FORM 5 PROJECT WORK FOR ADDITIONAL MATHEMATHICS 2009 by mizie0o0
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Additional Mathematics 2009
PROJECT WORK FORADDITIONAL MATHEMATHICS
2009
Circles In Our Daily Life
Name : Muhamad Termizi Bin Muhamad
Kelas : 5 Sains Teknologi Kejuruteraan
K/P : 920102-07-5241
Guru Pembimbing : Sir Mohd Nasir Bin
Sekolah : Sekolah Menengah Kebangsaan Seri Samudera
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Additional Mathematics 2009
No. Content Page1. Acknowledgement 22. Objective 33. Introduction 44. Part 1 6
5. Part 2a 136. Part 2b 157. Part 3 17
First of all, I would like to say Alhamdulillah, for giving me the strength
and health to do this project work until it done.
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Additional Mathematics 2009
Not forgotten to my parents for providing everything, such as money,
to buy anything that are related to this project work and their advise, which
is the most needed for this project. Internet, books, computers and all that as
my source to complete this project. They also supported me and encouraged
me to complete this task so that I will not procrastinate in doing it.
Then I would like to thank my teacher, Sir Mohd Nasir Bin …… for
guiding me and my friends throughout this project. We had some difficulties
in doing this task, but he taught us patiently until we knew what to do. He
tried and tried to teach us until we understand what we supposed to do with
the project work.
Last but not least, my friends who were doing this project with me and
sharing our ideas. They were helpful that when we combined and discussed
together, we had this task done.
The aims carrying out this project work are:
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i. To apply and adapt a variety of problem-solving strategies to solve
problems;
ii. To improve thinking skills;
iii. To promote effective mathematical communication;
iv. To develop mathematical knowledge through problem solving in a
way that increases students’ interest and confidence;
v. To use the language of mathematics to express mathematical ideas
precisely;
vi. To provide learning environment that stimulates and enhances
effective learning;
vii.To develop positive attitude towards mathematics.
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Additional Mathematics 2009
A circle is a simple shape of Euclidean geometry consisting of those
points in a plane which are the same distance from a given point called the
centre. The common distance of the points of a circle from its center is called
its radius. A diameter is a line segment whose endpoints lie on the circle and
which passes through the centre of the circle. The length of a diameter is
twice the length of the radius. A circle is never a polygon because it has no
sides or vertices.
Circles are simple closed curves which divide the plane into two
regions, an interior and an exterior. In everyday use the term "circle" may be
used interchangeably to refer to either the boundary of the figure (known as
the perimeter) or to the whole figure including its interior, but in strict
technical usage "circle" refers to the perimeter while the interior of the circle
is called a disk. The circumference of a circle is the perimeter of the circle
(especially when referring to its length).
A circle is a special ellipse in which the two foci are coincident. Circles
are conic sections attained when a right circular cone is intersected with a
plane perpendicular to the axis of the cone.
The circle has been known since before the beginning of recorded
history. It is the basis for the wheel, which, with related inventions such as
gears, makes much of modern civilization possible. In mathematics, the
study of the circle has helped inspire the development of geometry and
calculus.
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Additional Mathematics 2009
Early science, particularly geometry and Astrology and astronomy, was
connected to the divine for most medieval scholars, and many believed that
there was something intrinsically "divine" or "perfect" that could be found in
circles.
Some highlights in the history of the circle are:
• 1700 BC – The Rhind papyrus gives a method to find the area of a
circular field. The result corresponds to 256/81 as an approximate
value of π.
•
300 BC – Book 3 of Euclid's Elements deals with the properties of
circles.
• 1880 – Lindemann proves that π is transcendental, effectively settling
the millennia-old problem of squaring the circle.
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Additional Mathematics 2009
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Additional Mathematics 2009
Definition
In Euclidean plane geometry, π is defined as the ratio of a circle's
circumference to its
diameter:
The ratio C/d is constant, regardless of a circle's size. For example, if a
circle has twice the diameter d of another circle it will also have twice the
circumference C, preserving the ratio C/d .
Area of the circle = π × area of the shaded square:
Alternatively π can be also defined as the ratio of a circle's area (A) to
the area of a square whose side is equal to the radius:
These definitions depend on results of Euclidean geometry, such as the
fact that all circles are similar. This can be considered a problem when π
occurs in areas of mathematics that otherwise do not involve geometry. For
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this reason, mathematicians often prefer to define π without reference to
geometry, instead selecting one of its analytic properties as a definition. A
common choice is to define π as twice the smallest positive x for which
cos( x ) = 0. The formulas below illustrate other (equivalent) definitions.
