• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

The Fencing Problem.

Extracts from this document...

Introduction

Fencing Problem – Math’s Coursework

Fencing Problem – Math’s Coursework

A farmer has exactly 1000 meters of fencing and wants to fence of a plot of level land. She is not concerned about the shape of the plot but it must have a perimeter of 1000 m. She wishes to fence of a plot of land that contains the maximum area. I am going to investigate which shape is best for this and why.

I am going to start by investigating the different rectangles; all that have a perimeter of 1000 meters. Below are 2 rectangles (not drawn to scale) showing how different shapes with the same perimeter can have different areas.

http://biketrader@www.

...read more.

Middle

370

130

48100

380

120

45600

390

110

42900

400

100

40000

410

90

36900

420

80

33600

430

70

30100

440

60

26400

450

50

22500

460

40

18400

470

30

14100

480

20

9600

490

10

4900

Using this table I can draw a graph of height against area. This is on the next sheet.

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.

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. I am only going to use isosceles triangles because if I know the base 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 two. This means that I can say that:

Side = (1000 – 200) / 2 = 400

To work out the area I need to know the height of the triangle. Tow ork out the height I can use Pythagoras’ Theorem.

...read more.

Conclusion

As you can see from the graph, the line straightens out as the number of side’s increases. Because I am increasing the 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 number of sides, so I cannot find the area using the equation for the other shapes. I can find out the area by using π. To work out the circumference of the cir le the equation is πd. I can rearrange this so that diameter equals circumference/π. From that I can work out the area using the πr² equation.

DIAMETER = 1000 / π = 318.310

RADIUS = 318.310 / 2 = 159.155

AREA = π × 159.155² = 79577.472m²

From this I have concluded that a circle has the largest area when using a similar circumference. This means that the farmer should use a circle for her plot of land so that she can gain the maximum area

What impression did William Woodruff give of the visitors from America? How does the author use language to show the different reaction s of the family to the visitors?

487

204

152

488

...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 Coursework - The Fencing Problem

    I can work out the vertical height of the right-angled triangle by using the formula '100 tan 54'. This gives me the result 137.6. So, the total area of the right angled triangle is 6880 m�, and the total area of the isoclines triangle is 13760 m�.

  2. Fencing problem.

    = the total perimeter of the scalene triangle � 2 Area of a scalene triangle (s) = 357 m + 349 m + 294 m � 2 Area of a scalene triangle (s) = 1000 � 2 Area of a scalene triangle (s)

  1. The Fencing Problem

    These results confirm that 22360.68 is the largest area. Additionally, the triangle it resembles is indeed isosceles. With this information I can deduce that the reason being for the isosceles having the largest area is owing to the fact that it has the largest height. Therefore, I can predict that isosceles triangles will prove to have greater areas overall than that of scalene triangles.

  2. Fencing Problem

    This should narrow my results down further, to give me a more accurate result. Base Side A Side B Perimeter S Area (m�) 332 334 334 1000 500 48111.37 332.5 333.75 333.75 1000 500 48112.07 333 333.5 333.5 1000 500 48112.45 333.5 333.25 333.25 1000 500 48112.50 334 333 333

  1. The Fencing Problem

    59088.47 90 200.00 60000.00 100 196.96 59088.47 110 187.94 56381.56 120 173.21 51961.52 130 153.21 45962.67 140 128.56 38567.26 150 100.00 30000.00 160 68.40 20521.21 170 34.73 10418.89 These results show that a parallelogram has the greatest area when angle x is 90?

  2. Fencing Problem

    I have completed my research on scalene triangles; I will start a new investigation on pentagons. In this investigation I am going to be using trigonometry, to find the area of the whole pentagon, I am going to split it up into 5 smaller triangles.

  1. t shape t toal

    22 23 24 25 Again using the traditional method, we get 75 (8 + 13 + 17 + 18 + 19), using the formula: t = 5x - 7g t = (5 x 8) - (7 X 5) t = 40 - 35 t = 5 5x - 7g cannot be used to find the T-total (t)

  2. The Fencing Problem

    Each shape has an n number of equilateral triangles. This is what a general triangle looks like for each shape: To find the area of this triangle, I need to split it into two to make two right-angles triangles. Then I can work out the area of one of the

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