Download Using the following problem solving approach

Using the following problem solving approach (outlined below) - How would you
determine the total number of segments in a hexagon without actually counting them?
1. Understand the problem
2. Devise a plan
3. Carry out the plan
4. Examine the solution
Solution:
A counting technique that can do this is called combinations. To count the number of
line segments in an n-gon (both diagonals and sides), one would compute a
"combination of n vertices taken 2 at a time." A shorter way of saying this is "n
choose 2." Two at a time since each line segment is determined by 2 vertices.
n
The formula for this is C 2n , which is sometimes also written as   .
 2
C 2n = n!/(2!*(n-2)!).
"n!" is read "n factorial." It equals n(n-1)(n-2)*...*3* 2* 1.
Therefore, the number of line segments in an n-gon is [n(n-1)(n-2)* ...3*
2* 1]/[(2*1)*(n-2)* ...3* 2* 1] which reduces to n(n-1)/2.
In case of a hexagon, we get n = 6.
 n  6
6!
720
Therefore,      

 15 (Answer)
 2   2  2!(6  2)! 2 * 24
The solution for the number of sections as obtained above can be verified by actual
counting how many new segments can be drawn at each vertex and then adding up the
numbers from each vertex.
Let's look at hexagon ABCDEF first.
A
B
F
C
E
D
From vertex A, we can draw 5 new segments. (AB, AC, AD, AE, AF)
From vertex B, we can draw 4 new segments. (BC, BD, BE, BF)
From vertex C, we can draw 3 new segments. (CD, CE, CF)
From vertex D, we can draw 2 new segments. (DE, DF)
From vertex E, we can draw 1 new segment. (EF)
From vertex F, we can draw 0 new segments.
The total number of segments (including diagonals) is 1 + 2 + 3 + 4 + 5 or
15.
Notice that this is the sum of the first 5 counting numbers and that the highest number
of segments is one less than the number of vertices, 6.
From algebra, we know that
n(n  1)
n(n  1)  2n n[( n  1)  2] n(n  1)


 (n  1)   n  1  2  n 
2
2
2
For a hexagon, plug in n = 6, which then gives the total number of segments
6(6  1) 6 * 5 30


 15 , which is the same result as
(including diagonals) =
2
2
2
obtained above.