# The Fencing Problem.

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Introduction

The Fencing Problem

By Alexa Glick

10 Bruce

Aim

To make a fence that is 1000m long. The area inside the fence has to have the maximum area. I am investigating which shape would give this.

Hypothesis

I think that the area of a shape will always increase as the number of sides increase.

Method

I am going to start investigating different shape rectangles, all with a perimeter of 1000m. Below are four rectangles (which are not drawn to scale) this is showing how different size shapes with the same perimeter can have different areas.

300m 499m

1m

200m

250m

498 m

250m

2m

By looking at these rectangles, it is shown that any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m X 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side.

Middle

9600

490

10

4900

500

0

0

Using this formula I can draw a graph of base length (x) against area.

As you can see, the graph has formed a parabola, According to the table and the graph; the rectangle with a base of 250m has the greatest area. This shape is also called a square, or a regular quadrilateral. Because I only measured to the nearest 10m, I cannot tell whether the graph is true, and does not go up just to the sides of 250m. I will work out the results using 249m, 249.5 and 249.75.

Base (m) | Height (m) | Area (m²) |

249 | 251 | 62499 |

249.5 | 250.5 | 62499.75 |

249.75 | 250.25 | 6249993.75 |

250 | 250 | 62500 |

250.25 | 249.75 | 62499.9375 |

250.5 | 249.5 | 62499.75 |

251 | 249 | 62499 |

All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas.

Now that I have found that a square has the greatest area of the rectangles group, I am going to find the triangle with the largest area. Because in any scalene or right-angled triangle, there is more than 1 variable, there are countless combinations, so I am only going to use isosceles triangles. This is because if know the base length, then I can work out the other 2 lengths, because they are the same. If the base is 200m long then I can subtract that from 1000 and divide it by 2. This means that I can say that.

Side = (1000 – 200) ÷ 2 = 400. This can be rearranged to the following formula.

Conclusion

No. Of sides | Area (m²) |

8 | 75444.174 |

9 | 76318.817 |

10 | 76942.088 |

As you can see, the larger the number sides on the shape, the larger the area is. This pattern has carried on going for all of the shapes that I have investigated, so I am going to investigate shapes with the following amount of sides:

20, 50, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000.

Below is a table showing the results that I got.

No. Of sides | Area (m²) |

20 | 78921.894 |

50 | 79472.724 |

100 | 79551.290 |

200 | 79570.926 |

500 | 79576.424 |

1000 | 79577.210 |

2000 | 79577.406 |

5000 | 79577.461 |

10000 | 79577.469 |

20000 | 79577.471 |

50000 | 79577.471 |

100000 | 79577.471 |

Here is a graph showing the No. Of sides against the Area.

As you can see form the graphs, the line straightens out as the number of side’s increases. Because I am increasing the number of sides by large amounts and they are not changing I am going to see what the result is for a circle. Circles have an infinite amount of sides, so I cannot find the area by using the equation that I have used to find the other amount of sides out. I can find out the area of a circle by using Ө (pie). I have to work out the circumference of a circle. I can rearrange this so that the diameter is circumference/Ө. From that I can work out the area using the Өr 2 equation.

1000/Ө = 318.310

318.310/2 = 159.155

Ө X 159.1552 = 79577.472m2

If I place this point on my graph it is at the same place, area wise, as the last results on my graph were. From this I conclude that a circle has the largest area when using a similar circumference.

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

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