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

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


   

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:
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
n is the number of sides
:
D is the number of diagonals
:
D=n(n-3)/2
:
35=n%5E2-3n%2F2
:
n%5E2-3n-70=0
:
throw out the negativesystem%28n=10%2Cn=-7%29
:
highlight%28n=10%29sides so a decagon
:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation an%5E2%2Bbn%2Bc=0 (in our case 1n%5E2%2B-3n%2B-70+=+0) has the following solutons:

n%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-3%29%5E2-4%2A1%2A-70=289.

Discriminant d=289 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--3%2B-sqrt%28+289+%29%29%2F2%5Ca.

n%5B1%5D+=+%28-%28-3%29%2Bsqrt%28+289+%29%29%2F2%5C1+=+10
n%5B2%5D+=+%28-%28-3%29-sqrt%28+289+%29%29%2F2%5C1+=+-7

Quadratic expression 1n%5E2%2B-3n%2B-70 can be factored:
1n%5E2%2B-3n%2B-70+=+1%28n-10%29%2A%28n--7%29
Again, the answer is: 10, -7. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-3%2Ax%2B-70+%29

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A polygon has 35 diagonals. How many sides does it have?
------------
You can figure out these big number problems by looking at small number examples.
------------
If 4 sides you get 2 diagonals
If 5 sides you get 6 diagonals
If 6 sides you get ? diagonals
---------------
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...?
------------
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.
--------------
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.
---------------
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.
----------------
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.
====================================
Hope this helps.
Cheers,
Stan H.