ACKNOWLEDGMENT

This Additional Mathematics Project Work plays a role in giving students a chance to apply the knowledge of Additional Mathematics learnt in the school syllabus.

I Raakesh a/l Tharmanathan from 5 Science 1, chose to do Additional Mathematics Project Work 1.

Though, I faced many challenges to complete this project work, i successfully managed to fulfil the requirements of this project work with the help of close reliances.

Therefore, I would like to take this opportunity to thank, Pn. Ong Lin Lin, my Additional Mathematics subject teacher in school. Without her guidance and tutelage, it would have been impossible for me to complete this project work.

Besides that, I would love to thank my parents. First of all, for chauffeuring me around to my teammates residence to complete my project work. Secondly, for their continues support that they showered upon me.

Last but not the least, I would like to thank my project work teammates, Sathish Kumar, Khairunnisa, Heng Jiao Faye and Nur Husna Qadirah for relentlessly working with me during the month period given for this project work.

HYPOTHESIS

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Polygons play a vital role in mathematics, many industries and human life.

PART 1

Polygons are evident in all architecture. They provide variation and charm in buildings. When applied to manufactured articles such as printed fabrics, wallpapers and tile flooring, polygons enhance the beauty of the structure itself.

a) Collect six pictures. You may use a camera to take the pictures or get them from magazines, newspapers, internet or any other resources.

Tile flooring

It is made up of a central hexagon surrounded by squares, equilateral triangles, and rhombi.

Construction

The Pentagon, headquarters of the United States Department of Defense is made up of pentagon shape.

Wallpaper

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It is made up of several hexagons combined together.

Nature

Bee's use the hexagon in their hives because it allows the most efficient use of space. Circles in a grid create spaces, and no polygon with more than six sides will be interlocking.

Printed Fabrics

It is made up of several polygons such as squares, triangles and quadrilaterals.

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Handcrafts

The hexagon is sometimes used to create the illusion of a cube by connecting every other vertex to the center, forming three diamonds, and shading each diamond differently.

b) Give the definition of polygon and write a brief history of it.

1The word "polygon" derives from the Greek πολύς (polús) "much", "many" and γωνία (gōnía) "corner" or "angle". In geometry a polygon ( pronounce as ̍ p ɒ l ɪ ɡ ɒ n ) is a flat shape consisting of straight lines that are joined to form a closed chain or circuit.

Polygons have been known since ancient times. The regular polygons were known to the ancient Greeks, and the pentagram, a non-convex regular polygon (star polygon), appears on the vase of Aristophonus, Caere, dated to the 7th century B.C. Non-convex polygons in general were not systematically studied until the 14th century by Thomas Bradwardine.

In 1952, Shephard generalized the idea of polygons to the complex plane, where each real dimension is accompanied by an imaginary one, to create complex polygons.

A polygon is traditionally a plane figure that is bounded by a closed path, composed of a finite sequence of straight line segments (i.e., by a closed polygonal chain). These segments are called its edges or sides, and the points where two edges meet are the polygon's vertices (singular: vertex) or corners.

An n-gon is a polygon with n sides. The interior of the polygon is sometimes called its body. A polygon is a 2-dimensional example of the more general polytope in any number of dimensions.

The basic geometrical notion has been adapted in various ways to suit particular purposes. Mathematicians are often concerned only with the closed polygonal chain and with simple polygons which do not self-intersect, and may define a polygon accordingly. Geometrically two edges meeting at a corner are required

1 Wikipedia

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to form an angle that is not straight (180°); otherwise, the line segments will be considered parts of a single edge – however mathematically, such corners may sometimes be allowed. In fields relating to computation, the term polygon has taken on a slightly altered meaning derived from the way the shape is stored and manipulated in computer graphics (image generation).

2According to syllabus, the Form 1 mathematics textbook and the dictionary.com, defines polygon as an enclosed figure on a plane that is bounded by three or more straight sides. From wikipedia, polygon is defined as a flat shape consisting of straight lines that are joined to form a closed chain or circuit.

