• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1818

Fencing Prblem

Extracts from this document...

Introduction

Introduction

A Farmer has a plot of land. She has only 1000m of fencing. She wants to fence off a plot of land. She is not concerned about the shape but the perimeter has to be exactly 1000m. The fenced off area has to be the maximum contained area. I am going to investigate the shape or shapes that could be used to fence in the maximum area using exactly 1000 metres of fencing each time.

Firstly, I am going to investigate 3 - sided shapes - triangles, 4 – sided shapes-quadrilaterals and 5 sided shapes-pentagons. From there on I am going to gather the results and see if I need to investigate any further.

TRIANGLES

In first part of my coursework I am going to investigate different types of triangles.

Firstly I will look at Isosceles triangles.

The formula for triangles:

There are two ways of working out the area. You will get the same answer with either formula.

  • Area = ½ x base x height
  • Area = base x height

2

Triangle 1

                             450m                        450m

    100m

I divide the Isosceles triangle

450 m2 - 50m2 = Height m2

202,500 – 2,500 = 200,000

Height = square root of 200,000

Height = 447.21 to two decimal places.

To calculate the area of this Isosceles triangle we calculate:

Area = ½ x base x height

Area = ½ x 100 x 447.21

Area = 22,361 m2 (meter squared) to the nearest whole number.

Triangle 2

I chose a different size isosceles triangle to continue my investigation

                             425m                        425m

   150m

...read more.

Middle

354

44,250

Triangle 5

293

293

42,924.5

In the above results, I have found out that the equilateral Triangle had the largest area. I also noticed that there is a pattern. The pattern is that as the base increases and the height decreases, the area also increases.


QUADRILATERALS

I am going to investigate different quadrilaterals making sure that the perimeter is 1000 metres at all times.

The formula for the quadrilaterals:

  • Perimeter = length + width + length + width.
  • Area = length x width.

My first shape is going to be a square:

                                       250m

A square has 4 sides. The 4 sides of the square are all equal. Therefore that means that all the edges will have the length of 250m due to the perimeter being 1000m.

Area = length x width.

Area = 250 meters x 250 meters

Area = 62,5002 (meters squared)

The next 5 shapes are going to be different types of rectangles. They will be numbered for further investigation:

(1)                                450m

                        50m                

(2)                400m  

                  100m

(3)        300m

                        200m                

(4)        350m

                                150m

(5)        280m

        220m


Results of the Quadrilaterals.

Shape

Length(m)

Width(m)

Area(m2)

Square

250

250

62500

Rectangle 1

450

50

22250

Rectangle 2

400

100

40000

Rectangle 3

300

200

60000

Rectangle 4

350

150

52500

Rectangle 5

280

220

61600

In the above results, I have found out that the Square had the largest area.

                                                      200 cm      

To find out the area of a pentagon, I am going to divide the pentagon into 5 isosceles triangles.

Each length is equal to 200 cm as 1000

   5

                  = 200

                   hyp                  hyp

                                  h        

                        200cm

To find out the hypotenuse, I am going to use the tangent formula

Tan = Opp

         Adj  

Tan 36 =100

...read more.

Conclusion

Shape

Base(m)

Height(m)

Area(m2)

Triangle 1

100

447

22,350

Triangle 2

150

418

31,350

Triangle 3

200

387

38,700

Triangle 4

250

354

44,250

Triangle6

300

316

47,400

Equilateral triangle

333.33

289

48,166

Triangle7

350

274

47,950

Triangle8

400

224

44,800

I then put these values in a graph, shown below. I realised I had reached the peak of the area as I reached the area of the equilateral triangle.

image00.png

The following table consists of all the measurements to the quadrilaterals above.

Shape

Length(m)

Width(m)

Area(m2)

Square

250

250

62500

Rectangle 1

450

50

22250

Rectangle 2

400

100

40000

Rectangle 3

300

200

60000

Rectangle 4

350

150

52500

Rectangle 5

280

220

61600

Bearing the calculations in mind, I then put the quadrilaterals in order and compared the base of the quadrilaterals with the Area. I than calculated the area for a few more quadrilaterals and tabulated them below in order and compared the base of the quadrilaterals with the Area.

Base (M)

Height(M)

Area(M)

 100

400

40000

 150

350

52500

 200

300

60000

 220

280

61600

 250

250

62500

 280

220

61600

 300

200

60000

 350

150

52500

 400

100

40000

I then put these values in a graph, shown below. I realised I had reached the peak of the area as I reached the area of the Square.

image01.png

I put the maximum results together to compare them.

The Circle gave me a coverage area of 79,382.34 metres squared. The Square gave me a maximum coverage area of 62,5002 (metres squared). The Triangle gave me a maximum coverage area of 48,1662 (metres squared).

Comparing all the values investigated so far in comparison the Circle covers the largest farm area.  Therefore I would recommend the Farmer to build her fence in a circle, as she will cover the area of 79,382.342 metres squared.

...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. Maths Fence Length Investigation

    Below is the formula and area when using a base of 200m. H2 = h2 - a2 H2 = 4002 - 1002 H2 = 150000 H = 387.298 1/2 X 200 X 387.298 = 38729.833m. Below is a table of result for isosceles triangles from a base with 10m to a base with 500m.

  2. Borders Investigation

    We know that the formula is cubic, as the differences become constant on the third row. We now work out the coefficient of the in the formula by dividing this constant number by six, and this gives us a result of .

  1. History Coursework: Local Study, Stanton Drew Stone Circles

    This circle lies in ruins and on private land. The average distance between the stones is 6.9m, again this measurement is close to the other measurements but because the south west circle is is ruins this measurement is again probably false from the original size between the stones because the

  2. Geography Investigation: Residential Areas

    I need to use a photographic form of collecting evidence otherwise I have no proof of what the area looks like and whether it actually has a problem or not. With the pictures I will annotate them to show problems or the good things about the residential areas.

  1. Maths Fencing Coursework

    52500 = 65625 300 275 200 225 Area= 200 X 300 Area= 225 X 275 = 60000 = 84375 250 250 Area= 250 X 250 = 62500 From these results, I have realised that the maximum area for a rectangle is square.

  2. The Fencing Problem

    quadrilateral with a perimeter of 1000m using either the base or the height. I will test my formulae on three shapes to make sure that they are correct. Test one: The first test will be on a square, the square sides are all 250m in length and it has an area of 62,500m�.

  1. To investigate the areas of different shapes when they are joined together on square ...

    the formula works, as it was able to predict the area of that shape before I counted it. I am now going to test shapes with 2 dots on the inside. Shapes with 2 dots inside: - Shape No. of dots inside No.

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

    This could mean that regular shapes yield larger areas but we do not have enough evidence to prove this yet we will carry on with the investigation. Although to make absolutely sure the maximum area is 25,000cm2 I will test values near to the maximum.

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