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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, A3A1 ∪ A2, A2∪ A3A1∪ A2∪ A3 Table of Results for 5, 6 and 7 transversals: 5 10 A1, A2, A3, A4A1 ∪ A2, A2∪ A3, A3 ∪ A4A1∪ A2∪ A3 , A2∪ A3∪ A4A1∪ A2∪ A3  ∪A4 6 15 A1, A2, A3, A4, A5A1 ∪ A2, A2∪ A3, A3 ∪ A4, A4∪ A5A1∪ A2∪ A3 , A2∪ A3∪ A4, A3∪ A4∪ A5A1∪ A2∪ A3  ∪A4,  A2∪ A3∪ A4  ∪A5A1∪ A2∪ A3  ∪A4 ∪A5 7 21 A1, A2, A3, A4, A5, A6A1 ∪ A2, A2∪ A3, A3 ∪ A4, A4∪ A5, A5∪A6A1∪ A2∪ A3 , A2∪ A3∪ A4, A3∪ A4∪ A5,  A4∪A5 ∪A6A1∪ A2∪ A3  ∪A4,  A2∪ A3∪ A4  ∪A5,  A3∪ A4∪ A5  ∪A6A1∪ A2∪ A3  ∪A4 ∪A5,  A2∪ A3∪ A4  ∪A5 ∪A6A1∪ 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

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 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,
• 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

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, Test of validity for m = 4, n = 3 • 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. 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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