Project Work Addmaths

A d d i t i o n a l M a t h e m a t i c s | 1

ACKNOWLEDGEMENT

In the name of Allah, the most Compassionate and the most Merciful. Alhamdulillah,

from the bless of Allah, the most gracious, I had done the Additional Mathematics’s project

work as an SPM assignment and task. Praise and gratitude I like to humbly express to Allah

S.W.T our Merciful Creator, Who gave me strength, time and ability to finish this project.

Hopefully, this project will help me to increase my interest and confidence through Additional

Mathematic and bring a booster for me to get A1 for my SPM. A very special thanks to my

beloved Additional Mathematic teacher, Puan Saadiah bt Abu Bakar for her support, brilliant

ideas, and also her encouragements in completing this project. Her committed and professional

skills help me to complete this project work accordingly and successfully. Then, I would like to

dedicate my sincere thanks towards my beloved family, especially my father, Prof. Dr. Husin

bin Wagiran, my mother, Pn. Norsiah Osman and also my sister, Dr. Noor Huzaimah bt. Husin

for the great support and contribution. To my beloved friends who will sit for the SPM this year

together, your support and help in regarding ideas are very meaningfull. Thanks a lot. Lastly,

thank you very much to everybody that involved in this project. May god bless all of you. Thank

you.


CONTENT

BIL TOPICS PAGES1 Introduction 32 Part 1 – General Aspect

Picture related to circles and semicircles

History of Pi

4

3 Part 2 – Task Specification Design of garden Conjecture

8

4 Part 3 – Problem Solving Table and generalisations Problem solving

10

5 Part 4 - Conclusion 23


INTRODUCTION

Every student taking Additional Mathematics for SPM is required to carry out a project work in

form 5. For this year, the SPM candidates need to implement a project work related to circles

and part of a circle.

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 student’ 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.


Part 1GENERAL ASPECT

Pictures related to circles and semicircles

History of Pi

(a) There are a lot of things around us related to circles or part of a circle. For example :


Double sided tape Car’s tyre Plate

Alarm clock spectacle Parabola mirror

Eye on Malaysia buttons Power switch

b) (i) Pi or is a mathematical constant related to circles.

http://rds.yahoo.com/_ylt=A0S020uWhPpJt2sAI1eJzbkF;_ylu=X3oDMTBpZTByOGFiBHBvcwMyBHNlYwNzcgR2dGlkAw--/SIG=1ive6ajqt/EXP=1241241110/**http%3A/images.search.yahoo.com/images/view%3Fback=http%253A%252F%252Fimages.search.yahoo.com%252Fsearch%252Fimages%253Fp%253Dcircle%252Bswitch%2526ei%253D%2526iscqry%253D%2526fr%253D%26w=381%26h=381%26imgurl=static.flickr.com%252F3%252F4093568_d066985f80.jpg%26rurl=http%253A%252F%252Fwww.flickr.com%252Fphotos%252Feecue%252F4093568%252F%26size=30k%26name=apple%2Bpower%2Bswit...%26p=circle%2Bswitch%26oid=5231ca7207c86cba%26fusr=eecue%26hurl=http%253A%252F%252Fwww.flickr.com%252Fphotos%252Feecue%252F%26no=2%26tt=2733%26sigr=11bloo1v4%26sigi=11a2tv12e%26sigb=12c123mg9%26sigh=1138lmvgk


By definition, pi is the ratio of the circumference of a circle to its diameter. Pi is always the

same number, no matter which circle you use to compute it. For the sake of usefulness people

often need to approximate pi. For many purposes you can use 3.14159, which is really pretty

good, but if you want a better approximation you can use a computer to get it. Here's pi to

many more digits: 3.14159265358979323846. The area of a circle is pi times the square of the

length of the radius, or "pi r squared":

A = r2

(ii) History of Pi,

Pi is a very old number. We know that the Egyptians and the Babylonians knew about the

existence of the constant ratio pi, although they didn't know its value nearly as well as we do

today. They had figured out that it was a little bigger than 3; the Babylonians had an

approximation of 3 1/8 (3.125), and the Egyptians had a somewhat worse approximation of

4*(8/9)^2 (about 3.160484), which is slightly less accurate and much harder to work with. For

more, see A History of Pi by Petr Beckman (Dorset Press). The modern symbol for pi [ ] was

first used in our modern sense in 1706 by William Jones, who 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(16/5^3 - 4/239^3) + ... = 3.14159... = (see

A History of Mathematical Notation by Florian Cajori). Pi (rather than some other Greek letter

like Alpha or Omega) was chosen as the letter to represent the number 3.141592... because

the letter [ ] in Greek, pronounced like our letter 'p', stands for 'perimeter'.

