# The Fencing Problem

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Introduction

7/7/02The Fencing Problem Clare Dutton

Question: A farmer has exactly 1000 metres of fencing and wants to fence of a plot of level land.

She is not concerned about the shape of the plot, but it must have a perimeter of 1000m. So it could be anything with a perimeter (or circumference) of 1000m.

She wishes to fence off the plot of land, which contains the maximum area.

Investigate the shape, or shapes, that could be used to fence in the maximum area using exactly 1000m of fencing each time.

Background work

Before I look for the answer, I will do some trial investigations to determine a good way of investigating which shape gives the maximum area.

Do regular or irregular shapes give a larger area?

Triangle

I know that the maximum area for an equilateral triangle (regular) is

48112.5224 m² (4.dp), by using ½ absinc.

1 side: 1000÷3=333⅓

Area (A) = ½ absinc

A= (333⅓ × 333⅓ × sin60) ÷2

A=48112.522432468813709095731708496

A=48112.5224 m² (4.dp)

An example of an irregular triangle is an isosceles triangle. To find the area of one it must be divided in two and Pythagoras’s theorem and ½ b×h must be used.

e.g.

Base=50m Other 2 sides= (1000-50) ÷2 =475m b²=h²-a² b²=475²-25² b²=225625-625 b²=225000 b=√225000 b=474.341649m | Area=½b×h A= (474.341649×50) ÷2 A=11858.54123m² |

Here are some more examples:

Base | Other 2 Sides | Height (b²=h²-a²) | Area=½b×h |

100 | 450 | 447.21359549995800 | 22360.679774997900 |

150 | 425 | 418.33001326703800 | 31374.750995027800 |

200 | 400 | 387.29833462074200 | 38729.833462074100 |

250 | 375 | 353.55339059327400 | 44194.173824159200 |

300 | 350 | 316.22776601683800 | 47434.164902525700 |

350 | 325 | 273.86127875258300 | 47925.723781702000 |

400 | 300 | 223.60679774997900 | 44721.359549995800 |

450 | 275 | 158.11388300841900 | 35575.623676894300 |

499 | 251 | 22.36067977499790 | 5578.989603861980 |

500 | 250 | 0 | 0 |

When the size of the base reaches 500m it is no longer possible to create an isosceles triangle because they would be inverted. The largest isosceles triangle is one with a base of 332m. This is because it is closest to the side length of an equilateral triangle. This is shown here:

Base | Other 2 Sides | Height (b²=h²-a²) | Area=½b×h |

320 | 340 | 300.00000000000000 | 48000.000000000000 |

330 | 335 | 291.54759474226500 | 48105.353132473700 |

332 | 334 | 289.82753492378900 | 48111.370797348900 |

333 | 334 | 288.96366553599800 | 48112.450311743600 |

333.3333333 | 333 | 288.67513459510200 | 48112.522432468800 |

Middle

This shows that a regular shape has a larger area than an irregular shape.

Quadrilateral

I know that a square (regular quadrilateral) has a maximum area of 62500m².

1 side: 1000÷4=250m 250×250=62500 A=62500 m² |

An example of an irregular quadrilateral is a rectangle.

Their areas can be found by using the formula: xy

e.g.

y=5m x=(1000-2y)/2 x=495m A=x × y A= 495×5 A=2475 m² |

Here are some other examples:
| As you can see in this table the rectangle with the largest area is the one with 260m and 240m as side lengths. These measurements are very close to those of the square. Even closer to the square’s measurements are 249m and 251m giving a maximum area of 62499m². |

The square gives the larger area and so we can see that a regular shape is better than an irregular shape. Another quadrilateral is a trapezium but this does not have a shape similar to the square or rectangle and so cannot have a larger area.

