• 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
• Level: GCSE
• Subject: Maths
• Word count: 2243

# The Fencing Problem.

Extracts from this document...

Introduction

The Fencing Problem

By Alexa Glick

10 Bruce

Aim

To make a fence that is 1000m long. The area inside the fence has to have the maximum area. I am investigating which shape would give this.

Hypothesis

I think that the area of a shape will always increase as the number of sides increase.

Method

I am going to start investigating different shape rectangles, all with a perimeter of 1000m. Below are four rectangles (which are not drawn to scale) this is showing how different size shapes with the same perimeter can have different areas.

300m                                                                                             499m

1m

200m

250m

498 m

250m

2m

By looking at these rectangles, it is shown that any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m X 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side.

Middle

9600

490

10

4900

500

0

0

Using this formula I can draw a graph of base length (x) against area.

As you can see, the graph has formed a parabola, According to the table and the graph; the rectangle with a base of 250m has the greatest area. This shape is also called a square, or a regular quadrilateral. Because I only measured to the nearest 10m, I cannot tell whether the graph is true, and does not go up just to the sides of 250m. I will work out the results using 249m, 249.5 and 249.75.

 Base (m) Height (m) Area (m²) 249 251 62499 249.5 250.5 62499.75 249.75 250.25 6249993.75 250 250 62500 250.25 249.75 62499.9375 250.5 249.5 62499.75 251 249 62499

All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas.

Now that I have found that a square has the greatest area of the rectangles group, I am going to find the triangle with the largest area. Because in any scalene or right-angled triangle, there is more than 1 variable, there are countless combinations, so I am only going to use isosceles triangles. This is because if know the base length, then I can work out the other 2 lengths, because they are the same. If the base is 200m long then I can subtract that from 1000 and divide it by 2. This means that I can say that.

Side = (1000 – 200) ÷ 2 = 400. This can be rearranged to the following formula.

Conclusion

 No. Of sides Area (m²) 8 75444.174 9 76318.817 10 76942.088

As you can see, the larger the number sides on the shape, the larger the area is. This pattern has carried on going for all of the shapes that I have investigated, so I am going to investigate shapes with the following amount of sides:

20, 50, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000.

Below is a table showing the results that I got.

 No. Of sides Area (m²) 20 78921.894 50 79472.724 100 79551.290 200 79570.926 500 79576.424 1000 79577.210 2000 79577.406 5000 79577.461 10000 79577.469 20000 79577.471 50000 79577.471 100000 79577.471

Here is a graph showing the No. Of sides against the Area.

As you can see form the graphs, the line straightens out as the number of side’s increases. Because I am increasing the number of sides by large amounts and they are not changing I am going to see what the result is for a circle. Circles have an infinite amount of sides, so I cannot find the area by using the equation that I have used to find the other amount of sides out. I can find out the area of a circle by using Ө (pie). I have to work out the circumference of a circle. I can rearrange this so that the diameter is circumference/Ө. From that I can work out the area using the Өr 2 equation.

1000/Ө = 318.310

318.310/2 = 159.155

Ө X 159.1552 = 79577.472m2

If I place this point on my graph it is at the same place, area wise, as the last results on my graph were. From this I conclude that a circle has the largest area when using a similar circumference.

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

the Circle * Circumference = x Diameter * I know my perimeter must be 1000 m so I will replace circumference by

2. ## Fencing problem.

I have discovered the area of six different polygons. These polygons that have been discovered are all regular. The results have been shown below in a tabulated form: Name of regular polygon Area (m2) Pentagon 68800 Hexagon 72217.2 Heptagon 74322.5 Octagon 75450.4 Nonagon 76392 Decagon 76950 I have represented the

1. ## The Fencing Problem

and the resulting value is the area. Area = {1/2[(1000 � 8) x h]} x 8 125 125 125 125 125 125 125 125 h 125 62.5 Regular Polygons - Decagon As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan)

2. ## t shape t toal

If we start with a grid of 9 by 9. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

1. ## Geography Investigation: Residential Areas

With this I can then begin to calculate values for my scatter graph. Table 12 Average Area Rating Street (distance (miles)) Average Area Rating 1 (0.4) 1.761 2 (0.8) 3.303 3 (1) 2.931 4 (1.4) 2.81 5 (1.8) 3.16 6 (2.2)

2. ## The fencing problemThere is a need to make a fence that is 1000m long. ...

62499.75 251 249 62499 All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas. Now that I have found that a square has the greatest area of the

1. ## Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

It would analyse what I thought on aspects of the area. I would rate each of these aspects from 1 to 5 with 1 being the lowest score and 5 the highest. I carried this survey out in 10 different places across the study area to give my opinions on Sutton Harbour as a whole.

2. ## Beyond Pythagoras

I got the first equation which was 3, 4, 5 and I times each number by 2. I times each number by to because then it will make the first number even and I checked if it worked. Here is my working out to see if I changed them into

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