These straight sides meet at a point called a vertex. A vertex is therefore the meeting point of the two sides of a polygon. The line that joins two non-adjacent vertices of a polygon is called a diagonal. For example, look at the diagram below:

Side Side

Diagonal Vertex

Types of polygon:

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Today a polygon is more usually understood in terms of sides as shown in the table above.

Historical image of polygon(1699):

c) There are various methods of finding the area of triangle. State four different methods.

3Various methods of finding the area of triangle may be used in practice, depending on what is known about the triangle. The following is a selection of frequently used formulae for the area of a triangle.

3 Wikipedia

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Calculating the area T of a triangle is an elementary problem encountered often in many different situations. The best known and simplest formula is :

Where b is the length of the base of the triangle, and h is the height or altitude of the triangle. The term "base" denotes any side, and "height" denotes the length of a perpendicular from the vertex opposite the side onto the line containing the side itself.

Although simple, this formula is only useful if the height can be readily found. For example, the surveyor of a triangular field measures the length of each side, and can find the area from his results without having to construct a "height".

Using trigonometry

Applying trigonometry to find the altitude h.

The height of a triangle can be found through the application of trigonometry.Knowing SAS: Using the labels in the image on the right, the altitude is h = a sin . Substituting this in the formula derived above :

The area of the triangle can be expressed as :

Where α is the interior angle at A, β is the interior angle at B, is the interior angle at C and c is the line AB.

Furthermore, since sin α = sin (π − α) = sin (β + ), and similarly for the other two angles :

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Knowing AAS :

and analogously if the known side is a or c.

Knowing ASA :

and analogously if the known side is b or c.

Using Heron's formula

The shape of the triangle is determined by the lengths of the sides alone. Therefore the area can also be derived from the lengths of the sides. By Heron's formula :

Where the formula below is the semiperimeter, or half of the triangle's perimeter.

Three equivalent ways of writing Heron's formula are :

Using vectors

The area of a parallelogram embedded in a three-dimensional Euclidean space can be calculated using vectors. Let vectors AB and AC point respectively from A to B and from A to C. The area of parallelogram ABDC is then :

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which is the magnitude of the cross product of vectors AB and AC. The area of triangle ABC is half of this,

The area of triangle ABC can also be expressed in terms of dot products as follows :

In two-dimensional Euclidean space, expressing vector AB as a free vector in Cartesian space equal to (x1,y1) and AC as (x2,y2), this can be rewritten as :

Using coordinates

If vertex A is located at the origin (0, 0) of a Cartesian coordinate system and the coordinates of the other two vertices are given by B = (xB, yB) and C = (xC, yC), then the area can be computed as 1⁄2 times the absolute value of the determinant :

For three general vertices, the equation is :

Which can be written as :

If the points are labeled sequentially in the counterclockwise direction, the above determinant expressions are positive and the absolute value signs can

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be omitted. The above formula is known as the shoelace formula or the surveyor's formula.

If we locate the vertices in the complex plane and denote them in counterclockwise sequence as a = xA + yAi, b = xB + yBi, and c = xC + yCi, and

denote their complex conjugates as , , and , then the formula :

Is equivalent to the shoelace formula.

In three dimensions, the area of a general triangle A = (xA, yA, zA), B = (xB, yB, zB) and C = (xC, yC, zC) is the Pythagorean sum of the areas of the respective projections on the three principal planes (i.e. x = 0, y = 0 and z = 0) :

PART 2

A farmer wishes to build a herb garden on a piece of land. Diagram 1 shows the shape of that garden, where one of its sides is 100 m in length. The garden has to

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be fenced with a 300 m fence. The cost of fencing the garden is RM 20.00 per metre. (The diagram below is not drawn to scale.)

Diagram 1

a) Calculate the cost needed to fence the herb garden.

