• 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. 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. The Fencing Problem

    By rearranging this equation you can work out the height of the triangle. Hypotenuse � - Base � = Height � VHeight � = Height In this case the equation is: 166.666�� - 83.333�� = 20,833.33�� V20,833.33� = 144.333m I can now enter the height of the equilateral triangle, as

  1. The Fencing Problem

    h 333 167 10) h 332.75 167.25 11) h 332.5 167.5 12) h 332.25 167.75 13) h 332 168 14) h 331.75 168.25 ^ Once more, we have discovered a higher area between the gaps. Judging by the measurements, I can predict that the largest area would be that of an equilateral triangle. Base (m) Sloping Height (m)

  2. Koch Snowflake Math Portfolio

    Calculation for values of N6, l6, P6 and A6 : General Formula for Nn: a rn a= 3 r= 4 n= 6 Therefore: 3 (4)6 = 12288 General Formula for ln: a rn a= 1 r= 1/3 n= 6 Therefore: 1 (1/3)6 = 1/729 General formula for Pn: Pn =

  1. Fencing Problem

    To find height we use: Tan X =__L__ 2 x H However, X is: X = _360_ 2 x n Substituting L and X, we get: Tan 180 = 1000 n 2Hn Rearranging the above formula to make H the subject: H = ___500___ n tan 180 n A = (1/2 x L x H)

  2. Fencing problem.

    I shall now calculate the height by drawing a line near the edge of the trapezium to form a right-angled triangle. Now to find the value of (BF) height, I shall use Pythagoras's theorem. The following are the steps that have been taken into consideration to find the height of the trapezium above: BC2 = BF2 + FD2 (200)

  1. Fencing Problem

    1000 500 48112.23 334.5 332.75 332.75 1000 500 48111.64 335 332.5 332.5 1000 500 48110.71 After testing my fourth table of values I noticed the exact same pattern, and I also notice the area had increased quite fairly since I began narrowing down my results from my first table.

  2. Geography Investigation: Residential Areas

    rating and this will give me the estimated average area rating for Cyprus Road. 4.291 + 0.79 = 5.1 (2 s.f.) It is impossible to get over 5 on the average area rating and this means that I expect Cyprus Road to receive a high 4 as an average for their area rating.

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