Sekolah Menengah Kebangsaan

Additional Mathematics Project2009

Title : application of pi

Name : Class : Teacher’s name :

CONTENT

1.Content………………………………………………………………………………….2

2.Preface…………………………………………………………………………………..3

3.Acknowledgement……………………………………………………………………...4

4.Introduction…………………………………………………………………………….5

5. Definition of circles ………………………………………………………………...…6

6. Part 1 Pictures of circle or part of a circle…………………………………………..…7 Definition for Pi or π ……………………………………………………………………………....8 – 9 History of Pi or π ………………………………………………………………10

7. Part 2 Relation between lengths of arcs PQR, PAB and BCR……….…………11 - 12 Relation between lengths of arcs PQR, PAB BCD and DER and its tabulate

findings……………………………………………………………………..13 - 14 Generalization…..………………………………………………………………15 Showing of the generalization is true…………………………………….15 - 16

8. Part 3 Expression of y in terms of π and x……………………………………....17 - 18 Diameters of two fish ponds…………………………………………………...19 Reduction of non-linear equation to simple linear form…………………….20 Graph……………………………………………………………………………21 Two methods to determine the area of flower pot……………….………22 - 23 Diameters of the flower beds……………………………………………...24 - 25

9. Conclusion……………………………………………………………………………26

10.Reference…………………………………………………………………………….27

PREFACE

This project is prepared by the students of class 5 Atom (2009) and it is based on the Additional Mathematics textbooks, internet search and reference book.

Additional Mathematics is an elective subject in secondary school. Each student who takes this subject has to carry out a project work on the given tasks. The project work for the year 2009 is about Circle by using the principle of Pi (π).

The aim of doing this project is to improve the skills in using mathematics for students. Working on this also gives a chance for students to apply their skills on what they had learnt to solve an assigned project. Therefore, every student stands a chance to improve their thinking skills, usage of languages and grammar as well as mathematics skills throughout the project.

After doing this project, the student will be able to master and understand more on the application of Additional Mathematics that they learnt in their school syllabus. The student can also learn some values during the completion of the project such as to learn how to work together or to be cooperative, improving their communication skills, responsibility and also not to give up easily on the task given.

Acknowledgement

My name is_____________. I am thankful that this additional mathematics project can be done just in time. For this, I would like to seize the opportunity to express my sincere gratitude for those who had been helping me during my work.

First and foremost, I would like to say a big thank you to my additional mathematics teacher, Miss Tiong Ping for giving me information about my project work. On the other hand, I would also like to thank my dear principle, Madam Fung for giving me the permission to carry out this project.

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

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

INTRODUCTION

Additional Mathematics is one of the compulsory subjects for SPM science stream candidates. All of the students would have to carry out a project work based on a topic given and must be submitted in three weeks time.

The objective of carrying out this project is to apply and adapt a variety of problem-solving strategies that we had learnt to solve the problems. Moreover, our thinking skills can be improved. It also promotes effective mathematical communication. Our confidence and interest towards mathematics will be increase through solving various types of problems.

Besides, the aim of doing this project is to use the language of mathematics to express mathematical ideas precisely. This also stimulates and enhances effective earning. During the project, we are able to develop our positive attitude towards mathematics. This makes the lesson to be more fun, useful and meaningful.

Definition of circle

In our daily life, there are lots of round substances. We call those as circles.

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.

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.

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.

Part 1(a)

There are a lot of things around us related to circles or part of circle. Here are some pictures of 5 such objects.

Diagram 1.1 : Pizza Slide (sector) Diagram 1.2 : Wheel of bicycle

Diagram 1.3 : Volleyball Diagram 1.4 : Circular Waterfronts

Diagram 1.5 : Thyco crater on the moon surface

Part 1(b) : Pi or π Story

Definition of Pi or π

Pi or π is a mathematical constant whose value is the ratio of any

circle's circumference to its diameter in Euclidean space; this is the

same value as the ratio of a circle's area to the square of its radius. It is

approximately equal to 3.14159 in the usual decimal notation. π is one of

the most important mathematical and physical constants: many

formulae from mathematics, science, and engineering involve π.

