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First of all, I would like to say Alhamdulillah to Allah for giving me
the strength and health to do this project work until it done.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,
Mdm. Far idah Bt Md. Yatim forguiding 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. she 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.
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The aims carrying out this project are :
i. To apply and adapt a variety of problem-solving
strategies to solve problems;
ii. To improve thinking skillsiii. 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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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 f igure 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,
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the study of the circle has helped inspire the development of geometry
and calculus.
Early science, particularly geometry and Astrology and astronomy,
wasconnected to the divine for most medieval scholars, and many
believed thatthere was something intrinsically "divine" or "perfect" that
could be found incircles.
One of highlights in the history of the circle is:
* 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 n.
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PART 1
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Wheel Of Bicycle Circles on water surface school Park
Fish pond Round table at school compound
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Definition
In Euclidean plane geometry, n is defined as the ratio of a
circle's circumference to its diameter:
The ratioCUis constant, regardless of a circle's size. For example, if
a circle has twice the diameterdof another circle it will also have twice
the circumference C, preserving the ratioc/d. Area of the circle = n x area
of the shaded square:
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Alternatively n 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
n occurs in areas of mathematics that otherwise do not involve
geometry. For this reason, mathematicians often prefer to define n
without reference togeometry, instead selecting one of its analytic
properties as a definition. Acommon choice is to define n as twice the
smallest positive x for which cos(x) = 0.
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HISTORY
The ancient Babylonians calculated the area of a circle by taking
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. The Greek letter n, often spelled out piin text, was adopted for
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the number from the Greek word forperimeter"nepiueTpoc.", first by
William Jones in 1707, and popularized by Leonhard Euler in 1737. The
constant is occasionally also referred to as the circular constant, Archimedes'
constant (not to be confused with an Archimedes number), or Ludolph's
number (from a German mathematician whose e fforts to calculate more of
its digits became famous).The name of the Greek letter n is pi, and this
spelling is commonly used in typographical contexts when the Greek letter is
not available, or its usage could be problematic. It is not normally
capitalised (II) even at the beginning of a sentence. When referring to
this constant, the symbol n is always pronounced like "pie" in English,
which is the conventional English pronunciation of the Greek letter . In
Greek, the name of this letter is pronounced /pi/.The constant is
named "n" because "n" is the first letter of the Greek words nepupepeia
(periphery) and nepiueTpoc. (perimeter), probably referring to its use in
the formula to find the circumference, or perimeter, of a circle, n is
Unicode character U+03C0 ("Greek small letter pi").
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PART 2
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(a) Diagram 1 shows a semicircle PQRof diameter 10. Semicircles PAB
and BCRof diameterdiand d2repectively are inscribed in the semicircle
PQRsuch that the sum ofdiand d2isequal to 10 cm.
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TABLE 1
From the Table 1 we know that the length of arc PQRis not affected by
the different in di and d2 in PABand BCRrespectively. The relation
between the length of arcs PQR , PABand BCRis that the length of arc
PQRis equal to the sum of the length of arcs PABand BCR, which is we
can get the equation:
LENGTH OF ARC PQR = LENGTH OF ARC PAB + LENGTH OF ARC
d1(cm) d2(cm)
Length of arc
PQR in terms
of (cm)
Length of arc
PAB in terms
of (cm)
Length of arc
BCR in terms
of (cm)
1 9 5 0.5 4.5
2 8 5 1.0 4.0
3 7 5 1.5 3.5
4 6 5 2.0 3.0
5 5 5 2.5 2.5 6 4 5 3.0 2.0
7 3 5 3.5 1.5
8 2 5 4.0 1.0
9 1 5 4.5 0.5
10 0 5 5.0 0.0
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(b) Diagram 2 shows a semicircle PQR of diameter 10cm. Semicircles
PAB,BCD and DER of diameter d1 , d2 and d3 respectively are inscribed
in the semicircle PQR such that the sum of d1 , d2 and d3 is equal to 10 cm.
