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Mathematics 18 Online
OpenStudy (ujjwal):

The total number of diagonals in a decagon is.. I need a general relation which will hold true for all polygons!

OpenStudy (accessdenied):

I would draw a diagram in this case and just count the number of unique diagonals from each point. As for the relation, you may need to consider a few more cases or otherwise use intuition on how these diagonals are counted.

OpenStudy (ujjwal):

Ah, it says the answer is 110. I am scared to count.

OpenStudy (accessdenied):

Hmm... that's a lot for a decagon, actually. I'm only getting 35.

OpenStudy (saifoo.khan):

access is right.

OpenStudy (accessdenied):

110, which is 11*10 is like, 11 for each point, and there's not even that many points to go to, let alone having unique points. D:

OpenStudy (ujjwal):

good point!

OpenStudy (ujjwal):

Are you sure, It's only 35. ?

OpenStudy (anonymous):

You could think of it like this. There are 10 points in a decagon, and you need 2 to make a diagonal. So how many ways are there to choose 2 distinct points from 10? Thats C(10,2) which is 45. Now take away the 10 sides of the decagon. That leaves you with 35.

OpenStudy (anonymous):

Or you could think of it like this. Lay 10 points around in a circle and start at one of them. Draw a line from that starting point to each of the other points. You will draw 9 lines. Then go the another point and draw lines to the other points that arent connected already. You will draw 8 lines. Then 7, 6, so on. This gives:\[9+8+7+\cdots +2+1 = 45\] Again subtract 10 for the sides to leave the diagonals.

OpenStudy (ujjwal):

@joemath314159 Does this method hold true for other polygons like pentagon and hexagon?

OpenStudy (anonymous):

It should. Say for a hexagon. There are 6 points. C(6,2) = 15. take away 6 for the sides, so there are 9 diagonals.

OpenStudy (ujjwal):

|dw:1339804971418:dw| we have 4 diagonals here. But c(5,2)-5=5

OpenStudy (accessdenied):

|dw:1339805080366:dw|

OpenStudy (anonymous):

|dw:1339805085317:dw|

OpenStudy (ujjwal):

Oh, i missed one. Thanks @joemath314159 and @AccessDenied .

OpenStudy (accessdenied):

You're welcome. Using the combination idea, you can make / simplify a generalized formula too. :D

OpenStudy (ujjwal):

yeah, the one involving the idea of combination is very easy and i understood it.

OpenStudy (anonymous):

The logical argument is whats important. How do you make a diagonal in a polygon? You have to choose 2 points and connect them. So the geometric problem of drawing diagonals becomes a combinatorial problem instead, which requires no pictures or things of the like to solve. This is why math is great :)

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