Transcript
N a me :
Class : 5 Kr Kreatif
School
: SMK Bandar Baru Sungai Lo Long
Teac Teach her : Pu Puan Haya Hayati ti Aini Aini bt. Ahm Ahmad
Title : ADDITIONAL MATHEMATICS PROJECT WORK 1/ 2010
Index
No.
Contents
Page
1.
Acknowledgement
3
2.
Objectives
4
3.
Introduction
5
4.
Exploration
6
5.
Procedure and findings
7 – 10
6.
Further Exploration
11 – 17
7.
Conclusion
18
8.
Reflection
19
Acknowledgement
First First of all, I would would like to say thank thank my teacher teacher Puan Hayati Hayati who had had gave us some some guides on how how to complete complete this paperwo paperwork. rk. Next to my fell fellow ow team teamma mate tes s who who had had held held on each each othe otherr firm firmly ly unti untill we had had comp comple lete ted d this this pape paperw rwor ork. k. Fina Finall lly y to my paren parents ts who who had had alwa always ys encourage me to put in some effort to complete such big task and also allowing me to visit each other's house to discuss about this paperwork.
While I was conducting this paperwork, I have gained consciousness in many other things in my my life. Completing this paperwork paperwork in a team had had gav gave me a chan chance ce to know know each each other ther and and broa roaden den my view on Math Mathem emat atic ics. s. I now now know know the the corr correc ectt way way to appl apply y mathe athema mati tic c knowledge in my daily life to solve a lot of problems.
Beside Besides s that, that, I also also have have learn learntt to accept accept other other ideas ideas from from differ different ent people people to make out out all the possibl possible e results. results. Then, Then, I know which is the best after after those compar comparison ison made. made. This way, way, I can save a part of time time and materials needed to construct any of the objects. For an example, a bridge construction. construction. We use the same same method to find out all the things we needed to complete it in the most beautiful and cost saving way.
Objectives
The aims of carrying out this project are : To learn the way to apply the formulas of mathematics in our daily life accurately. •
•
To widen our prospective view on mathematics.
•
To improve our mastering skills.
To use the the correct correct language language to express express our mathematical mathematical ideas ideas properly. •
•
•
To develop positive attitude towards mathematics to improve our way of thinking
INTRODUCTION
AREA
is a quantity expressing expressing the two-dimensional two-dimensional size of a defined part of a surface surface,, typically a region bounded by a closed curve curve.. The surface area of a 3-dimensional solid is the total area of the exposed surface, such as the sum of the areas of the exposed sides of a polyhedron polyhedron.. Area is an important invariant in the differential geometry of surfaces. surfaces. Area
(base X heigh height), t), The formulae related for this subject are: ½ (base
½ radi radius us
2
X θ (in radian),
VOLUME
is how much three-dimensio three-dimensional nal space a substance substance (solid, liquid, liquid, gas, or plasma) plasma) or shape occupies or contains, often quantified numerically using the SI derived unit, unit , the cubic metre. metre. The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces. Volume
Three Three dimens dimension ional al mathem mathematic atical al shapes shapes are also also assign assigned ed volume volumes. s. Volumes olumes of some some simple simple shapes, such as regular, straight-edged, and circular shapes can be easily calculated using arithmetic formulas.. The volumes of more complicated shapes can be calculated by integral calculus if a formulas formul formulaa exists exists for the shape' shape'ss bounda boundary ry.. One-di One-dimen mensio sional nal figure figuress (such (such as lines lines)) and and two twodimensional shapes (such as squares squares)) are assigned zero volume in the three-dimensional space. The volume of a solid (whether regularly or irregularly shaped) can be determined by fluid displacement.. Displacement of liquid can also be used to determine the volume of a gas. The displacement combined volume of two substances is usually greater than the volume of one of the substances. However, sometimes one substance dissolves in the other and the combined volume is not additive additive.. In differential geometry, geometry, volume is expressed by means of the volume form, form, and is an important global Riemannian invariant invariant.. In thermodynamics thermodynamics,, volume is a fundamental parameter, and is a conjugate variable to pressure to pressure.. The formula related to this subject are: Area X (length/ base/ height) History of Area
Heron (or Hero) of Alexandria (first century B.C.) is best known for having discovered the formula for calculating the area of a triangle. In Heron's formula, as it is called, the sides of a triangle are labeled "a," "b," and "c"; V is equal to half the perimeter. Arab scholars who studied the mathematics of the ancient Greeks claimed that this formula was known earlier to Greek mathematician Archimedes (ca. 287-212 B.C). The oldest written record of the formula, however, exists in Heron's Metrica. History of Volume The formulae for volume were discovered independently by many people over centuries of time. Not one individual person 'discovered' how to calculate the volume of a sphere, cube, rhomboid, etc. The formulae for calculating the volume of an object were arrived at many many years ago by many people in many civilisations. No single person can be regarded as the first to describe how to calculate volume.