(iii) About Pi


Pi is an infinite decimal. Unlike numbers such as 3, 9.876, and 4.5, which have finitely many

nonzero numbers to the right of the decimal place, pi has infinitely many numbers to the right of

the decimal point.

If you write pi down in decimal form, the numbers to the right of the 0 never repeat in a pattern.

Some infinite decimals do have patterns - for instance, the infinite decimal .3333333... has all

3's to the right of the decimal point, and in the number .123456789123456789123456789... the

sequence 123456789 is repeated. However, although many mathematicians have tried to find

it, no repeating pattern for pi has been discovered - in fact, in 1768 Johann Lambert proved that

there cannot be any such repeating pattern.

As a number that cannot be written as a repeating decimal or a finite decimal (you can never

get to the end of it) pi is irrational: it cannot be written as a fraction (the ratio of two integers).


Part 2TASK

SPECIFICATION Design of garden

Conjecture

This year, all SPM candidates whose taking additional mathematics is required to carry out a

project to design a beautiful garden. Basically, this garden is using the concept based on

semicircle shape. It contents of given several area of flower plot and fish ponds. The aspiration


of this garden is to get a combination of flora and fauna blended with soft elements and hard

elements such as water, flowers, rocks and marbles and to get an amazing garden with a

minimal cost.

Conjecture of the garden

The plan is designed as shown in diagram (i). The garden is predicted as a semicircle PQR

of 10cm diameter and semicircles PAB and BCR of several diameters in the semicircle

PQR as a fish ponds such that the sum of PB and BR is equal to 10cm. The adjustment of

the measurement is generalized approximately the sum of arc PAB and BCR is equal to

arc PQR. The generalisation is then wrote as :

PAB + BCR = PQR

A

B

C

P R10 cm

Q

Diagram (i)


Part 3PROBLEM SOLVING

Tables and generalisations

Problem solving

(a)


Solution :

Using the formula ,

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 4.52 8 5 1.0 4.03 7 5 1.5 3.54 6 5 2.0 3.05 5 5 2.5 2.56 4 5 3.0 2.07 3 5 3.5 1.58 2 5 4.0 1.09 1 5 4.5 0.5

We can see that whatever the diameter it is, show that the sum of length of arc PAB and length of arc BCR is equal to the length of arc PQR.

The relation between the length of arcs PQR, PAB, and BCR is

Length of arc PAB + length of arc BCR = length of arc PQR

(b)(i)

A

C

Q

P B Rd1 d210cm

Q

E


Solution :

Using the formula ,

d1

(cm)d2

(cm)d3

(cm)Length of arc PQR in terms

of (cm)


(cm)

Length of arc BCD in terms of

(cm)

Length of arc DER in terms of

(cm)1 2 7 5 0.5 1.0 3.52 2 6 5 1.0 1.0 3.03 2 5 5 1.5 1.0 2.54 2 4 5 2.0 1.0 2.05 2 3 5 2.5 1.0 1.56 2 2 5 3.0 1.0 1.07 2 1 5 3.5 1.0 0.58 1 1 5 4.0 0.5 0.51 1 8 5 0.5 0.5 4.02 1 7 5 1.0 0.5 3.52 2 6 5 1.0 1.0 3.02 3 5 5 1.0 1.5 2.52 4 4 5 1.0 2.0 2.02 5 3 5 1.0 2.5 1.52 6 2 5 1.0 3.0 1.02 7 1 5 1.0 3.5 0.51 8 1 5 0.5 4.0 0.5

The relation between the length of arcs PQR, PAB, BCD and DER is

Length of arc PQR = Length of arc PAB + length of arc BCD + length of arc DER

Based on the finding in (a) and (b), the generalizations about the length of arc of the outer semicircle and the length of arcs of the inner semicircles is

Length of arc0 = n length of arci , where n = 2,3,4...

