Name : khalida hanim bt. Che othman

I/C no. : 920118-03-5500

Form : 5 bestari

Teacher’s name : Mrs. Hjh. Wan marina bt.

Wan omar

Project Work For Additional Mathematics 2009

Content

Acknowledgement 3

Objectives 4

Introduction 5 - 6

Part 1 7 - 14

Part of circle 7

Definition of ‘Pi’ 8 - 9

History of ‘Pi’ 9 - 14

Part 2 15 - 20

Part 3 21 - 26

Conclusion 27

Reference 28

2 | P a g e

Project Work For Additional Mathematics 2009

Acknowledgement

First of all, I would like to say Alhamdulillah, for giving me the strength and health to do this project work.

Not forgotten 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 etc. 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, Mrs. Hjh Wan Marina Wan Omar for guiding me and my friends throughout this project. We had some difficulties in doing this task, but she 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.

3 | P a g e

Project Work For Additional Mathematics 2009

Objectives

The aims of carrying out this project work are;

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.

4 | P a g e

Project Work For Additional Mathematics 2009

Introduction

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.

5 | P a g e

Project Work For Additional Mathematics 2009

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.

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.

6 | P a g e

Project Work For Additional Mathematics 2009

Part 1 There are a lot of things around us related to circles or parts of a circles. We need to play with circles in order to complete some of the problems involving circles. In this project I will use the principles of circle to make our life easier.

Wheel of bicycle Clock

Old telephone Plate of food

7 | P a g e

Project Work For Additional Mathematics 2009

Paper cup

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

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

8 | P a g e

Project Work For Additional Mathematics 2009

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 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.

History

The history of π parallels the development of mathematics as a whole. Some authors divide progress into three periods: the ancient period during which π was studied geometrically, the classical era following the development of calculus in Europe around the 17th century, and the age of digital computers.

Geometrical period

That the ratio of the circumference to the diameter of a circle is the same for all circles, and that it is slightly more than 3, was known to ancient Egyptian, Babylonian, Indian and Greek geometers. The earliest known approximations date from around 1900 BC; they are 25/8 (Babylonia) and 256/81 (Egypt), both within 1% of the true value. The Indian text Shatapatha Brahmana gives π as 339/108 ≈ 3.139. The Hebrew Bible appears to suggest, in the Book of Kings, that π = 3, which is notably worse than other estimates available at the time of writing (600 BC). The interpretation of the passage is disputed, as some believe the ratio of 3:1 is of an interior circumference to an exterior diameter of a thinly walled basin, which could indeed be an accurate ratio, depending on the thickness of the walls (See: Biblical value of π).

9 | P a g e

Project Work For Additional Mathematics 2009

Archimedes (287–212 BC) was the first to estimate π rigorously. He realized that its magnitude can be bounded from below and above by inscribing circles in regular polygons and calculating the outer and inner polygons' respective perimeters:

By using the equivalent of 96-sided polygons, he proved that 223/71 < π < 22/7. Taking the average of these values yields 3.1419.

In the following centuries further development took place in India and China. Around AD 265, the Wei Kingdom mathematician Liu Hui provided a simple and rigorous iterative algorithm to calculate π to any degree of accuracy. He himself carried through the calculation to a 3072-gon and obtained an approximate value for π of 3.1416.

Later, Liu Hui invented a quick method of calculating π and obtained an approximate value of 3.1416 with only a 96-gon, by taking advantage of the fact that the difference in area of successive polygons forms a geometric series with a factor of 4.

Around 480, the Chinese mathematician Zu Chongzhi demonstrated that π ≈ 355/113, and showed that 3.1415926 < π < 3.1415927 using Liu Hui's algorithm applied to a 12288-gon. This value was the most accurate approximation of π available for the next 900 years.

Classical period

10 | P a g e

Project Work For Additional Mathematics 2009

Until the second millennium, π was known to fewer than 10 decimal digits. The next major advance in π studies came with the development of calculus, and in particular the discovery of infinite series which in principle permit calculating π to any desired accuracy by adding sufficiently many terms. Around 1400, Madhava of Sangamagrama found the first known such series:

This is now known as the Madhava–Leibniz series or Gregory-Leibniz series since it was rediscovered by James Gregory and Gottfried Leibniz in the 17th century. Unfortunately, the rate of convergence is too slow to calculate many digits in practice; about 4,000 terms must be summed to improve upon Archimedes' estimate. However, by transforming the series into

Madhava was able to calculate π as 3.14159265359, correct to 11 decimal places. The record was beaten in 1424 by the Persian mathematician, Jamshīd al-Kāshī, who determined 16 decimals of π.

