• 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
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  • Level: GCSE
  • Subject: Maths
  • Word count: 2073

Investigate different shapes of guttering for newly built houses.

Extracts from this document...

Introduction

Introduction

For my maths coursework I was required to investigate different shapes of guttering for newly built houses.

The purpose of guttering is to catch as much water running of the roof as possible.  The guttering for newly built houses needs to have as large an area as possible so it can hold as much rainwater as possible.  For the purpose of this investigation the material used for the production of the guttering will have a fixed width of 30cm.

I will be investigating 6 different shapes:

  • Triangle
  • Rectangle
  • Square
  • Semi-circle
  • Half-octagon
  • Trapezium

Task

Guttering

A firm has been asked to make guttering for newly built houses.

Investigate


Plan

To investigate which type of shape will hold the maximum amount of water, I will be calculating the area of the cross-section of each different type of guttering.  If possible, I will be changing the variables until I find the maximum area for each shape, the variables being length of sides and size of angles.  I will then compare the highest values of each shape and pick the one with the maximum area.


Formulas

Triangle

                                                Area = ½ a×b×sin c

Rectangle

                                                Area = a×b

Square

                                                Area = l×h

Semi-circle

                                                Area = ½×∏ ×r²

Half-octagon

                                                Area = divide the shape

                                                into triangles, work out

                                                the area of each triangle

                                                and then add them

                                                together

Trapezium

                                                Area = ½ (a+b)×h

Triangle

...read more.

Middle

×b

                                                = 7×16

                                                = 112cm²

4)                                                a = 8   b = 14

                                                Area = a×b

                                                = 8×14

                                                = 112cm²

5)                                                 a = 9  b = 12

                                                Area = a×b

                                                = 9×12

                                                = 108cm²

6)                                                a = 11  b = 8

                                                Area = a×b

                                                = 11×8

                                                = 88cm²

When I changed the sides to 9cm, 12cm, 9cm the area began to decrease.  Because rectangles 3 and 4 give the same area this indicates that the best lengths to use must be halfway between them.  I will now try to prove this:

7)                                                a = 7.5  b = 15

                                                Area = a×b

                                                = 7.5×15

                                                = 112.5cm²

This rectangle has proved my theory. The rectangle with side’s 7.5cm, 15cm, and 7.5cm gives the maximum area.

Rectangle results table

Length × height

Area cm²

2×26

52 cm²

5×20

100 cm²

7×16

112 cm²

8×14

112 cm²

9×12

108 cm²

11×8

88 cm²

7.5×15

112.5 cm²

Observations

Decreasing the length of the bottom side to a certain point, in this case 14cm, will increase the area of the guttering, but after this point the area begins to decrease again.

Differentiation

I will now use differentiation to ensure that my answer is the maximum value.

image00.png         Let the sides of the rectangle = x

image00.png        Hence, the base is 30 – 2x

                                        Area = x (30 – 2x)

                                        A = 30x¹-2x²

Differentiating with respect to x

dA

dx  = 30 - 4x¹

If the area is to be a maximum then:        dA

                                                                dx = 0

                                        30 – 4x = 0

                                        4x = 30

                                        x = 7.5

To check if it is a maximum value:

d²A

                                dx²          must be less than o

dA

                                dx  = 30 – 4x¹

d²A

                                dx²          = - 4

d²A

                                dx²            0         i.e. it’s the maximum

                                                        Maximum area = 112.5cm²

...read more.

Conclusion


4i)                                Height                Area of trapezium

a²= h²- b²          Area = ½ (a+b)×height

                                    = 8²- 2²        = 0.5×(18+14)×7.75

                                a²= 60                = 124cm²

                                a  = 60

                                    = 7.75cm

4ii)                                Height                Area of trapezium

a²= h²- b²          Area = ½ (a+b)×height

                                    = 8²- 3²        = 0.5×(20+14)×7.42

                                a²= 55                = 126.14cm²

                                a  = 55

                                    = 7.42cm

4iii)                                Height                Area of trapezium

a²= h²- b²          Area = ½ (a+b)×height

                                    = 8²- 4²        = 0.5×(22+14)×6.93

                                a²= 48                = 124.74cm²

                                a  = 48

                                    = 6.93cm

4iv)                                Height                Area of trapezium

a²= h²- b²          Area = ½ (a+b)×height

                                    = 8²- 5²        = 0.5×(24+14)×6.24

                                a²= 39                = 118.56cm²

                                a  = 39

                                    = 6.24cm

Conclusion

After working through many different trapeziums, I found that the best trapezium was 4ii as it had the largest area at 126.14cm².

