ACKNOWLEDGEMENT

This project could not have been completed with such amazing accuracy of answers and information without the help of a dedicated teacher, cooperative and loyal friends and of course loving parents.

First of all I would like to thank my additional math teacher, Puan Goh who has helped me in making this project a success. Without her wisdom and her guidance this project would be a complete failure. Her dedication in seeing her student achieving success is unbelievable. Puan Goh has taken time out of her busy schedule to guide me in completing this project with the least amount of errors. Her patience in explaining every question with such detail shows her utter most passion to educate the generation that will one day become the leaders of tomorrow.

Secondly, all the riches of this world are nothing compared to the people who I so proudly call as my friends. The cooperation between my group members Choo Siew Yen, Khong Shook Wai, Lau Hui Pin, Law Sui Yng, Ng Suet Ying and Vanessa Ong was something that could be compared to the likeness of cooperation that was present between historical figures such as the Wright brothers, the inventors of the world's first successful airplane. I would like to express my gratitude as this project could not have been completed without their help.

Thirdly, I would like to thank my parents for being so supportive of me in carrying out this project. Their help of providing the materials required in completing this Additional Mathematics program is deeply appreciated. They also provided me transport to the binding shop to bind my project. Furthermore, they allowed me to use the computer to complete this project.

In a nutshell I would like to thank all who've helped me in making this project a success.

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OBJECTIVE

To learn how to apply 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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INTRODUCTION

The arch is an incredible architectural discovery, dating back to ancient times but still in wide use today, as, up until the 19th century, it was the only known method for roofing a building without the use of beams. It comes in many shapes-semicircular (Roman), segmental (less than half a circle), or pointed (Gothic). The arch developed from the post and lintel or possibly the corbel, which is similar in shape and principle to the arch. Efforts to build corbelled roofs with smaller units and less weight could have eventually led to the discovery of the arch.

Arches are made of wedge-shaped blocks, called voussoirs, set with their narrow side toward the opening so that they lock together. The topmost voussoir is called the keystone. Once locked into place, the arch cannot collapse under any amount of weight and the only danger is of the voussoirs crumbling under the pressure. To keep this from happening, most arches require support from other arches, walls, or buttresses.

The arch has been found in many different cultures, as early as Mesopotamia. The Egyptians used it in tombs and vaults but never for monumental architecture, such as temples. They apparently thought it unsuited to this purpose. The Greeks also used the arch solely for practical constructions, but many of the principles they developed were later exploited by the Romans.

Overall, it was not until the time of the Etruscans that the arch was used in any kind of monument. The best example of this is the Porta Augusta, where the arch is combined with Greek architectural ideas. The Romans borrowed this combination and used it over and over again, but its invention belongs solely to the Etruscans.

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The Romans took many great strides in the development of the arch. While they borrowed many techniques from earlier races, the Romans invented the idea of setting an arch on top of two tall pedestals to span a walkway such as a public highway. The outer wall of the Colosseum appears composed almost completely of arches. Here we see examples of the barrel vault and the more complicated groined vault, both developed by the Romans from the basic arch. The Romans also used arches for common purposes, such as in the building of bridges and aqueducts.

Arches continued to be used in Medieval times, especially in cathedrals, (above, second from right), where they helped support the great weight of the stone ceilings, especially when walls were weakened by the presence of many windows. It is here that buttresses were often used to support the arches. Sometimes called "flying buttresses" because of their height, buttresses are a simple construction of a stone pillar with a "bridge" at the top that joins onto the arch or walls of the building, giving extra support to the construction. Arches were also often found in long rows in cathedrals to help support each other. It is about this time that the pointed arch began to be developed, as an alternative to the traditional rounded arch. This pointed or Gothic arch became very prevalent in the architecture of the time.

