a polygon has 65 diagonals .find the number of sides othe polygon?
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no of diagonals = n(n-3) /2
65= n(n-3)/2
130 = n^2 -3n
n^2 -3n -130 = 0
n^2 -13n +10 n -130 = 0
n ( n-13)+10(n-13)=0
(n-13)(n+10)=0
n= 13,-10
-10 sides are not possible so answer is 13 sides
65= n(n-3)/2
130 = n^2 -3n
n^2 -3n -130 = 0
n^2 -13n +10 n -130 = 0
n ( n-13)+10(n-13)=0
(n-13)(n+10)=0
n= 13,-10
-10 sides are not possible so answer is 13 sides
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Let the number of sides of the polygon be n
now, the formula of for the number of diagonal is= n(n-3)/2
A/Q. n(n-3)/2 =65
n^2- 3n= 130
n^2-3n-130=0
....now see in the pic.
now, the formula of for the number of diagonal is= n(n-3)/2
A/Q. n(n-3)/2 =65
n^2- 3n= 130
n^2-3n-130=0
....now see in the pic.
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