π is an irrational number, which means that its value cannot be

expressed exactly as a fraction m/n, where m and n are integers.

Consequently, its decimal representation never ends or repeats. It is also

a transcendental number, which means that no finite sequence of

algebraic operations on integers (powers, roots, sums, etc.) can be equal

to its value; proving this was a late achievement in mathematical history

and a significant result of 19th century German mathematics.

Throughout the history of mathematics, there has been much effort to

determine π more accurately and to understand its nature; fascination

with the number has even carried over into non-mathematical culture.

Circumference = π × diameter

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

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:

History of Pi or π

The ancient Babylonians calculated the area of a circle by taking 3 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.

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

Archimedes’ method—but because his book has been lost, little is known of his work. He calculated the value of the ratio of the circumference of a circle to its diameter to be 355/113. To compute this accuracy for pi, he must have started with 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 by William Jones in 1706, use of the symbol was popularized by Euler, who adopted it in 1737.

An 18th century French mathematician named Georges Buffon devised a way to calculate pi based on probability.

Conjecture

The conjecture that I made for the task given is written : The length of arc of the outer semicircle is equal to the sum of the lengths of arcs of

any number of the inner semicircles.

PB

C

R

Q

A

d1 d2

10 cm

Part 2(a)

Diagram 2.1

Diagram 2.1 shows a semicircle PQR of diameter 10cm. Semicircles PAB and BCR of diameter d1 and d2 respectively are inscribed in PQR 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 arc PQR, PAB, and BCR.

By using the 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

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 π

9 1 5 π 9/2 π ½ π

Table 2.1

Based on table 2.1 , we know that the length of arc PQR is not affected by the different in d1 and d2 in PAB and BCR respectively. The relationship 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, so we can get this equation:

SPQR = SPAB + SBCR

To prove this equation ,

(a) Let d1 = 3 and d2 = 7 , SPQR = 1/2 π(10) = 5 π SPAB + SBCR = ½ π(3) + ½ π(7)

= 3/2 π + 7/2 π = 10/2 π = 5π

= SPQR

(b) Let d1 = 4 and d2 = 6 , SPQR = 1/2 π(10) = 5 π SPAB + SBCR = ½ π(4) + ½ π(6)

= 2 π + 3 π = 5 π

P R

Q

AC

d1 d2

10 cm

B d3D

E

= SPQR

(c) Let d1 = 5 and d2 = 5 ,

SPQR = 1/2 π(10) = 5 π SPAB + SBCR = ½ π(5) + ½ π(5)

= 5/2 π + 5/2 π= 10/2 π

= 5π = SPQR

Therefore , SPAB + SBCR = SPQR

The equation is proven.

Part 2(b)(i)

Diagram 2.2

Diagram 2.2 shows a semicircle PQR of 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.

By using the formula , Arc of semicircle = ½πd.

d1

(cm)

d2

(cm)

d3

(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)

Length of arc DER in terms of π (cm)

1 2 7 5π ½ 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

Table 2.2

Based on table 2.2 , we know that the length of arc PQR is not affected by the different in d1, d2 and d3 in PAB, BCR and DER respectively. The relationship between the length of arcs PQR , PAB , DER and BCR is that the length of arc PQR is equal to the sum of the length of arcs PAB, BCR and DER , so we can get this equation:

SPQR = SPAB + SBCD + SDER

To prove that SPQR = SPAB + SBCD + SDER ,

(a) Let d1 = 2, d2 = 5 and d3 = 3

SPQR = 1/2 π(10) = 5 π SPAB + SBCD + SDER

= ½ π(2) + ½ π(5)+ ½ π(3)= π + 5/2 π + 3/2 π= 5 π = SPQR

(b) Let d1 = 2, d2 = 4 and d3 = 4

SPQR = 1/2 π(10) = 5 π

SPAB + SBCD + SDER

= ½ π(2) + ½ π(4)+ ½ π(4)= π + 2 π + 2 π= 5 π = SPQR

(c) Let d1 = 2, d2 = 2 and d3 = 6

SPQR = 1/2 π(10) = 5 π SPAB + SBCD + SDER

= ½ π(2) + ½ π(2)+ ½ π(6)= π + π + 3 π= 5 π = SPQR

Therefore , SPAB + SBCD + SDER = SPQR

The equation is proven.