Unique to architecture was the Islamic arch, found about the same time in the Middle East. Many advances were made in the arch by this culture as well. While the pointed arch was used here, the Muslims also developed a horseshoe-shaped arch and "stacked arches," an arch built above an arch. It is believed that the "stacked arch" idea developed by accident, when a builder was forced to use columns too short for his purpose and so stacked them on top of each other, with arches holding the stacked columns together. Islamic arches can be found in mosques throughout the Middle East.

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Moral Values

From this project, we learn how to work together as a single team. We obtained knowledge, joy, and the importance of working together. It is through this project that we have learnt how to work together to achieve a common goal while tolerating and giving in to each other.

We learnt the importance of being hardworking and why it is crucial that we maintain determination while achieving our goals. Therefore, through this task we have all become a better student and shall carry forth this value to imply in our lives in the near future.

Among the values attained is discipline and punctuality. When working together with other individuals, it is important that we stick to our promises and agreements such as meeting times and dates. We also learned to not let down out teachers and parents who placed high hopes in us.

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CONJECTURES

Structure 4 costs the minimum to construct.

Structure 1 is the strongest and most stable structure.

When k approaches the value of 4, the shape of the concrete structure will be a rectangle.

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PROCEDURE AND FINDINGS

The diagram below shows the gate of an art gallery. A concrete structure is built at the upper part of the gate and the words 'ART GALLERY' is written on it. The top of the concrete structure is flat whereas the bottom is parabolic in shape. The concrete structure is supported by two vertical pillars at both ends.

The distance between the two pillars is 4 metres and the height of the pillar is 5 metres. The height of the concrete structure is 1 metre. The shortest distance from point A of the concrete structure to point B, that is the highest point on the parabolic shape, is 0.5 metres.

(a) The parabolic shape of the concrete structure can be represented by various functions depending on the point of reference. Based on different points of reference, obtain at least three different functions which can be used to represent the curve of this concrete structure.

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Solution :

Function 1 :

x-4 -2 2 4

y

-2

2

4

6

y = -.125x2

+ 4.5

y InterceptLocal Maximum( 0 , 4.5 )

B xy

Maximum point (0, 4.5) and pass through point (2, 4)y=a(x−b)2+ cb = 0, c = 4.5y=a(x−0)2 + 4.5y=ax2 + 4.5 --- (1)Substitute (2, 4) into (1)4=a(2)2 + 4.54a = -0.5a = -0.125∴ y=−0.125x2 + 4.5

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Function 2 :

x (meter)-3 -2 -1 1 2 3 4

y (meter)

-2

-1

1

2

y = -0.125x2

+ 0.5

y InterceptLocal Maximum( 0 , 0.5 )

Maximum point (0, 0.5) and pass through point (2, 0)y=a(x−b)2 + cb = 0, c = 0.5y=a(x−0)2 + 0.5y=ax2 + 0.5 --- (2)Substitute (2, 0) into (2)0=a(2)2 + 0.54a = -0.5a = -0.125

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∴y=−0.125x2 + 0.5Function 3 :

x (meter)-4 -2 2 4 6 8 10

y (meter)

-4

-2

2

4

6

8

10

y = -.125(x – 2)2

+ 4.5

Maximum point (2, 4.5)

Maximum point (2, 4.5) and pass through point (0, 4)y=a(x−b)2 + cb = 2, c = 4.5y=a(x−2)2 + 4.5 --- (3)Substitute (0, 4) into (3)4=a(0−2)2 + 4.54a = -0.5a = -0.125∴ y=−0.125(x−2)2 + 4.5

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(b) The front surface of this concrete structure will be painted before the words 'ART GALLERY' is written on it. Find the area to be painted.

Solution :

x (meter)-3 -2 -1 1 2 3 4

y (meter)

-2

-1

1

2

y = -0.125x2

+ 0.5

y InterceptLocal Maximum( 0 , 0.5 )

Area to be painted= Area of rectangle – Area under the curve= 4 x 1 – 2 ∫

0

2

(−0.125 x2+0.5 )dx

= 4 – 2[−0.125 x33

+¿ 0.5 x ¿¿02= 4 – 2 (2

3 - 0 )

= 223

m2Page | 11

FURTHER EXPLORATION

(a) You are given four different shapes of concrete structures as shown in the diagrams below. All the structures have the same thickness of 40 cm and are symmetrical.

