# The Fencing problem.

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Introduction

The Fencing problem

I am making an investigation to find out how shape affects area. In my investigation I am going to see how area is affected when I change the shapes of different shapes all with the perimeter of 1000m. I am going to experiment with different side lengths and different numbers of sides. I will record my results in graphs and tables and come to a conclusion.

Hypothesis

My hypothesis is that as the number of sides rises, so will the areas.

## Results

Here are my results for 4-sided shapes.

A rectangle, with the width of 25m and the length of 475m.

Area = 25m x 475m

Area = 11875m

A rectangle with the length of 50m and the width of 450m

Area = 50m x 450m

Area = 22500m

A rectangle with the length of 425m and the width of 75m

Area = 75m x 425m

Area = 31875m

## The fencing problem

By Siân Salkeld

A rectangle with the length of 400m and the width of 100m.

Area = 400m x 100m

Area = 40000

A rectangle with the length of 375m and the width 125m

Middle

Base= 200m

Sides= 400m

Height² = 400²-200²/2

= 160 000 - 100²

= 160 000 – 10 000

= 150000

Height = √150000

Height =387.29833462074168851792653997824

Area = ½ base times height

Area = 100 x 387.29833462074168851792653997824

Area to 1dp = 38729.8

base = 300m

sides = 350m

Height² = 350² - 150²

= 122500 – 22500

= 100000

Height = √100000

= 316.22776601683793319988935444327

Area = 150 x 316.22776601683793319988935444327

Area to 1dp = 47434.2

Base = 400m

Sides = 300m

Height² = 300 ² - 200²

Height = 90000 – 40000

Height = 50000

Height = √50000

Height =223.60679774997896964091736687313

Area = 200 x 223.60679774997896964091736687313

Area to 1dp = 44721.4

A triangle with the base of 400 is smaller than a triangle with the base of 300 so I will work out the area of a triangle with te base of 350

Base = 350

Sides = 325

Height² = 325²– 17²

= 105625 – 30625

=75000

=√75000

height =273.8612787525830567284848914004

area = 175 x 273.8612787525830567284848914004

area to 1dp = 47925.7

Using the knowledge that a square was the biggest 4 sided shape I am going to work out the area of an equalateral triangle

Base = 333.333

Sides = 333.333

Area = ½ base x height

Height² = 333.3² - 166.7²

= 111110.888889 - 27788.89

=83321.998889

=√83321.998889

= 288.65550209375881564039836747977

Area = 166.7 x 288.65550209375881564039836747977

Area = 48118.9

To prove this is the biggest I can show you the area of a triangle with the base of 333.

Conclusion

#### Pentagon

The pentagon has 5 sides and therefore each side is 1000/5. Which means each side are 200m. The angle at the top of the triangle is 360 divided by 5. Each angle is 72˚.

Height =?

Tan 72/2 =200/2/?

##### Tan36 = 200/?

? = 100/tan 72

? = 137.6

Area = 5(137.6 x 100)

Area = 5 x 13760

Area = 68800

#### Hexagon

Base = 1000/6

= 166.666

Angle = 360 / 6

= 60

Tan 60/2 = 166.666/?

= Tan 30= 83.333/?

? = 83.333/Tan 30

? = 144.338

Area of triangle = 144.338 x 83.333

= 12028 x 6

Area of hexagon = 72168

#### Heptagon

Base = 1000/7

= 142.9

Angle = 360 /7

= 51.4

Tan 51.4/2 =142.9/2/?

Tan 25.7=71.4/?

? = 71.4/tan 25.7

? = 148.4

Area of triangle = 148.4 x 71.4

= 10595.76

Area of heptagon = 10595.76 x 7

= 74170.32

Decagon

Base = 1000/10

= 100

Angle = 360/10

= 36

Tan 36/2 = 100/2/?

Tan18 = 50/?

? = 50/tan18

? =153.9

Area of triangle = 153.9 x 50

=7695

Area of decagon = 7695 x 10

= 76950

#### Circle

The equation for working out the area of a circle is: -

Circumference = π x diameter²

Circumference = 1000m

1000m/π = diameter (d)

D = 318.3

Radius(r) = 318.3/2

R = 159.15

Area = π r ²

Area = π x 159.52

Area = 791572.5

Number of sides | Area |

3 | 48118 |

4 | 62500 |

5 | 68800 |

6 | 72168 |

7 | 74170.32 |

10 | 76950 |

Infinite | 791572.5 |

Evaluation

###### My theory

My hypothesis was correct and this is now my theory , as the number of sides increases the area of the shape increases

If I repeated my investigation I could improve by investigating further and getting more results to prove my theory.

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

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