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d1(cm) d2(cm) d3(cm)
Length ofarc PQR
in terms
of (cm)
Length ofarc PAB
in terms
of (cm)
Length ofarc BCD
in terms
of (cm)
Length ofarc DER
in terms
of (cm)
1 1 8 5 0.5 0.5 4.0
1 2 7 5 0.5 1.0 3.5
1 3 6 5 0.5 1.5 3.0
1 4 5 5 0.5 2.0 2.5
1 5 4 5 0.5 2.5 2.0
1 6 3 5 0.5 3.0 1.5
1 7 2 5 0.5 3.5 1.0 1 8 1 5 0.5 4.0 0.5
2 1 7 5 1.0 0.5 3.5
2 2 6 5 1.0 1.0 3.0
2 3 5 5 1.0 1.5 2.5
2 4 4 5 1.0 2.0 2.0
2 5 3 5 1.0 2.5 1.5
2 6 2 5 1.0 3.0 1.0
2 7 1 5 1.0 3.5 0.5
3 1 6 5 1.5 0.5 3.0
3 2 5 5 1.5 1.0 2.5 3 3 4 5 1.5 1.5 2.0
3 4 3 5 1.5 2.0 1.5
3 5 2 5 1.5 2.5 1.0
3 6 1 5 1.5 3.0 0.5
4 1 5 5 2.0 0.5 2.5
4 2 4 5 2.0 1.0 2.0
4 3 3 5 2.0 1.5 1.5
4 4 2 5 2.0 2.0 1.0
4 5 1 5 2.0 2.5 0.5
5 1 4 5 2.5 0.5 2.0 5 2 3 5 2.5 1.0 1.5
5 3 2 5 2.5 1.5 1.0
5 4 1 5 2.5 2.0 0.5
6 1 3 5 3.0 0.5 1.5
6 2 2 5 3.0 1.0 1.0
6 3 1 5 3.0 1.5 0.5
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7 1 2 5 3.5 0.5 1.0
7 2 1 5 3.5 1.0 0.5
8 1 1 5 4.0 0.5 0.5
TABLE 2
( i ) From the Table 2, we can say that the relation between the lengths of
arcs PQR, PAB, BCD and DER based on the table is the sum of all the
lengths of arcs of the inner semicircles which is PAB, BCD and DER
respectively will equal to the length of arc of the outer semicircle, which is
PQR.
Length of arc PQR = Length of arc PAB + Length of arc BCD + Length of
arc CDR
( ii ) Base on the findings in the table in (a) and (b) above, we conclude that:
The length of the arc of the outer semicircle = the sum of the length of arcs
of the inner semicircles for n inner semicircles where n = 2, 3, 4
Or
(s out) = n (s in), n = 2, 3, 4, ......
where,
s in = length of arc of inner semicircle
s out = length of arc of outer semicircle
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(c ) For different values of diameters of the outer semicircle, show that thegeneralization statedin b (ii) is still true.
The length of arc of the outer semicircle
The sum of the length of arcs of the inner semicircles
Factorise /2
Substitute
We get,
where d is any positive real number.
We can see that
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Assume the diameter of outer semicircle is 30cm and 4 semicircles are inscribed inthe 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 5 4 3 3
12 3 5 10 15 6 3/2 5/2 5
14 8 4 4 15 7 4 2 2
15 5 3 7 15 15/2 5/2 3/2 7/2
Let d1=10, d2=8, d3=6, d4=6, SABC= 5 + 4 + 3 + 3
15 = 5 + 4 + 3 + 3
15 = 15
The diameter of the outer semicircle,
10cm = 1cm + 1cm + 8cm
The length of arc of the outer semicircle,
0.5 + 0.5 + 4.0 = 5
The sum of the length of arcs of the inner semicircles
Factorise /2
(1cm + 1cm + 8cm) =5
In conclusion, we can conclude that
The length of the arc of the outer semicircle is equal to the sum of the length
of arcs of any number of the inner semicircles. This is true for any value of
the diameter of the semicircle.
