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  • Level: GCSE
  • Subject: Maths
  • Word count: 2123

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

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Maths Coursework

For this investigation, I will be looking at different shapes, and the areas the different shapes give. The exact question is:

A farmer has exactly 1000m 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 1000m. She wishes to fence of the area of land, which contains the maximum area.

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

I will show the working in the form of formulas, putting results in tables, and then transfer the tables into graphs. Once this is completed, I will draw up a conclusion.



My prediction is that as the number of sides increase, as will the area. I think this as the area of a rectangle, or any other quadrilateral, will have a bigger area than a triangle when using the same perimeter. I have no reason not to believe that this pattern of increasing sides/increasing area will continue.


I am going to start investigating different shape rectangles, all which have a perimeter of 1000m. Below are 4 rectangles (not to scale) showing how different shapes with the same perimeter can have different areas.  

In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length.

...read more.


 Because the regular rectangle was the largest before, I added 333.3 as a base length. This is the length of the base of a regular triangle. It is also an equilateral triangle.

The regular triangle seems to have the largest area out of all the areas but to make sure I am going to find out the area for values just around 333.

Base (m)

Side (m)

Height (m)

Area (m2)

























 This has proved that once again, the regular shape has the largest area.

Because the last 2 shapes have had the largest areas when they are regular, I am going to use regular shapes from now on. This would also be a lot easier as many of the other shapes have millions of different variables.

Because there are 5 sides, I can divide it up into 5 segments. Each segment is an isosceles triangle, with the top angle being 720. This is because it is a fifth of 360. This means I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 540 each. Because every isosceles triangle can be split into 2 equal right-angled triangles, I can work out the area of the triangle, using trigonometry. I also know that each side is 200m long, so the base of the triangle is 100m.

Length of side =   1000    =  200image09.png


...read more.


20, 50, 100, 200, 500, 1000

On the following page is a table showing the results that I got.

No. of sides

Area (m2)













 The pattern remains the same as the number of sides goes up. You can clearly see that as the number of sides increases, the area does to. This is the same from a three- sided shape, all the way up to 1000. If this is the case, then what would the area be of a shape that at an infinite number of sides?

 The final shape I am going to look at is the circle. Circles have an infinite amount of sides, so I cannot find the area by using the equation that I have used to find the other amount of sides out. I can find out the area of a circle by using π r2. But we don’t know the radius yet, so we have to work out the diameter. The circumference of a circle can be found using the equation is π d. I can rearrange this so that the diameter is circumference/ π. From that I can work out the area using the π r2 equation.


We know that the circumference has to be 1000, as that’s the amount of fencing given.

1000/ π = 318.310

Now we have the diameter of the circle. To find the radius, we simply divide by two, as the radius is half the diameter.

318.310/2 = 159.155

So we know have the radius. To find the area of a circle, you multiply π, by the radius squared.

π X 159.1552 = 79577.472m2

...read more.

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