• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15
16. 16
16
17. 17
17
18. 18
18
• 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))

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Fencing Problem essays

1. ## Fencing problem.

I have shown this beneath: Area of trapezium = 1/2 � (a + b) � h Area of trapezium = 1/2 � (190m + 310m) � 135m Area of trapezium = 1/2 � 500m � 135m Area of trapezium = 33750m2 So the area of the Gun Shape = Area

2. ## The Fencing Problem

and the resulting value is the area. Area = {1/2[(1000 � 15) x h]} x 15 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 h 66.66666667 33.33333334 Regular Polygons - Icosagon (20-sided-shape) As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan)

1. ## t shape t toal

I will now rotate the t-shape from its original position 90? anti-clockwise and translating it on various grids. I will then collect data and assess this algebraically. Data from 4 by 4 grid Data from 5 by 5 grid Data from 6 by 6 grid The formula like the formula for the t-shape rotated at 90?

2. ## t shape t toal

If we start with a Vertical translation we can work out a general formula for any size grid. (Vertically) I started to look at a 9 by 9 grid. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

1. ## Geography Investigation: Residential Areas

Questionnaire Having the occupant's view of the area I can find the average of the whole on area and compare it to other residential areas around Basingstoke. 5) Questionnaire I will know if the resident owns their own home or if it is privately rented - this will enable me to conclude the hypothesis.

2. ## A farmer has exactly 1000m of fencing and wants to fence off a plot ...

Hero was a Greek mathematician. He found a way of calculating the area of a scalene triangle if all the sides are known. Here are a few scalenes I've worked out the area of S is all the sides of the triangle divided by two which is the same as the perimeter divided by two.

1. ## The Fencing Problem My investigation is about a farmer who has exactly 1000 ...

400 x 100 40000 200 300 300 x 200 60000 225 275 275 x 225 61875 240 260 260 x 240 62400 All of the shapes that will be formed from using the lengths and widths in my table are all still rectangles as the sides and angles are still

2. ## The Fencing Problem. My aim is to determine which shape will give me ...

as in the triangle, which has 3 sides I received a total of 48,126m2, when I used the square I had an area of 62,500m2 and when I used the pentagon the area increased from the area which I received from the square, and the area I got was 68823m2.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to