# A Builder has to make drains from a sheet of plastic measuring 2m x 50cms. He finds the semi-circle produces the best drain. Prove This.

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Introduction

PROBLEM:

A Builder has to make drains from a sheet of plastic measuring 2m x 50cms. He finds the semi-circle produces the best drain. Prove This.

INTRODUCTION:

I will hope to prove in this coursework that a semi-circle will make the best drain for a builder. I will show this by calculating the area and volume of the semi-circle and other shapes. Hopefully I will find that the semi-circle will give the best area which will mean that it gives a bigger volume and will therefore hold more.

1.

2.

PROBLEM: Which way would be the best to bend the plastic?

SOLUTION: I would probably choose diagram 2 because it is the longest shape when bent and we would not need as many drains. The reason for me not choosing diagram 1 is because I would need to many pieces of drain to go around the whole house and it would hang to low and look unsightly.

WHICH SHAPES WILL I EXPLORE?

- Semi-Circle

This Shape is in 2D as if you were looking at it from straight on.

This is how it looks in 3D and how it will look on a building.

- Rectangle

This is the shape in 2D as if you were looking at it straight on.

Middle

25cm

54.265cm

175

25cm

27.136cm

*BOLD indicates turning point.

## Trapezium

By looking at the trapezium I can see that it is made up of two triangles and a rectangle.

We found B as 25cm because when we looked at the findings from the rectangle 25cm gave the best area for our shape.

To find the area for the trapezium I am not going to change the lengths of the side instead I will change the angle θ. The values for the angles can range from 1 to 89. For calculating each of my angle sizes I will go up in steps of 5.

To find the area of the trapezium I will use the formula:

Area of trapezium = ½ the sum of the parallel sides by the

Perpendicular height.

## Working Out For Trapezium

To find X we use the equation:

Sin45 = X/12.5

X = 12.5 Sin45

X = 8.8388cms

To find Y we use this equation:

Y = 2X+25

= 2(8.8388)+25

= 42.6776

To find H we use the equation:

Cos45 = H/12.5

H = 12.5 Cos45

H = 8.8388cms

Conclusion

Area of test-tube = Area of Semi-circle + area of rectangle

= 366.897 + 15.287

= 382.184

Height of Rectangle (side) cm | Circum. Of semicircle Cm | Radius of the semicircle Cm | Area of semicircle Cm2 | ## DiameterCm | Area Cm2 | Total Area Cm2 |

1 | 15.29 | 7.645 | 367.04 | 30.58 | 30.58 | 397.62 |

2 | 14.65 | 7.325 | 336.96 | 58.6 | 117.2 | 454.16 |

4 | 13.38 | 6.69 | 281.07 | 107.04 | 428.16 | 709.23 |

6 | 12.10 | 6.05 | 229.86 | 145.2 | 871.2 | 1101.06 |

8 | 10.83 | 5.415 | 184.14 | 173.28 | 1386.24 | 1570.38 |

10 | 9.55 | 4.775 | 143.19 | 191 | 1910 | 2053.19 |

12 | 8.28 | 4.14 | 107.64 | 198.72 | 2384.64 | 2492.28 |

14 | 7.01 | 3.505 | 77.15 | 196.28 | 2747.92 | 2825.07 |

16 | 5.73 | 2.865 | 51.55 | 183.36 | 2933.76 | 2985.31 |

18 | 4.46 | 2.23 | 31.23 | 160.56 | 2890.08 | 2921.31 |

20 | 3.18 | 1.59 | 15.88 | 127.2 | 2544 | 2559.88 |

22 | 1.91 | 0.955 | 5.73 | 84.04 | 1848.88 | 1854.61 |

24 | 0.64 | 0.32 | 0.64 | 30.72 | 737.28 | 737.92 |

## Conclusion

Now that I have investigated all five shapes I can now display my findings in order in the table below. The shapes are listed from the one with the largest area/volume to the one with the smallest. The one with the largest area/volume is the best shape for a drain.

## Stage | Area |

1 | 392.5 |

5 | 382.184 |

2 | 312 |

4 | 299.09 |

3 | 289.389 |

I can now see that the best drain is the semi-circle as it has the greatest area and volume.

## Extension

If I had more time I would have been able to investigate after shapes and went on to investigate about different sizes of plastic and how they effect the area and volume of the drain.

Evaluation

I found this project very rewarding and I thoroughly enjoyed it. I was pleased when at the end I found that the semi-circle was the shape that gave the greatest area and volume.

Although I enjoyed most of the project I found some parts quite difficult, for example the area of the trapezium had a lot of complex working out to do.

There are parts of the project that I could have made more accurate by rounding to a higher decimal place or using the Π button on the calculator.

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

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