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Additional Mathematics 2009
History
The ancient Babylonians calculated the area of a circle by taking 3
times the square of its radius, which gave a value of pi = 3. One Babylonian
tablet (ca. 1900–1680 BC) indicates a value of 3.125 for pi, which is a closer
approximation.
In the Egyptian Rhind Papyrus (ca.1650 BC), there is evidence that the
Egyptians calculated the area of a circle by a formula that gave the
approximate value of 3.1605 for pi.
The ancient cultures mentioned above found their approximations by
measurement. The first calculation of pi was done by Archimedes of
Syracuse (287–212 BC), one of the greatest mathematicians of the ancient
world. Archimedes approximated the area of a circle by using the
Pythagorean Theorem to find the areas of two regular polygons: the polygon
inscribed within the circle and the polygon within which the circle was
circumscribed. Since the actual area of the circle lies between the areas of
the inscribed and circumscribed polygons, the areas of the polygons gave
upper and lower bounds for the area of the circle. Archimedes knew that he
had not found the value of pi but only an approximation within those limits.
In this way, Archimedes showed that pi is between 3 1/7 and 3 10/71.
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Additional Mathematics 2009
A similar approach was used by Zu Chongzhi (429–501), a brilliant
Chinese mathematician and astronomer. Zu Chongzhi would not have been
familiar with Archimedes’ method—but because his book has been lost, little
is known of his work. He calculated the value of the ratio of the
circumference of a circle to its diameter to be 355/113. To compute this
accuracy for pi, he must have started with an inscribed regular 24,576-gon
and performed lengthy calculations involving hundreds of square roots
carried out to 9 decimal places.
Mathematicians began using the Greek letter π in the 1700s.
Introduced by William Jones in 1706, use of the symbol was popularized by
Euler, who adopted it in 1737.
An 18th century French mathematician named Georges Buffon devised a way
to calculate pi based on probability.
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Additional Mathematics 2009
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c : Assume the diameter of outer semicircle is 30cm and 4 semicircles are
inscribed in the outer semicircle such that the sum of d1(APQ), d2(QRS),
d3(STU), d4(UVC) is equal to 30cm.
d1 d2 d3 d4 SABC SAPQ SQRS SSTU SUVC
10 8 6 6 15 π 5 π 4 π 3 π 3 π
12 3 5 10 15 π 6 π 3/2 π 5/2 π 5 π14 8 4 4 15 π 7 π 4 π 2 π 2 π
15 5 3 7 15 π 15/2 π 5/2 π 3/2 π 7/2 π
let d1=10, d2=8, d3=6, d4=6, SABC = SAPQ + SQRS + SSTU +
SUVC
15 π = 5 π + 4 π + 3 π + 3 π
15 π = 15 π
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a. Area of flower plot = y m2
y = (25/2) π - (1/2(x/2)2π + 1/2((10-x )/2)2 π)
= (25/2) π - (1/2(x/2)2π + 1/2((100-20x+x2)/4) π)
= (25/2) π - (x2/8 π + ((100 - 20x + x2)/8) π)
= (25/2) π - (x2π + 100π – 20x π + x2π )/8
= (25/2) π - ( 2x2– 20x + 100)/8) π
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= (25/2) π - (( x2 – 10x + 50)/4)
= (25/2 - (x2 - 10x + 50)/4) π
y = ((10x – x2)/4) π
b. y = 16.5 m2
16.5 = ((10x – x2)/4) π
66 = (10x - x2) 22/7
66(7/22) = 10x – x2
0 = x2 - 10x + 21
0 = (x-7)(x – 3)
x = 7 , x = 3
c. y = ((10x – x2)/4) π
y/x = (10/4 - x/4) π
x 1 2 3 4 5 6 7
y/x 7.1 6.3 5.5 4.7 3.9 3.1 2.4
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3.0
4.0
5.0
6.0
7.0
8.0
0 1 2 3 4 5 6 7
X
Y/x
Additional Mathematics 2009
When x = 4.5 , y/x = 4.3
Area of flower plot = y/x * x
= 4.3 * 4.5
= 19.35m2
d. Differentiation method
dy/dx = ((10x-x
2
)/4) π
= ( 10/4 – 2x/4) π
0 = 5/2 π – x/2 π
5/2 π = x/2 π
x = 5
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