Final results table

Shape

Maximum Area

Triangle

112.5cm²

Rectangle

112.5 cm²

Square

110 cm²

Semi-circle

143.20 cm²

Half an octagon

135.75 cm²

Trapezium

126.14 cm²

Conclusion

After investigation of the chosen shapes and analysis of the results obtained, I have come to the conclusion that the most suitable shape for the guttering of the newly built houses would be the semi-circle.  It had the maximum area of 143.2cm².  The square was the worst as it only had an area of 100cm².  

The rest of the shapes were positioned as follows:

  • Semi-circle                143.2 cm²
  • Half-octagon                135.75 cm²
  • Trapezium                 126.14 cm²
  • Triangle                        112.5 cm²
  • Rectangle                112.5 cm²
  • Square                        100 cm²

(These findings contrast with some older metal guttering, which were made to a rectangular shape.  This suggests that advances in the production of guttering have made it more cost-effective to use the more efficient semi-circle shape.)

...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. Investigating different shapes of gutters.

    and the angle ( C ) L-2x I found it best to split the trapezium up to find the area. y L-2x y x h h x C C L-2x I have used the letter y to label the top part until I find its formula.

  2. To investigate the isoperimetric quotient (IQ) of plane shapes using the calculation shown below.

    This is shown below: The results of my investigation are shown on the table below Shape IQ General Formula Equilateral triangle 0.604 Square 0.7

  1. A length of guttering is made from a rectangular sheet of plastic, 20cm wide. ...

    5 2.347358 4.414737964 14.69472 24.69472 12.34736 54.51035 29 10 5 2.424048 4.373098536 14.8481 24.8481 12.42405 54.33159 21.5 10 5 1.832506 4.65208784 13.66501 23.66501 11.83251 55.04586 2nd Method We already had the angle the base and the side but we had to find out side x which was the extension at

  2. Geography Investigation: Residential Areas

    There is a lot there and so I need to use a technique that is going to break it down as much as possible so I can see if there is any kind of correlation that shows terraced houses are being faded out and the production of post 1900 semi-detached Victorian houses are being developed.

  1. GCSE Maths Coursework Growing Shapes

    100 -4 The formula for number of shapes added = 4n - 4 Check Number of shapes added = 4n - 4 = 4 � 3 - 4 = 8 Number of outer vertices Pattern no. (n) No. of outer vertices 1 4 2 8 3 12 4 16 5

  2. HL type 1 portfolio on the koch snowflake

    converging increase in the value of the geometric series of the conjecture. It is evident that in the fourth stage although there is an increase in area the increase itself is lesser than the any former increase. Therefore the predictions that and are true and, Indeed Hence Verified.

  1. When the area of the base is the same as the area of the ...

    2x20x4=160cm 3cm 18cm 972cm 324cm 3x18x4=216cm 4cm 16cm 1,024cm 256cm 4x16x4=256cm 5cm 14cm 980cm 196cm 5x14x4=280cm 6cm 12cm 864cm 144cm 6x12x4=288cm Height: Base length: Volume: Base Area: Area of 4 sides: 7cm 10cm 700cm 100cm 7x10x4=280cm 8cm 8cm 512cm 64cm 8x8x4=256cm 9cm 6cm 324cm 36cm 9x6x4=216cm 10cm 4cm 160cm 16cm

  2. The best shape of guttering

    B23*E23 A23 B23 C23 D23-10 0.5*C24*C24*SIN(D24) B24*E24 A24 B24 C24 D24-10 0.5*C25*C25*SIN(D25) B25*E25 A25 B25 C25 D25-10 0.5*C26*C26*SIN(D26) B26*E26 A26 B26 C26 D26-10 0.5*C27*C27*SIN(D27) B27*E27 A27 B27 C27 D27-10 0.5*C28*C28*SIN(D28) B28*E28 A28 B28 C28 D28-10 0.5*C29*C29*SIN(D29) B29*E29 A29 B29 C29 D29-10 0.5*C30*C30*SIN(D30) B30*E30 A30 B30 C30 D30-10 0.5*C31*C31*SIN(D31)

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