Part 2(b)(ii)

Base on the findings in the table in (a) and (b) above, it can be concluded that: The length of the arc of the outer semicircle is equal to the sum of the length of arcs of any number, n of the inner semicircles.

Souter = S2 + S3 + S4 + ……. + Sn, where n = 2,3,4,...

Part 2(c)

Diagram 2.3

Diagram 2.3 shows a big semicircle with n number of small inner circle. From diagram 2.3, we can see that:

And the length of arc of the outer semicircle:

The sum of the length of arcs of the inner semicircles:

Then, substitute the equation below:

into the equation of the sum of the length of arcs of the inner semicircles.

As a result, we will get the equation as shown in below :

Now, assume the diameter of outer semicircle is 20cm and 4 semicircles are inscribed in the outer semicircle such that the sum of d1, d2, d3, d4 is equal to 20cm.

Let d1 =4, d2 =6, d3 =3, d4 =7

Sin = ½ π (4+6+3+7) Sin = ½ π (20)

As mentioned above, Sout = ½ πd Sin = ½ π (20)

where 20 is the diameter, d as stated above.

Sout = ½ πdSin = ½ πd

∴Sin = Sout

From the example above, we can see that it carried only one conclusion which is:SINNER = SOUTER

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.

(S out) = ∑ n (S in), which n = 2, 3, 4, ...... andS in represents length of arc of inner semicircle,S out represents length of arc of outer semicircle.

This is true for any value of the diameter of the semicircle. In other words, for different values of diameters of the outer semicircle, it shows that the generalisation stated in b (ii) is still true.

Part 3(a)

The Mathematics Society is given a task to design a garden to beautify the school by using the design as shown in Diagram 3.1.The shaded region will be planted with flowers and the two inner semicircles are fish ponds.

Diagram 3.1

a) The area of the flower plot is y m2 and the diameter of one of the fish

ponds is x m. Express y in terms of π and x.

Part 3(b)

Find the diameters of the two fish ponds if the arc of the flower plot is 16.5 m2 (Use π = 22/7),Given y = 16.5 cm,

Part 3(c)Reduce the non-linear equation obtained in (a) to simple linear form and hence, plot a straight line graph. Using the straight line graph, determine the area of the flower plot if the diameter of one of the fish ponds is 4.5 m. By using linear law,y=5x/2 (π) – x2/4 (π)

Change it to linear form of Y = mX + c, y/x = (-x/4 + 5/2) π

x 0 1 2 3 4 5 6 7y/x7.9 7.1 6.3 5.5 4.7 3.9 3.1 2.4

Thus, a graph of y/x against x is plotted.

From the graph, when x = 4.5, y/x = 4.3 Area of flower pot = y/x x x= 4.3 × 4.5 = 19.35 m2

Part 3(d)The cost of constructing the fish ponds is higher than that of the flower plot. Use two methods to determine the area of the flower plot such that the cost of constructing the garden is minimum. We need to get the largest value of y so that the cost of constructing the garden is minimum,

Method 1: Differentiation

y = -πx^2/4 + 5πx/2dy/dx = -πx/2 + 5π/2(d^2)y/dx^2 = -π/2 <--- y has a maximum value.

