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Type I - Parallels and Parallelograms

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Introduction

Parallels and Parallelograms Math Portfolio

Introduction:

This investigation aims at finding a relationship between the numbers of horizontal parallel lines and the transversals. When these lines intersect they form parallelograms. The aim of this investigation is examine and determine a general statement for transversals and horizontal lines and how they affect the number of parallelograms formed within the figure. A diagram of a parallelogram and a transversal is shown below.

image00.png

These lines represent two transversals; however they are supposed to intersect with a horizontal parallel line.

image01.png

These lines represent two horizontal parallel lines. They are intersected with a number of transversals.

When there is one horizontal line and two transversals, this makes one parallelogram, A1, as shown below:

image09.png

This diagram shows that when there are two horizontal lines and two transversals, one parallelogram is formed. This parallelogram is called A1. However, when another transversal is added to the same diagram and same pair of horizontal lines, it looks like this:

image20.png

This figure shows that when another transversal is added to a pair of horizontal parallel lines, three parallelograms are formed. Although the third parallelogram is not labeled, it is clear that the total of A1image02.pngimage02.png A2 makes a big parallelogram, A3. So therefore, there is A1, A2, and A3. Another transversal added to this figure is shown below:image22.png

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Middle

3, A4, A5 and A6. A7 is shown because A1 can combine with A2, A1image02.pngimage02.png A2. Then A1image02.pngimage02.png A2image02.pngimage02.png A3 show another parallelogram, A8. After this A1image02.pngimage02.png A2image02.pngimage02.png A3image02.pngimage02.png A4 show that there is another parallelogram, A9. A10 is shown by the combination of A1image02.pngimage02.png A2image02.pngimage02.png A3image02.pngimage02.png A4image02.pngimage02.png A5. Then A11 is shown because the combination of  A1image02.pngimage02.png A2image02.pngimage02.png A3image02.pngimage02.png A4image02.pngimage02.png A5image02.pngimage02.png A6. The twelfth parallelogram, A12, is shown by combining A2image02.pngimage02.png A3. A13 is shown by combining A2image02.pngimage02.png A3image02.pngimage02.png A4. A14 is shown by combining A2image02.pngimage02.png A3image02.pngimage02.png A4image02.pngimage02.png A5. A15 is then shown when combining, A2image02.pngimage02.png A3image02.pngimage02.png A4image02.pngimage02.png A5image02.pngimage02.png A6. A16 is formed because of the combination of, A3image02.pngimage02.png A4. A17 is therefore shown by the combination of A3image02.pngimage02.png A4image02.pngimage02.png A5. Then after this, A18 is shown when combining, A3image02.pngimage02.png A4image02.pngimage02.png A5image02.pngimage02.png A6. By combining A4image02.pngimage02.png A5, this shows another parallelogram, A19. A20 is shown when combining A4image02.pngimage02.png A5image02.pngimage02.png A6. Lastly, the twenty first parallelogram is shown when combining A5image02.pngimage02.png 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.

Number of Transversals

Number of Parallelograms

2

1

3

3

4

6

5

10

6

15

7

21

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.

image03.png

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Conclusion

2

1

2

3

3

3

3

9

4

6

4

18

5

10

5

30

6

15

6

45

7

21

7

63

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 image10.pngimage10.png. To simplify this the formula can be written as, image11.pngimage11.png. Testing this formula is shown below:

When N =2, N = image12.pngimage12.png = image13.pngimage13.png = 3 Parallelagrams, this matches the value in the table.

When N =3, N =  image14.pngimage14.png = image15.pngimage15.png = 9 Parallelagrams. This also matches the actual value.

When N = 7, N = image16.pngimage16.png = image17.pngimage17.png = 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  image18.pngimage18.png.

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.  

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This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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