(i) Given that the cost to construct 1 cubic metre of concrete is RM840.00, determine whichstructure will cost the minimum to construct.

(ii) As the president of the Arts Club, you are given the opportunity to decide on the shape of the gate to be constructed. Which shape would you choose? Explain and elaborate on your reasons for choosing the shape.

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Solution :

(i) Structure 1 :

Area = 2 23m2

Volume = Area x Thickness = 2 2

3m2 x 0.4m

= 1615m3

Cost = 1615m3 x RM840

= RM896

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Structure 2 :

Area = Area of Rectangle – Area of Triangle = 1m x 4m - 1

2 x 4m x 0.5m

= 4m2 - 1m2 = 3m2Volume = Area x Thickness = 3m2 x 0.4m = 1.2m3Cost = 1.2m3 x RM840 = RM1008

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Structure 3 :

Area = Area of Rectangle – Area of Trapezium = 1m x 4m – (4m+1m)

2 x 0.5m

= 4m2−54m2

= 2.75m2Volume = Area x Thickness = 2.75m2 x 0.4m = 1.1m3Cost = 1.1m3 x RM840

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= RM924

Structure 4 :

Area = Area of Rectangle – Area of Trapezium = 1m x 4m – (2m+4m)

2 x 0.5m

= 4m2−1.5m2 = 2.5m2Volume = Area x Thickness = 2.5m2 x 0.4m = 1m3 Cost = 1m3 x RM840

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= RM840∴ Structure 4 will cost the minimum to construct, that is RM840.

(ii) As the president of the Arts Club, I will choose structure 1 which is an arch. 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.

Next, the parabolic arch is known to be the theoretically strongest form. The parabolic arch

carry all horizontal thrust to the foundation and so do not need additional elements. The arch

provides a structure which eliminates tensile stresses in spanning an open space. All the forces are

resolved into compressive stresses.

This is useful because several of the available building materials can strongly resist

compression but are very weak when tension, shear or torsional stress is applied to them. By using

the arch configuration, significant spans can be achieved. This is because all the compressive forces

hold it together in a state of equilibrium. This even applies to frictionless surfaces.

Lastly, arches have a high aesthetic value. For example, the Arch of Constantine in Italy and

The Arc de Triomphe in Paris. These arches are known for their high aesthetic value and people from

all around the world travel just to see it. Therefore, structure 1 which is an arch will act as a better

looking and more beautiful gate compared to the other structures.

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(b) The following questions refer to the concrete structure in the diagram below. If the value of k

increases with a common difference of 0.25 m;

(i) complete Table 1 by finding the values of k and the corresponding areas of the concrete structure to be painted.(ii) observe the values of the area to be painted from Table 1. Do you see any pattern? Discuss.

Solution :

(i)

k (m) Area to be painted (m2)0.00 4 - 12

(4)(0.5) = 30.25 4 - 12

(4+0.25)(0.5) = 2.93750.50 4 - 12

(4+0.5)(0.5) = 2.8750.75 4 - 12

(4+0.75)(0.5) = 2.8125Page | 18

1.00 4 - 12

(4+1)(0.5) = 2.751.25 4 - 12

(4+1.25)(0.5) = 2.68751.50 4 - 12

(4+1.5)(0.5) = 2.6251.75 4 - 12

(4+1.75)(0.5) = 2.56252.00 4 - 12

(4+2)(0.5) = 2.5Table 1

(ii)There is a pattern in the area to be painted.

