# Investigate different shapes of guttering for newly built houses.

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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

Middle

= 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.

Let the sides of the rectangle = x

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

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.)

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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