How do you find the no. of sides of a polygon if you know the no. of diagonals?

Question

If the number of diagonals in a polygon is 434, how many sides does it have? What formula do you use?

Answer 1

There are nC2 sides/diagonals to any polygon. nC2 is:

n! / (n-2)! (2)! = n(n-1)/2

Now what you have is the diagonals so:

diagonals + sides = nC2

434 + n = n(n-1)/2

868 + 2n = n² - n

868 = n² - 3n

n² - 3n - 868 = 0

(n - 31)(n + 28) = 0

n = 31 is the only answer (n = -28 is not possible).

Answer 2

No. of diagonal = n(n – 3)/2 where n = number of sides.

434 = n(n – 3)/2

n(n – 3) = 868

n² – 3n – 868 = 0

(n – 31)(n + 28) = 0

n = 31 or n = – 28

so number of sides = 31

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Answer 3

Use permutation / combination

nC2 = number of lines between 2 points

(n is the number of sides, so diagonals = total number of lines - number of sides)

nC2 - n = 434

n!/[(n-2)!2!] - n = 434

n^2 - 3n = 868

Solving quadratic equation, you get, n = 31

The number of sides = 31

Regards,

Aroop