In this figure, there are six transversals added to the pair of horizontal parallel lines. With six transversals, there are fifteen parallelograms within this diagram. The figure firstly obviously shows five parallelograms, A1, A2, A3, A4, and A5. A6 is formed when you combine A1 with A2, A1 A2. Then when you combine A1 with A2 and A3 (A1 A2 A3), A7 is formed. Also A1 A2 A3 A4 create another parallelogram, A8. Another parallelogram is formed when A1 A2 A3 A4 A5 are combined. This makes A9. When A2 A3 are combined this makes A10. Also when A2 A3 A4 are combined, this makes another parallelogram, therefore A11 is created. A2 A3 A4 A5, when these are combined, this makes A12. Then when, A3 combines with A4, A3 A4, this makes A13. A14 is shown when A3 A4 A5. Lastly, the fifteenth parallelogram formed is when A4 is combined with A5, A4 A5.
With seven transversals, there are twenty one parallelograms within in this figure. Firstly there are six obvious parallelograms listed, A1, A2, A3, A4, A5 and A6. A7 is shown because A1 can combine with A2, A1 A2. Then A1 A2 A3 show another parallelogram, A8. After this A1 A2 A3 A4 show that there is another parallelogram, A9. A10 is shown by the combination of A1 A2 A3 A4 A5. Then A11 is shown because the combination of A1 A2 A3 A4 A5 A6. The twelfth parallelogram, A12, is shown by combining A2 A3. A13 is shown by combining A2 A3 A4. A14 is shown by combining A2 A3 A4 A5. A15 is then shown when combining, A2 A3 A4 A5 A6. A16 is formed because of the combination of, A3 A4. A17 is therefore shown by the combination of A3 A4 A5. Then after this, A18 is shown when combining, A3 A4 A5 A6. By combining A4 A5, this shows another parallelogram, A19. A20 is shown when combining A4 A5 A6. Lastly, the twenty first parallelogram is shown when combining A5 A6.
The table below is the results found when adding a transversal each time to a pair of horizontal parallel lines, resulting in an amount of parallelograms.
Using technology, a graph of this table was made, to find a relationship between the number of transversals and the number of parallelograms made. The graph is shown below.
This graph shows that the general function has to be exponential or quadratic, because of the curve. However, when graphing this on the calculator, an exponential formula does not fit the graph as well as a quadratic formula, using regressions. The exponential graph increases much faster. Obviously in a quadratic there are negative values; therefore, the x value domain must be greater than or equal to two. Using technology, a general statement and formula was found. This was found by using a quadratic regression, when listing all the points. The general formula and statement is y = 0.5x2 -0.5x. To make sure this function is valid; numbers can be substituted into this function. Also when looking at the table and analyzing this regression, since the formula is quadratic, when squaring each of these numbers, they must be subtracted by the value of themselves and then divided by two. Therefore, another statement without technology is , N is the number of transversals. When using this formula, the numbers of parallelograms are found for each transversal.
Firstly substituting 2 into this function would be the first number, because of the domain, x has to be greater than or equal to 2 because the least number of transversals needed to form a parallelogram is 2. Therefore, y = (0.5) × (2)2 – (0.5) × (2). Y = 2 – 1 = 1 parallelogram.
Substituting 3 into this function, y = (0.5) × (3)2 – (0.5) × (3). This equals, y = 4.5 – 1.5 = 3 parallelograms.
Substituting 4 into this function, y = (0.5) × (4)2 – (0.5) × (4). Y = 8 – 2 = 6 parallelograms. Obviously these numbers follow the table of values and these numbers are following the pattern of parallelograms.
Extending this diagram, adding a horizontal parallel line to make three parallel horizontal lines will form many more parallelograms. Also a relationship between these new values and the other values can be shown, resulting in a general statement for all cases. Shown below, this is a figure with two transversals and three horizontal parallel lines.
In this diagram, there are three different parallelograms. Compared to the pair of horizontal parallel lines, instead of having one parallelogram, when another horizontal line is added this figure, this creates three parallelograms. Firstly, A1 and A2 are shown; however, when combining A1 A2, this forms another parallelogram, known as A3.
This figure represents another transversal added to the three horizontal parallel lines. Within this figure, there are nine parallelograms shown. Firstly there are A1, A2, A3, and A4. A5 is shown when combining A1 with A2. A1 A3, this makes another parallelogram, forming A6. Then A7 is formed when combining A3 A4. A8 is formed when combining A2 A4. Lastly, A9 is formed when combining all of these together, A1 A2 A3 A4.
In this figure, there are a total of eighteen parallelograms within this figure. Firstly there are A1, A2, A3, A4, A5, and A6. Also A7 is formed when combining A1 A2. A8 is formed when combining A1 A3 A5. A9 is formed when combining A1 A2 A3 A4. A10 is formed when combining, A3 A5. A11 is formed when combining A3 A4 A5 A6. A12 is formed when combining A2 A4 A6. A13 is formed when combining A3 A4. A14 is formed when A5 A6. A15 is formed when combining A2 A4. A16 is formed when combining A4 A6. Also when combining A1 A3, this forms the seventeenth parallelogram, A17. Lastly, when combining all of these together this forms A18, A1 A2 A3 A4 A5 A6.
Clearly when analysing these values, when comparing them to the two horizontal parallel lines, all of these values are multiplied by three. The table below shows this easier.
This table shows that the number of parallelograms for 2 horizontal parallel lines are multiplied by three for them to equal the vaules for parallelograms for 3 horizontal parallel lines. Therefore, for three horizontal parallel lines the general statement and formula is the same as the first formula multiplied by three. Therefore, the formula for three horizontal lines is . To simplify this the formula can be written as, . Testing this formula is shown below:
When N =2, N = = = 3 Parallelagrams, this matches the value in the table.
When N =3, N = = = 9 Parallelagrams. This also matches the actual value.
When N = 7, N = = = 63 Parallelegrams, this shows that the formula is correct because this is the actual value as well.
To further extend the results, by looking at the horizontal lines added each time, it is noticable that because these are all parallel lines, they have the same formula except different variables. When flipping the diagram, it appears to be the same. Therefore, the transversals act the same as as the horizontal parallel lines. Moreover, the formula for the overall formula dealing with the variables M, horizontal lines, and N, transversals is .
The limitation of this formula and general statement are non existant. These formulas are completely correct, if the correct domain is being used. The domain for this formula is that N and M have to be greater than or equal to 2. However, for a quadratic regrission the limitation is that the x values must be always positive and never negative.