Could use inductive proof too. First, check for base case of n = 3. In a triangle there are 0 diagonals. And (3)(3-3)/2 = 3*0/2 = 0, so initial condition works. Assume n sided polygon and add another vertex. The vertex can draw a diagonal to all of the original vertices except the two it borders so that is adding (n-2). It also allows a diagonal to be drawn between the two vertices it borders as they no longer share an edge, so that is adding 1. Diagonals for n+1 = Diagonals for n + (n-2) + 1 D = (n)(n-3)/2 + (n-2) + 1 D = (n2 - 3n) / 2 + (n-2) + 1 D = [n2 -3n + 2n - 4 + 2] /2 D = [n2 - n - 2] /2 D = [(n+1)(n-2)] /2 D = [(n+1)(n+1 - 3)] /2 And there we have that if true for the n case it is true for n+1, so true for all n >= 3
Thank you sir for this video. I didn't know how this formula made so I wanted to know its proof and your video explain everything that I wanted to know.
Yo wait I have an idea, how about you do one vertice, so for example. Every vertices has 5 diagonals, why not put that on every vertices? (Wont work on Concave polygons though)
@@alexisgiovanebinas6032 Sure, i'll try to be as clear as possible. To make a diagonal you need to take 2 vertex of a polygon, these vertex don't have to be contiguous (indeed if they are you'll make a side). Then i thought of all the possible ways you could arrange N vertex in groups of 2 and since the order isn't important (Taking vertex 1 and 4 would be the same as taking vertex 4 and 1) i thought of combinations. I applied the formula to a simple polygon like the pentagon and it didn't work at fist cuz i considered also the groups of contiguous vertex which, indeed, are the same as the number of Sides so i just subtracted N. And there you are
A new proof If n point are given Then total no. Of line passing through that. Is nc2 but we also counted sided along with diagonal so substract no. Of sides 'n' Then no of diagonal =nc2-n Which simplifies into n(n-3)/2
Well consider a polygon with n sides_ since a diagonal is a line going through two vertices therefore # diagonals=nC2 - n since sides also follows from the definition
Geometrically: the definition of nC2 is: given n distinct noncollinear points how many lines can we make each passing through two points..if we suppose these points are the vertices of a polygon. nC2= #diagonals + #sides.
It is not a formal proof, but it does contain all the necessary ingredients to make it an informal one. I preferred the informal one in this case because the target audience for this particular video was middle schoolers and some high schoolers. It is often the case that too much formality obscures the simplicity and fundamental nature of the proof--I thought my target audience would most benefit from a more intuitive explanation.
( What i understood ) (n-3) will get all possible diagonal of one vertex and then we have to multiple it by n again because we need to get all diagonals from all vertex(s) so n×n-3 but we have to do -3 part first and then multiply so n(n-3) and this will get us all possible diagonals but counted from both sides so we don't want to count 1 diagonal 2 times so divide the answer of n(n-3) by 2 so this forms the formula n(n-3/2). Hope you have understood
![Megan Thee Stallion - Bigger In Texas [Official Video]](http://i.ytimg.com/vi/QxTkZTLWdo4/mqdefault.jpg)
Could use inductive proof too.
First, check for base case of n = 3. In a triangle there are 0 diagonals. And (3)(3-3)/2 = 3*0/2 = 0, so initial condition works.
Assume n sided polygon and add another vertex. The vertex can draw a diagonal to all of the original vertices except the two it borders so that is adding (n-2). It also allows a diagonal to be drawn between the two vertices it borders as they no longer share an edge, so that is adding 1.
Diagonals for n+1 = Diagonals for n + (n-2) + 1
D = (n)(n-3)/2 + (n-2) + 1
D = (n2 - 3n) / 2 + (n-2) + 1
D = [n2 -3n + 2n - 4 + 2] /2
D = [n2 - n - 2] /2
D = [(n+1)(n-2)] /2
D = [(n+1)(n+1 - 3)] /2
And there we have that if true for the n case it is true for n+1, so true for all n >= 3
Tony James wot
i dont understand
this helps a lot, thanks
Tanks for confusing me more😅
I love the way you say diagonals. Btw, great video!
Outstanding mind blowing awesome sir 👌👌🙏🙏
I would be very pleased if you don't stop making videos like this 😄
This video was extremely helpful and I can actually explain this too using your method because I understood it! THANK YOU!
Number of pairs of points - number of sides
= (n choose 2) - n
= n * (n - 1)/2 - n
= (n^2 - n - 2n) / 2
= n * (n - 3) / 2
@@Ash12345p Each pair of points forms either a side or a diagnoal. If they're adjacent it's a side, otherwise a diagonal.
Will Bishop help
Wow such a great explanation of the formula 👏 ❤
thanks for purging me from this problem😂, love from India!!!!
A big thumbs-up 😊
Tqsm ...math is becoming easy by your grace ..
God bless u ...
Thanks for helping.😊😊
A Superb Explanation !
