Add Maths Complete

7/28/2019 Add Maths Complete

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ACKNOWLEDGEMENT

First of all, I wish to express gratitude to the Almighty for his guidance and

also giving me the strength to complete this project work.

Not forgetting my parents who provided everything, such as money, to buy

materials that are related to this project work and their never ending support which

was most needed for this project.

Then I would like to thank my teacher, Puan Hafizuriah, for guiding me and

my friends throughout this project. We had some difficulties in doing this task, but

she taught us patiently until we knew what to do. She gave her very best in

teaching us until we understood what we were supposed to do with the project

work.

Last but not least, my friends who were doing this project with me and

sharing their ideas. They were very helpful that when we combined and discussed

together, we had this task done.


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OBJECTIVE

The aims of carrying out this project work are to enable students to:

(a) apply mathematics to everyday situations and appreciate the importance

and the beauty of mathematics in everyday life.

(b) improve problem-solving skills, thinking skills, reasoning and mathematical

communication.

(c) develop positive attitude and personalities and instil mathematical values

such as accuracy, confidence and systemic reasoning.

(d) stimulate learning environment that enhances effective learning inquiry-base

and team work.

(e) develop mathematical knowledge in a way which increases students

interest and confidence.


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Title :Build a fencing surrounding the vegetable nursery in order to obtain the

maximum planting area in turn to produce the highest yields of crops by

using Herons method

INTRODUCTION

Triangles are shapes which are unique in their own way. With them, the

world has been able to see structures made of many types of triangles. For

example, the Egyptian pyramids are those found at Giza, on the outskirts ofCairo.

Several of the Giza pyramids are counted among the largest structures ever built.

The Pyramid of Khufu at Giza is the largest Egyptian pyramid. It is the only one of

the Seven Wonders of the Ancient World still in existence.

.
http://en.wikipedia.org/wiki/Giza_pyramid_complexhttp://en.wikipedia.org/wiki/Cairo,_Egypthttp://en.wikipedia.org/wiki/Pyramid_of_Khufuhttp://en.wikipedia.org/wiki/Seven_Wonders_of_the_Ancient_Worldhttp://en.wikipedia.org/wiki/Seven_Wonders_of_the_Ancient_Worldhttp://en.wikipedia.org/wiki/Giza_pyramid_complexhttp://en.wikipedia.org/wiki/Cairo,_Egypthttp://en.wikipedia.org/wiki/Pyramid_of_Khufuhttp://en.wikipedia.org/wiki/Seven_Wonders_of_the_Ancient_World


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A triangle has three angles and three sides. All the angles always add up to

180.There are three special names given to triangles that tell how many

sides (or angles) are equal.

There can be 3, 2 or no equal sides/angles:

Equilateral Triangle

Three equal sidesThree equal angles, always 60

Isosceles Triangle

Two equal sidesTwo equal angles

Scalene Triangle

No equal sidesNo equal angles

Triangles can also have names that tell you what type of angle is inside:

Acute Triangle

All angles are less than 90

Right Triangle

Has a right angle (90)

Obtuse Triangle

Has an angle more than 90


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Triangles have been a great contribution to the world today. It has given

shape to a building, as a toy for kids to learn, and even gives a name to the most

dangerous part of the sea, the Bermuda Triangle.

PART I


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History

The formula is credited to Heron (or Hero) of Alexandria, and a proof can be foundin his book, Metrica, written c. A.D. 60. It has been suggested thatArchimedesknew the formula over two centuries earlier, and since Metrica is a collection ofthe mathematical knowledge available in the ancient world, it is possible that the

formula predates the reference given in that work.A formula equivalent to Heron's namely:

, where

was discovered by the Chinese independently of the Greeks. It was published inShushu Jiuzhang(Mathematical Treatise in Nine Sections), written by QinJiushao and published in A.D. 1247.

Proof

A modern proof, which uses algebra and is quite unlike the one provided by Heron(in his book Metrica), follows. Let a, b, cbe the sides of the triangle andA, B, Cthe angles opposite those sides. We have

by the law of cosines. From this proof get the algebraic statement:

The altitude of the triangle on base a has length bsin(C), and it follows
http://en.wikipedia.org/wiki/Hero_of_Alexandriahttp://en.wikipedia.org/wiki/Archimedeshttp://en.wikipedia.org/wiki/Mathematical_Treatise_in_Nine_Sectionshttp://en.wikipedia.org/wiki/Qin_Jiushaohttp://en.wikipedia.org/wiki/Qin_Jiushaohttp://en.wikipedia.org/wiki/Algebrahttp://en.wikipedia.org/wiki/Anglehttp://en.wikipedia.org/wiki/Law_of_cosineshttp://en.wikipedia.org/wiki/Altitude_(triangle)http://en.wikipedia.org/wiki/Hero_of_Alexandriahttp://en.wikipedia.org/wiki/Archimedeshttp://en.wikipedia.org/wiki/Mathematical_Treatise_in_Nine_Sectionshttp://en.wikipedia.org/wiki/Qin_Jiushaohttp://en.wikipedia.org/wiki/Qin_Jiushaohttp://en.wikipedia.org/wiki/Algebrahttp://en.wikipedia.org/wiki/Anglehttp://en.wikipedia.org/wiki/Law_of_cosineshttp://en.wikipedia.org/wiki/Altitude_(triangle)


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The difference of two squares factorization was used in two different steps.