Exploration
Procedure and Findings 6
Solution: (a)
Function 1
Maximum point (0, 4.5) and pass through point (2,4)
Y= a
+c
Let b = 0 and c = 4.5 into Y= a Y=a Y=
+c
+ 4 .5 + 4.5……………… (1)
Substitute (2, 4) into (1) 4=
+ 4.5
4a = -0.5 A= -0.125
Therefore, Y =
+ 4.5
Function 2
7
Maximum point (0, 0.5) and pass through (2, 0)
Y=
+c
Let b= 0 and c = 0.5
Y=
Y=
+ 0.5 + 0.5............... (2)
Substitute (2, 0) into (2) 0=
+ 0.5
4a = -0.5 A = -0.125
Therefore, Y=
+ 0.5
Function 3
8
Maximum point (2, 4.5) and pass through (0, 4)
Y=
+c
Let b = 2 and c = 4.5 Y=
+ 4.5 …………… (3)
Substitute (0, 4) into (3) 4=
+ 4.5
4a = -0.5 A= -0.125
Therefore, Y =
+ 4.5
Function 4 A = Area of sector – Area of triangle = ½ r2 θ - ½ (base x height) 9
= ½ r2 θ – ½ (4 x length) Therefore, Therefore, A = ½ r 2 θ – (2 x length) a) Given Given base base = 4m Height = 1 m – 0.5 m = 0.5 m r2 = (r-h)2 + (½ )2 = (r- 0.5)2 – (4/2)2 = r2 – r + 0.25 + 4 r2 = 4.25……………………………2 4.25……………………………2 A = ½ r2 θ – (2 x length)…1 l ength)…1 Substitute 2 into 1 A1 = ½ (4.25)2 θ – ½ (r –h)(2) = 9.03125 θ – 2( 4.25 – 0.5) = 9.03125 θ – 3.75
sin θ = 2/4.25
= 9.03125(0.491 rad.) – 3.75
= 0.471 therefore, θ = 2806’
= 4.431 – 3.75 = 0.68
= (28.1/180 x ) rad.
Shaded Area = A – A1
= 0.491 rad.
