new add maths project work

8/9/2019 New Add Maths Project Work

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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

new add maths project work

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