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-Content-

No. Contents Page

1 Introduction 3 - 42 Objective 5

3 Acknowledge 6

4 Part 1 7 - 12

5 Part 2 13 - 18

7 Part 3 19- 23

8 References 24

-Introduction-

A circle is a simple shape ofEuclidean geometry consisting of those points in a plane which

are the same distance from a given point called thecentre. The common distance of the points of

a circle from its center is called its radius. A diameteris a line segment whose endpoints lie on

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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 apolygon because it has no sides orvertices.

Circles are simple closed curves which divide the plane into two regions, an interiorand an

exterior. In everyday use the term "circle" may be used interchangeably to refer to either the

boundary of the figure (known as theperimeter) 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 asgears, makes much of modern civilization

possible. In mathematics, the study of the circle has helped inspire the development of geometry

and calculus.

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 .[1]

300 BC Book 3 ofEuclid's Elements deals with the properties of circles.

1880 Lindemann proves that is transcendental, effectively settling the millennia-old

problem ofsquaring the circle.[2]

http://en.wikipedia.org/wiki/Polygonhttp://en.wikipedia.org/wiki/Vertex_(geometry)http://en.wikipedia.org/wiki/Curvehttp://en.wikipedia.org/wiki/Plane_(mathematics)http://en.wikipedia.org/wiki/Interior_(topology)http://en.wikipedia.org/wiki/Perimeterhttp://en.wikipedia.org/wiki/Disk_(mathematics)http://en.wikipedia.org/wiki/Circumferencehttp://en.wikipedia.org/wiki/Ellipsehttp://en.wikipedia.org/wiki/Focus_(geometry)http://en.wikipedia.org/wiki/Conic_sectionhttp://en.wikipedia.org/wiki/Conical_surfacehttp://en.wikipedia.org/wiki/Wheelhttp://en.wikipedia.org/wiki/Wheelhttp://en.wikipedia.org/wiki/Gearhttp://en.wikipedia.org/wiki/Gearhttp://en.wikipedia.org/wiki/Sciencehttp://en.wikipedia.org/wiki/Sciencehttp://en.wikipedia.org/wiki/Geometryhttp://en.wikipedia.org/wiki/Astrology_and_astronomyhttp://en.wikipedia.org/wiki/Astrology_and_astronomyhttp://en.wikipedia.org/wiki/History_of_science_in_the_Middle_Ageshttp://en.wikipedia.org/wiki/Rhind_papyrushttp://en.wikipedia.org/wiki/Circle#cite_note-0http://en.wikipedia.org/wiki/Euclid's_Elementshttp://en.wikipedia.org/wiki/Ferdinand_von_Lindemannhttp://en.wikipedia.org/wiki/Squaring_the_circlehttp://en.wikipedia.org/wiki/Circle#cite_note-1http://en.wikipedia.org/wiki/Polygonhttp://en.wikipedia.org/wiki/Vertex_(geometry)http://en.wikipedia.org/wiki/Curvehttp://en.wikipedia.org/wiki/Plane_(mathematics)http://en.wikipedia.org/wiki/Interior_(topology)http://en.wikipedia.org/wiki/Perimeterhttp://en.wikipedia.org/wiki/Disk_(mathematics)http://en.wikipedia.org/wiki/Circumferencehttp://en.wikipedia.org/wiki/Ellipsehttp://en.wikipedia.org/wiki/Focus_(geometry)http://en.wikipedia.org/wiki/Conic_sectionhttp://en.wikipedia.org/wiki/Conical_surfacehttp://en.wikipedia.org/wiki/Wheelhttp://en.wikipedia.org/wiki/Gearhttp://en.wikipedia.org/wiki/Sciencehttp://en.wikipedia.org/wiki/Geometryhttp://en.wikipedia.org/wiki/Astrology_and_astronomyhttp://en.wikipedia.org/wiki/History_of_science_in_the_Middle_Ageshttp://en.wikipedia.org/wiki/Rhind_papyrushttp://en.wikipedia.org/wiki/Circle#cite_note-0http://en.wikipedia.org/wiki/Euclid's_Elementshttp://en.wikipedia.org/wiki/Ferdinand_von_Lindemannhttp://en.wikipedia.org/wiki/Squaring_the_circlehttp://en.wikipedia.org/wiki/Circle#cite_note-1

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Objectives

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The aims of carrying out this project work are:

1. To apply and adapt a variety of problem-solving strategies to solve

problems;

2. To improve thinking skills;

3. To promote effective mathematical communication;

4. To develop mathematical knowledge through problem solving in a way

that increase students interest and confidence;

5. To use the language of mathematics to express mathematical ideas

precisely;

6. To provide learning environment that stimulates and enhances effective

learning;

7. To develop positive attitude toward mathematics.

Acknowledgement

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First of all I would like to say Alhamdulillah for giving the strength and

health to do this project work.

Not forgotten my parents for providing me everything such as money to

buy anything that are related to this project work and their advice which is the

most needed for this project. Internet , books , computers and all that . 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 Asmalia Jaafar for guiding

me and my friends throughout this project. We had some difficulties in doing this

task , but she taught us patiently until we know 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

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PART 1

SYMBOLS

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Wheel of a bicycle Circles on water surface School park

Fish pond Round table at school compound

Before I continue the task, first, we do have to know what dopi() related to a circle.

Definition

In Euclidean plane geometry, is defined as the ratio of a circle's circumference to its

http://en.wikipedia.org/wiki/Euclidean_geometryhttp://en.wikipedia.org/wiki/Euclidean_geometryhttp://en.wikipedia.org/wiki/Ratiohttp://en.wikipedia.org/wiki/Circlehttp://en.wikipedia.org/wiki/Circumferencehttp://en.wikipedia.org/wiki/Euclidean_geometryhttp://en.wikipedia.org/wiki/Ratiohttp://en.wikipedia.org/wiki/Circlehttp://en.wikipedia.org/wiki/Circumference

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diameter:

The ratioC

/d is 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 ratio C/d.