Amazing explanation sir
Thank you sir for this video. I didn't know how this formula made so I wanted to know its proof and your video explain everything that I wanted to know.
Thank you sooooooo much sir , you made me understand everything😅
God bless ya
Of course! I'm glad it helped. =)
Great help sir💕
Thanks for helping us
Yo wait I have an idea, how about you do one vertice, so for example. Every vertices has 5 diagonals, why not put that on every vertices?
(Wont work on Concave polygons though)
Perfect explanation 👍
[C(n,2)= n!/2*(n-2)!]-n
The -N at the end excludes the possibility of taking 2 contiguous vertex
wow! can you explain how you come up with that?
@@alexisgiovanebinas6032 Sure, i'll try to be as clear as possible.
To make a diagonal you need to take 2 vertex of a polygon, these vertex don't have to be contiguous (indeed if they are you'll make a side).
Then i thought of all the possible ways you could arrange N vertex in groups of 2 and since the order isn't important (Taking vertex 1 and 4 would be the same as taking vertex 4 and 1) i thought of combinations.
I applied the formula to a simple polygon like the pentagon and it didn't work at fist cuz i considered also the groups of contiguous vertex which, indeed, are the same as the number of Sides so i just subtracted N. And there you are
@@Yperrr thanks a lot. Appreciated it!
Super vedio 👍
Yeah very helpful but it would be confusing for some so you should have written the formula like N x N-3/2 where n is equal to number of sides
i was completely confused in class.. this video explained everything perfectly! Thankyou so much!!!
Really loving your solution........
Thank you so much sir...
Really helped me a lot.
THANK YOU SO MUCH! I understand!
Very good explanation
Thank you for the explanation
It was really really helpful.. thanks for uploading this video... I was searching for this since a long time
The number of diagonals from a single vertex is n-3 .pls explain this concept
Love the explanation ❤
Thank youu for such a simple explanation !😃
Thank u brother it helped me a lot
GOD BLESS YOU
THANK YOU SO MUCH
I'm going to use this
A new proof
If n point are given
Then total no. Of line passing through that. Is nc2 but we also counted sided along with diagonal so substract no. Of sides 'n'
Then no of diagonal =nc2-n
Which simplifies into n(n-3)/2
Very very much thanks to make this video.
thanks bro for the explanation
thanks..very nice explanation
Well consider a polygon with n sides_ since a diagonal is a line going through two vertices therefore
# diagonals=nC2 - n since sides also follows from the definition
Geometrically: the definition of nC2 is: given n distinct noncollinear points how many lines can we make each passing through two points..if we suppose these points are the vertices of a polygon. nC2= #diagonals + #sides.
Thank SIR 😊😊😊
I DON'T KNOW THAT TRICK
thank u sir so much sir
Are you indian or chinese who lives in america? This really helps me!
Thq soooo much for ur help
Thank you so much❤❤❤❤❤❤
Helped alot thanks
what happens when n is odd or the polygon has an odd number of sides
thank you, this was really helpful
I'm glad it helped! =)
Raghad Ayman heeh
Thanks
thanks!
Thanks you for understanding a question 🥰🥰🥰
Thanks mate
Wow👏👏👏👏👏
IN A SIMPLY WAY YOU UNDERSTAND
Thanks sir ❤️
Thanks 👍
Thank you sir
I had proved this many years ago when I was 13 years old .
very very thank you
U helped me so much thank you!:)
Thank you for your video sir😅😅😅😅😊
Thank you😘
Super
thank you
Thank u sir it really worked and I understand each and evrey thing's
👌🕺
By using n(n-3)/2 is very easy to do
Why not using combination method to prove? Seem like easier(in my opinion)
You didn't prove it though; you just tried examples and generalised it. I was expecting a proof by induction.
It is not a formal proof, but it does contain all the necessary ingredients to make it an informal one. I preferred the informal one in this case because the target audience for this particular video was middle schoolers and some high schoolers. It is often the case that too much formality obscures the simplicity and fundamental nature of the proof--I thought my target audience would most benefit from a more intuitive explanation.
Ps. Pls post more videos😂😂
nC2 - n
😘👌video
S=(n-2) where is come -2
Thank you sooo much
My pleasure! =)
Sir for covex polygon
What about a septagon
thankk youu
why do you multiply n by (n-3)? thank you!
( What i understood )
(n-3) will get all possible diagonal of one vertex and then we have to multiple it by n again because we need to get all diagonals from all vertex(s) so n×n-3 but we have to do -3 part first and then multiply so n(n-3) and this will get us all possible diagonals but counted from both sides so we don't want to count 1 diagonal 2 times so divide the answer of n(n-3) by 2 so this forms the formula n(n-3/2). Hope you have understood
N means number of sides :)
Budget khan academy
I think you r a Chinese by your accent.......
not integer, variable
okay
Thanks bro for the explanation
Thanks
Thank you sir
nC2- n