The first major European contribution since Archimedes was made by the German mathematician Ludolph van Ceulen (1540–1610), who used a geometric method to compute 35 decimals of π. He was so proud of the calculation, which required the greater part of his life, that he had the digits engraved into his tombstone.

Around the same time, the methods of calculus and determination of infinite series and products for geometrical quantities began to emerge in Europe. The first such representation was the Viète's formula,

found by François Viète in 1593. Another famous result is Wallis' product,

by John Wallis in 1655. Isaac Newton himself derived a series for π and calculated 15 digits, although he later confessed: "I am ashamed to tell you to how many figures I carried these computations, having no other business at the time."

In 1706 John Machin was the first to compute 100 decimals of π, using the formula

11 | P a g e

Project Work For Additional Mathematics 2009

with

Formulas of this type, now known as Machin-like formulas, were used to set several successive records and remained the best known method for calculating π well into the age of computers. A remarkable record was set by the calculating prodigy Zacharias Dase, who in 1844 employed a Machin-like formula to calculate 200 decimals of π in his head at the behest of Gauss. The best value at the end of the 19th century was due to William Shanks, who took 15 years to calculate π with 707 digits, although due to a mistake only the first 527 were correct. (To avoid such errors, modern record calculations of any kind are often performed twice, with two different formulas. If the results are the same, they are likely to be correct.)

Theoretical advances in the 18th century led to insights about π's nature that could not be achieved through numerical calculation alone. Johann Heinrich Lambert proved the irrationality of π in 1761, and Adrien-Marie Legendre also proved in 1794 π2 to be irrational. When Leonhard Euler in 1735 solved the famous Basel problem – finding the exact value of

which is π2/6, he established a deep connection between π and the prime numbers. Both Legendre and Leonhard Euler speculated that π might be transcendental, which was finally proved in 1882 by Ferdinand von Lindemann.

William Jones' book A New Introduction to Mathematics from 1706 is said to be the first use of the Greek letter π for this constant, but the notation became particularly popular after Leonhard Euler adopted it in 1737. He wrote:

There are various other ways of finding the Lengths or Areas of particular Curve Lines, or Planes, which may very much facilitate the Practice; as for instance, in the Circle, the Diameter is to the Circumference as 1 to (16/5 − 4/239) − 1/3(16/53 − 4/2393) + ... = 3.14159... = π}}

See also: history of mathematical notation

Computation in the computer age

The advent of digital computers in the 20th century led to an increased rate of new π calculation records. John von Neumann et. al. used ENIAC to compute 2037 digits of π in 1949, a calculation that took 70 hours. [33][34] Additional thousands of decimal places were obtained in the following decades, with the million-digit milestone passed in 1973. Progress was not only due to

12 | P a g e

Project Work For Additional Mathematics 2009

faster hardware, but also new algorithms. One of the most significant developments was the discovery of the fast Fourier transform (FFT) in the 1960s, which allows computers to perform arithmetic on extremely large numbers quickly.

In the beginning of the 20th century, the Indian mathematician Srinivasa Ramanujan found many new formulas for π, some remarkable for their elegance and mathematical depth. One of his formulas is the series,

and the related one found by the Chudnovsky brothers in 1987,

which deliver 14 digits per term. The Chudnovskys used this formula to set several π computing records in the end of the 1980s, including the first calculation of over one billion (1,011,196,691) decimals in 1989. It remains the formula of choice for π calculating software that runs on personal computers, as opposed to the supercomputers used to set modern records.

Whereas series typically increase the accuracy with a fixed amount for each added term, there exist iterative algorithms that multiply the number of correct digits at each step, with the downside that each step generally requires an expensive calculation. A breakthrough was made in 1975, when Richard Brent and Eugene Salamin independently discovered the Brent–Salamin algorithm, which uses only arithmetic to double the number of correct digits at each step. The algorithm consists of setting

and iterating

until an and bn are close enough. Then the estimate for π is given by

13 | P a g e

Project Work For Additional Mathematics 2009

Using this scheme, 25 iterations suffice to reach 45 million correct decimals. A similar algorithm that quadruples the accuracy in each step has been found by Jonathan and Peter Borwein. The methods have been used by Yasumasa Kanada and team to set most of the π calculation records since 1980, up to a calculation of 206,158,430,000 decimals of π in 1999. The current record is 1,241,100,000,000 decimals, set by Kanada and team in 2002. Although most of Kanada's previous records were set using the Brent-Salamin algorithm, the 2002 calculation made use of two Machin-like formulas that were slower but crucially reduced memory consumption. The calculation was performed on a 64-node Hitachi supercomputer with 1 terabyte of main memory, capable of carrying out 2 trillion operations per second.

