• 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
• Level: GCSE
• Subject: Maths
• Word count: 2470

# The coursework problem set to us is to find the shape of a gutter that generates the maximum flow of water and that can contain the most water in it. The plastic used for the gutter can be no more than 30cm in width.

Extracts from this document...

Introduction

Maths Coursework

Investigation

The coursework problem set to us is to find the shape of a gutter that generates the maximum flow of water and that can contain the most water in it. The plastic used for the gutter can be no more than 30cm in width.

##### Plan

To find the gutter that produces the biggest flow of water I must investigate different shapes of gutters to find which generates the largest volume. To do this I will use a variety of different shapes. I will firstly investigate a rectangle shaped gutter changing the length of the sides to form the largest volume. This is the simplest shape with a small amount of sides. The second I will investigate will be an isosceles triangle shaped gutter. The third shape I will investigate will be a trapezium shaped gutter and lastly I will investigate a semicircular shaped gutter. In calculating the trapezium and the triangle I will be using angles and length. This will take many trigonometry calculations. So rather than using a pen and paper methods to calculate these I will use spreadsheets typing in formulas to calculates answers which will only involve myself typing in the values of the shape.

## Spread sheets for shapes that change in angle

Side : In this box the length of the side of the shape is put.

Angle : Into this box the angle that the side forms with the perpendicular height is put.

Middle

15

So the triangle and the rectangle doubled formed the same shape but were divided in different ways.

## Trapezium Shaped gutters

My plan is to change the angle, width and length of the trapezium. This is because the angle in a triangle that creates the maximum area is 45°, I predict that this will be the angle that creates the largest area in the trapezium. If we look at the trapezium we see that it is basically a triangle divided into 2 and putting them on either side of a rectangle. Contrarily to the triangle and the rectangles area being 112.5cm2  I predict that the trapeziums area will be 112.52

sin

cos

b

As before doing every calculation would take to long so firstly I will change the angle 10° at a  time then when I have narrowed it down to less points I will change the angle 1° at a time so I can find the maximum area. When working out the area I need to consider the fact that there are 3 shapes 2 triangles and a rectangle.  I will work out the area for each shape then add it together and because the trapezium is symmetrical I only need to work out the area of 1 triangle then times it by 2.

 Side angle Height top Base area 10 0 10 0 10 100 10 10 9.848078 1.736482 10 115.5818 10 20 9.396926 3.420201 10 126.1086 10 30 8.660254 5 10 129.9038 10 40 7.660444 6.427876 10 125.8448 10 50 6.427876 7.660444 10 113.5191 10 60 5 8.660254 10 93.30127 10 70 3.420201 9.396926 10 66.34139 10 80 1.736482 9.848078 10 34.46582 10 90 6.1316 10 10 1.2314

Conclusion

° apart. The maximum area here was identical to the maximum area of the rectangle, a reason for this was discovered later.

I then went on to investigate the maximum area of the trapezium. This investigation required very much more variation, in the form of both angles and lengths. I found the sides that produced the maximum area for this shape were all equal at 10cm each; the angle at which the sides fell was 120°. I noticed a quality shared by both the triangle with the maximum area and the hexagon with the maximum area; if the shapes were doubled they would produce regular polygons. The triangle produced a square and the trapezium produced a hexagon. The doubled triangle led me to notice another connection between investigated shapes.                       The doubled triangle was identical to the doubled rectangle, they were both 15x15cm. I remembered the fact that their areas were identical, this was because they were half the same rectangle.

The circle can be described, as having 1 side, in a true circle this is the case. A shape that has an unimaginable number of sides has the appearance of a circle but it is not. A circle must have a curved side. The polygon with sides can have millions of sides but the semi circle has the largest area as it has only 1 side. So from the start you could predict the semicircle to hold the most water as it has a curve and a curve will always create a larger area than a many sided shape even if it appears to look like a curve.

Stephen Kyle

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

# Related GCSE Fencing Problem essays

1. ## Investigating different shapes of gutters.

= Area of circle = = = Area of semicircle = = = If I use 21cm for L the area would = (Throughout the investigation I will round my numbers to two = decimal places) = x = side of rectangle x x L -2x The general case for the area of a rectangle is A= x(L - 2x)

2. ## Fencing Problem

These can all be seen from the tables. Adding to that, as you increase the sides, the area is tending towards a certain number. This can be seen on the graph. To find that certain number I am going to calculate the area of a circle.

1. ## Fencing Problem

* I will use Sin once again as I have my hypotenuse and am trying to figure out my opposite i.e. base. * Sin 45 x 100 = 70.71067812 * Now that I have the base and height of my triangle I can figure out the area of the two triangles.

2. ## fencing problem part 2/8

This can only be achieved if the 'abcdCos2? ' part of the formula is a low as possible because this number is taken away from the number to be square rooted. Therefore abcdCos2 ? should equal zero. To make abcdCos2 ?

1. ## Fencing Problem

The radian formula would be: 250000/n X 1/(tan (?/n) The only part changed about this formula compared to the general formula is that the radian formula has got ? added to it. Small Angle Theory Sin O is roughly the same as theta O Tan O is roughly the same as theta.

2. ## t shape t toal

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 The Total amount of numbers inside the T shape is 11+12+13+8+18=62 The 11 at the left of the T-Shape will be called the T-Number.

1. ## t shape t toal

3 x 7 = 21 Or g x 7 = 21 Grid size (g) Number that you take away Gained by 3 x 3 21 3 x 7 4 x 4 28 4 x 7 5 x 5 35 5 x 7 6 x 6 42 6 x 7 7

2. ## t shape t toal

2 + 3 + 4 + 9 + 15 = 33 T + (T-6) + (T-11) + (T-12) + (T-13) = T-total Yes the formula does I will now simplify the formula. T + (T-6) + (T-11) + (T-12) + (T-13)

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