# The fencing problem.

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Introduction

Maths Coursework

Yr 10

Maths Coursework

By Anonymous

Table Of Contents

Introduction

The fencing problem

Investigating the relationship between the length of sides and the area

Investigating Rectangles

Investigating Triangles

Investigating the relationship between the number of sides and the area

Pentagons

Hexagons

Heptagons

Octagons

Nonagons

Decagons

Results Table

Conclusion

Introduction

The fencing problem

A farmer has exactly 1000m of fencing. She wishes to fence off a piece of land but does not know what shape will give her the largest area. I will investigate different shapes to see if I can find out what the maximum area of land can be fenced off with a given length of fencing.

Investigating the relationship between the length of sides and the area

Investigating Rectangles

I will start off this section by testing a variety of different sized rectangles. First I will try rectangles increasing or decreasing in 50’s, then in 10’s and finally in 5’s. Then I will draw a table of the rectangles and finally draw a graph.

PREDICTION: I think that none of the rectangles will have the same area because I have not used the same measurements for the length and breadth twice.

a = l x w

Here are the rectangles and areas increasing and decreasing by 50.

l (m) | w (m) | a (m²) | |||

1 | 50 | 450 | l x w | 50 x 450 | 22500 |

2 | 100 | 400 | l x w | 100 x 400 | 40000 |

3 | 150 | 350 | l x w | 150 x 350 | 52500 |

4 | 200 | 300 | l x w | 200 x 300 | 60000 |

5 | 250 | 250 | l x w | 250 x 250 | 62500 |

6 | 300 | 200 | l x w | 300 x 200 | 60000 |

7 | 350 | 150 | l x w | 350 x 150 | 52500 |

8 | 400 | 100 | l x w | 400 x 100 | 40000 |

9 | 450 | 50 | l x w | 450 x 50 | 22500 |

Here are the rectangles and areas increasing and decreasing by 10.

l (m) | w (m) | a (m²) | |||

1 | 260 | 240 | l x w | 260 x 240 | 62400 |

2 | 320 | 180 | l x w | 320 x 180 | 57600 |

3 | 490 | 10 | l x w | 490 x 10 | 49000 |

4 | 460 | 40 | l x w | 460 x 40 | 18400 |

5 | 310 | 190 | l x w | 310 x 190 | 58900 |

Here are the rectangles and areas increasing and decreasing by 5.

l (m) | w (m) | a (m²) | |||

1 | 345 | 155 | l x w | 345 x 155 | 53475 |

2 | 205 | 295 | l x w | 205 x 295 | 60475 |

3 | 325 | 175 | l x w | 325 x 175 | 56875 |

4 | 495 | 5 | l x w | 495 x 5 | 2475 |

5 | 425 | 75 | l x w | 425 x 75 | 31875 |

The point of this was to see which rectangle has the biggest area and whether I could see any relationships. The following table shows the area of all the rectangles in order of smallest to largest. I have also calculated the difference in metres between the length and width.

Length (m) | Width (m) | Area (m²) | Difference between length and width (m) |

5 | 495 | 2475 | 490 |

10 | 490 | 4900 | 480 |

40 | 460 | 18400 | 420 |

50 | 450 | 22500 | 400 |

75 | 425 | 31875 | 350 |

100 | 400 | 40000 | 300 |

150 | 350 | 52500 | 200 |

155 | 345 | 53475 | 190 |

175 | 325 | 56875 | 150 |

180 | 320 | 57600 | 140 |

190 | 310 | 58900 | 120 |

200 | 300 | 60000 | 100 |

205 | 295 | 60475 | 90 |

240 | 260 | 62400 | 20 |

250 | 250 | 62500 | 0 |

Middle

area = 33541 m²

- Equilateral

area = 48160 m²

4. Right-angled

area = 41625 m²

Base (m) | Side 1 (m) | Side 2 (m) | Area (m²) |

450 | 200 | 350 | 33541 |

200 | 400 | 400 | 38729 |

250 | 333 | 417 | 41625 |

333.3 | 333.3 | 333.3 | 48160 |

Again the shape with the largest area is the one where all the sides are the same length. Therefore my prediction was correct.

Investigating the relationship between the number of sides and the area

I have now proved that regular shapes have the largest area. Now I will try to find a relationship between regular shapes with up to ten sides.

Pentagons

I will start by drawing and finding the area of a regular pentagon and giving the answer to two decimal places. I will find the area by using the SIN rule.

This is the SIN Rule

I will first split the shape into ten right-angled triangles, then find the angles, then height. By then I will I will have enough information to use the formula:

All I need to do then is multiply by ten then I will have the area for the pentagon.

c = 137.64m This is the height of the right-angled triangle in the regular pentagon.

The area of the right-angled triangle can now be calculated.

= = 6882.5m²

Therefore the area of the regular pentagon is 6882.5 x 10 = 68825m²

Hexagons

Conclusion

number of sides = n

total length = 1000

length of sides =

I had to split the shape up into right-angled triangles and found that there was a relationship between the number of sides on the shape and the number of right-angles in the shape.

number of right-angles = 2n

The area of a right-angled triangle is

The base (b) is half of the length of the sides

There for the base (b) = x =

The height (h) is calculated by

Where c = height (h)

Angle B is calculated by dividing the number of degrees in a circle (360) by the number of right-angled triangles in the shape.

Angle B =

Angle C = 180 – A - B

Angle C = 180 – 90 -

I can now substitute the information I have gained into the formula to calculate the height (h).

I now have enough information to use the formula to calculate the area of a right-angled triangle.

area =

I now have a formula that will calculate the area the right-angled triangle. To find the area of the shape all I need to do is multiply this by the number of right-angled triangles in the shape.

Therefore the area of a shape with n sides can be calculated using the formula

I have produce a graph using microsoft excel which shows the results of the formula for shapes with between 20 and 1000 sides.

©2002 Anonymous Page of

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

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