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Rathdown School
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Design considerations: Only fold in half Maximize volume Which way to fold? Material costs Installation costs Maintenance costs
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Lesson Topic # of lesson periods 1Right angled triangles and Pythagoras’ theorem2 2Finding the length of a side in a right–angled triangle1 3Using trigonometry to solve practical problems2 42 5The Sine Rule2 6The Cosine Rule2 7Nets of 3 dimensional shapes2 8Volume of prisms2 9 To optimize the design of a rain gutter using one fold 2 × 35 min. (#9 = research lesson) 10 To optimize the design of a rain gutter using multiple folds 8 Area of triangle = ab sin C 1212
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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
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Posing the task Students will work in groups of 3 or 4 They will be given a piece of cardboard and asked to design a gutter by folding the cardboard in half 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
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θ θ 10·5 cm Volume = Length × Cross Sectional Area
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θ Area of triangle = ab sin C 1212 Area of triangle = (10·5)(10·5)sin θ 1212 Area of triangle = 55·125sin θ 10·5 cm
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θ Area of triangle = base × height 1212 y x (2x) 2 = 10·5 2 + 10·5 2 – 2(10·5)(10·5)cos θ 10·5 2 – x 2
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ϕ 10·5 cm 10·5sin ϕ 10·5cos ϕ
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30°60°90°120°150° 180° 0 150 300 450 600 750 900 1050 1200 1350 1500 1650 1800 Angle (degrees) Volume (cm 3 )
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Maximum volume was 1653·75 cm 3 Maximum occurs when angle is 90° 55·125 is a constant and sin θ is a variable
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(21– 2x) A = 0·5(10·5) 2 V = 55·125× 30 V = 1653·75 cm 3 4 sides 10·5 cm V= x(21– 2x) x 5·25 cm
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A = 0·5(7) 2 sin60° × 3 7 cm 60° V = 63·65× 30 V = 1910 cm 3 6 sides
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A = 0·5(6·86) 2 sin45° × 4 V = 66·55× 30 V = 1997 cm 3 5·25 cm 45° 8 sides
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4·2 cm 10 sides A = 0·5(6·8) 2 sin36° × 5 V = 67·86 × 30 V = 2036 cm 3 36°
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3·5 cm 30° 12 sides A = 0·5(6·76) 2 sin30° × 6 V = 65·54 × 30 V = 2056 cm 3
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3 cm 14 sides A = 0·5(6·76) 2 sin30° × 6 V = 69·02 × 30 V = 2071 cm 3 25·7°
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2·625 cm 22·5° 16 sides A = 0·5(6·73) 2 sin22·5° × 8 V = 69·27 × 30 V = 2078 cm 3
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4681012 14 1500 1600 1700 1800 1900 16 2000 2100 Sides of half polygon Volume (cm 3 )
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