• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 3964

The Fencing Problem

Extracts from this document...

Introduction

Maths Coursework: The Fencing Problem

There is a need 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.

Triangles: Scalene

image00.png

The diagram above is not to scale. Instead of having the perimeter to 1000m, only in this diagram, I have made the perimeters of the shape to 10, only to make this part of the investigation easier to understand. We know that the base of all the shapes is 2. The lengths for the equilateral triangle are 4 on each side. This part of the investigation is to explain why the triangle with the longest height cannot have the same base. The tallest triangle also has a perimeter of 10. One of the sides for the tallest triangle is 5, which is understandable. However the other side is 3. This is literally impossible to be because if this triangle was drawn to scale, then the side that is 3 will not end up reaching the base. The isosceles triangle has a side of 4 and it looks shorter than the side of 3. The only way the higher triangle will reach is if the base is shortened. So in the formula ‘h x b ÷ 2’, in the case of the higher triangle, the height will be longer but the base will be shorter.

Looking at this diagram, there is no need to draw out tables to find out whether or not a scalene triangle is bigger than an equilateral or an isosceles in terms of area. I have made it so that the base is the same width for all triangles. The lines that are going from top to bottom on each triangle represent the height.

...read more.

Middle

image07.png

My two diagrams have shown that the rectangle is the shape that will have the largest area no matter what the height or width of the other quadrilateral is.

Quadrilaterals: The Rectangle

image08.png

This is a basic example of a rectangle with a perimeter of 1000m. This is the formula to calculate the area of a rectangle: height x width

height x width = area

200 x 300 = 60000m²

Therefore, the area of the triangle above is 60000m². Now I must do this for lots of rectangles using the same method as I did for triangles; fill out some tables zooming in on each table every time, and then showing the results on a chart. I’m starting by going up by 50m each time:  

BASE(m)

HEIGHT(m)

AREA(m²)

50

450

22500

100

400

40000

150

350

52500

200

300

60000

250

250

62500

300

200

60000

Already the area has started to decrease somewhere around the 200m-300m spot so I will zoom in to find a closer answer:

BASE(m)

HEIGHT(m)

AREA(m²)

210

290

60900

220

280

61600

230

270

62100

240

260

62400

250

250

62500

260

240

62400

The area is biggest around the 240m-260m spot so I shall zoom in again:

BASE(m)

HEIGHT(m)

AREA(m²)

241

259

62419

242

258

62436

243

257

62451

244

256

62464

245

255

62475

246

254

62484

247

253

62491

248

252

62496

249

251

62499

250

250

62500

251

249

62499

I have to zoom in one last time to finally figure out which rectangle has the largest area. This time I am going up in 0.1 of a metre:

BASE(m)

HEIGHT(m)

AREA(m²)

249.1

250.9

62499.19

249.2

250.8

62499.36

249.3

250.7

62499.51

249.4

250.6

62499.64

249.5

250.5

62499.75

249.6

250.4

62499.84

249.7

250.3

62499.91

249.8

250.2

62499.96

249.9

250.1

62499.99

250

250

62500

250.1

249.9

62499.99

In conclusion, I have investigated that the rectangle with the largest area (that has a perimeter of 1000m), has a base length of 250m and a height length of 250m. On the next page is a line graph showing the relationship between the base and the area of the rectangles. It also shows that the rectangle with the base of 250m will have the largest area:

image29.png

Conclusion:

Just like the equilateral, coincidentally, this shape happens to be a square; which is a regular polygon.

...read more.

Conclusion

Comparing the formula for the area of a circle with the area of n

There is an interesting relationship between the two formulas. The formula for the area of n is:

                                                    A=     250,000

n tan (180)

         n

The formula for the area of a circle that has a 1000m perimeter is:

A= πr² = π x 100,000

                    4π²

= 250,000

   π

There is a relationship between the denominators. Let’s take 5 as an example for n. the denominator for the first equation would then be:

5 tan (180)

        5

The answer to this is:

3.149591887

The value of π goes on forever in terms of decimal places but on a calculator, it is:

3.141592654

As you can see the values are very close between the two denominators. Now I shall try substituting n with a lower number than 5. I will try using 3:

3 tan (180)

         3

The answer is:

3.144466757

As you can see the answer is getting closer to π but it isn’t going lower. Below is a table showing that as the numbers are decreasing, the value is getting closer to π but never reaching it. It is acceptable to even use numbers like 0.1 to represent n because we know that a circle doesn’t have 0.1 sides, it has an infinite number of sides:

n

n tan (180)

         n

5

3.149591887

4

3.146706537

3

3.144466757

2

3.142869254

1

3.141911687

0.1

3.141595844

Even when n is 0.1, the value doesn’t reach the same as π. We can then say that π is a limit which cannot be reached but can get closer and closer to being reached.

The final part of this investigation will be a graph showing the ‘shaded’ part of the circle against the number of sides. To find the area of the shaded part, I have to subtract the area of the regular polygon from the area of the circle:

image28.png

In conclusion, my investigation has shown that a farmer should have a regular circular fence as the shape of the plot.

...read more.

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

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Maths investigation - The Fencing Problem

    = 68794.7 Hexagon Using trigonometry 1/2 x 6 x 166 2/3 x 166 2/3 x sin 60 Area = 72168.8 Septagon Each side = 142.9 142.9/sin 51.4 = Y/sin 64.3 Y = 164.8 Area = 7x( 1/2 x 164.8 x 164.8 x sin 51.4)

  2. Fencing problem.

    The trigonometry function TAN shall be used to find the height, which can also be known as length of opposite. TAN � = Opposite � Adjacent TAN 600 = Opposite � 83.35m Opposite = TAN 600 � 83.35 Opposite = 144.4m = Height.

  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. Fencing Problem

    * 5000 m� + 23284.27125 m� = 28284.27125 m� * I will follow this exact method to find the area of other parallelograms, but change certain values, (angles a & b, side length, base, height), in order to find the largest area possible of a parallelogram with a perimeter of

  1. t shape t toal

    in with the data for a T-shape with a T-number of 69. t = 5x - 7g t = (5 x 69) - ( 7 x 10) t = 345 - 70 t = 275 This proves that you have to use -5g for an upward translation of 1.

  2. Geography Investigation: Residential Areas

    an actual summary of how Basingstoke changes from CBD to the outskirts, or in theory, the high class housing area. As well as the above geographical reason for doing my investigation, I also would like to investigate and find out how the areas in Basingstoke do change, people's opinions, property

  1. Fencing Problem

    I will multiply this number by 9 because the 9 segments are all the same size. Therefore I will get the overall area of the nonagon. Nonagon Area Area: Ex. Angle = 360� = 40� 9 Int. Angle = 180�-40�=140� = 140��2=70� Tan = opp adj Tan 70 = h 55.55 2.75 x 55.55 = h 152.8(1d.p)

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

    far all my graphs prove that my statement was true and I have also realised that the increase between the shapes have decreased. Before the increase between the equilateral triangle and the square was 14,376 compared to the difference of the hexagon and heptagon, which is 1993.526.

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