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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_World7/28/2019 Add Maths Complete
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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)7/28/2019 Add Maths Complete
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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_squares7/28/2019 Add Maths Complete
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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_theorem7/28/2019 Add Maths Complete
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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._Robbins7/28/2019 Add Maths Complete
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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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