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Content

1)Content2)Acknowledgement3)Introduction4)Part1

-Pictures of circle or part of a circle-Definition for Pi or π-History of Pi or π

5)Part2-Relation between lengths of arcs PQR, PAB and BCR-Relation between lengths of arcs PQR, PAB BCD and DER and tabulate its

findings-Generalization-Showing of the generalization is true

6)Part3-Expression of y in terms of π and x-Diameters of two fish ponds-Reduction of non-linear equation to simple linear form-Graph-Two methods to determine the area of flower pot-Diameters of the flower beds

7)Question papers

8)Conclusion

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Acknowledgement

My name is Kumaresen Nair.I am thankful that this additional mathematics

project can be done just in time. For this, I would like to seize the opportunity toexpress my sincere gratitude for those who had been helping me during mywork.

First and foremost, I would like to say a big thank you to my additionalmathematics teacher, Mr.Yap Tian Key for giving me information about myproject work. On the other hand, I would also like to thank my dear principle,Mr.Haji Misman Bin Selamat for giving me the permission to carry outthis project.

 Also, I would like to thank my parents. They had brought me the thingsthat I needed during the project work was going on. Not only that, they alsoprovided me with the nice suggestion on my project work so that I had not meetthe dead end throughout this process.

Lastly, I would like to say thank you to my friends and the modern accessin our daily life. All of my relevant information came from my friends and theinternet. I managed to use all these access in our daily life, such as: computer tofinish my additional mathematics project.

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Definition of circle

 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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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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There are a lot of things around us related to circles or parts of a circles.

a) Pictures of 5 such objects is collected. The internet is used to get the pictures.

ring

bangle

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Clock

Mercury cell

  Pizza

We can conclude that,

There are a lot of things around us related to circles or parts of a circles

b) pi or π is a mathematical constant related to circles.π is defined and a brief history of π is stated.

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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 .

 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 thatall circles are similar . This can be considered a problem when π occurs in areas

of mathematics that otherwise do not involve geometry. For this reason,mathematicians often prefer to define π without reference to geometry, insteadselecting one of its analytic properties as a definition. A common choice is todefine π as twice the smallest positive x for which cos( x ) = 0.[6] The formulasbelow illustrate other (equivalent) definitions.

History

Pi has been known for almost 4000 years—but even if we calculated the number of seconds in those 4000 years and calculated pi to that number of places, wewould still only be approximating its actual value. Here’s a brief history of finding

 pi :

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The ancient Babylonians calculated the area of a circle by taking 3 times thesquare 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 theEgyptians calculated the area of a circle by a formula that gave the approximatevalue of 3.1605 for  pi .

The ancient cultures mentioned above found their approximations bymeasurement. 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 PythagoreanTheorem to find the areas of two regular polygons: the polygon inscribed withinthe circle and the polygon within which the circle was circumscribed . Since theactual area of the circle lies between the areas of the inscribed and

circumscribed polygons, the areas of the polygons gave upper and lower boundsfor 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 showedthat pi is between 3 1/7 and 3 10/71.

 A similar approach was used by Zu Chongzhi (429–501), a brilliant Chinesemathematician and astronomer. Zu Chongzhi would not have been familiar with

 Archimedes’ method—but because his book has been lost, little is known of hiswork. He calculated the value of the ratio of the circumference of a circle to itsdiameter to be 355/113. To compute this accuracy for  pi , he must have startedwith 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 byWilliam Jones in 1706, use of the symbol was popularized by Euler, who adoptedit in 1737.

 An 18th century French mathematician named Georges Buffon devised a way tocalculate pi based on probability. You can try it yourself at the Exploratoriumexhibit Throwing Pi .

We can conclude that, pi or π is very useful for daily life.

 

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 A semicircle PQR of diameter 10 cm

a) is showed in diagram 1.

PAB =d1

BCR=d2

d1 + d2 = 10

Strategy of saving question part 2 (a) :

I determine the relation between length of arc PQR , PAB, and BCR by using

various values of d1 and the corresponding values of d2.

Using formula: Arc of semicircle = ½πd

d1

(cm)

d2

(cm)

Length of arc PQR 

in terms of π (cm)

Length of arc PAB

in terms of π (cm)

Length of arc BCR 

in terms of π (cm)

1 9 5 π ½ π 9/2 π

2 8 5 π π 4 π

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3 7 5 π 3/2 π 7/2 π

4 6 5 π 2 π 3 π

5 5 5 π 5/2π 5/2 π

6 4 5 π 3 π 2 π

7 3 5 π 7/2 π 3/2 π

8 2 5 π 4 π π9 1 5 π 9/2 π ½ π

10 0 5 π 5 π 0

SPQR  = SPAB + SBCR 

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

  5π = ½ π(3) + ½ π(7)

5π = 3/2 π + 7/2 π

5π = 10/2 π

5π = 5 π

We can conclude that

Length of arc PQR = Length of 

arc PAB + Length of arc BCR

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b) A semicircle PQR of diameter 10 cm is showed in diagram 2 .

