• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Mathematics SL Parellels and Parallelograms. This task will consider the number of parallelograms formed by intersecting m horizontal parallel lines with n parallel transversals; we are to deduce a formula that will satisfy the above.

Extracts from this document...

Introduction

Jia-Der Ju Wang

Parallels and Parallelograms

Mathematics Coursework

This task will consider the number of parallelograms formed by intersecting m horizontal parallel lines with n parallel transversals; we are to deduce a formula that will satisfy the above.

Methodology

  1. We started out the investigation with a pair of horizontal parallel lines and a pair of parallel transversals. One parallelogram (A1) is formed (shown in Figure 1)
  1. A third parallel transversal is added to the diagram as shown in Figure 2. Three parallelograms are formed: A1, A2, and A1A2
  1. When a fourth transversal is added to Figure 2 (Figure 3), six parallelograms are formed. A1, A2, A3, A1A2, A2A3, A1A3
  1. Figure 4 has 5 transversals cutting the pair of horizontal parallels, forming ten parallelograms. A1, A2, A3, A4, A1A2, A2A3, A3A4, A1A3, A2A4, A1A4
  1. A sixth transversal was added to Figure 5, forming 15 parallelograms shown in Figure 6. A1, A2, A3, A4, A5, A1A2, A2A3, A3A4, A4A5, A1A3, A2A4, A3A5, A1A4, A2A5, A1A5
  1. When a seventh transversal is added, twenty-one parallelograms are formed (Figure 7). A1, A2, A3
...read more.

Middle

image06.png

image07.png

image08.pngimage07.pngimage09.png

Solve the two new equations as simultaneous equations

image10.png

Answer:  image11.png

Replace the values in the following formula image02.png

image03.png

But when I tried out the formula I found out that the answers didn’t match:

U2 = 3 (3 +1)  = 6

          2

U3 = 4 (4 +1)  = 10

          2

The results we had have moved one term so instead of adding 1 to “n”, we need to subtract 1 to “n” so the values can match. Now our final formula is:

Un = n (n -1)

          2

Once again I tested the above:

U2 = 2 (2 -1) = 1

          2

U3 = 3 (3 -1) = 3

          2

  1. In order to find the general formula for the parallelograms formed by m horizontal parallel lines intersected by n parallel transversals; I decided to further the investigation by considering the number of parallelograms formed by three horizontal parallel lines intersected by a pair of parallel transversals (Figure 8). Three parallelograms were formed: A1, A2, and A1A2
  1. When a third transversal was added to the above figure, nine parallelograms were formed (Figure 9). A1, A2, A3, A4, A1A2, A3A4, A1A3, A2A4, A1A4
  1. One more transversal was added to Figure 9, to form 18 parallelograms (figure 10). A1, A2, A3, A4, A5, A6, A1A2, A3A4
...read more.

Conclusion

  1. When I compared all the formulas I got, I found the relationship which led me to find the general formula.

2 horizontal lines:1n (n -1)

                          2

3 horizontal lines: 3(n (n -1))

                          2

4 horizontal lines:6(n (n -1))

                          2

5 horizontal lines:10(n (n -1))

                                                  2

The number and sequence repeat the formula (n (n -1)) and multiplies by the first term.                                                                  2

  1. From this realization I was able to find that the final formula for calculating the number of parallelograms formed when m horizontal parallel lines are intersected by n parallel transversals:

m(m -1)   x   n(n -1)

    2                    2

  1. To test the validity of the formula I tested it against previously counted parallelograms (Figure 10), the intersection of 4 transversals with 3 horizontal parallel lines should form 18 parallelograms:

Using the formula:

m(m -1)   x   n(n -1)

    2                    2

3(3 -1)   x    4(4 -1)

    2                  2

3(2)  x  4(3)

  2          2

6  x  12

 2      2

3  x  6

= 18 parallelograms

Limitations

It is very difficult to test the validity of the formula when there are lots of parallel transversals intersecting lots of horizontal parallels, because even though we will get a number, to prove there are so many parallelograms is confusing and difficult; therefore we can only assume that the answer might be right.


image04.png


...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Math SL Circle Portfolio. The aim of this task is to investigate positions ...

    Limitations in the general statement, , (n= ) is that ? 0.5. The value 0.5 will only work for the case when r=1 and values are changed.

  2. How many pieces? In this study, the maximum number of parts obtained by n ...

    n=1 * Check n = 1: (1/2)(1) 2 + (1/2)(1) + 1 = 2 ? * Assume n = k k ? (1/2)k 2 + (1/2)k + 1 = (2 + 4 + 7 + 11 + 16 + .....+k) k=1 * Assume n = k + 1 k+1 ? (1/2)(k +1) 2 + (1/2)(k + 1)

  1. This essay will examine theoretical and experimental probability in relation to the Korean card ...

    As player 1 already has one of those two cards, for player 1 to get that hand, player 1 must get the other card to match the hand which is 1/18. As player 1 has the highest hand in the whole game, the card that player 2 has does not

  2. MATH IB SL INT ASS1 - Pascal's Triangle

    The parameter a in our functions is always 1. The parameter b in our functions is always -n, thus we can write b=-n. The parameter c in our functions is always the number of the numerator, thus we can say c=Xn or c = 0.5n² + 0.5 (which is the general formula for Xn)

  1. Math IA Type 1 Circles. The aim of this task is to investigate ...

    The general statement gives us OP?= 2. From the diagram, Figure 9, we can see that OP? does exist and it does equal 2; therefore the general statement is still valid. Figure 9. Another value of OP will be tested, OP= . Figure 10 In Figure 9, circle C3 cannot be drawn, because there is no point

  2. Mathematic SL IA -Gold medal height (scored 16 out of 20)

    ?a? is greater than 0, and ?a? is lesser than 0 respectively. However, no matter what value do x and y have, the gradient ?a? is the same in a linear function. Figure 7 the graph of winning gold medal heights from year 1932 to 1980 Figure 7 has the

  1. Math SL Fish Production IA

    However, a linear function does not have enough variables and since the graph does not have a direct linear relationship, a more suitable function would be a polynomial equation. A quartic function is suitable for this set of data because there are 5 variables in a quartic function.

  2. Maths IA. In this task I am asked to investigate the positions of ...

    This triangle has been constructed by the dots (midpoints) of the circles in the diagram above. Side?s A to P and O to P are both the same, they are 2. I know this because points A and O are both the radiuses of P making the lengths of those lines equal.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work