• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15
  16. 16
  17. 17
  18. 18
  • Level: GCSE
  • Subject: Maths
  • Word count: 1962

The fencing problem.

Extracts from this document...


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.




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

1000÷4 = 250



          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:




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

...read more.


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





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.




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


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

...read more.


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.


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. Geography Investigation: Residential Areas

    my hypothesis which when it comes to wrapping up my coursework as a whole I know exactly what I found from this hypothesis. Table 5 Type of housing Street Terrace Detached Semi-Detached Flat Maisonette Sarum Hill 10 0 0 0 0 Bounty Road 4 2 2 1 1 Penrith Road

  2. The Fencing Problem

    Area = 8 (.5 ( (1000 ( 8)) ( ((500 ( 8) ( (tan (180 ( 8)) = 75,444.173m� Diagram Regular 9 Sided Polygon: A nine-sided regular polygon is called a regular nonagon Area = 9 (.5 ( (1000 ( 9)) ( ((500 ( 9) ( (tan (180 ( 9))

  1. The Fencing Problem

    x h]} x 100 Regular Polygons - Chiliagon (1000-sided-shape) As you can see, I have not incorporated a diagram of a Hectogon because of its close resemblance to a circle (at this scale factor). However, the calculations of the base and height of the triangles are shown, and of course the calculations.

  2. t shape t toal

    p = number of translations x = T-number g = grid size t = T-total Combinations of translations (diagonal) For diagonal translation across a grid a combination of horizontal and vertical translations are used, therefore I predict that if I combine my 2 found equations for horizontal and vertical movement's, I can generate a general formula for diagonal translations.

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

    This has happened so far, but to support my prediction, I will look at the regular hexagon, a 6-sided shape. To find the area of my hexagon, I will use the same method as I did for my pentagon. First I will find out the length of each side: 1000/n

  2. Fencing Problem

    * Height tan36 = 100 * Height = 100 * Height = 137.638192 * Now that I have the height of my right-angled triangle I can figure out the area of the triangle. I will follow the formula; 1/2 base x height.

  1. Investigating different shapes to see which gives the biggest perimeter

    If I substitute the values from the pentagon into the formula I get, h = 100/tan 36�. So height is equal to 137.64m. I can now work out the area of the isosceles using the formula 1/2 base x height: Area = 1/2 base x height = (1/2 x 200)

  2. Koch Snowflake Math Portfolio

    = unit� A1 = unit� A2 = unit� A3 = unit� A4 = unit2 Hence, as can be seen by the aforementioned values, as n gets larger, there is a geometric increase in the perimeter and area.

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