# SOLUTION: A polygon has 35 diagonals. How many sides does it have?

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> SOLUTION: A polygon has 35 diagonals. How many sides does it have?      Log On

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Question 173411This question is from textbook Introductory Algebra
: A polygon has 35 diagonals. How many sides does it have?This question is from textbook Introductory Algebra

Found 2 solutions by Mathtut, stanbon:
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n is the number of sides
:
D is the number of diagonals
:
D=n(n-3)/2
:

:

:
throw out the negative
:
sides so a decagon
:
 Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=289 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 10, -7. Here's your graph:

You can put this solution on YOUR website!
A polygon has 35 diagonals. How many sides does it have?
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You can figure out these big number problems by looking at small number examples.
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If 4 sides you get 2 diagonals
If 5 sides you get 6 diagonals
If 6 sides you get ? diagonals
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Let's see.
If you have 4 sides you have 4 vertices.
There are 4C2 = 4*3/1*2 = 6 pairs.
Four of those pairs are used to make sides.
That leaves 2 pairs for diagonals. Hmmm...?
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Similarly:
If you have 5 sides you have 5C2 = 10 pairs of vertices.
5 of the pairs are used for sides, leaving 5 pairs for diagonals.
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Let's try 6 sides.
You have 6C2 = 15 pairs of vertices.
6 of the pairs are used for sides, leaving 9 pairs for vertices.
Draw the six-sided polygon and see if it has 9 diagonls.
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If all of this is correct try it on 35 sides.
You have 35C2 = 595 pairs
35 of the pairs are used for sides, leaving 560 for diagonals.
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How about a polygon with "n" sides.
You have nC2 pairs of vertices
n of the pairs are used for sides, leaving nC2 - n pairs for diagonals.
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Hope this helps.
Cheers,
Stan H.