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(52779)  (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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