An important recent development was the Bailey–Borwein–Plouffe formula (BBP formula), discovered by Simon Plouffe and named after the authors of the paper in which the formula was first published, David H. Bailey, Peter Borwein, and Plouffe. The formula,

is remarkable because it allows extracting any individual hexadecimal or binary digit of π without calculating all the preceding ones. Between 1998 and 2000, the distributed computing project PiHex used a modification of the BBP formula due to Fabrice Bellard to compute the quadrillionth (1,000,000,000,000,000:th) bit of π, which turned out to be 0.

In 2006, Simon Plouffe, using the integer relation algorithm PSLQ, found a series of beautiful formulas. Let q = eπ, then

and others of form,

where q = eπ, k is an odd number, and a, b, c are rational numbers. If k is of the form 4m + 3, then the formula has the particularly simple form,

14 | P a g e

Project Work For Additional Mathematics 2009

for some rational number p where the denominator is a highly factorable number, though no rigorous proof has yet been given.

Part 2

(a) Diagram 1 shows a semicircle PQR of diameter 10. Semicircles PAB and BCR of diameter d1

and d2 repectively are inscribed in the semicircle PQR such that the sum of d1 and d2 is equal to 10 cm.

Complete Table 1 by using various values of d1 and the corresponding values of d2. Hence, determine the relationship between the lengths of arcs PQR, PAB and BCR.

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 0.5 π 4.5 π

2 8 5 π 1.0 π 4.0 π

3 7 5 π 1.5 π 3.5 π

15 | P a g e

Project Work For Additional Mathematics 2009

4 6 5 π 2.0 π 3.0 π

5 5 5 π 2.5 π 2.5 π

6 4 5 π 3.0 π 2.0 π

7 3 5 π 3.5 π 1.5 π

8 2 5 π 4.0 π 1.0 π

9 1 5 π 4.5 π 0.5 π

From the Table 1 we know that the length of arc PQR is not affected by the different in d1

and d2 in PAB and BCR respectively. The relation between the length of arcs PQR , PAB

and BCR is that the length of arc PQR is equal to the sum of the length of arcs PAB and

BCR, which is we can get the equation:

SPQR = SPAB + SBCR

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

5π =

12 π(3) +

12 π(7)

5π =

32 π +

72 π

5π =

102 π

5π = 5 π

16 | P a g e

Project Work For Additional Mathematics 2009

(b) Diagram 2 shows a semicircle PQR of diameter 10 cm. 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.

(i) Using various values of d1 and d2 and the corresponding values of d3, determine the relation between the lengths of arcs PQR, PAB, BCD and DER. Tabulate your

findings.

d1(cm) d2(cm) d3(cm)

Length ofarc PQR interms of π

(cm)

Length ofarc PAB interms of π

(cm)

Length ofarc BCD interms of π

(cm)

Length ofarc DER interms of π

(cm)1 1 8 5 π 0.5 π 0.5 π 4.0 π1 2 7 5 π 0.5 π 1.0 π 3.5 π1 3 6 5 π 0.5 π 1.5 π 3.0 π1 4 5 5 π 0.5 π 2.0 π 2.5 π

17 | P a g e

Project Work For Additional Mathematics 2009

1 5 4 5 π 0.5 π 2.5 π 2.0 π1 6 3 5 π 0.5 π 3.0 π 1.5 π1 7 2 5 π 0.5 π 3.5 π 1.0 π1 8 1 5 π 0.5 π 4.0 π 0.5 π2 1 7 5 π 1.0 π 0.5 π 3.5 π2 2 6 5 π 1.0 π 1.0 π 3.0 π2 3 5 5 π 1.0 π 1.5 π 2.5 π2 4 4 5 π 1.0 π 2.0 π 2.0 π2 5 3 5 π 1.0 π 2.5 π 1.5 π2 6 2 5 π 1.0 π 3.0 π 1.0 π2 7 1 5 π 1.0 π 3.5 π 0.5 π3 1 6 5 π 1.5 π 0.5 π 3.0 π3 2 5 5 π 1.5 π 1.0 π 2.5 π3 3 4 5 π 1.5 π 1.5 π 2.0 π3 4 3 5 π 1.5 π 2.0 π 1.5 π3 5 2 5 π 1.5 π 2.5 π 1.0 π3 6 1 5 π 1.5 π 3.0 π 0.5 π4 1 5 5 π 2.0 π 0.5 π 2.5 π4 2 4 5 π 2.0 π 1.0 π 2.0 π4 3 3 5 π 2.0 π 1.5 π 1.5 π4 4 2 5 π 2.0 π 2.0 π 1.0