Area of the circle = area of the shaded square

Alternatively 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:[3][5]

These definitions depend on results of Euclidean geometry, such

as the fact that all circles aresimilar. This can be considered a

problem when occurs in areas of mathematics that otherwise do

not involve geometry. For this reason, mathematicians often prefer

to define without reference to geometry, instead selecting one of

itsanalytic properties as a definition. A common choice is to

define as twice the smallest positivex for which cos(x) = 0.[6] The formulas below illustrate

other (equivalent) definitions.

History

The ancient Babylonians calculated the area of a circle by taking 3 times the square of its

radius, which gave a value ofpi = 3. One Babylonian tablet (ca. 19001680 BC) indicates a

value of 3.125 forpi, which is a closer approximation.

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P

B

C

R

Q

A

d1 d2

10 cm

PART 2

Part 2 (a)

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Diagram 1 shows a semicirclePQR of diameter 10cm. SemicirclesPAB andBCR of diameter d1

and d2 respectively are inscribed inPQR such that the sum of d1 and d2 is equal to 10cm. By

using various values of d1 and corresponding values of d2, I determine the relation between

length of arcPQR,PAB, andBCR.

Using formula: Arc of semicircle = d

d1

(cm)

d2

(cm)

Length of arcPQR in

terms of (cm)

Length of arcPAB in

terms of (cm)

Length of arcBCR in

terms of (cm)

1 9 5 9/2

2 8 5 4

3 7 5 3/2 7/2

4 6 5 2 3

5 5 5 5/2 5/2

6 4 5 3 2

7 3 5 7/2 3/2

8 2 5 4

9 1 5 9/2

Table 1

From the Table 1 we know that the length of arcPQR is not affected by the different in d1 and d2

inPAB andBCR respectively. The relation between the length of arcsPQR ,PAB andBCR is

that the length of arcPQR is equal to the sum of the length of arcsPAB andBCR, which is we

can get the equation:

SPQR = SPAB + SBCR

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P R

Q

A

C

d1 d2

10

cm

Bd3

D

E

Let d1= 3, and d2 =7 SPQR = SPAB + SBCR

5 = (3) + (7)

5 = 3/2 + 7/2

5 = 10/2

5 = 5

The Arc length of sector

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d1 d2 d3 SPQR SPAB SBCD SDER1 2 7 5 1/2 7/2

2 2 6 5 3

2 3 5 5 3/2 5/2

2 4 4 5 2 2

2 5 3 5 5/2 3/2

SPQR = SPAB + SBCD + SDER

Let d1 = 2, d2 = 5, d3 = 3 SPQR = SPAB + SBCD + SDER

5 = + 5/2 + 3/2

5 = 5

The length of arc of sector

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15 = 15

The Arc length of sector

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Part 3

a. Area of flower plot = y m2

y = (25/2) - (1/2(x/2)2 + 1/2((10-x )/2)2 )

= (25/2) - (1/2(x/2)2 + 1/2((100-20x+x2)/4) )

= (25/2) - (x2

/8 + ((100 - 20x + x2

)/8) )

= (25/2) - (x2 + 100 20x + x2 )/8

= (25/2) - ( 2x2 20x + 100)/8)

= (25/2) - (( x2 10x + 50)/4)

= (25/2 - (x2 - 10x + 50)/4)

y = ((10x x2)/4)

b. y = 16.5 m2

16.5 = ((10x x2)/4)

66 = (10x - x2) 22/7

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2.0

3.0

4.0

5.0

6.0

7.0

8.0

0 1 2 3 4 5 6 7

X

Y/x

66(7/22) = 10x x2

0 = x2 - 10x + 21

0 = (x-7)(x 3)

x = 7 , x = 3

c. y = ((10x x2)/4)

y/x = (10/4 - x/4)

X 1 2 3 4 5 6 7

y/x 7.1 6.3 5.5 4.7 3.9 3.1 2.4

By using linear law method.

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When x = 4.5 , y/x = 4.3

Area of flower plot = y/x * x

= 4.3 * 4.5

= 19.35m2

Thus, the area of flower plot is equal to 19.35m2

d. Differentiation method

dy/dx = ((10x-x2)/4)

= ( 10/4 2x/4)

0 = 5/2 x/2

5/2 = x/2

x = 5

Completing square method

y = ((10x x2)/4)

= 5/2 - x2/4

= -1/4 (x2 10x)

y+ 52 = -1/4 (x 5)2

y = -1/4 (x - 5)2 - 25

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x 5 = 0

x = 5

e. n = 12, a = 30cm, S12 = 1000cm

S12 = n/2 (2a + (n 1)d

1000 = 12/2 ( 2(30) + (12 1)d)

1000 = 6 ( 60 + 11d)

1000 = 360 + 66d

1000 360 = 66d

640 = 66d

d = 9.697

Thus the diameter of the flower plot is 9.697cm

Tn (flower bed) Diameter

(cm)T1 30

T2 39.697

T3 49.394

T4 59.091

T5 68.788

T6 78.485

T7 88.182

T8 97.879

T9 107.576

T10 117.273T11 126.97

T12 136.667

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REFERENCES

Source via internet :

http://en.wikipedia.org/wiki/Pi

http://ualr.edu/lasmoller/pi.html

http://egyptonline.tripod.com/history.htm

http://www.gap-system.org/~history/HistTopics/Pi_through_the_ages.html

News papers :

News Straits Times

Metro

Berita Harian

Magazines :