• 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. Fencing problem.

    Firstly I shall find the area of the rectangle then the trapezium that the rectangle has been placed upon. Area of rectangle = Base � Height Area of rectangle = 325 � 25 Area of rectangle = 9375m2 Now I shall find the area of the trapezium.

  2. The Fencing Problem

    Area = {1/2[(1000 � 10000) x h]} x 10000 I have explored various regular polygons so far, and we can see now that as the number of sides is increased, the area is larger each time. I can now assume that a circle will finally give us the largest area

  1. Geography Investigation: Residential Areas

    This means the residential areas actually get in a sense better because the way people look after their properties changes and gets better, hence receiving a low Index of Decay rating. Also to conclude, I have found that residents own more homes away from the CBD - this is probably

  2. Maths GCSE Courswork

    The next isosceles triangle has two sides of length 320m and a base of 360m. Base of one triangle = 360 /2 = 180m Length A = c2 - b2 = a2 = 3202 - 1802 = 700002 = V70000 = 264.57m Area = base x height / 2 =

  1. Fencing Problem - Math's Coursework.

    Tow ork out the height I can use Pythagoras' Theorem. Below is the formula and area when using a base of 200m. H� = h� - a� H� = 400� - 100� H� = 160000 - 10000 H� = 150000 H = 387.298 1/2 � 200 � 387.298 = 38729.833m wweb ebw stebebud eeb ebnt ceb enebtral ebcoeb uk.

  2. An Investigation into the Varying Isoperimetric Quotients of Differing Shapes.

    Height =using tan 45=O/A = tan 45 = h/x/2 h = x/2 * tan 45 Area of 2 triangular segments = 1/2 * (base * height) = 1/2 (x* tan 45 * x/2) =1/2 * x * x/2*tan 45 = x2/4 * tan 45 Area of a square = 4*

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