SOLUTION: How many sides are there in a polygon if the number of sides equals the number of diagonals?

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Question 1164700: How many sides are there in a polygon if the number of sides equals the number of diagonals?
Found 3 solutions by Edwin McCravy, Theo, ikleyn:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!

0 sides and 0 diagonals.  In other words, the polygon is degenerated into a
single point, which has no sides and no diagonals.  In higher geometry courses
that is allowed.  But in lower level geometry it is not.  If you are in a
lower level of geometry, then the answer is "There are no such polygons,
because there must always be 2 fewer diagonals than sides".

Edwin

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
the number of diagonals is equal to the number of sides times (number of sides - 3) / 2.

when n = 3, the number of diagonals is equal to 3 * (3 - 3) / 2 = 0
when n = 4, the number of diagonals is equal to 4 * (4 - 3) / 2 = 1
when n = 5, the number of diagonals is equal to 5 * (5 - 3) / 2 = 5

your answer is that the number of sides equals the number of diagonals when the number of sides = 5.


Answer by ikleyn(52780)   (Show Source): You can put this solution on YOUR website!
.

n sides and    diagonals.


This quantities are equal (given)


    n = 


Simplify


    2n = n*(n-3)

    n^2 - 3n - 2n = 0

    n^2 - 5n = 0

    n*(n-5) = 0.


The roots are  n= 0  and  n= 5.


Of them, only n= 5 is interesting for us.


It is the problem's answer.


ANSWER.  5-sided polygon (pentagon).

Solved.



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