# Firstly I will investigate rectangles to find the maximum area of that shape.

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Introduction

Firstly I will investigate rectangles to find the maximum area of that shape.

The following table shows the formulas I used to obtain the results for the different rectangles.

Length (L) | Width (500-L) | Area L(500-L) |

(m) | (m) | (m²) |

10 | =(500-A3) | =A3*B3 |

20 | =(500-A4) | =A4*B4 |

30 | =(500-A5) | =A5*B5 |

40 | =(500-A6) | =A6*B6 |

50 | =(500-A7) | =A7*B7 |

100 | =(500-A8) | =A8*B8 |

150 | =(500-A9) | =A9*A9 |

200 | =(500-A10) | =A10*B10 |

250 | =(500-A11) | =A11*B11 |

Here are my results:

Length (L) | Width (500-L) | Area L(500-L) |

(m) | (m) | (m²) |

10 | 490 | 4900 |

20 | 480 | 9600 |

30 | 470 | 14100 |

40 | 460 | 18400 |

50 | 450 | 22500 |

100 | 400 | 40000 |

150 | 350 | 52500 |

200 | 300 | 60000 |

250 | 250 | 62500 |

From these results we can see that the largest area belongs to the rectangle with the dimensions of 250x250. Therefore the rectangle with the largest area is actually a square.

250m

250m x 250m = 62,500m²

250m

I will now investigate triangles; I will start with an equilateral triangle.

Middle

triangle. To do this I will split it into 2 right-angled

triangles and use Pythagoras theorem.

166.6m

a = 333.3² - 166.6²

a = 111111.1 – 27777.7

a = √83333.4111

a = 288.7m

To find the area of a triangle you need to multiply ½base x height.

Therefore area = 0.5 x 333.3 x 288.7

Area = 48116.7m

I will now investigate isosceles triangles

Firstly I will take a measurement for the base. As the other two sides have the same lengths, to obtain them I can subtract the base from 1000. Then to get one side I can divide the answer by two.

To obtain the area you need to again use Pythagoras. When investigating the triangles I discovered that the base must be below 500 otherwise it will not be correct due to the fact that the other sides would not reach a sufficient point.

The following table shows the formulas I used to obtain the results for the different isosceles triangles.

Base (b) | Length of one of the other sides | height² | height | Area |

(m) | 0.5(1000-B) | (m) | (m²) | |

50 | =0.5*(1000-A3) | =(B3*B3)-(A3*0.5)*(A3*0.5) | =SQRT(C3) | =0.5*A3*D3 |

100 | =0.5*(1000-A4) | =(B4*B4)-(A4*0.5)*(A4*0.5) | =SQRT(C4) | =0.5*A4*D4 |

150 | =0.5*(1000-A5) | =(B5*B5)-(A5*0.5)*(A5*0.5) | =SQRT(C5) | =0.5*A5*D5 |

200 | =0.5*(1000-A6) | =(B6*B6)-(A6*0.5)*(A6*0.5) | =SQRT(C6) | =0.5*A6*D6 |

250 | =0.5*(1000-A7) | =(B7*B7)-(A7*0.5)*(A7*0.5) | =SQRT(C7) | =0.5*A7*D7 |

300 | =0.5*(1000-A8) | =(B8*B8)-(A8*0.5)*(A8*0.5) | =SQRT(C8) | =0.5*A8*D8 |

350 | =0.5*(1000-A9) | =(B9*B9)-(A9*0.5)*(A9*0.5) | =SQRT(C9) | =0.5*A9*D9 |

400 | =0.5*(1000-A10) | =(B10*B10)-(A10*0.5)*(A10*0.5) | =SQRT(C10) | =0.5*A10*D10 |

450 | =0.5*(1000-A11) | =(B11*B11)-(A11*0.5)*(A11*0.5) | =SQRT(C11) | =0.5*A11*D11 |

Conclusion

The spreadsheet formulas and the results are on the next page.

I have discovered that the areas of regular polygons continue to increase but only very slightly. An immense number such as 1 million would be either very impractical or may probably be impossible.

I am now going to investigate the circle. Circles have an infinite number of sides, so I cannot find the area using the equation for the other shapes. I can find out the area by using pi. To work out the circumference of the circle the equation is π x diameter. I can rearrange this so that diameter equals the circumference/π. From that I can work out the area using the πr² equation.

Diameter = 1000 / π = 318.310

Radius = 318.310 / 2 = 159.155

Area = π × 159.155² = 79577.472m²

From this I have concluded that a circle has the largest area when using a similar circumference. This means that the farmer should use a circle for her plot of land so that she can gain the maximum 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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