• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 4001

Fencing Problem

Extracts from this document...

Introduction

Maths Coursework: Fencing Problem Investigate the shape, or shapes that could be used to fence in the maximum area using exactly 1000 meters of fencing each time. Introduction I will attempt to solve this problem by investigating different shapes with different numbers of sides to find the maximum area that can be gained using 1000 meters of fence. The perimeter of the shapes, no matter how many sides it has, must be 1000m. I will begin by considering three sided shapes then I will move on to other shapes with various number of sides e.g. rhombus, trapezium, parallelogram, pentagon, hexagon etc. Triangle I will look at different types of triangles these are: Right angle triangle, scalene, isosceles and equilateral. Right Angle Triangle Considering the fact that I have to use exactly 1000m to construct a right angle triangle, it would be mathematically incorrect to assume the values of the sides because Pythagoras Theorem states that a2 + b2 = c2 . Therefore, I will have to make a model and construct a right angle triangle with a perimeter of 1000m. I will use a scale of 1: 10,000 using a string. I managed to find that the base and height are 3.75cm & 2.00cm, using Pythagoras Theorem the hypotenuse is 4.25cm. This would all add up to 10cm. using the scale; the sides would be as shown on the diagram below. 425m 375m 200m I have found that the only right angle triangle that can be formed using 1000m as total length without any decimals would have the above lengths. Area of a triangle is: A = 1/2 x B x H A = 1/2 x 200 x 375 A = 37500 m2 However, I can also have other lengths on the sides but this would have decimals: a2 + b2 = c2 32 + 42 = 52 However, this would add up to 12 in total, which exceeds the normal perimeter by 2. ...read more.

Middle

When entering values for the length I could not have entered 0m or 500m because the shape would not be a rectangle then, it would have been a straight line. The areas in the table increases and then decreases. This proves that the areas of the rectangles get larger as the difference between the length and width get smaller. This type of curve is a quadratic graph, which comes from quadratic equations, as you will see next page. This can mathematically be explained. The formula for perimeter we deduced earlier was L + W = 500 If we make W the subject, it will look like this: W= 500- L Now putting this in the area formula which is A = L x W This will change to L x (500-L) Simplifying this would be 500L - L2 Rearranging this equation would look like this: -L2+ 500L making it a quadratic equation. Using the completing the square method this could be factorized to look like this: - (L2- 500L) - (L-250)2+ 62,500 This indicates that the maximum value obtainable for the area of a rectangle with a perimeter of 1000 will be 62,500 and it will fall under the length value of 250. Pentagon I am first going to look at an irregular pentagon, the house pentagon even when I know a regular pentagon will give me a larger area. I am doing this just to prove that regular shapes give larger areas. Now I am going to look at the regular pentagon of all sides = 200m. The perimeter of the regular pentagon is 1000m so the equation is: 5L = 1000 (where L = Length) This can be simplified as: L = 200 m I am using a regular pentagon to prove my statement; a regular pentagon has a bigger area then an irregular area both with 1000m, Even though we have seen that the rectangles and triangles on the previous pages were equilateral and produced the greatest area. ...read more.

Conclusion

This is also shown on the column on the table, which is the difference between the height and the radius. The difference gets smaller as n increases. To explain this even better, I have researched an A-level book to provide rigorous proof. In the A-Level book, it says applying Tan to a very small angle is the same as the angle itself. Therefore, when n comes closer to infinity thus PI/n becomes very small, Tan (PI/n) would be the same as PI/n. To test this we can have a look at the table. As n gets larger, the column PI/n becomes the closer or even same as Tan (PI/n). Tan Pi/n Pi/n Tan Pi/n Pi/n 0.727 0.628 0.246 0.242 0.577 0.524 0.228 0.224 0.482 0.449 0.213 0.209 0.414 0.393 0.199 0.196 0.364 0.349 0.187 0.185 0.325 0.314 0.176 0.175 0.294 0.286 0.167 0.165 0.268 0.262 0.158 0.157 After finding this out, I can compare the formulas to find the area of a circle area of an n-sided shape and try to find the connection. I have used radians because the area for the circle has PI in it and the area of the polygons has 180o in it. Therefore, I can change the 180o into PI so it is easier to compare the formula. Area of Circle Area Of n-sided Polygon Therefore, as you can see the formulas for the height of the triangle in the polygon when n tends to infinity and the radius of the circle are equal. In addition, the formulas for the area of the circle and the area of the polygon when n tend to infinity are equal. That is why the values are almost equal in the table. Conclusion I therefore conclude that the largest shape of 1000m perimeter the farmer can use is a circle of radius 159.15m that gives an area of 79577.47m2. ?? ?? ?? ?? Hamza Sharif Ms. Butt 10X1 - 1 - ...read more.

The above preview is unformatted text

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

    I have showed this on the graph by putting a line through the highest point (highest value). Base (m) Sloping Height (m) Perpendicular Height (m) Perimeter (m) Area (m�) 330 335 291.55 1000 48105.35 330.5 334.75 291.12 1000 48107.34 331 334.5 290.69 1000 48109.00 331.5 334.25 290.26 1000 48110.35 332

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

    20 78921.1894 2000 79577.4061 5000 79577.4611 20000 79577.4709 50000 79577.4714 From looking at my chart you can see that as the sides go up the area is still going up, just. This just about proves that my prediction is correct but there is one thing that will definitely prove that as the number of sides increase, so does the area.

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

    this point the line for the area of the base joins the line of the volume at point 1cm. There are some trends in the graph like the volume line increases its height when there is less cm cut off for example at 1cm it is at a big height

  2. Beyond Pythagoras

    30600cm2 1530cm 9 95cm 900cm 905cm 42750cm2 1900cm 10 105cm 1100cm 1105cm 57750cm2 2310cm Shortest side (a) The difference between the shortest side of each triangle is 10, which means the formula starts off with 10n. After finding out 10n, there is always 5 left so the formula for a is 10n + 5.

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

    simply divide 360� by 5, the 5 being the number of sides the pentagon has.

  2. Fencing problem.

    In a scalene triangle all three sides and all three angles are different. I shall not be using the area of a triangle formula during this part of the investigation. The following formula shall be used: Area of a scalene triangle = Vs (s-a)(s-b)(s-c)

  1. Beyond Pythagoras.

    b + c = 1st term = 4 + 4 + c = 8 = 8 + c = 8 = c = 8 - 8 = c = 0 Bellow are the figures I fount for a, b and c: a = 4 b = 4 c = 0

  2. Beyond Pythagoras.

    + ( 2n + 2n) = (2n + 2n + 1) Lets see a + b (2n + 1) + (2n + 2n) = 4n + 4n + 1 + 4n + 4n + 8n = 8n +4n + 8n + 4n + 1 so if a + b =

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