• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1348

A farmer has brought 1000 metres of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the biggest size. I will investigate

Extracts from this document...

Introduction

Maths Coursework

The fencing problem:

Aim:

A farmer has brought 1000 metres of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the biggest size. I will investigate different shapes the fencing can make to achieve the largest area.

Rectangles

I begin my investigation by starting with different shaped rectangles; I will change the value of the widths and lengths by going up in increments of 50m.

                        450mimage00.png

                                                                   50m

Area = 450 x 50 = 22,500m²

                        400mimage01.png

                                                          100m

Area = 400 x 100 = 40,000m²

350mimage08.png

150m

Area = 350 x 150 = 52,500m²

    300m

image17.png

200m

Area = 300 x 200 = 60,000m²

   250mimage19.png

      250m

Area = 250 x 250 = 62,500m²  * Becomes a square*

I have noticed that if you increase and decrease the lengths and widths by 50m the area will increase until it reaches the optimum length and width where the maximum area for the shape can be obtained, a square. From then onwards the lengths, widths and areas begin to repeat themselves in reverse order.

Here is a table to show how the area changes and how it begins to repeat itself after a square is reached.

...read more.

Middle

Area (m²)

480

260

10000

24000

400

300

223.6

44720

333.3

333.3

288.7

48111.9

300

350

316.2

47340

250

375

353.6

44200

180

410

400

36000

150

425

418.3

31372.5

50

475

474.3

11857.5

I have drawn a table so that I can see clearly which triangle had the biggest area:

As you can clearly see from my table, my results show that the equilateral triangle (or regular triangle) is able to obtain the largest area with a perimeter of 1000m.

Because of this pattern all shapes that Ii use now shall also be regular.

I will now investigate a regular pentagon; because there is five sides I can divide it into segments, each segment being an isosceles triangle with a top angle of 72° (one fifth of 360° is 72°). I can work out the other angles by subtracting 72 from 180 and then dividing the answer by two, that gives an answer of 54°.

Because every isosceles triangle can be split into 2 equal right angles triangles I can use trigonometry to work out the area of the triangle; I can then multiply the answer by 10 to give me the area of the full pentagon. I also know that each side is 200m long (1000 divided by 5), so the base of the triangle is 100m.

...read more.

Conclusion

Decagon

image12.png

Each of the ten isosceles triangles that make up this regular decagon look like this:

image10.png

Tan = Opposite

Adjacent

Tan 18º = 0.32

     0.32 = 50

                                                                         ?

     50 / Tan 18º = 153.88

     100 / 2 = 50

     50 x 153.88 = 7,694

                7,694 x 10 = 76,940

A = 76,940m²        

image13.png

image14.png

1000     /    Tan    360

           2 x 12                  2 x 12

        41.6    /    0.2679

        = 155.2819

        1000    x    1000     Tan    360

           2 x 12        2 x 12             2 x 12

        41.6    x    41.6    /    Tan 15

        = 6,459.7238

        1000    x    1000     Tan    360      x      12

           2 x 12        2 x 12             2 x 12

        41.6    x    41.6    /    Tan 15    x     12

        = 77,516.6853

A = 77,516.6853 m²

image15.png

image16.png

In order to show more clearly how the area is increasing with the number of sides I have drawn up a table to show sides against area:

No of sides

Area (m)

3

48111.9

4

62500

5

68800

6

72150

7

74120.2

8

75445

10

76940

12

77,516.7

20

78,914.1

As I have already come to the conclusion that as the number of side’s increases the area increases I am going to work out the area of a circle. Circles have infinite sides and should prove my conclusion to be correct.

*r = radius*

image18.png

c = 1000 = 2πr

      500 = πr

Area of a circle = πr²

πr² = 79,622m²

500 = π x Ans² = 79,577.47155

π

A = 79,577.5m2

From this I conclude that a circle has the largest area when given a perimeter of 1000m.

...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

    The next simplest shape is the 4-sided shape, namely a rectangle. I have constructed another formula linking the sides. From the diagram the area must equal -X� +500X Unlike in the previous example, this turns out to be a quadratic equation so I can plot it on a graph.

  2. A farmer has 1000 metres of fencing. She wants to use it to fence ...

    Answer = 11858.54cm2 Answer = 22360.68cm2 Answer = 31374.75cm2 Answer = 38729.83cm2 Answer =44194.17cm2 Answer =47434.16cm2 Answer =47925.72cm2 Answer =44721.36cm2 Answer =35575.62cm2 Table per 1000 a b base 475 475 50 11858.54 450 450 100 22360.68 425 425 150 31374.75 400 400 200 38729.83 375 375 250 44194.17 350 350

  1. When the area of the base is the same as the area of the ...

    I drew a net of a rectangle 18 by 8 and then cut off 0.5 etc of each of the 4 corners. I then measured the Area of the base, the area of the 4 sides and the volume. To measure the area of the base and the area of

  2. Investigating different shapes to see which gives the biggest perimeter

    To find out the height I must half the triangle to get a right-angled triangle from the isosceles and then I will use Pythagoras's theorem to work out the height. Normal Isosceles Triangle: When I cut the triangle in half, I get a right-angled triangle like this: I can now use Pythagoras's theorem to work out the height.

  1. the fencing problem

    I will start off by looking at isosceles triangles to find the largest area with the perimeter of 1000m. Isosceles triangles 400 � 2 = 200 base x Height H= 300� - 200� 2 300m 300m H= 90000- 40000 400 x 223.60 H= 50000 2 H= 50000 Area = 44720m

  2. The Fencing Problem.

    This will be =B2*C2 where B2 is the length and C2 is the width. This formula is the same as the one used previously to calculate the area of a rectangle. The formulas and headings will be entered in as shown in the table below.

  1. Maths Coursework - The Fencing Problem

    Again, like the square, they produce an area of 62500m�. The Trapezium The trapezium can be worked out in almost the same way the same way as the parallelogram using the 'vertical height = side sin 90' formula. Using the area formula ' area = 1/2 x (a + b)

  2. The Fencing Problem

    have one angle, I can work out the area: Area = 1/2 x b x c x Sin A Area = 1/2 x 333.3 x 333.3 x Sin (60) = 48,112.5cm2 I think that the above triangle will give the highest area (for triangles)

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