Proof using the Pythagorean theorem
http://en.wikipedia.org/wiki/Difference_of_two_squareshttp://en.wikipedia.org/wiki/Difference_of_two_squares


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Triangle with altitude h cutting base cinto d+ (c d).

Heron's original proof made use ofcyclic quadrilaterals, while other arguments appeal totrigonometry as above, or to the incenterand one excircle of the triangle. The followingargument reduces Heron's formula directly to the Pythagorean theorem using only elementarymeans.

We wish to prove The left-hand side equals

while the right-hand side equals

via the identity It therefore suffices to show

and

Substituting into the former,

as desired. Similarly, the latter expression becomes

Using the Pythagorean theorem twice, and allows us tosimplify the expression to

The result follows.

Numerical stability
http://en.wikipedia.org/wiki/Cyclic_quadrilateralhttp://en.wikipedia.org/wiki/Trigonometryhttp://en.wikipedia.org/wiki/Incenterhttp://en.wikipedia.org/wiki/Excirclehttp://en.wikipedia.org/wiki/Pythagorean_theoremhttp://en.wikipedia.org/wiki/File:Triangle_with_notations_3.svghttp://en.wikipedia.org/wiki/File:Triangle_with_notations_3.svghttp://en.wikipedia.org/wiki/Cyclic_quadrilateralhttp://en.wikipedia.org/wiki/Trigonometryhttp://en.wikipedia.org/wiki/Incenterhttp://en.wikipedia.org/wiki/Excirclehttp://en.wikipedia.org/wiki/Pythagorean_theorem


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Heron's formula as given above is numerically unstable for triangles with a verysmall angle. A stable alternative involves arranging the lengths of the sides so that

and computing

The brackets in the above formula are required in order to prevent numericalinstability in the evaluation.

Generalizations

Heron's formula is a special case ofBrahmagupta's formula for the area of acyclic quadrilateral. Heron's formula and Brahmagupta's formula are both special

cases ofBretschneider's formula for the area of a quadrilateral. Heron's formulacan be obtained from Brahmagupta's formula or Bretschneider's formula bysetting one of the sides of the quadrilateral to zero.

Heron's formula is also a special case of the formula for the area of a trapezoid ortrapezium based only on its sides. Heron's formula is obtained by setting thesmaller parallel side to zero.

Expressing Heron's formula with a CayleyMenger determinant in terms of the

squares of the distances between the three given vertices,

illustrates its similarity to Tartaglia's formula for the volume of a three-simplex.

Another generalization of Heron's formula to pentagons and hexagons inscribedin a circle was discovered by David P. Robbins.

Heron-type formula for the volume of a tetrahedron
http://en.wikipedia.org/wiki/Numerical_stabilityhttp://en.wikipedia.org/wiki/Brahmagupta's_formulahttp://en.wikipedia.org/wiki/Cyclic_quadrilateralhttp://en.wikipedia.org/wiki/Bretschneider's_formulahttp://en.wikipedia.org/wiki/Quadrilateralhttp://en.wikipedia.org/wiki/Trapezoid#Areahttp://en.wikipedia.org/wiki/Cayley%E2%80%93Menger_determinanthttp://en.wikipedia.org/wiki/Distancehttp://en.wikipedia.org/wiki/Tartaglia's_formulahttp://en.wikipedia.org/wiki/Volumehttp://en.wikipedia.org/wiki/Simplexhttp://en.wikipedia.org/wiki/David_P._Robbinshttp://en.wikipedia.org/wiki/Numerical_stabilityhttp://en.wikipedia.org/wiki/Brahmagupta's_formulahttp://en.wikipedia.org/wiki/Cyclic_quadrilateralhttp://en.wikipedia.org/wiki/Bretschneider's_formulahttp://en.wikipedia.org/wiki/Quadrilateralhttp://en.wikipedia.org/wiki/Trapezoid#Areahttp://en.wikipedia.org/wiki/Cayley%E2%80%93Menger_determinanthttp://en.wikipedia.org/wiki/Distancehttp://en.wikipedia.org/wiki/Tartaglia's_formulahttp://en.wikipedia.org/wiki/Volumehttp://en.wikipedia.org/wiki/Simplexhttp://en.wikipedia.org/wiki/David_P._Robbins


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IfU, V, W, u, v, ware lengths of edges of the tetrahedron (first three form atriangle; u opposite to Uand so on), then

where


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Part II

First of all, place the same triangle on a Certesian Plane. Label all the vertices of the triangle

with A, B and C, and state their coordinates.

