• 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

The Fencing Problem.

Extracts from this document...

Introduction

Prediction

Now that I have looked at the different triangles and quadrilaterals, I can predict that the shapes with the greater number of sides, will have the maximum area as the number of sides, will affect the calculation of the area of the shape of the plot of land. Therefore, the larger the number of sides in the shape the greater the area of land. In turn this will benefit the farmer as the maximum area of land is achieved with the minimum amount of fencing which in the long run will save the farmer money.

Further to my prediction, I think that a circle would give the greatest area of land because a circle is technically made

...read more.

Middle

              c²= 333⅓²- 166.67²

              c²=83 333⅓

               c= 288.68 metres

Area = ½bh

        =½ x333⅓ x 288.68

        =48 122.52 metre²

OR

Area = ½bc SinA

         =½ x 333⅓ x 333⅓ x sin 60º

         =48 122.52

For the isosceles triangle I will take an example of different bases and heights to find out which combination gives the maximum area.

Number 1

a=100 b=c=450image03.pngimage02.png

height   a²=b²+c²

50²=450²+c²

c²=450²-50²

c²= 200 00

c+ 447.21 meters

Area=½bh

=½x 100x 447.21

= 22 360.68 metres²image04.png

Number 2

a=200 b=c=400image02.png

height a²=b²+c²

100²=400²+c²

c²=400²-100²

c²=150 000

c=387.29 meters

Area=½bh

=½x200x387.29

=38729.83 metres²image05.png

Number 3

a=300 b=c=350image02.pngimage03.png

...read more.

Conclusion

Area(m²)

1

100

447.21

22 360.68

2

200

387.29

38 729.83

3

300

316.23

47 434.16

4

400

223.61

44 722

The isosceles triangle which has the maximum area out of the ones which I have investigated is number 3. Number 3 has a base of 300 meters and a height of 316.23 meters. It also has an area of 47 434.16 metres²

I am now going to investigate a right angle triangle.

Ratio of length of sides

a : b : c

3 : 4 : 5

1=1000= 83½m

      12

a = 3 x 83½= 250m

b = 4 x 83½= 333⅓m

c = 5 x 83½= 416⅔m

Angle A

  a      =    c

SinA     SinC

 250   =  416⅔

SinA  = Sin 90º

SinA x 416⅔= Sin 90º x 250

Sin A = Sin90º x 250

                  416⅔

SinA = 0.6

      A=36.87º

Area= ½bc SinA

       =½x333⅓x416⅔x Sin36.87º

       =41 666.77 m²

The last type of triangle I am going to invest is the scalene.

image08.png

image09.png

Number 1

a=400, b= 450 & c=150

CosA= b²+c²-a²

              2bc

CosA=450²+150²-400²

              2x450x150

CosA=  65 000

            135 000

CosA=0.48

A=61.22º

Area= ½bcSinA

=½x450x150xSin61.22º

=29 581.02

Number 2

a=350 b=450 c=200

CosA=b²+c²-a²

             2bc

CosA=450²+200²-350²

              2x450x200

CosA=120 000

           180 000

CosA=0.67

      A=48.19º

Area= ½bcSinA

=½x450x200xSin48.19º

=33 541.18m²

...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. Fencing problem.

    Firstly I will find the area of the triangle the followed by the rectangle that it is places on. This has been shown below: To find the area of the triangle above I am going to make point (F) between (EB).

  2. The Fencing Problem

    possible, while keeping to the 1000m perimeter (or circumference, in this case). Circle I will now aim to prove my prediction; I will use the conventional area formula (A = ?r�). To find out the radius, I will rearrange the formula for finding out the circumference (since we have a known variable already, C = 1000).

  1. Geography Investigation: Residential Areas

    In this hypothesis I am predicting that the outskirts of town will have more houses that are owned by the individual resident and that the property is not rented, may it be privately or rented from a housing association. For the initial enquiry into this hypothesis I will use all the residents I surveyed from every road.

  2. HL type 1 portfolio on the koch snowflake

    considering the values in terms of 6 decimal places both equal to 0.692200. Therefore we can conclude with the result that when n=17, . 6. Now we investigate what happens to the Perimeter and Area as 'n' gets very large.

  1. Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

    It means that late night drunks come down teats hill and expect to be able to use the footbridge to get to the Barbican. But it closes at night and this means that these people will now stay around Ruby's community and cause havoc until late into the night.

  2. Graphs of Sin x, Cos x; and Tan x

    Two sides and an angle opposite to one of the two sides. 2. One side and any two angles. Remember Try to use the formula with the unknown at the top of the fraction. We can use the sine rule to find the size of an angle or the length of a side.

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