# Looking for a polygon with given diagonals...

• Jun 10th 2010, 11:12 PM
quantumie
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

• Jun 11th 2010, 12:32 AM
sa-ri-ga-ma
Number of diagonalizes is 54.

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

$108 = n(n-3)$

$108 = n^2 - 3n$

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

Solve the quadratic to find n.
• Jun 11th 2010, 12:35 AM
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Another way to solve besides using the quadratic equation is to notice that

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

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.
• Jun 11th 2010, 01:53 AM
quantumie
Thank you for the replies...I really appreciate it..