4 5 1 5 π 2.0 π 2.5 π 0.5 π5 1 4 5 π 2.5 π 0.5 π 2.0 π5 2 3 5 π 2.5 π 1.0 π 1.5 π5 3 2 5 π 2.5 π 1.5 π 1.0 π5 4 1 5 π 2.5 π 2.0 π 0.5 π6 1 3 5 π 3.0 π 0.5 π 1.5 π6 2 2 5 π 3.0 π 1.0 π 1.0 π6 3 1 5 π 3.0 π 1.5 π 0.5 π7 1 2 5 π 3.5 π 0.5 π 1.0 π7 2 1 5 π 3.5 π 1.0 π 0.5 π8 1 1 5 π 4.0 π 0.5 π 0.5 π

SPQR = SPAB + SBCD + SDER

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

18 | P a g e

Project Work For Additional Mathematics 2009

5 π = π +

52 π +

32 π

5 π = 5 π

(ii) Based on your findings in (a) and (b), make generalization about the length of the arc of the outer semicircle and the lengths of arcs of the inner semicircles for n inner

semicircles where n = 2, 3, 4, …….

Souter = S1 + S2 + S3 + S4 + S5

(c) For different values of diameters of the outer semicircle, show that the generalization stated in b(ii) is still true.

The length of arc of the outer semicircle

= s

12 (2πr) =

12 (

2 πd2 ) =

πd2

The sum of the length of arc of the inner semicircles

Factorise

π2

Sin =

π2 (d1 + d2 + ……….. dn)

Substitute

d1 + d2 + ……… + dn = d

We get,

Sin =

π2 (d) =

πd2

Where d is any positive real number,

We can see that,

Sin = Sout

19 | P a g e

Project Work For Additional Mathematics 2009

Examples

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 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, SABC= 5π + 4π+ 3π+ 3π

15 π = 5π+ 4π+ 3π+ 3π

The diameter of the outer semicircle, d = d1 + d2 + …… + dn

10cm = 1cm + 1cm + 8cm

The length of arc of the outer semicircle, d1 + d2 + d3

0.5π + 0.5π + 4.0π = 5π

The sum of the length of arcs of the inner semicircles

Factorise π/2

(1cm + 1cm + 8cm) = 5π

In conclusion, we can conclude that,

20 | P a g e

Project Work For Additional Mathematics 2009

The length of the arc of the outer semicircle is equal to the sum of the length ofarcs of any number of the inner semicircles. This is true for any value of the diameter of the semicircle.

In other words, for different values of diameters of the outer semicircle, the generalisations stated in b (ii) is still true.

Part 3The 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.

(a) The area of the flower pot is y m2 and the diameter of one of the fish ponds is x m. Express y in terms of π and x.

Area of ADC

=

12

π (10

2 )2

=

252

π

Area of AEB

=

12

π ( x2

2 )

21 | P a g e

Project Work For Additional Mathematics 2009

=

12

π ( x2

4 )=

x2

8π

Area of BFC

=

12

π ( x2 )

2

=

12

π ( x2

4 )=

x2

8π

Area of the shaded region=Area of ADC – (Area of AEB + Area of BFC)

=

252

π−[ x2

8π+(25

2π−5 x

2π+ x2

8π )]

=

252

x−( x2

8π+25

2π−5x

2π+ x2

8π )

=

252

π− x2

8π−25

2π+ 5 x

2π− x2

8π

=− x2

4π+ 5 x

2π

(b) Find the diameters of two fish ponds if the area of the flower pot is 16.5m2. ( Use π=22

7 )Given y = 16.5

16 . 5=− x2

4π+ 5 x

2π

16 . 5=− x2

4π+ 5 x

2π

22 | P a g e

Project Work For Additional Mathematics 2009

16 . 5=− x2

4 (227 )+ 5 x

2 (227 )

To eliminate the π, divide all the terms by π.