At maximum point, (d^2)y/dx^2 = 0.-πx/2 + 5π/2 = 0πx/2 = 5π/2x = 5 mx = 5 m:

maximum value of y = -π(5^2)/4 + 5π(5)/2= 6.25π m^2

Method 2: Completing the Square

y = -πx^2/4 + 5πx/2= -π/4(x^2 - 10x)= -π/4(x^2 - 10x + 25 - 25)= -π/4[(x - 5)^2 - 25]= -π/4(x - 5)^2 + 25π/4

y is a n shape graph as a = -π/4.

Hence, it has a maximum value.When x = 5 m, maximum value of the graph = 6.25π m^2

Part 3(e)

The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society as shown in Diagram 3.2. The sum of the diameters of the semicircular flower beds is 10 m.

Diagram 3.2

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.

By using arithmetic progression,

(i) The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society. (n = 12)

(ii) The sum of the diameters of the semicircular flower beds is 10 m. (S12 = 10 m = 1000 cm)

(iii) The diameter of the smallest flower bed is 30 cm. (a = 30 cm)

(iv) The diameter of the flower beds are increased by a constant value successively. (d =?)

S12 = (n/2)[2a + (n - 1)d]1000 = (12/2)[2(30)+(12-1) d]1000 =6(60 + 11d)1000 = 360 + 66d

66d = 640d = 640/66 = 9.679

Since the first flower bed is 30 cm,

Hence the diameters of remaining 11 flower beds expressed in arithmetic progression are:

Tn (flower bed), givenTn = a + (n – 1)d, where

a = 30 and d = 9.697Diameter (cm)

T1 30.00T2 39.70T3 49.39T4 59.09T5 68.79T6 78.49T7 88.18T8 97.88T9 107.6

T10 117.3T11 127.0T12 136.7

Also, the AP can be written as :

30.00 cm, 39.70 cm, 49.39 cm, 59.09 cm, 68.79 cm, 78.49 cm, 88.18 cm, 97.88 cm, 107.60 cm, 117.30 cm, 127.00 cm, 136.70 cm

Conclusion of each tasks :Task 1 :

There are a lots of thing surrounding us have to do with circles. As examples, clock, coin, ball and so on. Based on the research about pi , we can conclude that pi does with circle. There is no specific value of pi. However, the approximate value of it is 3.14159. Not only that, pi is defined as the ratio of a circle's circumference to its diameter:

Also, pi 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:

Task 2 :

Conjecture is made during the process of task 2. It stated that the length of arc of the outer semicircle is equal to the sum of the lengths of arcs of any number of the inner semicircles. The conjecture brings to a generalization , which the length of the arc of the outer semicircle is equal to the sum of the length of arcs of any number, n of the inner semicircles.

Souter = S2 + S3 + S4 + ……. + Sn, where n = 2,3,4,...

This generalization still true when the values of the diameter of the outer semicircle are different. Therefore, the conjecture made is accepted.

Task 3 :

Any method of solving quadratic equation can be used in the task given. In order to solve this, I had tried the method of differentiation and completing the square to find out the minimum cost of constructing the garden. Also, there are also problems in finding the area of the fish pond, the diameter of it and the area of the flower at its certain length. I would also determine the common difference in order to determine the diameters of the remaining flower beds by applying Arithmetic Progression.

CONCLUSION

Pi (π) is a very useful mathematics related to circle in which it helps the mankind to solve many problems easily involving circle. We are able to know how we can use this unit to solve various problems involving objects that are circular 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 quite impressed with the usage of circle and its ways to help us in solving problems although 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 its principles when apply to solve the problem involving circles.

Furthermore, I am able to interpret carefully when handling such mind twisting 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 careful when handling such question mentioned. I am really appreciating the government as they gave us this opportunity to do this project in the process of understanding and learning deeply into circles. I would like to thanks my additional mathematics teacher as without his help, I would not be able to accomplish this project.

REFERENCE

Book :

Longman Essential Additional Mathematics SPM, PEARSON MALAYSIA SDN.BHD. Lan Foo Huat, Yong Kien Cheng

Websites :

1) http://en.wikipedia.org/wiki/Pi 2) http://images.google.com/ 3) http://en.wikipedia.org/wiki/Circle