• 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
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  • Level: GCSE
  • Subject: Maths
  • Word count: 1962

The fencing problem.

Extracts from this document...

Introduction

Math investigation

Aim -

A farmer has exactly 1000 meters of fencing, with it she wishes to fence off a plot of level land. She is not concerned about the plot, but it must have a perimeter of 100m.

I will be investigating different possible shapes, which could have the maximum area.

Hypothesis -

My hypothesis I predict that the shape that has the maximum area is a circle, as it as an infinite number of sides.

I will be starting my investigation with Quadrilaterals.

The first shape I will be investigating is a square, which is a regular shape with four sides that has all sides of equal length and equal angles.

                                                             Width

image00.png

                  Length

The perimeter has to be 1000 meters, therefore I will divide 1000 by four

1000÷4 = 250

                                                     250m

image00.png

          250m                                                                    250m

                                                       250m

To find the area of a square, I have to use the formula:

Length x width

Therefore, I will times 250 x 250

Area = length x width

Area = 250m x 250m

Area = 62500m²

A square is a regular shape and that is why there is only one possible area for a square, which = 62500m².

The next shape that I will be investigating, is a rectangle:

                                                            Length

image08.png

 Width

There are many possible areas for a rectangle with the perimeter of 1000 meters because the sides of a rectangle are different.

...read more.

Middle

The same theory applies to a Trapezium, as they are made up of diagonals and the heights will be less than the sides that are made up of hypotenuses.

This is because the hypotenuse is the longest side in a triangle and this is why the height is always going to be smaller.

To work out the length of a trapezium the following formula is needed:

 ½(a + b) x h

                                                a        

image10.pngimage09.png

image11.png

                                                 B

Area of a trapezium = ½ (a + b) h

Kites and Rhombuses

I do not need to investigate these shapes, both because if I take an equilateral kite, it becomes a Rhombus and a Rhombus is a square. The square has the biggest area of all the quadrilaterals. This shows that you do not have to investigate Kites and Rhombuses as they all fall under the categories of the square.

image12.pngimage13.png

image14.png

image16.png

Now I will investigate the areas of Triangles.

First I will try an equilateral triangle. It has all equal sides and the same angles.

There are three sides that all have to be the same length. The three lengths have to equal 1000 metres so you therefore have to divide 1000 by 3.

1000÷3 = 333.3333333 (to 7 d.p)

 333.3333333m                                  333.3333333m

333.3333333m

To find the area of a triangle you have to use the formula;

...read more.

Conclusion

My final investigation is the Circle:

Perimeter = 2Πrimage20.png

                                                        2Πr = 1000m

                                                        Πr = 1000 ÷ 2        

                                                        Πr = 500m

                                                        r = 500 ÷ Π        

                                                        r = 159.155

                                                AREA = Πr²

                                                         = Π (159.155) ²

                                                         = Π (25330.296)

Area of a Circle = 79577.47155m ²        

I have proven my hypothesis as the circle has the largest area with a perimeter of 1000m.

Radians

I am going to develop on the area of a circle as it is very similar to the area of a regular polygon which,

Area = 250000 ÷ (n x tan (180/n)

Radius = 1000 ÷ 2Π

From this I can see that I can get the formula of a regular polygon. This is how:

Place the above equation into a formula for a circle and you get

Area = Π x 500² / Π²

250000 / Π

Eventually the Polygon will turn into a circle, as a circle has an infinitve number of sides.

To develop on the idea of

250000/ n x tan (180/n) I am changing the formula into radians

Π / 180 is the method I found out for radians.

250000/nx tan (180/n) x (Π  /180)

From this I can see that 180 and 180 cancel and as tan doesn’t make a difference in the equation so I am left with

25000/ n(Π/n)

The n’s cancel out…

250000/ Π

Which shows again that the formula for a circle and polygon is very similar

I can also see that I can make an equation to use any perimeter.

If I take my original formula:

A = n(( ½ x 1000/n) x ½ x1000/n/Tan (180/n))

A = n(( ½ x p/n) x ½ x p/n/Tan (180/n))        

...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. The Fencing Problem

    ( ((500 ( 9) ( (tan (180 ( 9)) = 76,318.817m� Diagram Regular 10 Sided Polygon: A ten-sided regular polygon is called a regular decagon. Area = 10 (.5 ( (1000 ( 10)) ( ((500 ( 10) ( (tan (180 ( 10)) =76,942.088m� Diagram Regular 15 Sided Polygon A fifteen-sided regular polygon is called a regular.

  2. 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 =

  1. Geography Investigation: Residential Areas

    In today's society, that just isn't big enough for a couple who want to start a family etc, so they want to leave. So far my hypothesis had been proved correct with the two studies I have done so far, the final section of this hypothesis I am going to

  2. The Fencing Problem

    to find the height, and then substituting the values into the conventional triangle formula (A = 1/2bh). I then multiply the area of the triangle by n (the number of sides the polygon has) and the resulting value is the area.

  1. Fencing Problem

    * Then I add the method to find the area of the entire polygon to the formula to figure out the area of the Isosceles triangle. * This formula can be cancelled down to this much simpler and easier version to use.

  2. My investigation is about a farmer who has exactly 1000 metres of fencing and ...

    and I have the adjacent (adj), this means that I use tan because tan = opp/adj. I already know that my angle is 60� so all I need is to find half of my base which means I divide the existing base by 2: Base (b)/2 = 133 ?

  1. Fencing problem.

    100 Opposite = TAN 540 � 100 Opposite = 137.6m = Height. I shall now substitute the height into the formula below: Area of a triangle = 1/2 � Base � Height Area of a triangle = 1/2 � 200 � 137.6 = 13760m2 I have now found the area of one triangle.

  2. Investigate the shapes that could be used to fence in the maximum area using ...

    This means that the angle halves, making it 22.5�. We already know that the base of the triangle is a side, therefore making it 125, but as we are concentrating on half the triangle, that number has to be halved also, giving us 62.5.

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