• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1634

Guttering Investigation

Extracts from this document...


Guttering Coursework

Guttering Coursework

In my investigation I will be looking at guttering, where I will try and find the best type; that is the one which holds the greatest volume of water.

I was walking around my hometown of Antrim where I noticed how all the guttering was made into a semi circular open top shape. I wondered if this is because this is the shape which holds most water or is it because it is the cheapest to manufacture.

My hypothesis is that semi circular guttering will hold the greatest volume of water.

I will now investigate to see whether my hypothesis is true of whether there is another which can hold a larger volume of water. I will use a sheet of plastic and fold it into different shapes. I will find out the shapes’ cross sectional area rather than the volume as it is easier and the length will be the same so it is only the cross sectional area that varies.

I will fold a sheet of paper into different shapes to help me visualise the different shapes possible.

I will be investigating guttering which will have a perimeter of “B” cm.

...read more.


Each side will be length “y” and the base will be B-2y

Area = x(B-2x)

     A = Bx-2x2

Below is a sketch of the graph A= By-2y2

I realise from my maths knowledge that this is a quadratic expression. I also know that because it is a-x2.

   Its shape will be image17.png . The graph will cross the axis where A=0.

                                      ∴Put A=0                



x=0, B-2x=0



Therefore the graph will cross the x axis at the values of 0 and B/2. Below is the sketch of my graph.


Due to the graph’s symmetry, the maximum area occurs half way between the two points 0 and B/2. This is the point B/4.


Maximum area = B/2 x B/4

                         = (B2)/8 = 0.125B2

I can now prove this by using differentiation.


dA/dx = B- 4x

At maximum point dA/dx = 0




 To show this is maximum I will find d2A/dx2

d2A/dx2 = -4

Since d2A/dx2 is negative then the area is the maximum value when x =1/4. I have proved what I have found before. It is the same as triangular cross section but less that semi circular.

...read more.



If the perimeter is B and there are n sides then the length of each side is B/n  

I will divide the half regular polygon into n congruent triangles.


Area of 1 triangle = ½ bh

                             = ½ x B/n x h

                             = Bh/2n

Area of half regular polygon = n x bh/2n

                                                    = Bh/2

                                                    = ½ Bh

I would like h to be in terms of n (number of sides).

Looking at one triangle:


tan θ = B/h

h tan θ= B/2n

h=B/2ntan θ


θ = 180/2n


B/2ntan (90/n)

Area of ½ regular polygon= ½ Bh

                                          = ½ B x B/(2n tan) (90/n)

                                          = B2/ 4n tan (90/n)

I will double check my formula by using n=3 to find the area of a half regular hexagon which I already know from the cross sectional area of a trapezium.

A= B2/ (4(3) tan (90/3))

A= B2/(12tan 30)

A= 0.1443B2

This is the same area I got for the half regular hexagon so I know my formula works.

   I will now use Microsoft Excel to record my results will see how the area increases as the number of sides increase for half regular polygons.

As I thought the area does get close to that of the semi circle as the sides increase but never exceeds it.

CONCLUSION: My hypothesis was correct, and the semi-circular guttering is the type which can hold the most water.

...read more.

This student written piece of work is one of many that can be found in our GCSE Miscellaneous 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 Miscellaneous essays

  1. GCSE Maths Shape Coursework

    All the different shapes in the T=20 table work with this formula. Therefore I will test the formula with all the other tables below, so I can be sure that it does work with all numbers of triangles. If it works with all four of the numbers of triangles that

  2. Equable shapes

    LW = 2L + 2W LW-2L = 2W L(W-2) = 2W L = This is the formula to find L (the length) given W (the width) Test the formula Suppose the width of a rectangle is 42 L = L = L = 2.1 Area = 42 x 2.1 = 88.2 Perimeter = 2(42 + 2.1)

  1. Graphing Quadratic Functions

    9 -4 26 -11 -1 7 6 -7 23 -14 0 6 5 -8 22 -15 1 7 6 -7 23 -14 2 10 9 -4 26 -11 3 15 14 1 31 -6 With these results the below graph was given: Graph of y = (x-h)2 We can clearly

  2. Box Coursework

    5,5,8,6,0,1,3,0, 8 180 0 1 190 0 200 3 1 Modal height for boys is 150?h<160 Girls height Stem Leaf Frequency 130 2,0 2 140 9,5,5,9,7,7,8,8 7 150 5,5,3,0,1,3,9,7,6,8,9,8,3,0,9,2,9,2 18 160 2,3,2,5,2,2,5,5,2,5,2,1,7,1,2,0,2,0,2,3 21 170 0,2 2 180 0,0 2 The modal height for girls is 160?h<170 I have now worked

  1. Tubes. I was given a piece of card measuring 24 cm by 32cm, and ...

    As you can see from the chart above, the 8 * 8 * 8 dimensions give me the largest volume. This is not an isosceles triangle, but it is an equilateral triangle. To justify that the 8 * 8 * 8 dimensions have the largest volume this I have made the chart below.

  2. T-Total Maths coursework

    time I move the T one place to the right, this is because each digit increases by 1, making it 5 each time. From this I predict that for T-Number 27 the T-Total will be 65, I will now prove my theory.

  1. Layers investigation

    Therefore there are 9 different arrangements of cubes. To further test this theory, I used a 3 by 4 grid. In a 3 by 4 grid there are 12 squares. Of this 12 squares are possible empty squares and 11 are possible filled in squares.

  2. Maths Investigative task on perimeter of a rectangle and volume of shapes

    least when the length and the width are almost the same or the same. We can prove this as in the above example; the least perimeter is when the dimensions are 32 by 31.25, which is close to being a square of 32 by 32.

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