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Maths SL Portfolio - Parallels and Parallelograms

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Introduction

IB Standard Level Maths: Portfolio Piece 1

Parallels and Parallelograms

Table of Results for 4 transversals:

Transversals

Number of Parallelograms

Parallelograms

Diagram

4

6

A1, A2, A3

A1 ∪ A2, A2 A3

A1 A2 A3

image05.png

Table of Results for 5, 6 and 7 transversals:

5

10

A1, A2, A3, A4

A1 ∪ A2, A2 A3, A3 ∪ A4

A1 A2 A3 , A2 A3 A4

A1 A2 A3  A4

image06.png

6

15

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

A1 A2 A3  A4 A5

image08.png

7

21

A1, A2, A3, A4, A5, A6

A1 ∪ A2, A2 A3, A3 ∪ A4, A4 A5, A5A6

A1 A2 A3 , A2 A3 A4, A3 A4 A5,  A4A5 A6

A1 A2 A3  A4,  A2 A3 A4  A5,  A3 A4 A5  A6

A1 A2 A3  A4 A5,  A2 A3 A4  A5 A6

A1 A2 A3  A4 A5 A6

image09.png

 Let n = number of transversals and letp = number of parallelograms

Transversals (n)

Parallelograms (p)

2

1

3

3 (1 + 2)

4

6 (1 + 2 + 3)

5

10 (1 + 2 + 3 + 4)

6

15 (1 + 2 + 3 + 4 + 5)

7

21 (1 + 2 + 3 + 4 + 5 + 6)

n

1 + 2 + … + (n – 1)

Use of Technology:

Using the TI – 84 Plus, press STAT 1: Edit.

Type in L1, L2:         (2, 1)

                (3, 3)

                (4, 6)

                        …etc.

Using Quadreg, L1, L2,

...read more.

Middle

 A3 A4  A5 A6, A3 A4 A5  A6 A7,A4 A5 A6  A7 A8, A5 A6 A7  A8 A9

        = 5

  • A1 A2 A3  A4 A5 A6,
  • A2 A3 A4  A5 A6 A7,  
  • A3 A4 A5  A6 A7 A8,
  • A4 A5 A6  A7 A8 A9

        = 4

  • A1 A2 A3  A4 A5 A6 A7,
  • A2 A3 A4  A5 A6 A7 A8,
  • A3 A4 A5  A6 A7 A8 A9

        = 3

  • A1 A2 A3  A4 A5 A6 A7A8,  
  • A2 A3 A4  A5 A6 A7 A8A9

        = 2

  •  A2 A3  A4 A5 A6 A7A8 A9

        = 1

 45

p = sum of all integers from 1 to (10 – 1)

= sum of all integers from 1 to 9

= 1 + 2 +3 + 4 + 5 + 6+ 7 + 8 + 9

= 45

p = (n2 – n) ÷ 2

p= 102 – 10 ÷ 2

= 90 ÷ 2

= 45

 Let m = number of horizontal lines

image11.png

If there are three horizontal lines, intersecting two transversals (m = 3, n = 2) then p = 3. Similarly, if there are three transversals, and two horizontal lines, (m= 2, n = 3), then we also obtain p= 3.

Conclusion:

Hence, m horizontal lines and n transversals produce the same amount of parallelograms as n horizontal lines and m transversals.

General Statement:

If there are m horizontal lines, and two transversals, then p = sum of all integers from 1 to (m – 1). Note that this rule is identical to the above investigation of n transversals and two horizontal lines.

Test of Validity for m = 10, n = 2image12.png

We will now prove that if m = 10, where n = 2, we will get the same p value of 45 as example 1 above where m = 2 and n = 10.

e.g. 2) 10 horizontal lines, 2 transversals

Manual method:

  • A1,
  • A2,
  • A3,
  • A4,
  • A5,
  • A6,
  • A7,
  • A8,
  • A9

                = 9

  • A1 ∪ A2,
  • A2 A3,
  • A3 ∪ A4,
  • A4 A5,
  • A5A6,
  • A6A7,
  • A7A8,
  • A8A9

                = 8

  • A1 A2 A3,
  • A2 A3 A4,
  • A3 A4 A5,
  • A4 A5 A6,
  • A5 A6 A7,
  • A6 A7 A8,
  • A7 A8 A9,

                = 7

  • A1 A2 A3  A4,  
  • A2 A3 A4  A5,  
  • A3 A4 A5  A6,
  • A4 A5 A6  A7,
  • A5 A6 A7  A8,
  • A6 A7 A8  A9

                = 6

  • A1 A2 A3  A4 A5,  
  • A2 A3 A4  A5 A6,
  • A3 A4 A5  A6 A7,
  • A4 A5 A6  A7 A8,
  • A5 A6 A7  A8 A9

                = 5

  • A1 A2 A3  A4 A5 A6,  
  • A2 A3 A4  A5 A6 A7,  
  • A3 A4 A5  A6 A7 A8,
  • A4 A5 A6  A7 A8 A9
...read more.

Conclusion

m horizontal lines, and n transversals, the resultant value of p equals the product of p1and p2, where;

p1 = number of parallelograms for m horizontal lines and two transversals

p2 = number of parallelograms for 2 horizontal lines and n transversals.

Hence, for any diagram with m horizontal lines and n transversals,

image04.png

Test of validity for m = 4, n = 3

image13.png

  • A1,
  • A2,
  • A3,
  • A4,
  • A5,
  • A6

        = 6

  • A1 ∪ A2,
  • A2 A3,
  • A4 A5,
  • A5A6,
  • A1 A4,  
  • A2 A5,
  • A3 A6

        = 7

  • A1 A2 A3,
  • A4A5 A6

        = 2

  • A1 A2A4 A5,  
  • A2A3 A5 A6

        = 2

  • A1 A2 A3  A4 A5 A6

= 1

p=6 + 7 + 2 + 2 + 1

        = 18

p = ½ m (m – 1) x ½ n (n – 1)

        = ½ 4 (4 – 1) x ½ 3 (3 – 1)

        = 6 x 3

        = 18

Scope/limitations:

The formula will be valid for m, n ≥ 2. If either value were to be 1 or 0, it would be impossible to create any parallelograms.

image07.png

Explanation of generalisation:

A diagram with m horizontal lines and 2 transversals creates p1parallelograms.

A diagram with 2 horizontal lines and n transversals creates p2parallelograms.

It follows that if a diagram were created, with m horizontal lines, and n transversals, we would be able to fit p1parallelograms vertically and p2 parallelograms horizontally, giving us a total of p1 x p2 parallelograms.

 

...read more.

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