The trapezium (red) appears to have the same area as the square (black) if you take the overlap on the right and add it inside the square on the left. The perimeters are not the same though. Side x is longer than y and so the square needs a larger perimeter to make it 1000m. Therefore the square has a larger area than a trapezium. This will be the same with a rectangle and trapezium and a parallelogram. |

Here is the rectangle fitting into the square:

The rectangle overlaps the square on one side but leaves a gap on the other where the square is larger. You could argue that if you took the overlapping strip and placed it in the gap the areas would be the same. But there would still be a space (shown in blue) where the rectangle was smaller. This proves that irregular shapes do not have a larger area than regular ones. |

Conclusion

A=48112.5224 m² (4.dp)

4 sides

b=1000÷4=250 m

÷2=125 m

360÷4=90°

÷2=45°

Height of Triangle=125÷Tan45

h=125 m

Area of 1 triangle =½ b×h

A=½ × 250 × 125

A=15625 m²

(×4) Area of Square= 62500m²

5 sides: Pentagon

1 side: 1000 ÷5=200 m

÷2=100 m

Angle at centre point: 360 ÷5=72°

÷2=36°

Height=100/tan 36

Height=1376.38192

Area of triangle=½ b×h

Area of 1 triangle=137638.192m²

(×5) Area of Pentagon=688190.96m²

6 sides: Hexagon1 side: 1000÷6=166.6666667 m

÷2=83.33333335 m

1 angle: 360÷6= 60°

÷2= 30°

Height=83.33333335/ tan30

Height= 144.3375673 m

Area of triangle = ½ b×h

A=83.33333335×144.3375673 m²

(×6) Area of hexagon=12028.13061 m²

7 Sides: Heptagon

1000÷7=142.8571429 m

÷2=71.42857145 m

360÷7=51.42857143°

÷2=25.71428572°

h=71.42857145÷Tan 25.71428572

h=148.3229569 m

Area of triangle=½ b×h

A=10594.49692 m²

(×7) Area of Heptagon= 74161.478m²

8 Sides: Octagon

1000÷8=125 m

÷2=62.5 m

360÷8=45°

÷2=22.5°

h=62.5÷Tan 22.5

h=150.8883476 m

Area of triangle=h ×62.5

A=9430.5217525 m²

(×8) Area of Octagon= 75444.1738 m²

9 Sides: Nonagon

1000 ÷9=111.1111111 m

÷2=55.55555555 m

360 ÷9= 40°

÷2= 20°

h=55.55555555 ÷Tan 20

=152.6376344m

Area of triangle=152.6376344×55.55555555

A=8479.86858 m²

(×9) Area of Nonagon= 76318.81722 m²

10 sides: Decagon

1000÷10=100 m

÷2=50 m

360÷10=36°

÷2=18°

h=50÷Tan 18

h=153.8841769 m

Area of triangle= 153.8841769 × 50

A=7964.208845m²

(×10) Area of Decagon= 76942.08845 m²

11 sides: Undecagon

1000÷11=90.90909091 m

÷2=45.45454546 m

360÷11=32.72727273°

÷2=16.36363637°

h=45.45454546÷Tan 16.36363637

h=154.8039654 m

Area of triangle=½ b×h

A=7036.54388m²

(×11) Area of Undecagon= 77401.98268 m²

12 sides: Dodecagon

1000 ÷12=83.33333333 m

÷2=41.66666667 m

360 ÷12=30°

÷2=15°

h=41.66666667÷Tan 15

h=155.502117 m

Area of triangle=½ b×h

A=6479.254876 m²

(×12) Area of Dodecagon= 7751.05851 m²

13 sides: Tridecagon

1000÷13=76.92302692 m

÷2= 38.46153846 m

360÷13= 27.71428571°

÷2=13.84615385°

h=38.46153846÷Tan13.84615385

h=155.8620234 m

Area of triangle=½ b×h

A=5994.693207 m²

(×13) Area of Dodecagon=77931.01169m²

14 sides: Tetra-decagon

1000÷14=71.42857143 m

÷2=35.71428572 m

360÷14= 25.71428571°

÷2=12.85714286°

h=35.71428572÷Tan 12.85714286

h=156.4745095 m

Area of triangle=½ b×h

A=156.4745095×35.71428572 m²

(×14) Area of Tetra-decagon= 78237.25476 m²

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

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