How would you approach this question (1 Viewer)

[namburger]

If a polygon has n sides, how many diagonals does it have?

My friends used simple cases such as square,pentagon and hexagon. Then tried to find a relationship which was quite difficult
only one of them got the answer

can anyone post some working out/explanation or alternative approaches
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[Trebla]

A diagonal is defined as the line joining two non-adjacent vertices. So, given one vertex in an n-sided polygon, there are 2 adjacent vertices. So it cannot form a diagonal with those adjacent vertices. This leaves (n - 3) vertices which it can form diagonals with. (n minus itself and the two adjacent vertices)

Now that's the case for ONE vertex. There are n vertices in an n-sided polygon, so it appears there are n(n - 3) diagonals. BUT we've double counted the diagonals. A diagonal from point A to point B is exactly the same as a diagonal from point B to point A. Thus, total diagonals is n(n - 3)/2

 

[namburger]

A diagonal is defined as the line joining two non-adjacent vertices. So, given one vertex in an n-sided polygon, there are 2 adjacent vertices. So it cannot form a diagonal with those adjacent vertices. This leaves (n - 3) vertices which it can form diagonals with. (n minus itself and the two adjacent vertices)

Now that's the case for ONE vertex. There are n vertices in an n-sided polygon, so it appears there are n(n - 3) diagonals. BUT we've double counted the diagonals. A diagonal from point A to point B is exactly the same as a diagonal from point B to point A. Thus, total diagonals is n(n - 3)/2

haha thx, makes perfect sense

 

[aakash]

Another way to approach this:

for a diagonal we need 2 vertices, no. of ways do choose this = nC2
but in these we also chose adjacent vertices (ie we counted the sides as well which are n in total).
so required answer = nC2 - n
this reduces to n(n-3)/2...same as what Trebla got.