• 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. The Fencing Problem

    its base = its height. This type of shape is called a regular polygon. A polygon is a many sided shape and a regular polygon is a many sided shape with all angles and sides the same. The square is a regular polygon because its four sides are the same lengths and they all meet at 90(.

  2. Fencing problem.

    segment of my coursework below: I have shown the length of the sides representing the points shown on the above diagram below: AB = 380m BC = 45m CD = 70m DE = 135m EF = 190m FA = 180m To find the area of the shape I shall divide the Gun Shape into to halves.

  1. The Fencing Problem

    This will make it clearer for us to see that the circle is the highest, and also help us see that as the number of sides increase, the area becomes larger. General Formula for N-Gon (Polygon with n number of sides)

  2. Geography Investigation: Residential Areas

    However, we can see one thing and that is that the highest penalty point is for Sarum Hill which is located in the centre of town and with the lowest score of The Beaches, located in the district of Hatch Warren in Basingstoke.

  1. Maths GCSE Courswork

    base of 340m: Base of one triangle = 340 / 2 = 170m Length A = c2 - b2 = a2 = 3302 - 1702 = 800002 = V80000 = 282.84m Therefore, the height of the triangle is 282.843m. Area = base x height / 2 = 340 x 282.843

  2. Fencing Problem - Math's Coursework.

    Below is a table of results for isosceles triangles from the base with 10m to a base with 500m. Base (m) Side (m) Height (m) Area (m) 10 495 494.975 2474.875 20 490 489.898 4898.980 30 485 wwag agw stagagud eag agnt cag enagtral agcoag uk.

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