# Thread: Looking for a polygon with given diagonals...

1. ## Looking for a polygon with given diagonals...

The formula for getting the number of diagonals of a certain polygon is:

N (N-3) / 2

where n stands for the polygon..

But here's the question:

What kind of polygon has 54 diagonals?

I tried to manipulate the formula as in reciprocated it in this form:

54 divided by the difference of 54 and 3 then multiply the quotient by two

The answer I obtained is 2.11

2. Number of diagonalizes is 54.

$\displaystyle 54 = \frac{n(n-3)}{2}$

$\displaystyle 108 = n(n-3)$

$\displaystyle 108 = n^2 - 3n$

$\displaystyle n^2 - 3n - 108 = 0$

Solve the quadratic to find n.

3. Another way to solve besides using the quadratic equation is to notice that

$\displaystyle 108=2\cdot2\cdot3\cdot3\cdot3$

We are looking for two divisors of 108 whose product is 108 and whose difference is 3. Notice that 108=(4*3)*(3*3) and that 4*3 and 3*3 differ by 3.

4. Thank you for the replies...I really appreciate it..