# A farmer has brought 1000 metres of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the biggest size. I will investigate

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Introduction

Maths Coursework

The fencing problem:

Aim:

A farmer has brought 1000 metres of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the biggest size. I will investigate different shapes the fencing can make to achieve the largest area.

Rectangles

I begin my investigation by starting with different shaped rectangles; I will change the value of the widths and lengths by going up in increments of 50m.

450m

50m

Area = 450 x 50 = 22,500m²

400m

100m

Area = 400 x 100 = 40,000m²

350m

150m

Area = 350 x 150 = 52,500m²

300m

200m

Area = 300 x 200 = 60,000m²

250m

250m

Area = 250 x 250 = 62,500m² * Becomes a square*

I have noticed that if you increase and decrease the lengths and widths by 50m the area will increase until it reaches the optimum length and width where the maximum area for the shape can be obtained, a square. From then onwards the lengths, widths and areas begin to repeat themselves in reverse order.

Here is a table to show how the area changes and how it begins to repeat itself after a square is reached.

Middle

Area (m²)

480

260

10000

24000

400

300

223.6

44720

333.3

333.3

288.7

48111.9

300

350

316.2

47340

250

375

353.6

44200

180

410

400

36000

150

425

418.3

31372.5

50

475

474.3

11857.5

I have drawn a table so that I can see clearly which triangle had the biggest area:

As you can clearly see from my table, my results show that the equilateral triangle (or regular triangle) is able to obtain the largest area with a perimeter of 1000m.

Because of this pattern all shapes that Ii use now shall also be regular.

I will now investigate a regular pentagon; because there is five sides I can divide it into segments, each segment being an isosceles triangle with a top angle of 72° (one fifth of 360° is 72°). I can work out the other angles by subtracting 72 from 180 and then dividing the answer by two, that gives an answer of 54°.

Because every isosceles triangle can be split into 2 equal right angles triangles I can use trigonometry to work out the area of the triangle; I can then multiply the answer by 10 to give me the area of the full pentagon. I also know that each side is 200m long (1000 divided by 5), so the base of the triangle is 100m.

Conclusion

Decagon

Each of the ten isosceles triangles that make up this regular decagon look like this:

Tan = Opposite

Adjacent

Tan 18º = 0.32

0.32 = 50

?

50 / Tan 18º = 153.88

100 / 2 = 50

50 x 153.88 = 7,694

7,694 x 10 = 76,940

A = 76,940m²

1000 / Tan 360

2 x 12 2 x 12

41.6 / 0.2679

= 155.2819

1000 x 1000 Tan 360

2 x 12 2 x 12 2 x 12

41.6 x 41.6 / Tan 15

= 6,459.7238

1000 x 1000 Tan 360 x 12

2 x 12 2 x 12 2 x 12

41.6 x 41.6 / Tan 15 x 12

= 77,516.6853

A = 77,516.6853 m²

In order to show more clearly how the area is increasing with the number of sides I have drawn up a table to show sides against area:

No of sides | Area (m) | |

3 | 48111.9 | |

4 | 62500 | |

5 | 68800 | |

6 | 72150 | |

7 | 74120.2 | |

8 | 75445 | |

10 | 76940 | |

12 | 77,516.7 | |

20 | 78,914.1 |

As I have already come to the conclusion that as the number of side’s increases the area increases I am going to work out the area of a circle. Circles have infinite sides and should prove my conclusion to be correct.

*r = radius*

c = 1000 = 2πr

500 = πr

Area of a circle = πr²

πr² = 79,622m²

500 = π x Ans² = 79,577.47155

π

A = 79,577.5m2

From this I conclude that a circle has the largest area when given a perimeter of 1000m.

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

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