Area = ½ a×b×sin c
= 0.5 ×15×15×sin 100°
= 110.79cm²
C = 110°
Area = ½ a×b×sin c
= 0.5 ×15×15×sin 110°
= 105.7cm²
90º is the best angle to use, because up until this angle the area was increasing and after this angle the area begins to decrease again.
Triangle results table
Observations
Increasing the angle to a certain point, in this case 90º, will increase the area of the guttering but after this point the area begins to decrease again.
Triangle conclusion
After investigating different angles for the triangular guttering, I have found that the triangle with the sides of 15cm and an angle of 90º gives the maximum cross-sectional area for the triangle, which is 112.5cm²
Rectangle
The angles in the rectangle cannot be varied, however the only restrictions to the length of the sides are that the two opposing sides must be of equal length, and that the sum of the three lengths must equal 30cm.
To find the area of the rectangle I will be using the formula: Area = a×b
1) a = 2 b = 26
Area = a×b
= 2×26
= 52cm²
2) a = 5 b = 20
Area = a×b
= 5×20
= 100cm²
3) a = 7 b = 16
Area = a×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
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
Rectangle conclusion
After using a graph, differentiation and investigating the rectangle shaped guttering I have found that the rectangle with sides 7.5cm, 15cm, 7.5cm gives the maximum area of 112.5cm²
Square
I will only be able to investigate one square shape because there is nothing I can vary; each of the sides will have to be of equal length. Neither, can I change the angles as a square has a pre-requisite of four 90º angles.
To find the area of the square I will be using the formula:
Area = l×h
Length = 10 height = 10
Area = l×h
= 10×10
=100
Conclusion
There is only one type of square I can use; this is because nothing, sides nor angles can be varied. So the maximum area for the square shaped guttering is 100cm²
Semi-circle
I will only be able to investigate one semi-circle because there are no angles or sides I can vary.
To find the area of the semi-circle I will be using the formula: Area = ½ Π×r²
Firstly, I will have to find the radius, and to do this I will find the diameter and then divide it by two. I will then change the formula to find the circumference so I can find the diameter.
The formula to find the circumference is:
Circumference = Π×diameter
60 = Π×d
60÷Π = d
19.098593 = d
Diameter ÷ 2 = radius
19.098593÷ 2 = 9.5492965
So the area of the semi-circle is:
A =½Π×r²
=0.5×Π×9.5492965²
=143.2394462cm²
=143.2cm²
Conclusion
As nothing can be varied for the semi-circle the maximum area of the semi-circle is 143.2cm²
Half-octagon
I will be using half of a regular octagon. There is nothing to vary in a regular octagon as all the sides and angles are equal, so I will only be getting one value for the half-octagon shaped guttering.
To find the area of the cross-section of the half-octagon guttering I will be splitting it into triangles and working out the area of each triangle. I will then add them together to give me the complete area of the half octagon.
Each triangle will have equal areas, equal sides and equal angles so I can just multiply the area of one triangle by four.
Straight line = 180º
180º÷ 4 = 45º
I am going to half this triangle to give me an angle of 90º; this will make the calculation of the height easier.
I can now use trigonometry to find the height of this triangle.
To find the height:
Tan 22.5 = opp
adj
= 3.75
h
h = 3.75
tan 22.5
= 9.05cm
To find the area of this triangle:
Area = ½ base×height
Area = 0.5×7.5×9.05
=33.94cm²
Each of the four triangles, which make up the half-octagon, has an area of 33.94cm²
To find the area of the half-octagon:
Area = area of triangle×4
= 33.94×4
= 135.75cm²
Conclusion
Because the angles and the sides of the half octagon are all equal, and nothing can be varied, this means that there can only be one possible area for the half-octagon, which is 135.75cm²
Trapezium
For this shape I will be changing the lengths of the sides firstly and finding the best lengths to use. I will then be changing the size of the angles to find the maximum area for the trapezium.
To find the area of the trapezium I will be using the formula: Area = ½ (a+b)×height
I will first have to find the height by using Pythagoras’ theorem h²= a²+b². I will have to change the subject of this formula to find the height, so I will be using the formula
a²= h²- b²
1) Height Area of trapezium
a²= h²- b² Area = ½ (a+b)×height
= 5²- 1² = 0.5×(22+20)×4.89
a²= 24 = 102.69cm²
a = √24
= 4.89cm
2) Height Area of trapezium
a²= h²- b² Area = ½ (a+b)×height
= 6²- 1² = 0.5×(20+18)×5.92
a²= 35 = 112.1cm²
a = √35
= 5.92cm
3) Height Area of trapezium
a²= h²- b² Area = ½ (a+b)×height
= 7²- 1² = 0.5×(18+116)×6.39
a²= 48 = 117.81cm²
a = √48
= 6.93cm
4) Height Area of trapezium
a²= h²- b² Area = ½ (a+b)×height
= 8²- 1² = 0.5×(16+14)×7.94
a²= 63 = 119.10cm²
a = √63
= 7.94cm
5) Height Area of trapezium
a²= h²- b² Area = ½ (a+b)×height
= 9²- 1² = 0.5×(14+12)×8.94
a²= 80 = 116.22cm²
a = √80
= 8.94cm
6) Height Area of trapezium
a²= h²- b² Area = ½ (a+b)×height
=10²- 1² = 0.5×(12+10)×9.95
a²= 99 = 109.45cm²
a = √99
= 9.95cm
In this shape after I changed the lengths of the sides I found that the trapezium with sides 8cm, 14cm, 8cm, has the largest area. I will now use these measurements and change the size of the angle by changing the difference between the lengths of sides 'a' and 'b'.
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
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:
(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.)