Method 1:

Area triangle ABC = Area of trapezium OABC + Area of trapezium EBCD

Area of trapezium OACD

= (1+9) (6) + (9+5) (3) - (1+5) (9)= 30+21-27

= 24


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Method 2:

= 29.168

Area = (AB)(AC)sin BAC

= x 10 x 97 x sin29.168

= 2


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Method 3 :

Area ABC =Area of APQR [Area APB + Area ABC +Area ACR]

= 8 x 9 [ x 6 x 8 + x 3 x 4 + x 9 x 4 ]

= 72 48

= 24

A(0.1)

P

B(6,9)

C(9,5)

E D

R


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Method 4 : Finding area under the curve using integration

MAB =

Eq of AB: y = + 1

MBC =

Eq of BC: y-5 =

MAC =

Eq of AC =

Area of ABC

=


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= 24

Method 5 : Using Herons formula

From method 2,

AB = 10 BC = 5 AC =

S =

=

=

Area = ]

=

=

= 24


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Method 6 :

9y = 4x + 9

4x 9y + 9 = 0

a =4, b = -9, c = 9

Perpendicular distance from D to AC

BD =


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= |

= |- |

=

Area ABC = x AC X BD

= x x

= 24

Part III

i)


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Total surface area of pyramid

= 4( ) + (12)(12)

= 423.886

ii)

Area of triangle VBA

=

= 12= 69.9714

Vn2

= 102

+ 62

Vn2

= 136

Vn =

Vn = 11.6619


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Area = x 12 x

= 93.723

Total surface area

=AVAD + AVAB + AVBC + AVCD + A ABCD

= ( x 12 x 10) + ( x 12 x + ( x 12 x )+( x 10 x

12) + (12 x 12)

= 451.446

Which type of pyramid is more economically to build in terms ofcost?

The right pyramid, because the total surface area is minimum.

PART IV

Further exploration


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Total length for fencing = 80m

a) The height of =

= a-(1600-80a+a)

=

=

Area = x (80 2a) ( )

= (40-a)( )( )

= ( )(40-a)( )

82a

A

B C

a a


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b) ( )(40-a) ( ) > 0

( )(40-a) ( ) < 0

20 < a < 40

c) A = ( )(40-a) ( )

i) First method : using calculus

= ( )(40-a)[(( 2)]+ [- ]

= (40-a)[ ]- ]

= ][(40-a)-(2a-40)]

= ][80-3a]

= 0, = 0

2040


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80-3a = 0

a=

[-3]+(80-3a)[- 2)]

= -

When a = ,

= -

= -5.196 < 0

As a conclusion, to obtain the maximum planting area, a is

equal to


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That means that the shape of the planting area is an equilateral

triangle with sides equal to .

To obtain the maximum planting area, a =80

3;

1

2

2

80 80, 40 2 40 80 3

3 3

307.9201m

Area A

=

=

ii) SecondMethod :Tabulate Method

Table 1 :

a , 40(40 ) 2 40)Area A a a=

20 0

21 169.9412

22 227.684

23 263.3629

24 286.2167

25 300

26 306.7246

27 307.6361

28 303.5787

29 295.161

30 282.8427

31 266.9831

32 247.8709

33 225.7432

34 200.7984

35 173.2051

36 143.1084

37 110.6345

38 75.8946639 38.98718

40 0


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Table 2 :

a , 40(40 ) 2 40)Area A a a=

26.0 306.7246

26.1 307.061

26.2 307.3406

26.3 307.5642

26.4 307.732926.5 307.8474

26.6 307.9086

26.7 307.9173

26.8 307.8743

26.9 307.7803

27 307.6361

27.1 307.4425

27.2 307.227.3 306.9094

27.4 306.5712


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Table 3 :

a , 40(40 ) 2 40)Area A a a=

26.6 307.9086

26.61 307.9118

26.62 307.9145

26.63 307.916626.64 307.9183

26.65 307.9194

26.66 307.92

26.67 307.9201

26.68 307.9197

26.69 307.9187

26.7 307.9173

26.71 307.915326.72 307.9128

26.73 307.9098

26.74 307.9062

26.75 307.9022

26.76 307.8976

26.77 307.8925

From the tables above, the area is a maximum is 307.9201 m2whena = 26.67 m.


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CONCLUSION

I have done researches throughout the internet and discussed with a friend

who had helped me a lot in completing this project. Through the completion of this

project, I had learnt many skills and techniques.

This project really helped me to understand more about the uses of Heron

Formulae in our daily life. This project also helped expose the techniques of

application of additional mathematics in real life situations.

While conducting this project, I found a lot of information. Apart from that,

this project encourages students to work together and share their knowledge. It

also encourages student to gather information from the internet, improve thinking

skills and promote effective mathematical communication.


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Last but not least, I would like to propose that this project should be

continued because it brings a lot of benefits to students and also tests the

students understanding in Additional Mathematics.

Reflection

After spending countless hours, days

and nights to finish this

Additional Mathematics Project,

here is what I got to say:


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Completing this project makes me realize how important

Additional Mathematics is. Also, completing this project makesme realize

how fun and likable Additional Mathematics is.


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NAME : PURNASHANKRI MURUGAYAH

CLASS : 5 TERENGGANU

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