# Fencing problem

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Introduction

Maths Coursework SSMC Nicholas Thorburn

Fencing problem

I realise intuitively that a circle will cover the largest area, but it is hard to prove it. I decided to do this logically, according to the sheet, by looking at the area that can be covered by a rectangle. I decided to produce a table, and then a graph, showing varying lengths and breadths in a logical manner, decreasing one by 50, and increasing the other by 50.

These are my preliminary results:

Width (m) | 150 | 175 | 200 | 225 | 250 | 275 | 300 | 325 | 350 |

Length (m) | 350 | 325 | 300 | 275 | 250 | 225 | 200 | 175 | 150 |

AREA (m2) | 52500 | 56875 | 60000 | 61875 | 62500 | 61875 | 60000 | 56875 | 52500 |

I produced a graph showing area from this:

You can see that this graph is a parabola, so the highest point will indicate the greatest area, which is 62,500. I have put the base along the x axis because you can work out the other two

Middle

212.5

200

Length (m)

200

212.5

225

237.5

250

262.5

275

287.5

300

AREA (m2)

60000

61093.75

61875

62343.75

62500

62343.75

61875

61093.75

60000

Triangles

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.

I will then follow on by investigating the other type of triangle, scalene.

This is how I worked out the area:

b2 = a2 - c2

= 4502 - 502

= 200,000

= 447.2135955

= ½ X 25 X 447.2135955 = 5590.169944m

I produced a graph from this, below:

This again gives good evidence that that a regular sided shape (an equilateral triangle) covers more are than an isosceles triangle, and that scalene triangles cover less area than an isosceles triangle.

As you can see this graph is a-symmetrical.

Conclusion

Using SOHCAHTOA I can work out that I need to use Tangent.

O100

T = tan36

This has given me the length of H so I can work out the area.

Area = ½ X b X H = ½ x 100 X 137.638 = 6881.910

I now have the area of half of one of the 5 segments, so I simply multiply that number by 10 and I get the area of the shape

Area = 6881.910 X 10 = 68819.096m2.

I am going to work out the area of the 2 shapes using the same method as before.

Hexagon:

1000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3.

360 ÷ 6 = 60 ÷ 2 = 30

Area = ½ X b X H = ½ x 83 1/3 X 144.338 = 6014.065

6014.065 X 12 = 72168.784m2

Heptagon:

1000 ÷ 7 = 142.857 ÷ 2 = 71.429

360 ÷ 7 = 51.429 ÷ 2 = 25.714

Area = ½ X b X H = ½ X 71.429 X 148.323 = 5297.260

5297.260 X 14 = 74161.644m2

My prediction was right, so as the number of sides increases, the area increases. Below is a table of the number of sides against area. I have not gone far with this, or produced a graph, as all that proves is the area increase tails off, towards the higher-sided shapes.

No. of sides | Area (m2) |

3 | 48112.522 |

4 | 62500.000 |

5 | 68819.096 |

6 | 72168.784 |

7 | 74161.644 |

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

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