# The Fencing Problem.

Extracts from this document...

Introduction

## Charles Watson

## The Fencing Problem

## Introduction

## I have been given 1000 meters of fencing and my aim is to find out the maximum area inside.

## Prediction

I would predict that the more sides the shape has, then possibly the bigger the area it will have, although I have nothing to base this on, it will be what I am about to investigate.

Shapes:

I am going to start with the rectangle, I think this is a good starting block because I am able to vary the widths and lengths to see which has the bigger area. If I discover that the rectangles with equal sides i.e. square bring me the best result, then I will try to direct my investigation into furthering that particular theory.

## Rectangles

## Area = 40 000 m2

Area = 60 000 m2

Area = 62 500 m2

It appears that the square

Middle

In order to find out the area of the triangle, I need to multiply half base and height, half base = 400 ÷ 2 = 200,

∴ 200 x 170 = 34 000m

The Equilateral Triangle

I will work out the area by using half base multiplied by height, 333.3 ÷2 = 166.65,

∴166.65 x 270 = 44995.5

I will now try using a trapezium, however my investigation is going according to my prediction.

Trapezium _

I am using a trapezium because it is a four sided shape that I have not used yet.

I will split it up into 3 sections and find out the middle section and then do the calculations for the outer sections using half base times height:

200 x 200 = 40 000m 50 x 200 = 10 000m

50 x 200 = 10 000m

### Total = 60 000m

I will now carry on by looking at shapes with more sides, I will now look at a 6 sided shape:

## The Hexagon

I will work out the area in the same way that I went about finding out the area of the pentagon.

Conclusion

= 155.4

∴41.65 x 155.4 = 6474.1 x 12 = 77 688.9

I can now put all the shapes in a table and I notice a pattern forming,

Equilateral Triangle (3) 44995.5

Triangle (3)

Square (4) 625, 00

Trapezium (4)

Pentagon (5) 68, 562.50

Hexagon (6) 72, 168.79

Octagon (8) 75, 444.20

Dodecagon (12) 77, 688.90

I will now use a shape with infinite sides because I can see a trend growing, although I am not sure that I will get any where just using more shapes, so I am going to use:

## The Circle

Perimeter = 1000m

Diameter = 1000 ÷

=318.31 ÷ 2= 159.15

∴ r = 79, 577. 47

This proves my theory correct, as areas increased we grew more closer towards the circle. I have no real solid evidence to say why this is so, although I could suggest that because circles have no corners or straight edges, this increases the overall area of usable space.

This strong positive correlation on my graph shows the pattern which I have seen and predicted since the start, this proves that my results are correct and I do not think I should any anomalies as I worked out each shape using my own calculations.

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

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