# The Fencing Problem

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Introduction

THE FENCING PROBLEM

The problem:

A farmer has 1000m of fencing and wants to fence a plot of land. It can be any shape but must have a perimeter of no more than 1000m. I am going to investigate the effects of using different shapes to get the largest possible area using the 1000m of fence.

Triangles:

I will now investigate different types of triangles as they have the least number of sides. Isosceles and scalene triangles will be my starting point.

Hypothesis:

The triangle’s area will increase as the triangle’s height increases.

Below are examples of different triangles.

Here is an equilateral triangle, a triangle with all angles and sides the same:

Here is a right-angled triangle, a triangle with a right angle:

Middle

Quadrilaterals:

Next, I will investigate simple four sided shapes called quadrilaterals. I will investigate different rectangles and squares.

## Hypothesis

I predict that the closer the length and width values get to equalling each other, the larger the area will be.

Below are a square and a rectangle:

100 150

50 RECTANGLE 50

100 SQUARE 100

150

100

The formula to work out the area is:

Area = Length x Height

E.g.

100 x 100 = 10000

50 x 150 = 7500

This shows just how different areas of shapes can be, even with the same perimeter.

QUADRILATERALS | ||

Height | Width | Area |

20 | 480 | 9600 |

40 | 460 | 18400 |

60 | 440 | 26400 |

80 | 420 | 33600 |

100 | 400 | 40000 |

120 | 380 | 45600 |

140 | 360 | 50400 |

160 | 340 | 54400 |

180 | 320 | 57600 |

200 | 300 | 60000 |

220 | 280 | 61600 |

240 | 260 | 62400 |

260 | 240 | 62400 |

280 | 220 | 61600 |

300 | 200 | 60000 |

320 | 180 | 57600 |

340 | 160 | 54400 |

360 | 140 | 50400 |

380 | 120 | 45600 |

400 | 100 | 40000 |

420 | 80 | 33600 |

440 | 60 | 26400 |

460 | 40 | 18400 |

480 | 20 | 9600 |

Conclusion:

I have chosen different values of length and width and found that the biggest area is created when the lengths and widths of the sides are equal, at 250m. This quadrilateral, with all sides the same, is a square.

Therefore, the largest area for a four-sided shape is 62500m2.

Polygons:

The square is another regular shape, like the equilateral triangle, so from now on I will use regular polygons whilst finding out how the number of sides affects the area inside.

Hypothesis:

Conclusion

POLYGONS: | ||||||||

n | Side | Angle | Radiens | 1/2 side | 1/2 angle | Height | Triangle area | Total area |

5 | 200 | 72 | 0.628319 | 100 | 36 | 137.6382 | 13763.82 | 68819.1 |

6 | 166.6667 | 60 | 0.523599 | 83.33333 | 30 | 144.3376 | 12028.13 | 72168.78 |

7 | 142.8571 | 51.42857 | 0.448799 | 71.42857 | 25.71429 | 148.323 | 10594.5 | 74161.48 |

8 | 125 | 45 | 0.392699 | 62.5 | 22.5 | 150.8883 | 9430.522 | 75444.17 |

9 | 111.1111 | 40 | 0.349066 | 55.55556 | 20 | 152.6376 | 8479.869 | 76318.82 |

10 | 100 | 36 | 0.314159 | 50 | 18 | 153.8842 | 7694.209 | 76942.09 |

20 | 50 | 18 | 0.15708 | 25 | 9 | 157.8438 | 3946.095 | 78921.89 |

30 | 33.33333 | 12 | 0.10472 | 16.66667 | 6 | 158.5727 | 2642.879 | 79286.37 |

40 | 25 | 9 | 0.07854 | 12.5 | 4.5 | 158.8276 | 1985.344 | 79413.78 |

50 | 20 | 7.2 | 0.062832 | 10 | 3.6 | 158.9454 | 1589.454 | 79472.72 |

60 | 16.66667 | 6 | 0.05236 | 8.333333 | 3 | 159.0095 | 1325.079 | 79504.74 |

70 | 14.28571 | 5.142857 | 0.04488 | 7.142857 | 2.571429 | 159.0481 | 1136.058 | 79524.04 |

80 | 12.5 | 4.5 | 0.03927 | 6.25 | 2.25 | 159.0731 | 994.207 | 79536.56 |

90 | 11.11111 | 4 | 0.034907 | 5.555556 | 2 | 159.0903 | 883.835 | 79545.15 |

100 | 10 | 3.6 | 0.031416 | 5 | 1.8 | 159.1026 | 795.5129 | 79551.29 |

200 | 5 | 1.8 | 0.015708 | 2.5 | 0.9 | 159.1419 | 397.8546 | 79570.93 |

300 | 3.333333 | 1.2 | 0.010472 | 1.666667 | 0.6 | 159.1491 | 265.2485 | 79574.56 |

400 | 2.5 | 0.9 | 0.007854 | 1.25 | 0.45 | 159.1517 | 198.9396 | 79575.84 |

500 | 2 | 0.72 | 0.006283 | 1 | 0.36 | 159.1528 | 159.1528 | 79576.42 |

600 | 1.666667 | 0.6 | 0.005236 | 0.833333 | 0.3 | 159.1535 | 132.6279 | 79576.74 |

700 | 1.428571 | 0.514286 | 0.004488 | 0.714286 | 0.257143 | 159.1539 | 113.6813 | 79576.94 |

800 | 1.25 | 0.45 | 0.003927 | 0.625 | 0.225 | 159.1541 | 99.47133 | 79577.06 |

900 | 1.111111 | 0.4 | 0.003491 | 0.555556 | 0.2 | 159.1543 | 88.41905 | 79577.15 |

1000 | 1 | 0.36 | 0.003142 | 0.5 | 0.18 | 159.1544 | 79.57721 | 79577.21 |

As you can see, the area increases as the number of sides go up.

Finding the area of the circle:

The circumference is 1000 (as this is the length of fencing available)

From this we can find the diameter of a circle with a 1000m circumference

Circumference = π x diameter

Diameter = circumference / π

= 1000 / π

= 318.3

Radius = ½ diameter

= 159.15 r = 159.15 m c = 1000 m

Area = π x radius2

= π x 25328.7

= 79522.5 m2

Conclusion:

This proves my theory, as you can

see the circle has the largest area out

of all the shapes I have experimented with. This is because a circle has an infinite number of sides and if the area size goes up as the number of sides goes up, an infinite number of sides will obviously have the largest area.

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

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