• 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

GCSE Mathematics

The Fencing Problem

The Problem

        A farmer has 1000 meters of fencing and wants to fence off a plot of level land. She is not troubled about the shape of the plot but it must have a perimeter of 1000 m. She needs to fence of a plot of land that contains the maximum area. I am going to investigate which shape will give the biggest area and show my working.

To start of I will use the simplest shapes and find the largest possible area of each shape. A simple shape is a shape, which has an easy formula to finding its area. For example a rectangle has the formula of length x width to find its area.

Rectangle/Square

I am going to start by investigating rectangles, that all have a perimeter of 1000 meters. Below are 2 rectangles showing how different shapes with the same perimeter can have different areas.

image00.png

image04.png

Length x Width

Perimeter = 1000m          

...read more.

Middle

=(1000-A3*2)

=A3*B3

Using this table I made a graph on excel of one width (A) against Area (AxB). This is on the next page.

The line of results on the graph has made a plane curve (parabola). From the table and the graph, we can see that the rectangle with a Length (A) of 250m has the greatest area. This shape is called a square and is the regular shape of rectangles.

Triangle

I then found the triangle with the largest area. I only used isosceles triangles because if I know the base I can work out the other 2 lengths because they are equal.image06.pngimage05.png

image02.pngimage01.png

image03.png

For example

If B = 200

1 Side = (1000 – 200) / 2 = 400

To work out the area I need to work out H (the height). To work out H (height) I can use Pythagoras’ Theorem. The formula and area a triangle with B (base) of 200m is shown below.

H² = h² - a²

H² = 400² - 100²

H² = 160000 – 10000

H² = 150000

H = 387.298

½ × 200 × 387.298 = 38729.

...read more.

Conclusion

Then using SOH CAH TOA (trigonometry) you can work out that you need to use Tangent.

H = 100 tan54 = 137.638

O = 100

T = tan 36

This gives me the length of H so I can work out the area.

Area = ½ x b x H = ½ x 100 x 137.638 = 6881.910

Now I have the area of half of one of the segments, all you need to do is multiply that number by 10 and get the area of the shape.

Area = 6881.910 × 10 = 68819.096m²

So from this I can be sure that the more sides a shape has the larger its area will be I can also be sure that the regular of each shape gives the largest area. Using the same method of working out as before I then worked out the area of a regular Hexagon and Heptagon.

I used the same method as before to work out the area of the 2 shapes.

Hexagon

1000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3.

360 ÷ 6 = 60 ÷ 2 = 30

Area = ½ * b * H = ½ * 83 1/3 * 144.338 = 6014.065

6014.065 * 12 = 72168.784m2

Heptagon

1000 ÷ 7 = 142.857 ÷ 2 = 71.429m

360 ÷ 7 = 51.429 ÷ 2 = 25.714 degrees

Area = ½ * b * H = ½ * 71.429 * 148.323 = 5297.260

5297.260 * 14 = 74161.644m2

...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. t shape t toal

    We need now to combine the two equations. I only need to instance of 5x - 7g as only one T-shape is being translated: t = (5x - 7g) - b(5g) + (a5) a = number of translations horizontally b = number of translations vertically x = T-number g =

  2. The Fencing Problem

    This process can be used on any sized shape. 55 5/9 ( 8 = 444 4/9m 55 5/9 ( 6 = 333 1/3m 55 5/9 ( 4 = 222 2/9m The lengths of the sides are shown below Side A = 444.444m Side B = 333.333m Side C = 222.222m

  1. Fencing Problem

    The reason for this might be the fact that the difference between the values of the sides. As the difference becomes smaller, the area increases. We saw the same trend when looking at triangles. I will now look other parallelogram.

  2. Fencing problem.

    AC2 = AB2 + BC2 (416.67) 2 = (333.33) 2 + (250) 2 173608.9 = 111108.9 + 62500 173608.9 = 173608.9 Now it is certain that the triangle is a right hand triangle the area can be found. Area of triangle = 1/2 � Base � Height Area of triangle

  1. The Fencing Problem

    at this new table, we can see that the highest area has not varied. In any event, there is still a difference of 4 between the highest area and the other two areas surrounding it; a higher value could lie within that gap.

  2. Fencing Problem

    x (500 - 375) x (500 - 250) = 1953123000 m * 1953123000 m = 44194.17382 m� PERIMETER = 1000 m AREA = 44194.17382 m� * 1000 / 2 = 500 m * 500 x (500 - 350) x (500 - 350) x (500 - 300) = 2250000000 m * 2250000000 m = 47434.1649 m� PERIMETER = 1000

  1. The Fencing Problem

    336.25 293.684 48090.674 328.0 336.00 293.258 48094.241 328.5 335.75 292.831 48097.493 329.0 335.50 292.404 48100.430 329.5 335.25 291.976 48103.050 330.0 335.00 291.548 48105.353 330.5 334.75 291.119 48107.338 331.0 334.50 290.689 48109.003 331.5 334.25 290.259 48110.347 332.0 334.00 289.828 48111.371 332.5 333.75 289.396 48112.072 333.0 333.50 288.964 48112.450 333.5 333.25 288.531

  2. t shape t toal

    translation 2) enlargement 3) rotation and 4) reflection. I will be using two types of transformations firstly; these will be translation, which I have already been using, and rotation. I will also be using these transformations in combination. I will firstly start off by rotating the t-shape clockwise 90?

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