Using simple mathematics, to find the cost needed to fence the herb garden, given that 300 m has to be fenced with the cost of fencing the garden is RM 20.00 per metre : The cost needed to fence the herb garden = 300 m x RM 20.00 = RM 6000.00

Therefore, RM 6000.00 is needed to fence the herb garden.

b) Complete table 1 by using various values of p, the corresponding values of q and

θ.

p (m) q (m) θ (degree) Area (m²)50 150 Undefined Undefined55 145 28.25° 1887.3665 135 44.81° 3092.1375 125 53.13° 3749.9985 115 57.69° 4130.7695 105 59.75° 4308.38

100 100 60.00 4330.13105 95 59.75 4308.38115 85 57.69° 4130.76125 75 53.13° 3749.99135 65 44.81° 3092.13145 55 28.25° 1887.36150 50 Undefined Undefined

The values of p (m) 50 and p (m) 150 are not portrayed in the above table, because these values give a negatif value area (m²) and mathematically it is not correct to portray such negatif value area (m²).

Therefore, these values were omitted with the corresponding values of q (m), θ (degree) and area (m²) also omitted.

The values of p (m) and q (m) are inserted, while keeping in mind that the total perimeter of the herb garden needed to be fenced is 300 m and the given side is 100 m.

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Therefore, p m + q m + 100 m = 300 m

To find the (degree), the following formulae is used,

cos R = p + q - r 2pq

While using this formulae, it is noted that I have the values of all three sides which is the values of p m, q m and 100 m (given side).

Where,R = r = 100 m (given side)

For example, taking the values of p = 55 m and q = 145 m and r = 100 m :

cos R = p + q - r 2pqcos R = 55² + 145² - 100² 2(55)(145)cos R = 0.8809R = 28.25°

Whereas to find the Area (m²), the following formulae is used,

Area of pqr = ½ qp sin R

For example, taking the values of p = 55 m and q = 145 m and R = 28.25 :

Area of pqr = ½ qp sin R = ½ (145)(55) sin 28.25

= 1887.36 m

c) Based on your findings in (b), state the dimension of the herb garden so that the enclosed area is maximum.

Based on my findings in (b), the maximum enclosed area is 4330.13m. Hence, the dimension of the herb garden is as such, p = 100 m, q = 100m with = 60 to make sure dimension of the herb garden has the maximum enclosed area.

d)(i) Only certain values of p and of q are applicable in this case. State the range of

values of p and of q.

The range of values of p is :

50 m p 150 m

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It is therefore known that, that the value of p should not be less than 50 m or equal to 50 m and the value of p should not also be more than 150 m or equal to 150 m.

Whereas, the range of values of q is :

50 m q 150 m

It is therefore known that, that the value of q should not be less than 50 m or equal to 50 m and the value of q should not also be more than 150 m or equal to 150 m.

Therefore, the range of values of p and of q are the same.

(ii) By comparing the lengths of p, q and the given side, determine the relation between them.

By comparing the lengths of p, q and the given side, I came up with this relation that states :

When p (m) increases, then q (m) decreases, and when p(m) decreases, then q (m) increases.

(iii)Make generalisation about the lengths of sides of a triangle. State the name of the relevant theorem.

4The lengths of sides of a triangle for any given ABC, it is said that :

a = b + c - 2bc cos A

Therefore, the name of the relevant theorem based on the above generalisation is Cosine Rule. The cosine rule is also known as the law of cosine. The law of cosines is useful for computing the third side of a triangle when two sides and their enclosed angle are known, and in computing the angles of a triangle if all three sides are known.

PART 3

If the length of the fence remains the same 300 m, as stated in part 2 :

4 Additional Mathematics Reference Book

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a) Explore and suggest at least five various other shapes of the garden that can be constructed so that the enclosed area is maximum.

5Given the maximum enclosed area is 4330.13 m², thus to suggest five various other shapes of the garden that can be construsted, provided it is a regular polygon, can be found by the formula :

By definition, all sides of a regular polygon are equal in length. If you know the length of one of the sides, the area is given by the formula above.