16 . 5227

=− x2

4+5 x

2

5 .25=5x2

− x2

421=10 x−x2

x2−10 x+21=0

Factorise the equation to get the value of x.( x – 7 )( x – 3) = 0x = 7 or x = 3

∴x = 7 cm or x = 3 cm

(c) Reduce the non-linear equation obtained in (a) to simple linear form hence, plot a staright line graph. Using the straight line graph, determine the area of the flower pot if the diameter of one of the fish ponds is 4.5 m.

Linear Law

y= x2

4π+ 5 x

2π

Change it to linear form of Y = mX + Cyx=−π

x4+ 5 π

2

Y =

yx

X = x

m = −n

4

C =

5 π2

Thus, a graph of

yx against x was plotted and the line of best fit was drawn.

23 | P a g e

2.0

3.0

4.0

5.0

6.0

7.0

8.0 Y/x

2.0

3.0

4.0

5.0

6.0

7.0

8.0 Y/x

2.0

3.0

4.0

5.0

6.0

7.0

8.0 Y/x

Project Work For Additional Mathematics 2009

X = x 1.0cm 1.5 cm 2.0 cm 2.5 cm 3.0 cm 3.5 cm 4.0 cm 4.5 cm 5.0 cm

Y =

yx

7.069 6.676 6.283 5.890 5.498 5.105 4.712 4.320 3.927

Find the value of

yx when x = 4.5 m.

Then multiply

yx you get with 4.5 to get the actual value of y.

24 | P a g e

2.0

3.0

4.0

5.0

6.0

7.0

8.0 Y/x

2.0

3.0

4.0

5.0

6.0

7.0

8.0 Y/x

2.0

3.0

4.0

5.0

6.0

7.0

8.0 Y/x

Project Work For Additional Mathematics 2009

From the graph above, when the diameter of one of the fish pond is 4.5 m, the value of 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 m2

(d) The cost of constructing the fish ponds is higher than that of the flower pot. Use two methods to determine the area of the flower pot such that the cost of constructing the garden is minimum.

Method 1: Differentiation

y=− x2

4π+ 5 x

2π

dydx

=−πx2+ 5

2π

d2 ydx2

=−π2 y has maximum value

At maximum point,

d2 ydx2

=0

−πx2+ 5

2π=0

πx2=5

2π

x = 5cm

maximum value of y = − 52

4 π+

5 (5 )2

π

= 6.25 π m2

Method 2 : Completing The Square

25 | P a g e

Project Work For Additional Mathematics 2009

y=− x2

4π+ 5 x

2π

=− π4

( x2−10 x )

=− π4

( x2−10 x+25−25 )

=− π4

[ ( x−5 )2−25 ]

=− π4

( x−5 )2+25π4

y is a intersect shape graph as a=−π

4Hence, it has a maximum value.When x = 5m, maximum value of the graph = 6.25π m2

(e) The principal suggested an additional of 12 semicircular flower beds to design submitted by the Mathematics Society as shown in Diagram 4. The sum of the diameters of the semicircular flower beds is 10 cm.

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

S12 = ( n

2 ) [2 a+(n−1 ) d ]

10=(122 ) [2 (0 . 3 )+(12−1 ) d ]

= 6 (0 . 6+11d ) = 3.6 + 66d66d = 6.4

26 | P a g e

Project Work For Additional Mathematics 2009

d =

16165

Since the first flower bed is 0.3m,Hence the diameters of remaining 11 flower beds expressed in arithmeticprogression are:

131330

m ,163330

m,195330

m ,227330

m,259330

m ,291330

m,323330

m ,355330

m ,387330

m,419330

m,451330

m

ConclusionPart 1Not 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(π), we accept 3.142 or 227 as the best value of pi. The

circumference of the circle is proportional as pi(π) x diameter. If the circle has twice the diameter, d of another 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 of arc BCR. The length of arc for each semicircles can be obtained as

in length of arc = 12 (2πr). As in conclusion, outer semicircle is also

equal to the inner semicircles where Sin= Sout .

27 | P a g e

Project Work For Additional Mathematics 2009

Part 3In semicircle ABC(the shaded region), and the two semicircles which is AEB and BFC, the area of the shaded region semicircle ADC 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.

Reference

i. http://en.wikipedia.org/wiki/Piii. http://www.one-school.net/iii. Additional Mathematics Text Book Form 4 & 5

28 | P a g e