The area to be painted decreases as the k increases 0.25m and form a series of numbers :

3, 2.9375, 2.875, 2.8125, 2.75, 2.6875, 2.625, 2.5625, 2.5We can see that the difference between each term and the next term is the same.

2.9375 – 3 = -0.06252.875 - 2.9375 = -0.06252.8125 - 2.875 = -0.06252.75 - 2.8125 = -0.06252.6875 - 2.75 = -0.06252.625 - 2.6875 = -0.06252.5625 - 2.625 = -0.06252.5 - 2.5625 = -0.0625∴ We can deduce that this series of numbers is an Arithmetic Progression (AP), with a common difference, d = -0.0625In conclusion, when k increases 0.25m, the area to be painted decreases by −0.0625m2.

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(c) Express the area of the concrete structure to be painted in terms of k. Find the area as k approaches the value of 4 and predict the shape of the concrete structure.

Solution :

The area of the concrete structure to be painted

= 4 x 1 - 12

(4 + k)(0.5)

= 4 - 14

(4 + k)

= 4 – 1 - 14

k

= (3 - k4

)m2

When

k → 4

14

k → 1

Area → 2m2

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The shape of the concrete structure will be a rectangle with length 4m and breadth 0.5m, which may look like this :

CONCLUSION

a. In our daily lives, arches are often used in construction, and can often be seen in doorways or entrances. The parabolic shape of an arch can be calculated using the quadratic function.

Various points are substituted into the general formula f(x) = ax + bx + c and then solved

using completing the square. The function is rewritten as f(x) = a(x – b) + c, with (a, b) as vertex point. The equations obtained are y=−0.125x2 + 4.5, y=−0.125x2 + 0.5 and

y=−0.125(x−2)2 + 4.5. Since in all cases a < 0, negative parabolas are formed.

b. The area of the arch can be obtained by calculating the rectangular surface area of the structure. Then, integrate the function of the curve. The differences between these two values will be the area of the arch.

Further exploration

a. Since all four structures have the same thickness, it is the area of the arch will determine the cost needed. The area of the arch is obtained and then multiplied by the constant thickness of 0.4m for every structure. The shape of structure 4 needed the least money to build was chosen to be the shape of the gate to be constructed.

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b. When the value of k is increased by 0.25m each time, the area occupied by the arch also decreases by 0.0625m². The value obtained for two parallel sides of the structure is longer than the value of the other perpendicular pair of parallel sides. Hence the structure is a rectangle.

REFLECTION

While I was conducting this project, I have learnt more about Additional Mathematics and its

applications. First and foremost, I have learnt how to apply the principles of quadratic functions and

integration in our daily life. Next, I have also learnt how to relate one formula with another to get a

final answer needed. Through applying formulas, we can save materials used in all sorts of

productions.

Besides that, I have learnt something that will be very beneficial to me in the future. That is, I

will be better in using computer programs. I have learnt how to type out equations and display

workings correctly. I have also learnt how to draw graphs using a computer. I am now better in using

Microsoft Word which is widely used in the world.

Furthermore, I have gained many moral values through this project. I learnt that I must work

hard to finish a project within a given time frame. I have also learnt that co-operation and teamwork

is very important in completing a given task. With co-operation and teamwork, my team and I

managed to come up with answers quickly and were able to finish our project easily.

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Lastly, I learnt to be proactive by finding information and answers related to this project. I

learnt about responsibility in completing a project and that we should not give up halfway. Patience

is also needed to solve the questions correctly and efficiently.

REFERENCE

Internet :

http://forum.myhometuition.com/viewtopic.php?f=4&t=5

http://en.wikipedia.org/wiki/Arch

Books :

Integrated Curriculum for Secondary Schools: Additional Mathematics Form 5, Nur Niaga

Sdn. Bhd.

Focus Super Additional Mathematics SPM Revision Book, Penerbitan Pelangi Sdn. Bhd.

Human resources :

Puan Goh Siew Khimg, 5 Cengal Additional Mathematics Teacher.

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