• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 2247

The Fencing Problem

Extracts from this document...

Introduction

        Maths coursework        

Maths coursework

THE FENCING PROBLEM

Investigation

A farmer has exactly 1000 metres of fencing and wants to fence off a plot of level land.

She is not concerned about the shape of the plot, but it must have a perimeter (or circumference) of 1000m.

She wishes to fence off the plot of land, which contains the maximum area.

Therefore, my aim is to investigate the shape, or shapes that could be used to fence in the maximum area using exactly 1000 metres of fencing each time.

Rectangles

I will start with working out the area of rectangles as it is easy to find its area.

image10.jpgimage10.jpgimage10.jpgimage10.jpgimage10.jpgimage00.png

22,500m2           40,000m2            52,500m2            60,000m2        62,500m2

image10.jpgimage10.jpgimage10.jpgimage10.jpg

60,000m2           52,500m2             40,000m2           22,500m2

Table

Width / m

Height / m

Area / m2

50

450

22,500

100

400

40,000

150

350

52,500

200

300

60,000

250

250

62,500

300

200

60,000

350

150

52,500

400

100

40,000

450

50

22,500

Graph

image12.png

The table shows that the maximum area of a rectangle with a perimeter of 1000m is a 250 x 250 square.

Proof

In the graph, if we look at one point on either side of the highest point, they are clearly less than the maximum point, which proves that a square has the highest area.

Triangles

Next, I will look at the maximum area of a triangle with a perimeter of 1000m.

I will use the cosine rule to work out one of the angles:

Cos A = (b2 + c2 – a2) / (2bc)

Cos A = (2502 + 3502 – 4002) / (2 x 250 x 350)

...read more.

Middle

Cos A = 0.5 so A = 60o

Now that I have one angle, I can work out the area:

Area = ½ x b x c x Sin A

Area = ½ 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) because with the rectangles, the one with the highest area was the square, which had all 4 sides the same length. Similarly, with the above triangle, all three sides were the same length and this meant that the above triangle had a higher area than the first one. However, to make sure that my prediction is right, I will work out the area of a triangle very similar to the one above.

image05.png

I will use the cosine rule to work out one of the angles:

Cos A = (b2 + c2 – a2) / (2bc)

Cos A = (3402 + 326.62 – 333.32) / (2 x 340 x 326.6)

Cos A = 0.5006 so A = 59.96o

Now that I have one angle, I can work out the area:

Area = ½ x b x c x Sin A

Area = ½ x 340 x 326.6 x Sin (59.96) = 48,074.02cm2

This shows that shapes with all sides the same length have the highest formula.

Why do regular shapes have the biggest area?

All regular sided shapes have an n (n being the number of sides in that shape) number of equilateral triangles. Therefore, it is important to realise why an equilateral triangle has the biggest area compared to any other type of triangle. However,

...read more.

Conclusion

Circumference = 2πr

1000 = 2πr

1000 / 2π = r

159.1549431 = r

Now I can calculate the area:

Area = πr2

Area = 79,577.4715459477cm2.

This shows that the shape with the maximum is with a perimeter (or circumference) of 1000m is a circle. I firstly worked out that as the number of sides increased, the area increased as well but at a decreasing rate. The reason that I realised that a circle would give the maximum area was because as the number of sides increased, it would at one point reach an infinite number, which would be a circle. So therefore, a circle must have the maximum area because it has the most number of sides from any other shape.

Why a circle has the maximum area

image04.pngimage03.png

As we can see from the two pictures above, the triangle and octagon fit into the circle easily. However, there are many gaps left between the shape and the circle. This represents the extra are the circle has and answers why a circle has the maximum area.

Conclusion

Ultimately, I would recommend the farmer build a fence in the shape of a circle, with a circumference of 1000m and a maximum area of 79577.4715459477cm2. This will give her the maximum area with the length of fence that she has.

...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. The Fencing Problem

    =(A12*C12)/2 The table below is the same table as above but in normal view. Base (m) Sloping Height (m) Perpendicular Height (m) Perimeter (m) Area (m�) 50 475 474.34 1000 11858.54 100 450 447.21 1000 22360.68 150 425 418.33 1000 31374.75 200 400 387.30 1000 38729.83 250 375 353.55

  2. Math Coursework Fencing

    By drawing a line from every one of vertices to the centre of the pentagon, one can draw isosceles triangles. . The number of triangles formed, will always be equivalent to the number of sides of the polygon. An example is shown below in a regular pentagon with sides of

  1. My investigation is about a farmer who has exactly 1000 metres of fencing and ...

    by 2 to get right-angled triangles, so I can put 180/n to get a more simple equation: L/(2tan(180/n)) The 180/n is also in brackets because I am multiplying that answer by 2tan, and then the whole of that answer is will be divided into L.

  2. Investigate the shapes that could be used to fence in the maximum area using ...

    by 8, the number of sides an octagon has. 1000m/ 8 = 125m I will now look at one of the 8 segments to find out its area. The angle at the top of the triangle, round the outside, we know is 360�.

  1. The Fencing Problem.

    To find this I will divide 360 by the number of right-angled triangles (in this case 10). I can now tell the following about the triangle: - I can now use Trigonometry to find the height of the triangle. SOH CAH TOA I know what the opposite is and the angle, and I want to know what the adjacent is.

  2. Maths Coursework: The Fencing Problem

    4239.934289 x 2 = 8479.868579m2 (Area of isosceles triangle). 8479.868579 x 9 = 76318.81721m2 Area of Nonagon = 76318.81721m2 Regular Decagon Now I'll aim to find the area of a regular decagon within a circle, therefore each length of the decagon = 100m.

  1. Fencing problem.

    I shall be investigating many kinds of rectangles. * Trapezium: this is a quadrilateral with a single pair of parallel sides. * Parallelogram: this has two pairs of parallel sided with equal opposite angles. * Rhombus: this is the final shape that I shall be investigating is another quadrilateral with parallel sides and the diagonals bisect each other at 900.

  2. Graphs of Sin x, Cos x; and Tan x

    The equation of the graph is therefore y = 3cos2x�. Remember If the stretch is parallel to the x-axis with a scale factor of , then the transformation is f(bx). The sine rule We can use the sine rule when we are given: 1.

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