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

Borders - Fencing problem.

Extracts from this document...

Introduction

GCSE Coursework

Borders

Here are the first four examples of the shapes in this sequence:

image03.png



The sequence begins with a single white square, which is then surrounded by black squares to form the second shape. Each new cross is then formed by completely surrounding the previous cross with a border of black squares. In each new cross, the previous cross can be seen as the area of white squares in the centre.

In this investigation I will try to get algebraic formulas from the sequence, each expressing one property in terms of another (e.g. defining the area as the diameter squared). These formulas can then be checked and, proven, using a variety of maths skills.

Defining N

The variable that I will be using to refer to terms in this investigation will be n.

...read more.

Middle

0

0

1

4

2

8

3

12

4

16

5

20

From this table it becomes obvious to see that the formula for the perimeter of the shape is P = 4n

If you divide the perimeter of a cross by four to obtain a single side of the cross then you can see why this works. The length of one side is exactly the same as n. The length of one side must be multiplied by 4 to give the perimeter and for this reason P = 4n.

Area

To work out this formula I will draw a diagram that finds out the differences between consecutive terms, and then the differences between those differences.

     N:  0   1    2     3  

Area: 1    5   13   25 image00.pngimage01.pngimage00.pngimage01.pngimage00.pngimage01.png

            4     8  12

image02.pngimage02.png

44

Because the diagram needs to go down two levels the find

...read more.

Conclusion

of the cross. Therefore the 2n2 must be the rest of the circle. Because it is 2n2 then it must be possible to split in to two equal parts each part being n2. n2 must equal the part of the cross above or below the row of squares worked out by 2n+1.

image07.png

I will also try the formula out on the next cross to see if it works.

image09.png

The n value of this cross is 6.

I have counted the squares manually in the cross and there were 85 squares altogether.

So:

2n2 + 2n +1

2(6)2 + 2(6) + 1

72 + 12 + 1

= 85

Again the formula works. I wont draw out any more of the crosses because I think we have drawn enough. Now I will draw a table showing the area of the next few crosses in the sequence by using the formula to work it out.

n

2n2+2n+1

7

113

8

145

9

181

10

221

...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem 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 GCSE Fencing Problem essays

  1. Maths GCSE Courswork

    Base (m) Height (m) Area (m2) Perimeter (m) 300 400 223.6 44720 1000 310 380 244.95 46540.5 1000 320 360 264.57 47622.6 1000 330 340 282.84 48082.8 1000 333.33 333.33 288.67 48111.186 1000 340 320 300 48000 1000 350 300 316.22 47433 1000 360 280 331.66 46432.4 1000 370 260

  2. Fencing problem.

    100 � 153.9 = 7695m2 I have now found the area of one triangle. I shall multiply this figure by the number of triangles present within the decagon: Area of decagon = Area of one triangle � Number of triangles Area of decagon = 7695 � 10 = 76950m2 Results I have now completed the third segment of my coursework.

  1. Fencing Problem

    * 360� / 5 = 72� * I will now split each Isosceles triangle into 2 right-angled triangles and then calculate the new angle and side length. This will allow me to use trigonometry. * Now that I have my Right angled triangle (half of my isosceles triangle)

  2. Geography Investigation: Residential Areas

    d2 / n3-n) ?d2= 24 therefore 6?d2= 144 n = 6 therefore n3-n = 210 Rs = 1-(144/210) Rs = 0.32 (correct to 2 significant figures) Null Hypothesis: Rejected. I hypothesized that there was no correlation between the two variable statistics of intangible score and average time lived in the area.

  1. Fencing Problem - Math's Coursework.

    484.768 7271.520 40 480 479.583 9591.660 50 475 474.342 11858.550 60 470 469.042 14071.260 70 465 463.681 16228.835 80 460 458.258 18330.320 90 455 452.769 20374.605 100 450 447.214 22360.700 110 445 441.588 24287.340 120 440 435.890 26153.400 130 This coursework from www.studentcentral.co.uk (http://www.studentcentral.co.uk/coursework/essays/2307.html)

  2. The Fencing Problem

    I am trying to only use the base length and perimeter I will need to find the height length using the base and perimeter length. The equation for this is: Perimeter = 1000m 1000 = 2H + 2B All sides must add up to 1000m 500 = H + B

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