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  • Level: GCSE
  • Subject: Maths
  • Word count: 2200

Fencing investigation.

Extracts from this document...


About the Coursework

A farmer has exactly 1000 metres of fencing, with it she wishes 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. It could be any shape with a perimeter (or circumference) of 1000m.

        What she does wish to do is fence off the plot of land which contains the maximum area.


In this coursework, I am therefore going to investigate all of the different possible shapes that have a perimeter/circumference of 1000m, trying to eliminate any useless shapes by giving evidence of their uselessness and thoroughly investigating the more important, regular, larger area shapes.

I will be investigating all types of shapes: Triangles, Quadrilaterals, 5-sided polygons, 6-sided polygons, 10 and         11-sided polygons, larger shapes and also circles.

The triangles I will be investigating are: Equilaterals, Isosceles and Scalene.

The quadrilaterals I will be investigating are: Squares, Rectangles, Trapeziums and Parallelograms.

In the first section of the coursework, I will be investigating all of the 3 types of triangles (there is only one possibility for an Equilateral triangle).



           333.3m        333.3m



Area = ½ base x         

333.3m                333.3m




Pythagoras’ Theorem is then used to work out the perpendicular height of the triangle.

                333.32 = 166.62 + x2

                333.32 – 166.62 = x2

111111.1 – 27777.7 = x2

83333.3 = x2

√83333.3 = x

288.675m = x



        333.3m        333.3m



        Now that the height of the equilateral triangle has been solved, the next step is to work out the total area of the triangle using the simple formula.

                Area = ½ x base x height

                Area = ½ x 333.3 x 288.675

                Area = 166.

...read more.


        As we already have both the length and the breadth of the rectangle shown above, we can solve the area with no trouble whatsoever.

                Area = length x breadth

                Area = 400m x 100m

Area = 40,000m2




        200m        200m



Area = ½ (base + top)


        200m        200m

                        100m        200m        100m

If both sides of the trapezium are 200m long, the top is 200 and the base is 400, then if I split the trapezium into a rectangle and two triangles, then the base of each of those triangles will be 100m long and so therefore I can use Pythagoras to solve the height, therefore I can work out the area.

Pythagoras’ Theorem must now be used to solve the perpendicular height of this trapezium if the area is to be found.

2002 = 1002 + x2

                2002 – 1002 = x2

                30000 = x2

                √30000 = x

173.205m = x

Now that the perpendicular height of the trapezium has been solved, working out the area is an easy procedure.

Area = ½ (base + top) height

                Area = ½ (400 + 200) 173.205

                Area = ½ (600) 173.205

                Area = 300 x 173.205

Area = 51,961.52m2

I see no reason to bother with the trapeziums though because the length of the perpendicular height will always be shorter than the length of one of the sides. Therefore the area is always less than a similar rectangle.

As I explained above, I can split the shape into a rectangle and two right-angled triangles and those two triangles have their hypotenuse as the two sides of the trapezium, and in a right-angled triangle, the hypotenuse is always the longest side.



...read more.


Since 180o is the same as π in radians, my formula can also be written as:

250,000                                (π÷n is the same as 180÷n in

       n(tan(π÷n))                                       radians)

180 divided by n is the expression for the angle.

As ‘n’ increases, the angle gets smaller. When the angle is smaller, the tangent of a small angle is the same as the angle in radians, therefore there is no need for tan (in the case of very small angles, the tangent of the angle makes no difference to the angle itself, therefore as n increases and the angle gets small there is no need for tan. Therefore tan can be discounted and the formula will therefore be written as):



        Now the n’s can cancel out and I’m left with:



        which is the same as the area for a circle.


In the previous section:
I discovered that the circle had the largest area out of all the other shapes that I investigated.
I couldn’t use my formula for the circle because a circle does not have any sides, it is a curved shape.
End of Section 4

The More regular shapes are, the larger their areas will be.

        The more sides a shape has, the larger its area will be.

I have noticed that the more sides a shape has, the closer and closer the number on the bottom line of my formula gets to π (as explained in the previous page).

        In conclusion, the circle is the largest shape of the lot and the farmer should use a circular fence for her farm.

End of Coursework

...read more.

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