# Maths Fencing Coursework

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Introduction

Maths Coursework

The Fencing Problem

A farmer has brought 1000 metres of fencing to make a crop area. The farmer wants the fencing to be put in a shape where it cand hold the maximum area/size. I will investigate many shapes so he can achieve the largest area.

The farmer does not mind what shape is created so firstly I am going to investigate:

Squares, rectangles & quadililaterals.

After I will investigate: Triangles, polygons and circles.

RECTANGLES

I have collected information about rectangles that show me what lenght and width is needed to get the maximum area. I have organised all my data on a spreadsheet.

To work out the area I do Length x Width.

Length: On the spreasheet I typed three numbers that went up in acsending order in 25s, I then highlighted them and dragged it down to maximum it could go to.

Width: To work out the width In a column next to the length I would write for e.g. ( If I was on the 2B slot) : = (1000/20) –A3

Area: To work the area I have to type in the first area . Then to work ou the rest I highlight it and drag it down.

I am going work thes rectangles out to see wheather my graph is wrong or right.

Middle

Side a: 250.5

Side b: 250.5

Side c: 499

√(500*(500-250.5)* (500-250.5)* (500-499)) =5578.989604

Here is a equilateral Triangle example:

Side a: 333.333

Side b: 333.333

Side c: 333.333

√(500*(500-333.333)* (500-333.333)* (500-333.333)) =48112.285 cm ²

TRIANGLES

I have collected information about triangles that show me what type triangle has the maximum area. I have organised all my data on a spreadsheet.

To work out the area I use Hero’s rule, Pythagoras or another formula.

Scalene

I have used Heros rule to work out the area of this Scalene triangle to ensure that the spreadsheet data gets the same results.

S=500

Area= √(500*(500-3)* (500-498)* (500-499)) =704.9822693

These are the spreadsheet data results in 100’s.

a | B | c | s | s-a | s-b | s-c | Area |

2 | 498 | 499 | 500 | 498 | 2 | 1 | 705.6911506 |

101 | 400 | 499 | 500 | 399 | 100 | 1 | 4466.542287 |

201 | 300 | 499 | 500 | 299 | 200 | 1 | 5468.089246 |

301 | 200 | 499 | 500 | 199 | 300 | 1 | 5463.515352 |

401 | 100 | 499 | 500 | 99 | 400 | 1 | 4449.719092 |

As you can see these results are not very accurate.

I now have to go in 10’s for more accurate results.

a | B | c | s | s-a | s-b | s-c | Area |

151 | 350 | 499 | 500 | 349 | 150 | 1 | 5116.150897 |

161 | 340 | 499 | 500 | 339 | 160 | 1 | 5207.686627 |

171 | 330 | 499 | 500 | 329 | 170 | 1 | 5288.194399 |

181 | 320 | 499 | 500 | 319 | 180 | 1 | 5358.17133 |

191 | 310 | 499 | 500 | 309 | 190 | 1 | 5418.025471 |

201 | 300 | 499 | 500 | 299 | 200 | 1 | 5468.089246 |

211 | 290 | 499 | 500 | 289 | 210 | 1 | 5508.629594 |

221 | 280 | 499 | 500 | 279 | 220 | 1 | 5539.855594 |

231 | 270 | 499 | 500 | 269 | 230 | 1 | 5561.924127 |

241 | 260 | 499 | 500 | 259 | 240 | 1 | 5574.943946 |

251 | 250 | 499 | 500 | 249 | 250 | 1 | 5578.978401 |

261 | 240 | 499 | 500 | 239 | 260 | 1 | 5574.047004 |

These results show me that the maximum area for a scalene triangle may be the area of an isosceles triangle.

I will go in 0.5’s just to see if my prediction is right.

a | B | c | s | s-a | s-b | s-c | Area |

252 | 249 | 499 | 500 | 248 | 251 | 1 | 5578.888778 |

251.5 | 249.5 | 499 | 500 | 248.5 | 250.5 | 1 | 5578.944793 |

251 | 250 | 499 | 500 | 249 | 250 | 1 | 5578.978401 |

250.5 | 250.5 | 499 | 500 | 249.5 | 249.5 | 1 | 5578.989604 |

250 | 251 | 499 | 500 | 250 | 249 | 1 | 5578.978401 |

My prediction was right that the maximum area for a scalene triangle is an isosceles triangle.

Isosceles a²+b²=c²

I have used Pythagoras to work out the area of the Isosceles triangle.

Example:

0.5² + b² = 499.5²

0.25 + b = 249500.25

249500.25 - 0.25 =249500

b² =249500

b = √249500

b/ height= 499.4997497

=499.49 cm

Area = 499.49 x 1

2

Area= 249.745

Conclusion

N= number of sides

The angle of a shape is: 360

2N

The length of one side is: 1000

2N

Height of a triangle is length of a side divideded by the angle times tan.

1000 tan 360

2N 2N

= 1000

2N tan 180

N

I have found what the height is so I have to times the height with the base and divide by 2.

BASE HEIGHT

1000 X 1000 X 1I EIGHTER DIVED BY TWO OR TIMES BY 1/2

N 2N tan 180 2

N

= 1,000,000 X N1

2N² tan 180 X 2

N

= 1,000,000 X N = 250,000 X N

4N² tan 180 N² tan 180

N N

Shape | Sides | Maximum Area |

pentagon | 5 | 68819.09602 |

hexagon | 6 | 72168.78365 |

heptagon | 7 | 74161.47845 |

octagon | 8 | 75444.17382 |

enneagon | 9 | 76318.81721 |

decagon | 10 | 76942.08843 |

hendecagon | 11 | 77401.9827 |

dodecagon | 12 | 77751.05849 |

triskaidecagon | 13 | 78022.2978 |

tetrakaidecagon, tetradecagon | 14 | 78237.25478 |

Penta kaidecagon, pentadecagon | 15 | 78410.50182 |

hexakaidecagon, hexadecagon | 16 | 78552.17956 |

heptakaidecagon | 17 | 78669.52214 |

octakaidecagon | 18 | 78767.80305 |

enneakaidecagon | 19 | 78850.94024 |

icosagon | 20 | 78921.89393 |

From the results I have got I can see that the more sides the polygon has the higher the area, So I have decided to find out the area of a circle.

2пr= 1000 cm r = radius

пr=1000 п= 3.141592654…..

2

r = 500

п

п x (500 ) ² = 79577.47155

П

This shows me that the circle has the maximum area with a perimeter of 1000cm.

My prediction was wrong that the polygon with the most sides would have the highest area

But actually the shape with only one side (circle) has the maximum area this is something I have just learnt after doing the calculations.

To conclude I have examined many shapes such as triangles (equilateral, scalene & isosceles), quadrilaterals and many polygons. I have learnt that a one sided shape can hold the maximum area with a perimeter of 1000cm.

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

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