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• Level: GCSE
• Subject: Maths
• Word count: 1962

# The fencing problem.

Extracts from this document...

Introduction

## Aim -

A farmer has exactly 1000 meters of fencing, with it she wishes to fence off a plot of level land. She is not concerned about the plot, but it must have a perimeter of 100m.

I will be investigating different possible shapes, which could have the maximum area.

Hypothesis -

My hypothesis I predict that the shape that has the maximum area is a circle, as it as an infinite number of sides.

I will be starting my investigation with Quadrilaterals.

The first shape I will be investigating is a square, which is a regular shape with four sides that has all sides of equal length and equal angles.

Width

Length

The perimeter has to be 1000 meters, therefore I will divide 1000 by four

1000÷4 = 250

250m

250m                                                                    250m

250m

To find the area of a square, I have to use the formula:

Length x width

Therefore, I will times 250 x 250

Area = length x width

Area = 250m x 250m

Area = 62500m²

A square is a regular shape and that is why there is only one possible area for a square, which = 62500m².

The next shape that I will be investigating, is a rectangle:

Length

Width

There are many possible areas for a rectangle with the perimeter of 1000 meters because the sides of a rectangle are different.

Middle

The same theory applies to a Trapezium, as they are made up of diagonals and the heights will be less than the sides that are made up of hypotenuses.

This is because the hypotenuse is the longest side in a triangle and this is why the height is always going to be smaller.

To work out the length of a trapezium the following formula is needed:

½(a + b) x h

a

B

## Area of a trapezium = ½ (a + b) h

Kites and Rhombuses

I do not need to investigate these shapes, both because if I take an equilateral kite, it becomes a Rhombus and a Rhombus is a square. The square has the biggest area of all the quadrilaterals. This shows that you do not have to investigate Kites and Rhombuses as they all fall under the categories of the square.

Now I will investigate the areas of Triangles.

First I will try an equilateral triangle. It has all equal sides and the same angles.

There are three sides that all have to be the same length. The three lengths have to equal 1000 metres so you therefore have to divide 1000 by 3.

1000÷3 = 333.3333333 (to 7 d.p)

333.3333333m                                  333.3333333m

333.3333333m

To find the area of a triangle you have to use the formula;

Conclusion

My final investigation is the Circle:

Perimeter = 2Πr

2Πr = 1000m

Πr = 1000 ÷ 2

Πr = 500m

r = 500 ÷ Π

r = 159.155

AREA = Πr²

= Π (159.155) ²

= Π (25330.296)

Area of a Circle = 79577.47155m ²

I have proven my hypothesis as the circle has the largest area with a perimeter of 1000m.

I am going to develop on the area of a circle as it is very similar to the area of a regular polygon which,

Area = 250000 ÷ (n x tan (180/n)

From this I can see that I can get the formula of a regular polygon. This is how:

Place the above equation into a formula for a circle and you get

Area = Π x 500² / Π²

250000 / Π

Eventually the Polygon will turn into a circle, as a circle has an infinitve number of sides.

To develop on the idea of

250000/ n x tan (180/n) I am changing the formula into radians

Π / 180 is the method I found out for radians.

250000/nx tan (180/n) x (Π  /180)

From this I can see that 180 and 180 cancel and as tan doesn’t make a difference in the equation so I am left with

25000/ n(Π/n)

## The n’s cancel out…

250000/ Π

Which shows again that the formula for a circle and polygon is very similar

I can also see that I can make an equation to use any perimeter.

If I take my original formula:

A = n(( ½ x 1000/n) x ½ x1000/n/Tan (180/n))

A = n(( ½ x p/n) x ½ x p/n/Tan (180/n))

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