# The Fencing Problem.

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Introduction

MATHS COURSEWORK INVESTIGATION:-

THE FENCING PROBLEM

Planning:-

My coursework problem is to find out the highest possible area by using different kind of shapes with different sides and angles. The main aim is to use as many shapes possible but their perimeter must always add up to 1000m. So if it is a rectangle:-

(a+a^) + (b+b^) = 1000m

In case of isosceles triangle:-

b + (s+s^)=1000m

The shapes the I am going to use are:-

- Rectangles
- Isosceles Triangles
- 5 sided figure
- 6 sided figure
- 10 sided figure
- 20 sided figure

Rectangles:-

The first shape that I am going to use is a rectangle; I decided to use it because it is a simple shape to work with. A rectangle will look like this:

To work out the area of a rectangle you would use this formula:

Length x width= Xm2

This would apply to all rectangles and squares. This is the table of results for the areas of the rectangles that I used :-

Length | Width | Area |

25 | 475 | 11875 |

50 | 450 | 22500 |

75 | 425 | 31875 |

100 | 400 | 40000 |

125 | 375 | 46875 |

150 | 350 | 52500 |

175 | 325 | 56875 |

200 | 300 | 60000 |

225 | 275 | 61875 |

250 | 250 | 62500 |

275 | 225 | 61875 |

300 | 200 | 60000 |

325 | 175 | 56875 |

350 | 150 | 52500 |

375 | 125 | 46875 |

400 | 100 | 40000 |

425 | 75 | 31875 |

From the table I can see that if you would constantly increase the length 25m each time. Using rectangles as the shape we can see the areas are always increasing until the length and the width equal each other; this is because if you go beyond the point where the shape is a square the areas would start to repeat itself. This brings us to a conclusion that the highest area that we can get is when we have a square where the sides equal each other.

Middle

Length

Base

Area

100

450

22360.67

120

440

26153.39

140

430

29698.48

160

420

32984.80

180

410

36000

200

400

38729.83

220

390

41168.25

240

380

43266.61

260

370

45033.32

280

360

46432.75

300

350

47434.16

320

340

48000

340

330

48083.26

360

320

47623.52

380

310

46540.30

400

300

44721.36

Because when I did rectangles I found out that the highest area was when it was a square and not a rectangle, that is why I decided to use an a equilateral triangle this is the result that I got.

Length | Base | Area |

333.33 | 333.33 | 48112.9 |

From the first table I can see that the highest area was 48083.26m2because after that area the other areas started to repeat themselves, but I got this result before I used an a equilateral triangle which gave the area of 48112.9m2 this showed me that the again the highest area can only be achieved if you are using regular shapes (where the side/bases equal each other).

Other Shapes:-

The other shapes that I used were: 5 sided, 6 sided, 10 sided and 20 sided figures. I started by using a pentagon (5 sided shape). A pentagon will look similar to this.

To work out the area of a regular pentagon we would firstly divide the total perimeter (1000) by the number of sides, in this case by 5: 1000/5, this therefore gives us 5 different small triangles inside one shape. Each triangle will have a base of 200m. After getting the triangles than we can see that all of them are isosceles therefore we would use the same formula as we used before, with other triangles (1/2 x base x height) We already know that all the interior angles add up to 360, we must again divide 360 by the number of side (360/5) which gives us 72 degrees at the top of each triangle.

Conclusion

But we don’t know the height so we need to refer to SOH-CAH-TOA

To find out that we need to use TAN. The final calculation to get the height of the triangle would be:

25/tan 9= 157.84m

So the height is 157.84m. Finally we just use the normal triangle formula and we would see this calculation (25 x 157.84) x20= 78920m2

Conclusion:-

From the results that I got and the shapes that I used the highest are can be achieved when you have more sides. If the number of sides keeps on increasing the shapes will start to look more like a circle so I tried to use circle , this is the result that I got.

To find the area of the circle we would use this formula:

∏ x r2

But we don’t know the radius. To find out the radius I had to follow these steps:

- The circumference is 1000
- The formula to work out the circumference we have to use this formula: 2∏r

Than to work out the radius we would have to divide 1000 by 2∏

This would give us the radius being 159.15m.

Lastly we would just apply the normal formula of ∏r2. And we would get the area of 79577.47m2.

From the results I got, I can conclude that the highest area can be achieved when I used a circle.

This can be proved by increasing the number of sides of a shape, because if you keep on increasing the amount of sides it will look more and more like a circle.

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

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