Where,s = the length of any sideN = the number of sidestan = the tangent function calculated in degrees

Basically the formula above play a purpose in finding the length of one side, given the area 4330.13 m².

For example, tetragon shape : Maximum enclosed area = sN/4 tan (180/N) 4330.13 = s(4)/4 tan (180/4)s = 65.80 m

Below are five of the various other shapes of the garden that can be constructed so that the enclosed area is maximum.

Tetragon shape

Area of tetragon is s2, where s is the length of a side.

65.80 m

Maximum enclosed area = s approx.

= (65.80) approx.

= 4329.64 m approx.

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Pentagon shape

Area of pentagon is 1.72s2 approx., where s is the length of a side.

50.17 m

Maximum enclosed area of pentagon = 1.72s² approx.

= 1.72 (50.17)² approx .

= 4329.29 m² approx .

Hexagon shape

Area of hexagon is 2.598s2 approx., where s is the length of a side.

40.82 m

Maximum enclosed area of hexagon = 2.598s² approx.

= 2.598(40.82)² approx .

= 4328.97 m² approx .

Heptagon shape

Area of heptagon is 3.633s2 approx., where s is the length of a side.

34.52 m

Maximum enclosed area of heptagon = 3.633s² approx.

= 3.633(34.52)² approx .

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= 4329.19 m² approx .

Octagon shape

Area of octagon is 4.828s2 approx., where s is the length of a side.

29.95 m

Maximum enclosed area of octagon = 4.828s² approx.

= 4.828(29.95)² approx .

= 4330.72 m² approx .

b) Draw conclusion from your exploration in (a) if :

(i) The demand of herbs in the market has been increasing nowadays. Suggest three types of local herbs with their scientific names that the farmer can plant in the herb garden to meet the demand. Colllect pictures and information of these herbs.

6These are few herbs that the farmer can plant in the herb garden to meet the demand of herbs in the market :

Agrimony

Agrimonia eupatoria is the scientific name for agrimony. The name Agrimonia may have its origin in the Greek “agremone”. The species name eupatoria relates to Mithradates Eupator, King of Pontus, who is credited with introducing many herbal remedies.

Agrimony is a genus of 12-15 species of perennial herbaceous flowering plants in the family Rosaceae, native to the temperate regions of the Northern Hemisphere, with one species also in Africa. The species grow to

6 Wikipedia & Herbal Medicine

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between 0.5 m - 2 m tall, with interrupted pinnate leaves, and tiny yellow flowers borne on a single (usually unbranched) spike.

Uses of Agrimony : Used as food plants by the larvae. Used by the ancient Greeks to treat eye ailments

is made into brews to cure diarrhea and disorders of the gallbladder, liver and kidneys.

Anglo-Saxons made a solution from the leaves and seeds for healing wounds (this used continued through the Middle Ages and afterward, in a preparation called "musket-shot water") prescribed for athlete's foot.

To treat skin diseases, asthma, coughs, gynecological complaints as a gargling solution for sore throats and ulcers.

Acts as an astringent and bitter tonic. To treat urinary problems and urinary tract infections.

Tongkat Ali

Tongkat ali, a medicinal herb with the scientific name, Eurycoma longifolia Jack, is a popular herbal supplement with many purported benefits. According to the British Journal of Sports Medicine, the active constituents of tongkat ali appear to be eurycomalacton, eurycomanon, and eurycomanol.

Uses of Tongkat Ali : Ergogenic Aid :-

Ergogenic aid is a supplement that enhances the body's adaptive response to exercise. According to a study by S. Hamzah and A. Yusof in 2003 in "The British Journal of Sports Medicine," a tongkat ali extract in conjunction with a weight-training program, caused an increase in lean body mass and a loss of fat tissue relative to placebo. This supports the use of tongkat ali by athletes as an ergogenic aid.

