• 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. The Fencing Problem

    = 500 � n 3) To find out the interior angle A�C A�B = 360 � n A�C = 0.5 x (360 � n) A B = 180 � n 1000 � n 2 4) To acquire the perpendicular height; O Tan A�C = AC � h Tan ?

  2. Maths GCSE Courswork

    360 x 264.57 / 2 = 95245.2 / 2 = 47622.6m2 OR Heron's Formula Area = = 500 (500 - 320)(500 - 360)(500 - 320) = 2268000000 = V2268000000 = 47622.6m2 The area of this isosceles triangle is 47622.6m2. The next triangle has two sides of length 330m and a

  1. Fencing problem.

    I have discovered the area of six different polygons. These polygons that have been discovered are all regular. The results have been shown below in a tabulated form: Name of regular polygon Area (m2) Pentagon 68800 Hexagon 72217.2 Heptagon 74322.5 Octagon 75450.4 Nonagon 76392 Decagon 76950 I have represented the

  2. The Fencing Problem

    I will proceed to investigate the regular polygon with the largest area. Regular 3 Sided Polygon: A regular-sided polygon is called an equilateral triangle. Area = 3 (.5 ( (1000 ( 3)) ( ((500 ( 3) ( (tan (180 ( 3))

  1. Fencing Problem

    So I will do this calculation: * Sin 45 x 100 = 70.71067812 * Now that I have the height of the triangle I will figure out the base of the triangle so I can figure out the total area of the triangle.

  2. Geography Investigation: Residential Areas

    higher in the CBD and gets better as you work your way out from the centre. In theory, the newer, planned houses are on the outskirts of the town, thus they will have a lower index of decay. However, the nineteenth century, older, unplanned houses in the more central part of town should have a higher index of decay.

  1. 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.

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

    When I find the area of the segment to find the area of the square all I will have to do is times the area of the segment by eight because the segment is 1/8 of the square. O H h 90 45 A x/2 H= Hypotenuse O=Opposite A=Adjacent h=height Using trigonometry.

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