How many triangles can we form if we draw all the diagonals of a hexagon?
I thought that the answer is $\binom{6}{3}=20$ but this is not the right answer, why?
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$\begingroup$The problem is very unclear (see the comments). Here is one interpretation (which is probably not the one intended, but who knows?):
Drawing all 9 diagonals of a regular hexagon divides it into 24 regions, of which 6 are quadrilaterals, leaving 18 triangles.
$\endgroup$ 3 $\begingroup$Assuming a regular hexagon:
If you draw all diagonals of a regular hexagon you have $3 \cdot 6 = 18$ possible triangles, but 3 of those are the same (the equilateral triangles) so we have $18 - 3 = 15$ possible triangles.
$\endgroup$ $\begingroup$let me set of this numbers, where in every number corresponds with a number of sides of every polygon.. ( 3,4,5,6,7,8,9,10 ),,let me answer how many diagonal can be drawn from the fixed vertex?? and how many triangles are formed from this diagonal??
1.) Triangle = 3 sides, 0 diagonal, 1 triangle
2.) quadrilateral = 4 sides, 2 diagonal formed, 8 triangles formed
3.) Pentagon = 5 sides, 5 diagonal formed, 40 triangles formed
4.) hexagon = 6 sides, 9 diagonal formed, ????????? :))
$\endgroup$ $\begingroup$I can see 35 in a pentagon, by organising my triangles by the quantity of shapes each is constructed of:
10 triangles made of 1 shape. 10 triangles made of 2 shapes. 10 triangles made of 3 shapes. 5 triangles made of 5 shapes. Total of 35 triangles.
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