# 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 | |||

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 | |||

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 | |||

7 | 21 | 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 A1∪ A2∪ A3 ∪A4 ∪A5 ∪A6 |

→ 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,

Middle

= 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 ∪A7∪A8,
- A2∪ A3∪ A4 ∪A5 ∪A6 ∪A7 ∪A8∪A9

= 2

- ∪ A2∪ A3 ∪A4 ∪A5 ∪A6 ∪A7∪A8 ∪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

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 = 2

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,
- A5∪A6,
- A6∪A7,
- A7∪A8,
- A8∪A9

= 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

Conclusion

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,

Test of validity for m = 4, n = 3

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

= 6

- A1 ∪ A2,
- A2∪ A3,
- A4∪ A5,
- A5∪A6,
- A1 ∪A4,
- A2 ∪A5,
- A3 ∪A6

= 7

- A1∪ A2∪ A3,
- A4∪A5 ∪A6

= 2

- A1∪ A2∪A4 ∪A5,
- A2∪A3 ∪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.

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.

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

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