Add Maths Project

Name :

Class : 5 Kreatif

School : SMK Bandar Baru Sungai Long

Teacher : Puan Hayati Aini bt. Ahmad

Title : ADDITIONAL MATHEMATICS PROJECT WORK 1/ 2010

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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

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Acknowledgement

First of all, I would like to say thank my teacher Puan Hayati who had gave us some guides on how to complete this paperwork. Next to my fellow teammates who had held on each other firmly until we had completed this paperwork. Finally to my parents who had always 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 life. Completing this paperwork in a team had gave me a chance to know each other and broaden my view on Mathematics. I now know the correct way to apply mathematic knowledge in my daily life to solve a lot of problems.

Besides that, I also have learnt to accept other ideas from different people to make out all the possible results. Then, I know which is the best after those comparison made. This way, I can save a part of time and materials needed to construct any of the objects. For an example, a bridge construction. We use the same method to find out all the things we needed to complete it in the most beautiful and cost saving way.

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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 correct language to express our mathematical ideas properly.

To develop positive attitude towards mathematics

to improve our way of thinking

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INTRODUCTIONAREA

Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a region bounded by a closed 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. Area is an important invariant in the differential geometry of surfaces.

The formulae related for this subject are: ½ (base X height), ½ radius2

X θ (in radian),

VOLUME

Volume is how much three-dimensional space a substance (solid, liquid, gas, or plasma) or shape occupies or contains, often quantified numerically using the SI derived unit, the cubic 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.

Three dimensional mathematical shapes are also assigned volumes. Volumes of some 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 formula exists for the shape's boundary. One-dimensional figures (such as lines) and two-dimensional shapes (such as 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 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.

In differential geometry, volume is expressed by means of the volume form, and is an important global Riemannian invariant. In thermodynamics, volume is a fundamental parameter, and is a conjugate variable 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.

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Exploration

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Procedure and Findings

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 + c

Y = a + 4.5

Y = + 4.5……………… (1)

Substitute (2, 4) into (1)

4 = + 4.5

4a = -0.5

A= -0.125

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Therefore, Y = + 4.5

Function 2

Maximum point (0, 0.5) and pass through (2, 0)

Y = + c

Let b= 0 and c = 0.5

Y = + 0.5

Y= + 0.5............... (2)

Substitute (2, 0) into (2)

0 = + 0.5

4a = -0.5

A = -0.125

Therefore, Y= + 0.5

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Function 3

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

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Therefore, Y = + 4.5

Function 4

A = Area of sector – Area of triangle

= ½ r2 θ - ½ (base x height)

= ½ r2 θ – ½ (4 x length)

Therefore, A = ½ r2 θ – (2 x length)

a) Given 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

A = ½ r2 θ – (2 x length)…1

Substitute 2 into 1A1 = 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

= 4.431 – 3.75 therefore, θ = 2806’

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= 0.68 = (28.1/180 x ) rad.

Shaded Area = A – A1 = 0.491 rad.

= 4 – 0.68

= 3.32 m2

Further Explorationa) I) Structure 1

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

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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

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

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Structure 3

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

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Structure 4

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

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Structure choosing

As the president of the Arts Club, I will choose structure 1. This is because the cost needed is worthwhile. We can construct a rigid structure by maximizing our structure's performance by using only the smallest amount 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 higher pressure. Therefore, it lasts longer compared to other structures and needed less frequent maintenance.

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Furthermore, based on survey conducted, majority people felt uneasy when they saw sharp edges on most objects. Since from the beginning of time, a lot of people considered circular objects are the most perfect objects. In addition, the parabolic shape matches the standard of society nowadays.

b) I)

K(m)Area to be painted (m2)

(correct to 4 decimal places)0 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

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ii) By using K = 0 m, the structure may be like this:

When the value of K gradually increases, the area will gradually decreases.

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:

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C onclusion:

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 architecture, drawer and

designer. Through applying all these formulas, I can save 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.

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ReflectionBy 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 architecture, drawer and designer. Through applying all these formulas, I can save 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.

From this paperwork, I have learnt the correct way to display the workings correctly and arrange the calculations and topics 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. Only through teamwork, we can finish any task given by all means. 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.

After completing this paper, I have learnt how to co-operate with my teammates to complete this paperwork. 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 architecture, drawer and designer. Through applying all these formulas, I can save materials used in all sorts of production. At the same time, the cost will be reduced.

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Not only that, I know the history of these formulas and also able to determine the correct formula to be used in certain 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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