Rathdown School
Design considerations:
• Only fold in half
• Maximize volume
Which way to fold?
• Material costs
• Installation costs
• Maintenance costs
Lesson
Topic
# of lesson
periods
1
Right angled triangles and Pythagoras’ theorem
2
2
Finding the length of a side in a right–angled triangle
1
3
Using trigonometry to solve practical problems
1
Area of triangle = ab sin C
2
2
5
The Sine Rule
2
6
The Cosine Rule
2
7
Nets of 3 dimensional shapes
2
8
Volume of prisms
2
9
To optimize the design of a rain gutter using one fold
10
To optimize the design of a rain gutter using multiple
folds
4
2
2 × 35 min.
(#9 = research lesson)
8
Introduction

Use project management skills to develop best
options


Use mathematics to solve real world problems
Use recent topics such as trigonometry, and
volume of 3D shapes and nets, to investigate the
design of a rain gutter for a building which will
give the maximum flow
Posing the task


Students will work in groups of 3 or 4

Each group will be given an A3 size placemat,
graph paper and A4 size cardboard sheets

The task is to optimize the design of the gutter
to achieve the maximum flow of water.

Each group must prepare a graph to illustrate
their results

Each group will present their findings
They will be given a piece of cardboard and
asked to design a gutter by folding the
cardboard in half
Volume = Length × Cross Sectional Area
θ
10·5 cm
10·5 cm
θ
1
Area of triangle = ab sin C
2
1
Area of triangle = (10·5)(10·5)sin θ
2
Area of triangle = 55·125sin θ
10·5 cm
θ
x
1
Area of triangle = base × height
2
θ
10·52 – xy2
10·5 cm
(2x)2 = 10·52 + 10·52 – 2(10·5)(10·5)cos θ
10·5 cm
10·5sinϕ
ϕ
10·5cosϕ
Volume (cm3)
1800
1650
1500
1350
1200
1050
900
750
600
450
300
150
0
30°
60°
90°
120°
Angle (degrees)
150°
180°



Maximum volume was 1653·75 cm3
Maximum occurs when angle is 90°
55·125 is a constant and sinθ is a variable
4 sides
A = 0·5(10·5)2
V = 55·125× 30
V = 1653·75 cm3
V= x(21– 2x)
(21–
10·52x)
cm
5·25
x cm
6 sides
A = 0·5(7)2sin60° × 3
V = 63·65× 30
V = 1910 cm3
60°
7 cm
8 sides
A = 0·5(6·86)2sin45° × 4
V = 66·55× 30
V = 1997 cm3
45°
5·25 cm
10 sides
A = 0·5(6·8)2sin36° × 5
V = 67·86 × 30
V = 2036 cm3
36°
4·2 cm
12 sides
A = 0·5(6·76)2sin30° × 6
V = 65·54 × 30
V = 2056 cm3
30°
3·5 cm
14 sides
A = 0·5(6·76)2sin30° × 6
V = 69·02 × 30
V = 2071 cm3
25·7°
3 cm
16 sides
A = 0·5(6·73)2sin22·5° × 8
V = 69·27 × 30
V = 2078 cm3
22·5°
2·625 cm
2100
Volume (cm3)
2000
1900
1800
1700
1600
1500
4
6
8
10
12
Sides of half polygon
14
16
n