# Fencing Prblem

Extracts from this document...

Introduction

Introduction

A Farmer has a plot of land. She has only 1000m of fencing. She wants to fence off a plot of land. She is not concerned about the shape but the perimeter has to be exactly 1000m. The fenced off area has to be the maximum contained area. I am going to investigate the shape or shapes that could be used to fence in the maximum area using exactly 1000 metres of fencing each time.

Firstly, I am going to investigate 3 - sided shapes - triangles, 4 – sided shapes-quadrilaterals and 5 sided shapes-pentagons. From there on I am going to gather the results and see if I need to investigate any further.

TRIANGLES

In first part of my coursework I am going to investigate different types of triangles.

Firstly I will look at Isosceles triangles.

The formula for triangles:

There are two ways of working out the area. You will get the same answer with either formula.

- Area = ½ x base x height
- Area = base x height

2

Triangle 1

450m 450m

100m

I divide the Isosceles triangle

450 m2 - 50m2 = Height m2

202,500 – 2,500 = 200,000

Height = square root of 200,000

Height = 447.21 to two decimal places.

To calculate the area of this Isosceles triangle we calculate:

Area = ½ x base x height

Area = ½ x 100 x 447.21

Area = 22,361 m2 (meter squared) to the nearest whole number.

Triangle 2

I chose a different size isosceles triangle to continue my investigation

425m 425m

150m

Middle

354

44,250

Triangle 5

293

293

42,924.5

In the above results, I have found out that the equilateral Triangle had the largest area. I also noticed that there is a pattern. The pattern is that as the base increases and the height decreases, the area also increases.

QUADRILATERALS

I am going to investigate different quadrilaterals making sure that the perimeter is 1000 metres at all times.

The formula for the quadrilaterals:

- Perimeter = length + width + length + width.

- Area = length x width.

My first shape is going to be a square:

250m

A square has 4 sides. The 4 sides of the square are all equal. Therefore that means that all the edges will have the length of 250m due to the perimeter being 1000m.

Area = length x width.

Area = 250 meters x 250 meters

Area = 62,5002 (meters squared)

The next 5 shapes are going to be different types of rectangles. They will be numbered for further investigation:

(1) 450m

50m

(2) 400m

100m

(3) 300m

200m

(4) 350m

150m

(5) 280m

220m

Results of the Quadrilaterals.

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

Square | 250 | 250 | 62500 |

Rectangle 1 | 450 | 50 | 22250 |

Rectangle 2 | 400 | 100 | 40000 |

Rectangle 3 | 300 | 200 | 60000 |

Rectangle 4 | 350 | 150 | 52500 |

Rectangle 5 | 280 | 220 | 61600 |

In the above results, I have found out that the Square had the largest area.

200 cm

To find out the area of a pentagon, I am going to divide the pentagon into 5 isosceles triangles.

Each length is equal to 200 cm as 1000

5

= 200

hyp hyp

h

200cm

## To find out the hypotenuse, I am going to use the tangent formula

Tan = Opp

Adj

Tan 36 =100

Conclusion

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

Triangle 1 | 100 | 447 | 22,350 |

Triangle 2 | 150 | 418 | 31,350 |

Triangle 3 | 200 | 387 | 38,700 |

Triangle 4 | 250 | 354 | 44,250 |

Triangle6 | 300 | 316 | 47,400 |

Equilateral triangle | 333.33 | 289 | 48,166 |

Triangle7 | 350 | 274 | 47,950 |

Triangle8 | 400 | 224 | 44,800 |

I then put these values in a graph, shown below. I realised I had reached the peak of the area as I reached the area of the equilateral triangle.

The following table consists of all the measurements to the quadrilaterals above.

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

Square | 250 | 250 | 62500 |

Rectangle 1 | 450 | 50 | 22250 |

Rectangle 2 | 400 | 100 | 40000 |

Rectangle 3 | 300 | 200 | 60000 |

Rectangle 4 | 350 | 150 | 52500 |

Rectangle 5 | 280 | 220 | 61600 |

Bearing the calculations in mind, I then put the quadrilaterals in order and compared the base of the quadrilaterals with the Area. I than calculated the area for a few more quadrilaterals and tabulated them below in order and compared the base of the quadrilaterals with the Area.

Base (M) | Height(M) | Area(M) |

100 | 400 | 40000 |

150 | 350 | 52500 |

200 | 300 | 60000 |

220 | 280 | 61600 |

250 | 250 | 62500 |

280 | 220 | 61600 |

300 | 200 | 60000 |

350 | 150 | 52500 |

400 | 100 | 40000 |

I then put these values in a graph, shown below. I realised I had reached the peak of the area as I reached the area of the Square.

I put the maximum results together to compare them.

The Circle gave me a coverage area of 79,382.34 metres squared. The Square gave me a maximum coverage area of 62,5002 (metres squared). The Triangle gave me a maximum coverage area of 48,1662 (metres squared).

Comparing all the values investigated so far in comparison the Circle covers the largest farm area. Therefore I would recommend the Farmer to build her fence in a circle, as she will cover the area of 79,382.342 metres squared.

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

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