Libido Enhancer :-As tongkat ali is thought to increase levels of the male sex hormone testosterone, it may have utility as a libido-enhancer or general male sexual aid. According to a 2003 study by Ang HH, Ngai TH and Tan TH in "The Journal of Phytomedicine," when an extract of tongkat ali was administered to middle-aged rats, their sexual performance increased relative to the placebo group.

Anti-Cancer Agent :-Tongkat ali has also been studied as a possible cancer-fighting agent.According to a study by Tee TT and Azimahtol HL in 2005 in the

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journal, "Anticancer Research," an extract of tongkat ali exerted anti-proliferative effects against human breast cancer cells whereas in a 2003 study conducted at National Cheng Kung University and published in "The Journal of Natural Products," an extract of tongkat ali killed human lung cancer cells, as well as demonstrating anti-malarial effects.

Lemongrass

The scientific name of lemongrass is cymbopogon. It is a genus of about 55 species of grasses, (of which the type species is Cymbopogon citratus) native to warm temperate and tropical regions of the Old World and Oceania. It is a tall perennial grass. Lemongrass is native to India and tropical Asia. It is widely used as a herb. It has a subtle citrus flavor and can be dried and powdered, or used fresh. West-Indian lemon grass (Cymbopogon citratus), also known as serai in Malay, is assumed to have its origins in Malaysia. Lemon grass is also known as Gavati Chaha in the Marathi language.

Uses of Lemongrass : Used in teas (mostly in African countries, Democratic Republic of

the Congo and Latin American countries), soups and curries. Suitable for poultry, fish, beef, and seafood. Its oil which has anti-fungal properties is used as a pesticide and

a preservative (it is put on the ancient palm-leaf manuscripts found in India as a preservative which keeps the manuscripts dry so that the text is not lost to decay due to humidity).

Is used both as a medical herb and in perfumes (mostly in India by using cymbopogon citratus).

Is used as an addition to tea, and in preparations like 'kadha,' which is a traditional herbal 'soup' used against coughs and colds.

Has medicinal properties and is used extensively in Ayurvedic medicine to help with relieving cough and nasal congestion.

Pandan leaves

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Pandanus amaryllifolius is a tropical plant in the Pandanus (screwpine) genus, which is known commonly as pandan leaves. The plant is rare in the wild, but is widely cultivated. It is an upright, green plant with fan-shaped sprays of long, narrow, bladelike leaves and woody aerial roots. The plant is sterile, flowers only very rarely, and is propagated by cuttings. In Bangladesh it is called ketaki, along with the other variety of pandan there (Pandanus fascicularis). It is also known as pandan wangi in Indonesian, soon-mhway in Burmese, bai tooey in Thai, rampe in Sinhala, and lá dứa in Vietnamese. The leaves are used either fresh or dried, and are commercially available in frozen form in Asian grocery stores in nations where the plant does not grow. They have a nutty, botanical fragrance which enhances the flavor of Indonesian, Singaporean, Filipino, Malaysian, Thai, Bangladeshi, Vietnamese, Chinese, Sri Lankan, and Burmese foods.

Uses of Pandan leaves : Is used widely in Southeast Asian cooking as a flavoring. Is used to enhance the flavor of pulao, biryani and sweet

coconut rice pudding, payesh (in Bangladesh). The leaves can be steeped in coconut milk, which is then added

to the dish. Woven into a basket which is used as a pot for cooking rice. Gives white bread, jasmine rice and basmati rice (as well as

bread flowers Vallaris glabra) their typical smell. Have a repellent effect on cockroaches.

Misai Kucing

The scientific name for misai kucing is called as Orthosiphon stamineus. It is a traditional herb that is widely grown in tropical areas. There are two general species, known as Orthosiphon stamineus "purple" and Orthosiphon stamineus "white". Also known as Java tea, it was possibly introduced to the west in the early 20th century. The plant is a herbaceous shrub, with a

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unique flower. The flowers have fine filaments that resembles cat's whiskers- hence its Malay name, Misai Kuching [Cat’s whiskers]. In the wild, the plant can be seen growing at fringes of the jungle, in wastelands and along roadsides. It can grow to a height of 1.5 meters.

