• 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. Medicine and mathematics

    hour interval between doses> Dosage / mg Hour 1 2 3 4 5 Total Amount of penicillin in body 0 300 300 1 180 180 2 108 108 3 64.8 64.8 4 38.88 38.88 5

  2. The Fencing Problem

    334 289.83 1000 48111.37 332.5 333.75 289.40 1000 48112.07 333 333.5 288.96 1000 48112.45 333.5 333.25 288.53 1000 48112.50 334 333 288.10 1000 48112.23 334.5 332.75 287.66 1000 48111.64 335 332.5 287.23 1000 48110.71 335.5 332.25 286.79 1000 48109.46 336 332 286.36 1000 48107.88 336.5 331.75 285.92 1000 48105.97 Once again, the highest area has changed.

  1. Maths GCSE Courswork

    1000m Area = length x width = 125 x 375 = 46875m2 The next rectangle has a length of 150m and a width of 350m: Perimeter = 350 + 150 + 350 + 150 = 1000m Area = length x width = 150 x 350 = 52500m2 Below, I have

  2. Fencing problem.

    a right-angled triangle I shall consider taking the same procedures that have been taken into mind as before. I shall not need to use the Pythagoras theorem, as I shall know the height of the triangle. But to prove that the above triangle is a right-angled triangle I shall use the Pythagoras's theorem.

  1. 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: 333.333�� - 165.5� = 83333.333�� V83333.333�� = 288.675metres I can now enter the height of the equilateral triangle, as

  2. Fencing Problem

    Having to work with two different lengths i.e. L and B makes it difficult to find the area therefore if I can substitute L to B, I can easily work out the area. Looking back at the previous rearranging of the formula 2L + B = 1000, which is: L

  1. Fencing Problem

    I am also going to follow Hero's formula in order for me to find the largest possible area. * 1000 / 2 = 500 m * 500 x (500 - 475) x (500 - 475) x (500 - 50) = 140625000 * 140625000 = 11858.54123 PERIMETER = 1000 m AREA

  2. t shape t toal

    + (T+2) + (T+6) + (T+1) = T-total 5T + 7 = T-total Does this formula work for other t-shapes in the grid. 5T + 7 = T-total 5 ? 9 + 7 = T-total 52 = 52 Data from 5 by 5 grid 5 + 10 + 15 + 9 + 8 = 47 T + (T-3)

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