The Fencing Problem.

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:-

Rectangles
Isosceles Triangles
5 sided figure
6 sided figure
10 sided figure
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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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 kind of triangle would be called an a equilateral triangle.

Because when I did rectangles I found out that the highest area was when it was a square and not a rectangle, that is why I decided to use an a equilateral triangle this is the result that I got.

From the first table I can see that the highest area was 48083.26m2 because after that area the other areas started to repeat themselves, but I got this result before I used an a equilateral triangle which gave the area of 48112.9m2 this showed me that the again the highest area can only be achieved if you are using regular shapes (where the side/bases equal each other).

Other Shapes:-

The other shapes that I used were: 5 sided, 6 sided, 10 sided and 20 sided figures. I started by using a pentagon (5 sided shape). A pentagon will look similar to this.

To work out the area of a regular pentagon we would firstly divide the total perimeter (1000) by the number of sides, in this case by 5: 1000/5, this therefore gives us 5 different small triangles inside one shape. Each triangle will have a base of 200m. After getting the triangles than we can see that all of them are isosceles therefore we would use the same formula as we used before, with other triangles (1/2 x base x height) We already know that all the interior angles add up to 360, we must again divide 360 by the number of side (360/5) which gives us 72 degrees at the top of each triangle. Than we divide the angle at the top by 2: (72/2) which will give us 36 for each half of the triangle.

But we don’t know the height so we need to refer to SOH-CAH-TOA

To find out that we need to use TAN. The final calculation to get the height of the triangle would be:

100/tan 36= 137.64m

So the height is 137.64m. Finally we just use the normal triangle formula and we would see this calculation (100 x 137.64) x 5= 68820m2.

Hexagon:-

The next shape that I am going to use is a hexagon (6 sided figure). A regular hexagon will look similar to this.

To work out the area of a regular hexagon we would firstly divide the total perimeter (1000) by the number of sided, in this case by 6: 1000/6, this therefore gives us 6 different small triangles inside one shape. Each triangle will have a base of 166.67m. After getting the triangles than we can see that all of them are isosceles therefore we would use the same formula as we used before, with other triangles (1/2 x base x height) We already know that all the interior angles add up to 360, we must again divide 360 by the number of side (360/6) which gives us 60 degrees at the top of each triangle. Than we divide the angle at the top by 2: (60/2) which will give us 30 for each half of the triangle.

But we don’t know the height so we need to refer to SOH-CAH-TOA

To find out that we need to use TAN. The final calculation to get the height of the triangle would be:

83.33/tan 30= 144.33m

So the height is 144.33m. Finally we just use the normal triangle formula and we would see this calculation (83.33 x 144.33) x 6= 72162.11m2

Decagon:-

The next shape that I am going to use is a decagon (10 sided figure). A regular decagon will look similar to this.

To work out the area of a regular decagon we would firstly divide the total perimeter (1000) by the number of sides, in this case by 10: 1000/10, this therefore gives us 10 different small triangles inside one shape. Each triangle will have a base of 100m. After getting the triangles than we can see that all of them are isosceles therefore we would use the same formula as we used before, with other triangles (1/2 x base x height) We already know that all the interior angles add up to 360, we must again divide 360 by the number of side (360/10) which gives us 36 degrees at the top of each triangle. Than we divide the angle at the top by 2: (36/2) which will give us 18 for each half of the triangle

But we don’t know the height so we need to refer to SOH-CAH-TOA

To find out that we need to use TAN. The final calculation to get the height of the triangle would be:

50/tan 18= 153.88m

So the height is 153.88m. Finally we just use the normal triangle formula and we would see this calculation (50 x 153.88) x 10= 76940m2

Icosahedron:-

The next shape that I am going to use is an icosahedron (20 sided figure). A regular icosagon will look similar to this.

To work out the area of a regular icosahedron we would firstly divide the total perimeter (1000) by the number of sides, in this case by 20: 1000/20, this therefore gives us 20 different small triangles inside one shape. Each triangle will have a base of 50m. After getting the triangles than we can see that all of them are isosceles therefore we would use the same formula as we used before, with other triangles (1/2 x base x height) We already know that all the interior angles add up to 360, we must again divide 360 by the number of side (360/20) which gives us 18 degrees at the top of each triangle. Than we divide the angle at the top by 2: (18/2) which will give us 9 for each half of the triangle

But we don’t know the height so we need to refer to SOH-CAH-TOA

To find out that we need to use TAN. The final calculation to get the height of the triangle would be:

25/tan 9= 157.84m

So the height is 157.84m. Finally we just use the normal triangle formula and we would see this calculation (25 x 157.84) x 20= 78920m2

Conclusion:-

From the results that I got and the shapes that I used the highest are can be achieved when you have more sides. If the number of sides keeps on increasing the shapes will start to look more like a circle so I tried to use circle , this is the result that I got.

To find the area of the circle we would use this formula:

∏ x r2

But we don’t know the radius. To find out the radius I had to follow these steps:

The circumference is 1000
The formula to work out the circumference we have to use this formula: 2∏r

Than to work out the radius we would have to divide 1000 by 2∏

This would give us the radius being 159.15m.

Lastly we would just apply the normal formula of ∏r2. And we would get the area of 79577.47m2.

From the results I got, I can conclude that the highest area can be achieved when I used a circle.

This can be proved by increasing the number of sides of a shape, because if you keep on increasing the amount of sides it will look more and more like a circle.