Uses of Misai Kucing : Treats circulatory disorders. Consumed as a herbal tea. Flushing the kidneys and urinary tract. For treating ailments of the kidney and bladder stone. Relieves spasms of the smooth muscle in the walls of the

internal organs. Can reduce swelling. Acts as an anti-inflammatory agents that are not steroids. Enhance the functioning of the immune system and

reproductive system. Treats diabetes. Treats high blood pressure. Can protect cells from the damaging effects of free radicals. Provide a source of vitamin A. Prevents or reduces cancer [when cancerous lab rats were

treated with Misai Kucing, it was found that it slowed tumor development both in early and late stages of growth with no apparent side effects.

(ii) These herbs will be processed for marketing by a company. The design of the packaging plays the an important role in attracting customers. The company wishes to design an innovative and creative logo for the packaging. You are given the task of designing a logo to promote the product. Draw the logo on a piece of paper size A4. You must include at least one polygon shape in the logo.

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The logo is focused by the polygon shape of pentagon. Although, only two polygons are used, still the impact is strong to show that the Misai Kucing is the main focus in the product which attract people to see this simple yet tranquil piece of design. The rectangle shape is used to show the name of the product.

CONJECTURE

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MISAI KUCING

The hypothesis is accepted. Polygons are evident in mathematics. This can be seen in solving mathematical problem involving polygons. Besides that, polygons can also be seen in many industries. For example, the advertising industry makes a point in inserting polygons into advertisment banners, logos and etc. Furthermore, polygons are also evident in human life. From bedsheets to underwear, polygons are used in all kinds of fabrics, wallpapers, and tile flooring, polygons enhance the beauty of the structure itself.

FURTHER EXPLORATION

Heron’s Formula can be used to find the area of a triangle when the lengths of the three sides are known. There are many proofs of Heron’s Formula. Most can be categorised as algebraic, geometric, or trigonometric. Conduct a research using the internet to find out one proof of the Heron’s Formula in any one of the list categories.

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Below, is one of the proof of Heron’s Formula categorised under geometric and algebraic.

7Note: This proof was adapted from the outline of a proof on page 194 in the 6th edition of An Introduction to the History of Mathematics by Howard Eves.

Given triangle ABC, let the length of segment BC be a, the length of segment AC be b, and the length of segment AB be c.

Note the perimeter, p, of triangle ABC = a + b + c. Half of the perimeter is called the semiperimeter, s, and so for triangle ABC, s = (a + b + c)/ 2.

Let the point of intersection of the angles bisectors of angles A, B, and C be called point I. Construct segments AI, BI, and CI.

Next, let the point of intersection of the perpendicular to side AC through point I be called point D, the point of intersection of the perpendicular line to side BC through

7 A Geometric Proof of Heron's Formula by Shannon Umberger

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point I be called point E, and the point of intersection of the perpendicular line to side AB through point I be called point F.

By definition, point I is called the incenter of triangle ABC. The circle with point I as the center and that passes through points D, E, and F is called the incircle of triangle ABC.

Now, since segment IB is the angle bisector of angle B, then angles IBF and IBE are equal. Since angles IFB and IEB are right by construction, then they are equal, and thus angles BIF and BIE are equal (since all angles of a triangle sum to 180 degrees). Note that segments IF and IE are equal since both are radii of the same circle. Therefore, triangles BIF and BIE (in red) are congruent by SAS.

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Similarly, triangles CID and CIE (in green) are congruent by SAS.

Also similarly, triangles AIF and AID (in blue) are congruent by SAS.

Because corresponding sides of congruent triangles are congruent, it follows that segments BF and BE (in red) are equal, segments CD and CE (in green) are equal, and segments AD and AF (in blue) are equal.

