# The problem for this investigation is as follows: Farmer Jones has 1000 metres of fencing. What is the maximum area she can enclose?

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Introduction

Omar KadirMaths Coursework : Farmer Jones’ Fencing Problem26/02/03

Problem: The problem for this investigation is as follows:

Farmer Jones has 1000 metres of fencing. What is the maximum area she can enclose?

Mathematically, the problem can be interpreted to:

Which shape, with a perimeter of 1000 metres, has the largest area?

In order to go about answering the question, different types of shapes must be considered.

The original question of whether the shape will be a triangle, rectangle, polygon, or circle can be broken down into smaller questions. What will the lengths of the shapes be? If it is a triangle, will it be an Isosceles, an equilateral, or a Scalene? If it is a rectangle, what will the values of the width and length be? And if it is a polygon, will it be a regular or irregular shape?

I will take each shape separately, and investigate the different areas found within that particular shape. I will start with three-sided shapes and go on to four, five and six sided shapes, finishing with the circle.

Prediction:I predict that a circle with perimeter 1000metres will have the largest area.

I predict this because I also think the triangle will have a smaller area than the square (rectangle), which will be smaller than the pentagon, which will be smaller than the hexagon, and so on, until the final shape, the circle is reached.

Middle

260

100.000

24,000.00

490

255

70.711

17,324.20

The length of the base cannot go up to 500m,

because that would leave 250m for the other

two sides , therefore making it an ‘impossible triangle’,

as the two con

The results from the spreadsheet can be shown on a graph:

The formula to work out the area of any Isosceles triangle, by simply knowing the length of the base(x), can then be:

area = ½ base x height

a = ½ x ( c - b )

a = ½ x 1000-x - x

2 2

Equilateral Triangles:

There can only be one equilateral triangle with a perimeter of 1000 metres. Each one of the sides will have a value of 1000m divided by three (333.3m).

To find the area, we can use the same equation:

a = ½ b h

= ½ 333.3 x ( c - b )

= ½ 333.3 x 288.675

= 48,112.50 m2

We can now see that the triangle with the largest

area is the Equilateral trianngle, and not the Isoceles.

Scalene Triangles:

After looking into Isosceles triangles, and the equilateral triangle, I do not see it necessary to carry on any further with scalene triangles.

I noticed that as the sides of the triangle became closer in length, the area increased, and therefore it makes sense that the triangle with equal sides has the maximum area.

Investigating Rectangles

To find out the areas of the rectangles with a perimeter of 1000m, I used a spreadsheet again. By filling in the values for the width, the length and the area of each triangle can be worked out.

We know that the perimeter of the rectangle will be 1000m, and that one length plus one width will be 500m, all the time.

To find the length, we minus the width value from 500m. The formula used in the spreadsheet is:

=500-A2

From that, the area can easily be worked out, using a =wl, and on the spreadsheet using:

=B2*A2

Width(m) | Length(m) | Area(m ) |

20 | 480 | 9600 |

40 | 460 | 18400 |

60 | 440 | 26400 |

80 | 420 | 33600 |

100 | 400 | 40000 |

120 | 380 | 45600 |

140 | 360 | 50400 |

160 | 340 | 54400 |

180 | 320 | 57600 |

200 | 300 | 60000 |

220 | 280 | 61600 |

240 | 260 | 62400 |

250 | 250 | 62500 |

260 | 240 | 62400 |

280 | 220 | 61600 |

300 | 200 | 60000 |

320 | 180 | 57600 |

340 | 160 | 54400 |

360 | 140 | 50400 |

380 | 120 | 45600 |

400 | 100 | 40000 |

420 | 80 | 33600 |

440 | 60 | 26400 |

460 | 40 | 18400 |

480 | 20 | 9600 |

Conclusion

Area of triangle = ½ b x h

a = ½ x 100 x 100 After finding the area of one triangle, I must

TAN 36o multiply that figure by the number of triangles

a = 6,881.91 m2 found in the pentagon.

Area of Pentagon = no. of triangles x area of 1 triangle

a = 10 x 6881.91

a = 68,819.1 m2

Investigating Hexagons

I will use the same method for finding the area of a regular hexagon, as I did for finding the area of a regular pentagon.

Area of triangle = ½ b x h

a = ½ x 83.33 x 83.33

TAN 30o

a = 6,013.58 m2

Area of Hexagon = no. of triangles x area of 1 triangle

a = 12 x 6013.58

a = 72,163.01 m2

Investigating Octagons

Again, the same method is used.

Area of triangle = ½ b x h

a = ½ x 62.5 x 62.5

TAN 22.5o

a = 4715.26 m2

Area of Octagon = no. of triangles x area of 1 triangle

a = 16 x 4715.26

a = 75,444.17 m2

Throughout investigating these polygons, I have been using the same formula over and over again. I have noticed that within the formula, the one thing that kept changing was the number of triangles in the shape ( t ). From that, I can form this equation to work out the area of any polygon:

a = t ( ½ b xb/2 )

TAN 0

Investigating Circles

There can only be one circle with a perimeter of 1000 metres. To find the area of this circle, we first need to know the radius.

Radius = 1000

2

r = 159.15

Area = r2

a = 3.142 x 159.152

a = 79,572.53

Conclusion

My prediction is correct, and it has been proven. There is a definite relationship between the number of sides and the area, as this graph shows:

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

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