• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
• Level: GCSE
• Subject: Maths
• Word count: 1399

See how many squares would be needed in order to construct any cross-built up in the way described in the investigation.

Extracts from this document...

Introduction

Borders

Part 1-

Aim: -

My aim is to see how many squares would be needed in order to construct any cross-built up in the way described in the investigation.

Plan:-

• To use simple techniques first and begin with a simple number.
• Use examples, diagrams and label them.
• To record all results found in a table
• Extend investigation to 3-D

Method:-

To create a larger cross, surround the shaded squares with white squares. To begin my investigation I will start off with a simple number of squares.

Pattern 1- Number of shaded squares I have chosen is 1.

Total number of squares = 4 + 1

= 5

White squares = 4

By using the method stated in the investigation, I have constructed a larger cross. By using the same method I will again construct the next sequence of crosses.

Pattern 2-                                 Pattern 3- Total number of squares = 25

White squares = 12

Total number of squares = 13

White squares = 8

Pattern 4-

Total number = 41

White squares = 16

I have noticed that the squares going down are odd 1+3+1=5. 1+3+5+3+1=13. In each case 2 has been added to each pattern.1+3+5+7+5+3+1=25, like the previous pattern the addition has always been 2 to each digit.

Middle

So in result the general formula for this would be n x 4 or 4n. I will prove this formula to be correct by choosing a formula at random. I have chosen the number 10. 4n= 4 x 10= 40 white squares.

I am now going to search for the formula the formula for the dark squares. I will the information from the result table and differences (from pg2) I have found that the number of shaded squares equals to the total from the previous squares. To see this look at page 2. I have decided to draw a new table similar to that of the previous page.

Pattern            Total Squares           Shaded Squares             White Squares               Difference in

Squares

1                  5                          1                           4

4

2                 13          5                           8

8

3                 2513                           12

12

4                 41        25                            16

Throughout the investigation I will use the following symbols for the formulae, these will not change. n = pattern number, d= dark squares, w = white squares.

Trying for a formula for shaded squares

I have found out the general formula to find the number of shaded squares which is 2nsquared – 2n + 1, because this is a quadratic form of ax +bx+c, I found a second difference between the pattern of shaded squares.

Conclusion

Drawing simple diagrams at first can make it easier to find the answer for the investigation. Another good way is using different tables which hold a record of your findings take a part in finding an equation by recording the information on the table to justify your results. Using different techniques to tackle a problem is beneficial because of the reason that I drew a number of diagrams and a variety of assorted result tables to explain what I did. I also used different symbols and a plan to explain what I had done.

Equation 2n – 2n + 1 is very similar to equation 2n + 2n +1. I noticed this immediately after I had found the second equation. The difference between the two is the symbol before 2n. One is – (negative) and the other + (positive). This was because I had to add combine two equations together. The formula for white squares 4n and the formula for shaded squares 2n – 2n + 1, which in result gave 2n + 2n + 1.

The first equation I found was a linear equation, as 4n can be written as 4x. Equation 2 is a quadratic as it is in the form ax + bx+ c. the same with equation 3. I am planning to extend this investigation to 3D.

This student written piece of work is one of many that can be found in our GCSE T-Total 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

Related GCSE T-Total essays

1. T-Totals Investigation.

Could this be a pattern?) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

2. T-Shape investigation.

1 2 3 4 5 6 7 9 10 11 12 13 14 15 17 18 19 20 21 22 23 25 26 27 28 29 30 31 n 18 19 20 21 z 34 39 44 49 Right away I can see the t-total is increasing by 5.

1. T-total Investigation

These are the 3 different positions in which I put my 3by2 T on: Position 1: Position 2: 1 2 3 11 T-no 20 2 3 4 12 T-no 21 Position 3: 3 4 5 13 T-no 22 The bottom number of the T is the T-no.

2. have been asked to find out how many squares would be needed to make ...

To find these without drawing any more diagrams I will use my knowledge of the structure (e.g. 1 + 3 + 1). New orders Pattern Previous total New additions New totals 4 41 11 + 9 = 20 61 (20 + 41 = 61)

1. Black and white squares

45 6 61 146 20 60 41 86 7 85 231 24 84 61 147 1 Tn-2nd Wn-2 Bn-2 2 Tn-1st Wn-1 Bn-1 3 Tn Wn Bn After producing the formulas above, I have decided to look at the issue in a different angle, by adding the cumulative total number

2. Investigation in to How many tiles and borders is needed for each pattern

Anther formula I found was to work out the total tiles for each pattern For the nth term the formula is 2n +6n+5 Test The Formula: 1. Formula to calculate the number of borders is 4n+4 For example: when n=2 (4 2)+4 8+4=12 So then the 2nd pattern will have 12 borders.

1. Borders - Investigation into how many squares in total, grey and white inclusive, would ...

This gives the formula to find the number of white surrounding squares as: Un= 4n Following on from this, the total number of squares, with the total of white surrounding squares removed, gives the total of shaded squares for each width.

2. I am going to investigate how changing the number of tiles at the centre ...

B 10 14 18 22 26 +4 +4 +4 +4 B= 4N + 6 To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B.

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