= 4 – 0.68 = 3.32 m2
Further Exploration a) I) Structure 1
10
Using
3.142
Angle of triangle = 5308’ In radian = 5308’/1800 x = 0.927 rad. A1 / / Area of Sector = ½ r2 θ = ½ (4.5)2(0.927) = 9.386 m2 A2 / / Area of 2 triangle = 2 x ½ x 4 x 2 = 8 m2 Area of structure = (1 x 4) – (A1 – A2) = 4- 1.386 = 2.614 m2 Thickness = 0.4m Total volume
= 2.614 x 0.4 = 1.0456 m2
Cost = 1.0456 x RM 840.00 = RM 878.30
Structure 2
11
Area of trapezium = (L1+L2 / 2) (height) = (1.5/2)(2) = 1.5 m2 Total Area = 3m2 Thickness = 0.4 m Total volume = 3 x 0.4 = 1.2 m3 Cost = 1.2 x RM 840.00 = RM 1008.00
Structure 3 12
Total Area = 2 trapezium + rectangle = 2 (1.5/2 x 1.5) + (1 x 0.5) = 2.25 + 0.5 = 2.75 m2 Thickness = 0.4 m Total volume = 0.4 x 2.75 = 1.1 m3 Cost = 1.1 x RM 840.00 = RM 924.00
Structure 4 13
Total Area = 2 trapezium + rectangle = 2 (1.5/2 x 1) + (2 x 0.5) = 1.5 + 1 = 2.5 m2 Thickness = 0.4 m Total volume = 2.5 x 0.4 = 1 m3 Cost = 1 x RM 840.00 = RM 840.00
14
Structure choosing
As the president of of the Arts Club, I will choose structure structure 1. This is because the cost needed is worthwhile. worthwhile. We can construct construct a rigid structure structure by maximizing maximizing our structure's structure's performance by using using only the smallest amount of of materials needed.
Secondly, this structure is more stable compared to the others as the pressure exerted on it is distributed evenly to all parts of the structure because of its shape. In this way, it can withstand withstand higher pressure. Therefore, it lasts longer compared to other structures and needed less frequent maintenance. Furthermore, based on survey conducted, majority people felt uneasy when they saw sharp edges on most objects. objects. Since from the beginning of time, a lot of people considered circular objects are the most perfect perfect objects. In addition, the the parabolic shape matches the standard of society nowadays.
b) I) 15
0
Area to be painted (m2) (correct to 4 decimal places) 3.000
0.25
2.9375
0.50
2.8750
0.75
2.8125
1.00
2.7500
1.25
2.6875
1.50
2.6250
1.75
2.5625
2
2.5000
(m) K (m)
ii) By using
K =
0 m, the structure may be like this:
When the value of K gradually increases, the area will gradually decreases. 16
By using
K =
0.25 m, the structure may be like this:
c) When K approaches more to 4, the shape of concrete structure will be
something like :
Conclusion: 17
By doing this paperwork, I now learn that the knowledge of integration and area and even volume calculations can be used to solve daily problems p roblems such as in architecture, architecture, drawer and and designer. Through applying all all these formulas, I can save materials materials used in all sorts of production. At the same time, the cost will be reduced. Not only that, I know the history of these formulas and also able to determine the correct formula to be used in certain circumstances.
Reflection 18
By doing this paperwork, I now learn that the knowledge of integration and area and even volume calculations can be used to solve daily problems such as in architect architecture, ure, drawe drawerr and designer. designer. Through Through applyin applying g all these formulas, formulas, I can save materials used used in all sorts of production. At the same time, time, the cost will will be reduced. Not only that, I know the history of these formulas and also able to determine the correct formula to be used in certain circumstances.
From this paperwork, I have learnt the correct way to display the workings correctly and and arrange the calculations calculations and topics systematically. systematically. Through this, I have successfully completed this project along with my friends.
At the same time, I also have learnt that co-operation is very important for a team team.. Only Only thro throug ugh h team teamwo work rk,, we can fini finish sh any any task task given given by all mean means. s. Through teamwork, I also have learnt the correct way to listen to other members and to judge all the ideas we got after brainstorming through comparison based on calculations.
Afterr comp Afte comple leti ting ng this this pape paper, r, I have have lear learnt nt how how to co-o co-ope pera rate te with with my teammates to complete complete this paperwork. paperwork. By doing this paperwork, paperwork, I now learn learn that the knowledge of integration and area and even volume calculations can be used to solve daily daily problems problems such as in architec architecture, ture, drawer drawer and designer. designer. Through Through applying applying all these formulas, formulas, I can save materials materials used in all sorts of production production.. At the same time, the cost will be reduced.
Not only that, I know the history of these formulas and also able to determine the correct correct formula formula to be used in certain certain circumstances. circumstances. Now, I can finish any work associated with mathematics at a very fast and accurate rate.
I LOVE YOU ADDITIONAL MATHEMATIC
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