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In other words, for different values of diameters of the outer semicircle, the
generalisations stated in b (ii) is still true.
PART 3
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The Mathematics Society is given a task to design a garden to beautify their school
by using the design as shown in Diagram 3. The shaded region will be planted with
flowers and the two inner semicircles are fish pond.
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(a)
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(b)
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(c)
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Linear Law
y = - +
Change it to linear form of Y = mX + C.
= - +
Y =
X = x
m = -
C =
Thus, a graph of against x was plotted and the line of best fit was drewn.
X = x 1.0cm 1.5cm
2.0
cm
2.5
cm
3.0
cm
3.5
cm
4.0
cm
4.5
cm
5.0
cm
Y = 7.069 6.676 6.283 5.890 5.498 5.105 4.712 4.320 3.927
Find the value of when x = 4.5 m.
Then multiply y/x you get with 4.5 to get the actual value of y.
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From the graph above, when the diameter of one of the fish pond is 4.5 m,
the value of y/x is 4.35. Therefore, the area of the flower plot when the
diameter of one of the fish pond is 4.5 m is
4.35 m ( 4.5 m) = 19.575 m
2
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(d)
Method 1: Differentation
y = - +
= - +
= -
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Method 2: Completing the Square
y = - +
= - (x2 10x)
= - (x2 10x + 25 - 25)
= - [(x-5)2 25]
= - (x-5)2 + 25
y is a n shape graph as a = -
Hence, it has a maximum value.
When x = 5 m, maximum value of the graph = 6.25 m
2
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(e)
The principal suggested an additional of 12 semicircular flower beds to the
design submitted by the Mathematics Society. (n = 12)
The sum of the diameters of the semicircular flower beds is 10 m.
The diameter of the smallest flower bed is 30 cm. (a = 30 cm = 0.3 m)
The diameter of the flower beds are increased by a constant value
successively. (d = ?)
S12 = ( )[2a + (n - 1)d]
10 = ( )[2(0.3) + (12-1)d]
= 6(0.6 + 11d)
= 3.6 + 66d
66d = 6.4
d =
Since the first flower bed is 0.3 m,
Hence the diameters of remaining 11 flower beds expressed in arithmetic
progression are:
m, m, m, m, m, m, m, m, m, m, m
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CONCLUSION
Part 1
Not all objects surrounding us are related to circles. If all the
objects are circle, there would be no balance and stability. In our
daily life, we could related circles in objects. For example: a fan,
a ball or a wheel. In Pi(), we accept 3.142 or 22/7 as the best
value of pi. The circumference of the circle is proportional as
pi() x diameter. If the circle has twice the diameter, d ofanother circle, thus the circumference, C will also have twice of
its value, where preserving the ratio =Cid
Part 2
The relation between the length of arcs PQR, PAB and BCR where
the semicircles PQR is the outer semicircle while inner semicircle
PAB and BCR is Length of arc=PQR = Length of PAB + Length ofarc BCR. The length of arc for each semicircles can be obtained
as in length of arc = 1/2(2r). As in conclusion, outer semicircle is
also equal to the inner semicircles where Sin= Sout .
Part 3
In semicircle ABC(the shaded region), and the two semicircles
which is AEB and BFC, the area of the shaded region semicircleADC is written as in Area of shaded region ADC =Area of ADC
(Area of AEB + Area of BFC). When we plot a straight link graph
based on linear law, we may still obtained a linear graph because
Sin= Sout where the diameter has a constant value for a
semicircle.
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SMK TUN ISMAIL
Circles
AdditionalMathematics
Project Work 2009
Name : Hayyan Umairah BintiAmat Pejor
Class : 5 Arif
No.I/C : 920311-01-6432
Teachers Name : Faridah Bt. Md. Yatim
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CONTENT
NO. CONTENTS PAGE
1. Acknowledgement 1
2. Objective 2
3. Introduction 3 4
4. Part 1 5 105. Part 2 11 18
6. Part 3 19 27
7. Conclusion 28