The Fencing Problem.

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MATHS COURSEWORK INVESTIGATION:-

THE FENCING PROBLEM

Planning:-

                     My coursework problem is to find out the highest possible area by using different kind of shapes with different sides and angles. The main aim is to use as many shapes possible but their perimeter must always add up to 1000m. So if it is a rectangle:-

                   

       

 

 (a+a^) +  (b+b^) = 1000m  

 

In case of isosceles triangle:-

           

b + (s+s^)=1000m

The shapes the I am going to use are:-

  1. Rectangles
  2. Isosceles Triangles
  3. 5 sided figure
  4. 6 sided figure
  5. 10 sided figure
  6. 20 sided figure

Rectangles:-

                      The first shape that I am going to use is a rectangle; I decided to use it because it is a simple shape to work with. A rectangle will look like this:

 

To work out the area of a rectangle you would use this formula:

Length x width= Xm2

This would apply to all rectangles and squares. This is the table of results for the areas of the rectangles that I used :-

From the table I can see that if you would constantly increase the length 25m each time. Using rectangles as the shape we can see the areas are always increasing until the length and the width equal each other; this is because if you go beyond the point where the shape is a square the areas would start to repeat itself. This brings us to a conclusion that the highest area that we can get is when we have a square where the sides equal each other. The point where we are using a square using a is where the length and the width is 250m each this gives us a maximum area fir rectangles being at 62500m2.

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Isosceles Triangles:-

                                   The second shape that I used was a triangle. I decided to use it because it is quite a simple shape to use. A triangle will look similar to this:-

To work out the area of the triangle we would use this formula:

½ x base x height

This  formula works with every kind of triangles.

I predict that if we are to achieve the highest area than all the sides must be equal to each other, this ...

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