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  • Level: GCSE
  • Subject: Maths
  • Word count: 1698

Fencing Problem - Math's Coursework

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Introduction

Fencing Problem - Math’s Coursework

A farmer has exactly 1000 metres of fencing and wants to fence off a plot of level land. She is not concerned about the shape of the plot but it must have a perimeter of 1000 m. She wishes to fence off a plot of land that contains the maximum area. I am going to investigate which shape can provide her needs.

I am going to start by investigating the different rectangles; all that have a perimeter of 1000 meters. Below are 2 rectangles (not drawn to scale) showing how different shapes with the same perimeter can have different areas.

In a rectangle with a perimeter of 1000m, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m X 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side.

...read more.

Middle

480

20

9600

490

10

4900

500

0

0

Using this table I can draw a graph of height against area. This is on the next sheet.

As you can see, the graph has formed a parabola. According to the table and the graph, the rectangle with a base of 250m has the greatest area. This shape is also called a square, or a regular quadrilateral. Because I only measured to the nearest 10m, I cannot tell whether the graph is true, and does not go up just to the sides of 250m. I will work out the results using 249m, 249.5 and 249.75.

Base (m)

Height (m)

Area (m2)

249

251

62499

249.5

250.5

62499.75

24975

250.25

6249993.75

250

250

62500

250.25

249.75

62499.9375

250.5

249.5

62499.75

251

249

62499

Using this table I can draw a graph of height against area. This is on the next sheet

All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas.

Now that I have found that a square has the greatest area of the quadirateral group, I am going to find the triangle with the largest area.

...read more.

Conclusion

As you can see from the graph, the line straightens out as the number of side’s increases. Because I am increasing the sides by large amounts and they are not changing I am going to see what the result is for a 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 π. To work out the circumference of the circle the equation is πd. I can rearrange this so that diameter equals 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

My results:

Quadirateral: 62500m²

Triangle: 48107.689m²

Pentagon: 68819.096m²

Circle: 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.

...read more.

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