- Level: International Baccalaureate
- Subject: Maths
- Word count: 1592
Parallelograms. This investigation will focus on the number of parallelograms formed by intersecting lines also knows as transversals.
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Introduction
Parallels and Parallelogram
Internal Assessment
By: Olivia Bloch
This investigation will focus on the number of parallelograms formed by intersecting lines also knows as transversals. Figure 1 below shows a pair of horizontal parallel lines and a pair of parallel transversals. One parallelogram (A1) is formed. This figure with two transversals and two horizontal lines will be transformed by adding more transversals, and hence establishing the relationship between the number of transversals and parallelograms. The parallelogram is indicated in red.
Set Notation: A1 |
Figure 2 below demonstrates how adding one transversal can form 3 parallelograms, and these are indicated in red, blue, and green.
Set Notation: A1, A2 and A₁∪ A₂ |
Six parallelograms are then formed when adding a fourth parallel transversal, as seen in figure 3. The different parallelograms have been done using different colors, and the set notation illustrates the color it belongs to.
Set Notation: A1, A2, A3, A₁∪ A₂, A₂∪ A₃and A₁∪ A₂∪ A |
Set Notation: A1, A2, A3, A4, A₁∪ A₂, A₂∪ A₃, A3 ∪ A4 and A₁∪ A₂∪ A3 ∪ A4 |
Set Notation: A1, A2, A3, A4, A5, A₁∪ A₂, A₂∪ A₃, A3 ∪ A4, A4 ∪ A5, A₁∪ A₂ ∪ A3, A2∪ A3 ∪ A4, A3∪ A4 ∪ A5, A₁∪ A₂∪ A3 ∪ A4, A₂∪ A3 ∪ A4 ∪ A5 and A₁∪ A₂∪ A3 ∪ A4 ∪ A5 |
Set Notation: A1, A2, A3, A4, A5, A6, A₁∪ A₂, A₂∪ A₃, A3 ∪ A4, A4 ∪ A5, A5 ∪ A6, A₁∪ A₂ ∪ A3, A2∪ A3 ∪ A4, A3∪ A4 ∪ A5, A4∪ A5 ∪ A6, A₁∪ A₂∪ A3 ∪ A4, A₂∪ A3 ∪ A4 ∪ A5, A3∪ A4 ∪ A5 ∪ A6, A1∪ A2 ∪ A3 ∪ A4 ∪ A5, A2 ∪ A3 ∪ A4 ∪ A5 ∪ A6 and A₁∪ A₂∪ A3 ∪ A4 ∪ A5 ∪ A6 |
Middle
3
4
6
5
10
6
15
7
21
From this data we can generate a graph using excel, in order to determine the type of sequence we are dealing with.
The graph above shows an exponential growth; hence, it is a quadratic sequence. Moreover, the figure to the right shows the features of a quadratic equations; therefore, in order to determine the general statement of the sequence, we can use a calculator as shown below.
Hence the formula for this sequence is y = .5x2 - .5x
In order to determine the validity of this statement I have tested the formula with 2, 5, and 7 transversals.
y = .5x2 - .5x y = .5(2)2 - .5(2) y = 2 – 1 ∴ y = 1 | y = .5x2 - .5x y = .5(5)2 - .5(5) y = 12.5 – 2.5 ∴ y = 10 | y = .5x2 - .5x y = .5(7)2 - .5(7) y = 24.5 – 3.5 ∴ y = 21 |
As seen above the general statement was proven correct, as the outcome for all three of the numbers were those established earlier. |
After obtaining the general formula of the relation between the number of transversals and the number of parallelograms, the task is to consider the number of parallelograms formed by three horizontal parallel lines intersected byparallel transversals. This is illustrated in figure 7 below.
Set Notation: A1, A2, and A1∪ A2 |
Set Notation: A1, A2, A3, A4, A1∪ A2, A2∪ A4, A3∪ A1, A3∪ A4, and A1∪ A2∪ A3 ∪ A4 |
Set Notation: A1, A2, A3, A4, A5, A6, A1∪ A2, A3∪ A4, A5∪ A6, A3∪ A1, A3∪ |
Conclusion
Hence the general statement for the overall pattern is y = [n(n-1)/2] ∙ [m(m-1)/2] or y = [n(n-1) ∙ m(m-1)] / 4.
The values below were figured out by plotting in the equation in the spreadsheet (refer to attached spreadsheet)
Transversals (n) | Horizontal Line(s) (m) | Parallelogram(s) |
2 | 1 | 0 |
2 | 2 | 1 |
3 | 2 | 3 |
3 | 3 | 9 |
4 | 3 | 18 |
4 | 4 | 36 |
5 | 4 | 60 |
y = [n(n-1) ∙ m(m-1)] / 4 y = [2(2-1) ∙ 1(1-1)] / 4 y = (2 ∙ 0) / 4 ∴ y = 0 | y = [n(n-1) ∙ m(m-1)] / 4 y = [2(2-1) ∙ 2(2-1)] / 4 y = (2 ∙ 2) / 4 ∴ y = 1 | y = [n(n-1) ∙ m(m-1)] / 4 y = [3(3-1) ∙ 2(2-1)] / 4 y = (6 ∙ 2) / 4 ∴ y = 3 |
As seen above the general statement for the overall pattern was proven correct, as the outcome for all three of the numbers were those established earlier. |
A quadratic equation is a polynomial equation of the second degree; hence has specific rate patterns, which creates maximums and minimums. This particular pattern involves no negative x and y coordinates, because there is no such thing as negative lines, just like with a negative area, naturally also significant to this investigation. In conclusion to this statement both the domain and the range must be equal to or greater than zero.
Overall I think this investigation was an overall success, although it did take a lot longer than I has expected because of all the drawings and set notation. I did my best in including all the elements we were supposed to. Aside from this I feel like I was successful as I used numerous methods such as creating line regressions on my calculator, checking for quadratics by graphing, and validating my results when needed.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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