# Fencing Problem

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Introduction

Simran Singh Ghatore

11B

Mathematics

Fencing Problem

Aim-: A farmer has 100m of wire. She wishes to fence of an area of land into a shape, which will give her the maximum area.

- I have to measure and experiment with different shapes of 1000m circumference of perimeter to achieve the maximum area.

- I have to consider working from these shapes-:

- Quadrilaterals

- Triangles

- Pentagon

- Hexagon

- Circle

Prediction-: I predict that if the sides of a regular shape increase so will the area, coming to the conclusion of the circle holding the most area.

Square(l x w)*NOT TO SCALE

Rectangles (l x w)

In a rectangle, 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.

Length (m) | Width (m) | Area (m2) |

480 | 20 | 9600 |

425 | 50 | 22500 |

400 | 100 | 40000 |

350 | 150 | 52500 |

325 | 175 | 56875 |

300 | 200 | 60000 |

250 | 250 | 62500 |

Middle

Base (m) | Side (m) | Height (m) | Area (m2) |

480 | 260 | 10000 | 24000 |

400 | 300 | 223.607 | 44721.4 |

333.33 | 333.33 | 288.675 | 48112 |

300 | 350 | 316.228 | 47434.2 |

250 | 375 | 353.553 | 44194.125 |

180 | 410 | 400 | 36000 |

150 | 425 | 418.330 | 31374.75 |

50 | 475 | 474.342 | 11858.55 |

Graph

Conclusion

The regular triangle seems to have the largest area out of all the areas but to make sure I am going to find out the area for values just around 333.

Base (m) | Side (m) | Height (m) | Area (m2) |

333 | 333.5 | 288.964 | 48112.450 |

333.25 | 333.4 | 288.747 | 48112.518 |

333.3 | 333.4 | 288.704 | 48112.522 |

333.5 | 333.3 | 288.531 | 48112.504 |

333.75 | 333.1 | 288.314 | 48112.410 |

334 | 333.0 | 288.097 | 48112.233 |

This has proved that once again, the regular shape has the largest area.

Because the last 2 shapes have had the largest areas when they are regular, I am going to use regular shapes from now on.

Regular Pentagon

Because there are 5 sides, I can divide it into 5 segments. Each segment is an isosceles triangle, with a top angle of 720. This is because it is a fifth of 360. This means I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 540

Conclusion

I am going to apply the same method as before to solve the areas of these two regular shapes.

Hexagon:

1000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3

360 ÷ 6 = 60 ÷ 2 = 30

COS= Adjacent/Hypotenuse

Adjacent= 831/3 m andHypotenuse= h

= 831/3 / h

=83.33333335/COS60

=166.6666667 = 166.66666672 – 83.33333352

= √20833.33332

h = 144.338

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

TAN= Opposite/Adjacent

= x/71.4285 = h= 148.323

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

5297.260 X 14

= 74161.644m2

Octagon

TAN= Opposite/Adjacent

TAN 67.5o = h/62.5

h= 62.5xTan67.5 o

=Area= ½ x 125 x h

COS= Adjancent/Hypotenuse = 62.5/x

= 62.5/COS67.5=163.320m

TAN=62.5 x Tan67.5= 150.888m

Area= ½ x 150.888 x 125= 9430.3 x 8

=73444m2

My predictions were correct and as the number of side’s increases, the area increases. Below is a table of the number of sides against area:

No. of sides | Area (m2) |

3 | 48112.522 |

4 | 62500.000 |

5 | 68819.096 |

6 | 72168.784 |

7 | 74161.644 |

8 | 73444 |

Circle

I have already come to the conclusion that: as the number of side’s increases, the area increases. Circles have no sides and are infinite. This should prove my prediction correct.

C= 2 π x r

1000= 2 x 3.14 x r

1000= 6.28 x r

1000

6.28 = r

r= 159.23

Area = πr2

= 3.14 x 159.23 x 139.23

=79612.2m2

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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