By using the formula that states the area of a triangle equals one-half the base times the height, the area of triangle AIB = (1/ 2)*AB*IF,...

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...the area of triangle BIC = (1/ 2)*BC*IE,...

...and the area of triangle AIC = (1/ 2)*AC*ID.

Therefore, the area of triangle ABC is the sum of the areas of triangles AIB, BIC, and AIC.

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By substitution, the area of triangle ABC = (1/ 2)*AB*IF + (1/ 2)*BC*IE + (1/ 2)*AC*ID. Since segments IF, IE, and ID are equal (they are all radii of the same circle), it follows with a little algebra that the area of triangle ABC = (1/ 2)*ID*(AB + BC + AC). But AB + BC + AC = a + b + c, and so this sum is really the perimeter of triangle ABC; therefore, the area of triangle ABC = (1/ 2)*ID*p. Since (1/ 2)*p = s, the semiperimeter of triangle ABC, then the area of triangle ABC = ID*s.

Now, construct point G such that G lies on the same line as segment AC, and segment CG equals segment BE. Remember p = a + b + c = AB + BC + AC, and so by segment addition and substitution, p = (BE + CE) + (CD + AD) + (AF + BF) = (BF + BE) + (CE + CD) + (AD + AF). Furthermore, by substitution, p = 2*BE + 2*CD + 2*AD = 2(BE + CD + AD). Again, by substitution, p = 2(CG + CD + AD). But, AG = AD + DC + CG by segment addition, and so by substitution, p = 2*AG. Therefore, s = AG.

Thus, by substitution, the area of triangle ABC = ID*AG.

Construct a line perpendicular to segment AI through point I (in pink) and a line perpendicular to segment AG through point C (in light blue). Let the point of intersection of these two lines be called point H. Let the point of intersection of segment IH and segment AG be called point J. Construct segment AH (in yellow).

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Since angles AIH and ACH are right (by construction), then triangles AIH and ACH are right. Note the triangles share a common hypotenuse, segment AH. It follows that these triangles are inscribed in a common circle with segment AH as the diameter of the circle.

This fact means that quadrilateral AICH is cyclic, and so opposite angles AIC and AHC are supplementary.

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Since the sum of the angles at point I is 360 degrees, by angle addition, (angle BIF + angle BIE) + (angle CIE + angle CID) + (angle AID + angle AIF) = 360 degrees. By substitution, 2*(angle BIE) + 2*(angle CID) + 2*(angle AID) = 360 degrees, and so angle BIE + angle CID + angle AID = 180 degrees. But angle CID + angle AID = angle AIC by angle addition, thus angle BIE + angle AIC = 180 degrees, and so angles BIE and AIC are supplementary.

Since angles BIE and AHC are supplementary to the same angle, it follows that angle BIE = angle AHC.

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Because angles BIE and AHC are equal, and angles BEI and ACH are equal (both are right by construction), then triangles BIE and AHC are similar by AA similarity.

By substitution, AC/CG = AC/BE. Then, by the definition of similar triangles, AC/BE = HC/IE.

Now, because angles IJD and HJC are vertical angles, they are equal. Also, angles IDJ and HCJ are equal (both are right by construction). Therefore, triangles IJD and HJC are similar by AA similarity.

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By substitution, HC/IE = HC/ID. Then, by the definition of similar triangles, HC/ID = CJ/DJ.Thus, by transitivity, AC/CG = CJ/DJ.

The next few steps of the proof require a few algebra tricks.Since AC/CG = CJ/DJ, then AC/CG + 1 = CJ/DJ + 1.So AC/CG + CG/CG = CJ/DJ + DJ/DJ.This implies that (AC + CG)/CG = (CJ + DJ)/DJ.But by segment addition, AC + CG = AG, and CJ + DJ = CD.Therefore, by substitution, AG/CG = CD/DJ.Well, then, AG/CG * 1 = CD/DJ * 1.So AG/CG * AG/AG = CD/DJ * AD/AD.Then AG2/CG*AG = CD*AD/DJ*AD.

