• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1698

Fencing Problem - Math's Coursework

Extracts from this document...

Introduction

Fencing Problem - Math’s Coursework

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 of 1000 m. She wishes to fence off a plot of land that contains the maximum area. I am going to investigate which shape can provide her needs.

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.

In a rectangle with a perimeter of 1000m, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m X 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side.

...read more.

Middle

480

20

9600

490

10

4900

500

0

0

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, or a regular quadrilateral. Because I only measured to the nearest 10m, I cannot tell whether the graph is true, and does not go up just to the sides of 250m. I will work out the results using 249m, 249.5 and 249.75.

Base (m)

Height (m)

Area (m2)

249

251

62499

249.5

250.5

62499.75

24975

250.25

6249993.75

250

250

62500

250.25

249.75

62499.9375

250.5

249.5

62499.75

251

249

62499

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

All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas.

Now that I have found that a square has the greatest area of the quadirateral group, I am going to find the triangle with the largest area.

...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 circle 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

My results:

Quadirateral: 62500m²

Triangle: 48107.689m²

Pentagon: 68819.096m²

Circle: 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.

...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

    To find the height of the triangle I must now use Pythagorean theorem to find its height. This states that the square of the length of the hypotenuse of a right-angled triangle equals the sum of the squares of the lengths of the other two sides.

  2. The Fencing Problem

    h 335 165 2) h 334.75 165.25 3) h 334.5 165.5 4) h 334.25 165.75 5) h 334 166 6) h 333.75 166.25 7) h 333.5 166.5 8) h 333.25 166.75 9) h 333 167 10) h 332.75 167.25 11)

  1. Math Coursework Fencing

    Therefore this isosceles trapezium is a square. Shown geometrically: Transition Giving bigger area Parallelograms A parallelogram is a four-sided plane figure that has two sets of opposite parallel sides. Every parallelogram is a polygon, and more specifically a quadrilateral. The area of a parallelogram can be seen as twice the area of a triangle created by one of its diagonals.

  2. Fencing Problem

    that the closer the dimensions of the Isosceles triangles are to an Equilateral triangle the larger the area will be. The largest Isosceles triangle I have found with a perimeter of 1000 m has a base of 333.4 metres and sides of 333.3 metres, which is extremely close to the dimensions of an Equilateral triangle, which has sides of 333.3333333.

  1. GCSE Maths Coursework Growing Shapes

    Pattern no. (n) Triangles Added Triangles Added - 3n 1 0 -3 2 3 -3 3 6 -3 4 9 -3 5 12 -3 Formula for working out the number of triangles added = 3n-3 Check Triangles added = 3n-3 = 3 � 3 - 3 = 6 Width For

  2. Fencing - maths coursework

    I can see that the lengths of the sides get closer to each other the area increases. It looks like the greatest area will be when all the sides are exactly equal just like it was for the four-sided shapes.

  1. Koch Snowflake Math Portfolio

    The 's' is taken as 1(i.e. from above), and is therefore not taken into consideration. Verification: n = 0 = /4 n = 1 = =/3 n = 2 = = 10/27 Hence Proved. 4. When n = 4; i.

  2. A farmer has exactly 1000m of fencing and wants to fence off a plot ...

    30 170 300 27658.63337 200 480 320 1000 500 20 180 300 23237.90008 200 490 310 1000 500 10 190 300 16881.94302 The maximum value in scalene is identical to an isosceles! This could mean that the more irregular shapes are the smaller the area, or this could be put

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