• 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...


GCSE Coursework


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


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.














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.


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



Because the diagram needs to go down two levels the find

...read more.


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.


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


The n value of this cross is 6.

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


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.











...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

    the equilateral triangle which has all three sides of length 333.33m is the triangle which gives the maximum area, 48111.186m2, out of all the triangles investigated in this section. Rectangles The next set of shapes that I am going to be investigating is rectangles.

  2. Fencing problem.

    above data in the graph below: The graph above obviously indicates that decagon has the most coverage of area, this in other words means that it covers the most area therefore the 1000m fencing shall be used to most out of all the polygons.

  1. Fencing Problem

    I will use trigonometry again to figure out the height of the right-angled triangle. * I will use Tangent, as I have my opposite length to my angle and am trying to figure out my adjacent i.e. height. * Tan36 = 100 * I will change the subject or the formula to figure out the height.

  2. Geography Investigation: Residential Areas

    I would have expected to see a lower average time lived in the area, a lower intangible score. Although I personally cannot see a correlation in the two sets of data above I will use the statistical calculation called Spearman's rank which will give me a number between -1 and

  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

    Evidence of this is shown above using the first and second rectangles as examples. I have decided to work out a formula for quadrilaterals using the perimeter and the base length. The area of a quadrilateral is found using the equation Base ( Height Area = Base ( Height As

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