Now, since triangle AIJ is right (by construction), and D is a point on the hypotenuse, then DJ*AD = ID2 (by the geometric mean).

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So, by substitution, AG2/CG*AG = CD*AD/ID2.By cross multiplication, AG2*ID2 = CG*AG*CD*AD.It follows that AG*ID = [CG*AG*CD*AD]1/ 2.But, the area of triangle ABC = AG*ID.Thus, by transitivity, the area of triangle ABC = [CG*AG*CD*AD]1/ 2.

Finally, remember that segment AG = s.Note that segment CG = AG - AC by segment addition, and so CG = s - b by substitution.Also note that segment CD = AG - (AD + CG) by segment addition. So CD = AG - (AF + BF) by substitution, and so CD = s - c by substitution.And note that segment AD = AG - (CD + CG) by segment addition. So AD = AG - (CE + BE) by substitution, and so AD = s - a by substitution.

Therefore, by substitution, the area of triangle ABC = [s*(s - a)*(s - b)*(s - c)]1/ 2. QED.

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REFLECTION

While you were conducting the project, what have you learn? What moral values did you practice? Represent your opinions or feelings creatively through usage of symbols, illustrations, drawings, or even in a song.

While conducting this Additional Mathematics Project work, I have learnt to cooperate. Saying so I had to work in a team, thus my team and I learnt to cooperate with each other. We distributed the work in a very sophisticated manner and fairly.

Besides that, I have gained a wide mathematical knowledge from completing this project work. I have learnt to solve a question by using various methods. The knowledge gained from this project not only can it be applied for exam purposes,

but it can also be applied in the future for other reasons. I have also learnt that mathematics is used everywhere in daily life, from the most simple things like baking and decorating a cake, to designing and building monuments.

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At the very same time, the quote patience and perserverence is the key to success was very obvious in my completion of the project work. I have learnt to be more patient while striving to complete any kind of work.

This project work has made me more prepared to face the challenges of the outside world in getting things done perfectly. This is because the outside world is getting more competetive and those with better knowledge of solving problems stand higher chances to be successful.

Moreover, I practised certain moral values while doing this Additional Mathematics Project Work. I learnt to take responsibility doing this project work by finishing it on time.

Given the time constrain, I successfully completed the project work. Hence, I learnt to sacrifice and manage my precious time to focus on this project while preparing for my mid year examination.

Besides that, I practised to take responsibility. I took charge in steering my team to the right direction while doing this project work.

This project work had also taught me to be more confident when undertaking responsibilities. I also learned to be a more disciplined student who is punctual and independent.

Thanks to the polygons, human life have become condusive and lively.

BIBLIOGRAPHY

These are the following resources used in the Additional Mathematics Project Work :

1. The definition of polygon and its brief history was obtained form http://en.wikipedia.org/wiki/Polygon

2. For more in-depth definition of polygon and its brief history the information was from Form 1 Mathematics Textbook

3. The various methods of finding the area of a triangle was obtained from http://en.wikipedia.org/wiki/Triangle

4. To make generalisation about the lengths of sides of a triangle and to state the name of relevant theorem, information was obtained from SUCCESS, Additional Mathematics, SPM written by Wang Wei and Dr. Wong Sin Mong and published by OXFORD FAJAR

5. To explore and suggest at least five various other shapes of the garden that can be constructed so that the enclosed area is maximum, the necessary information was obtained from http://www.mathopenref.com/index.html

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6. To suggest three types of local herbs with their scientific names that the farmer can plant in the herb garden to meet the demand in the market, the information was obtained from http://en.wikipedia.org/wiki/Main_Page and from http://herbalmedicine.med-terms.net/?q=node/21

7. The information on Heron’s Formula in the FURTHER EXPLORATION was obtained from http://jwilson.coe.uga.edu/EMT668/EMAT6680.2000/Umberger/MATH7200/HeronFormulaProject/GeometricProof/geoproof.html

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