3)
5-9 (added 4)
so next you will add 5
9-14(added 5)
14-20(added six)

[dtweah]

Well guys I searched if this has been posted before, and didnt find it. Btw whats the best way to search if a question has been posted before or not. I simply put the whole question in the search, is there a smarter way I am missing?

Anway, here's the question






Now by taking general cases I was trying to solve and I came up with this equation where No. of diagonals of a polygon with n sides = (n-3)n/2
Its applicable here, but i want to know if its general, and if there is any other shorter way to solve the problem

A Polygon has 20 diagonals, how many sides does it have.

A 12
B 11
C 10
D 9
E 8

What is a shorter way? What can be shorter and surer than factoring
n^2 -3n -40=0
(n-8)(n+5)=0
n=8.
With formula this takes less than 15 seconds. There is no shorter way order than already knowing the answer!!

[Naruto]

Well guys I searched if this has been posted before, and didnt find it. Btw whats the best way to search if a question has been posted before or not. I simply put the whole question in the search, is there a smarter way I am missing?

Anway, here's the question






Now by taking general cases I was trying to solve and I came up with this equation where No. of diagonals of a polygon with n sides = (n-3)n/2
Its applicable here, but i want to know if its general, and if there is any other shorter way to solve the problem

A Polygon has 20 diagonals, how many sides does it have.

A 12
B 11
C 10
D 9
E 8

What is a shorter way? What can be shorter and surer than factoring
n^2 -3n -40=0
(n-8)(n+5)=0
n=8.
With formula this takes less than 15 seconds. There is no shorter way order than already knowing the answer!!

You misunderstood me, I meant is this formula genuine because i derived it based on eg. for triangel, quadrilateral, pentagon and hexagon. I wanted to know if there is another approach, because deriving that formula took me 4 mins.

[Vemuri]

I solved the question the same way as tohellandback did. But, your question is valid. Even I am curious to know if there is a relationship between sides & diagonals of a polygon. It will greatly help in answering the questions quickly.

[scoobydooby]

there indeed is a general formula to find out the number of sides given the number of diagonals. (dug out my high school math book)

the number of diagonals of a polygon of n sides is: nC2-n

[rah_pandey]

A formal proof of n^2-3n-40=0

for a n sided polygon lets choose one point say A
now there are n-1 points except A itself from which lines can be drawn to A but not all will be diagonals. 2 such lines that can be drawn are the sides passing from A and joining adjoining points

thus effectively n-3 points can be joined to form a diagonal from A thus n-3 diagonals are possible from any given point

for n points total no of such lines =n*(n-3)

but of these n*(n-3) there is overlap(by 2 times)

thus total no of diagonal =n*(n-3)/2

using for 20 diagonals

we get n(n-3)/2=20
=> n=8 or n=-5

[Vemuri]

there indeed is a general formula to find out the number of sides given the number of diagonals. (dug out my high school math book)

the number of diagonals of a polygon of n sides is: nC2-n

That's great Scoobydooby !!! Thanks a ton for providing the formula. Here goes my "thanks" to you

[Ian Stewart]

A formal proof of n^2-3n-40=0

There are a few good ways to prove the formula, and rah_pandey provided one above. An alternative proof - first starting with a simpler question:

* If you draw n dots on a page, how many different lines can be drawn connecting pairs of these dots? This is the same question as: If n people attend a party and all shake hands, how many handshakes take place? Since we need to choose 2 dots to make a line, and order doesn't matter, the answer is nC2.

* Now, looking at the n corners of an n-sided shape, we can draw nC2 different lines connecting pairs of these corners. All of these lines are diagonals *except* the n edges, so we have nC2 - n diagonals. That might look different from the formula in the post above, but it turns out to be equal if you expand.