PAB = d1BCD = d2

DER = d3

Strategy of saving question part 2 (b) (і) :

Determine the relation between length of arc PQR , PAB, BCD and DER by using

various values of d1 and d2 and the corresponding values of d3 .

d1 d2 d3 SPQR SPAB SBCD SDER

1 2 7 5π 1/2 π π 7/2 π

2 2 6 5π π π 3 π

2 3 5 5π π 3/2 π 5/2 π

2 4 4 5π π 2 π 2 π

2 5 3 5π π 5/2 π 3/2 π

SPQR = SPAB + SBCD + SDER

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

5 π = π + 5/2 π + 3/2 π5 π = 5 π

 Again, we can conclude that:

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Length of arc PQR = Length of arc PAB + Length of arc BCD + Length of arcCDR

Strategy of saving question part 2 (b) (іі) :

Based on my finding in (a) and (b), we can generalize about the length of the arc

of the outer semicirlcle and the lengths of arcs of the inner semicircles for n inner 

semicircle for n = 1,2,3,4,….

The length of arc of outer semicircle is = to the sum of the length of arc of inner 

semicircle for n = 1,2,3,4,….

Souter  = S1 + S2 + S3 + S4 + S5

Base on the findings in the table in(a) and (b) above, we 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.

Strategy of saving question part 2 (c) :

The generalize stated in b (іі) is still rtue for different values of diameters of the

outer semicircle.

 Assume the diameter of outer semicircle is 30cm and 4 semicircles are inscribed

in the outer semicircle such that the sum of d 1(APQ), d2(QRS), d3(STU), d4(UVC)

is equal to 30cm.

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d1 d2 d3 d4 S ABC S APQ 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 π

14 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, S ABC = S APQ + SQRS + SSTU + SUVC

15 π = 5 π + 4 π + 3 π + 3 π

15 π = 15 π

 As a result, 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 thediameter of the semicircle.In other words, for different values of diameters of the outer semicircle, show thatthe generalizations stated in b (ii) is still true.

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 A task is given to the mathematics society to design a garden to beautify theschool by using the design as shown in Diagram 3. the shaded region will beplanted with flowers and the two inner semicircles are fish ponds.

Strategy of saving question part 3 (a) :

a) the area of the flower plot = y mthe diameter of one of the fish ponds = x m

 Area of flower plot = y m2

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

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= (25/2) π - (1/2(x/2)2 π + 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

= (25/2) π - (( x2 – 10x + 50)/4)

= (25/2 - (x2 - 10x + 50)/4) π

y = ((10x – x2)/4) π

We can conclude that,

The area of flower plot,y = the area of semicircles – area of fish pond, which canbe express in the form y = ((10x – x2)/4) π

Strategy of saving question part 3 (b) :

 Area of the flower plot =16.5 m

2

, find the diameter of the two fish ponds.

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

 Again we can conclude that,

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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

When the area of the flower plot, y = 16.5 m2, the diameter of the fish ponds is

7m and 3m.

Strategy of saving question part 3 (c) :

The non-linear equation obtained in (a) is reduced to simple linear form and astraight line graph is plotted. The diameter of one of the fish ponds = 4.5m,determine the area of the flower plot.

.

y = ((10x – x2)/4) π

y/x = (10/4 - x/4) π

x 1 2 3 4 5 6 7y/x 7.1 6.3 5.5 4.7 3.9 3.1 2.4

 

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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

We can conclude that ,When x = 4.5 , y/x = 4.3, the area of flower plot is19.35m2

Strategy of saving question part 3 (d) :

The cost of constructing the fish ponds is higher than that of the flower plot.Determine the area of the flower plot using two methods such that the cost of constructing the garden is minimum.

Differentiation method

dy/dx = ((10x-x2)/4) π

= ( 10/4 – 2x/4) π

0 = 5/2 π – x/2 π

5/2 π = x/2 π

x = 5

Completing square method

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y = ((10x – x2)/4) π

= 5/2 π - x2/4 π

= -1/4 π (x2 – 10x)

y+ 52 = -1/4 π (x – 5)2 

y = -1/4 π (x - 5)2 - 25

x – 5 = 0

x = 5

We can conclude that,

By using differentiation method and completing square method can minimum the

cost of constructing the fish ponds.

 An additional of 12 semicircular flower beds is suggested by the principal to thedesign submitted by the Mathematics Society as shown in diagram 4. Sum of diameter of the semicircular flower beds = 10m.

Strategy of saving question part 3 (e) :

The diameter of the smallest flower bed = 30cm and the diameter of the flower beds are increased by a constant value successively.Determine the diameter of the remaining flower beds.

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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

We can conclude that,The diameter of flower bed willincrease by 9.697.

Tn(flower bed) Diameter(cm)

T1 30

T2 39.697

T3 49.394

T4 59.091

T5 68.788T6 78.485

T7 88.182

T8 97.879

T9 107.576

T10 117.273

T11 126.97

T12 136.667

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CONCLUSION

Pi (π) is a very useful mathematics related to circle in which it helps themankind to solve many problems easily involving circle. We are able to know

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how we can use this unit to solve various problems involving objects that arecircular in shape of even part of a circle shape.

Besides, in this project work we need to use a lot of mathematical concept

in order to get the answer. This makes me understand more about other mathematical concept besides Pi (π). So, after doing this project, I am quiteimpressed with the usage of circle and its ways to help us in solving problemsalthough there are some errors occur. Besides that, I also learnt many things for this which I can never find them in the textbook or reference book or even in our school syllabus. I am doing many researches to understand its usage and itsprinciples when apply to solve the problem involving circles.

Furthermore, I am able to interpret carefully when handling such mindtwisting problem that is in Part 3. This experience that I gain from this project

work can makes me apply to other subjects so that it will make me more carefulwhen handling such question mentioned. I am really appreciating the governmentas they gave us this opportunity to do this project in the process of understandingand learning deeply into circles. I would like to thanks my additional mathematicsteacher as without his help, I would not be able to accomplish this project.