(c)

10cm

AC

P D Rd1 d2B d3


Solution :

Using the formula ,

d1 (cm) d2 (cm)Length of arc

PQR in terms of (cm)


(cm)

Length of arc BCR in terms of

(cm)1 11 6 0.5 5.52 10 6 1.0 5.03 9 6 1.5 4.54 8 6 2.0 4.05 7 6 2.5 3.56 6 6 3.0 3.07 5 6 3.5 2.58 4 6 4.0 2.09 3 6 4.5 1.510 2 6 5.0 1.011 1 6 5.5 0.5

It is shown that the sum of the length of arc of the inner semicircle is equal to the length of arc of the outer semicircle.

Length of arc PQR = Length of arc PAB + length of arc BCR

It is show that the generalizations stated in b(ii) is still true.

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

A

C

Q

P B Rd1 d212cm

E

F

D

A B C

Flower plot

Fish ponds

10 mχ


(a) Solution :

y =

2y =

8y =

=

=

4y =

y =


(b) Solution :y = 16.5 m2

16.5 =

66 =

462 =

21 =

= 0

= 0

3 , 7

The diameter of the two fish ponds are 3m and 7m if the area of the flower plot is 16.5m2

(c)

y =


4y =

28y =

14y =

y =

Graph for non-linear equation :

Form a linear equation by using non-linear equation above,

y = ---------(1)

(1) : =

Graph is plotted using the following table :

Graph y/x against x is plotted by using scale of 1cm to 1 unit on the x-axis and 2cm to 1 unit on the y-axis

x 1 2 3 4 5 6 7 8 9y 7.07 12.6 16.5 18.9 19.6 18.9 16.5 12.6 7.07

x 1 2 3 4 5 6 7 8 9y/x 7.07 6.30 5.50 4.73 3.92 3.15 2.36 1.58 0.79


From the straight line graph, show that when the diameter of one of the fish ponds is 4.5m, the

area of the flower plot is 4.3 m.

Therefore, the area of the flower plot is

8

7

98765432

1

1

2

3

4

5

6

4.3


=

= 4.5

y = 19.35m2

Refer to the graph, the area of the flower plot is 19. 35m2.

(d) The cost of constructing the fish ponds is higher than that of the flower plot. Therefore, we

can determine the area of the flower plot such that the cost of constructing the garden is

minimum by using differentiation and completing the square method.


i) Using differentiation :

y =

Let : = = 14

= = 0

=

=

y is minimum when = 0,

=

0 =

0 =

=

5 =

Therefore, the area of flower plot is

y =

= 19.6 m2

ii) Using completing the square :

y =


y =

y =

y =

The equation above show that y has minimum value where 5

Therefore, the area of flower plot is :

y =

= 19.6 m2

Show that we can determine the area of the flower plot by using differentiation and

completing the square method.

(e) The principle suggested an additional of 12 semicircular flower beds to the design submitted

by the Mathematics Society as shown in diagram below. The sum of the diameters of the

semicircular flower beds is 10m. The diameter of the smallest flower bed is 30cm and the

diameter of the flower beds are increased by a constant value successively.


Given that :

Smallest diameter, a = 30cm

Number of flower beds = 12

Sum of all twelve diameter of the flower beds, S12 = 1000cm

S12 = 1000

1000 =

1000 =

640 =

9.7 =

The diameter of the remaining flower bed is determined by using the following formula :

Table below shows the diameter of the remaining flower beds :

Flower beds Diameter,cmFlower bed1 30 + (0) 9.7 30.0Flower bed2 30 + (1) 9.7 39.7Flower bed3 30 + (2) 9.7 49.4Flower bed4 30 + (3) 9.7 59.1Flower bed5 30 + (4) 9.7 68.8Flower bed6 30 + (5) 9.7 78.5Flower bed7 30 + (6) 9.7 88.2

E

F

A C

Flower plot

Fish ponds

10 mSmallest flower bed


Flower bed8 30 + (7) 9.7 97.9Flower bed9 30 + (8) 9.7 107.5Flower bed10 30 + (9) 9.7 117.2Flower bed11 30 + (10) 9.7 127.0Flower bed12 30 + (11) 9.7 136.7

Total diameter 1000

The table above shows the diameter of each flower bed in increasing order starts from the first

flower bed of 30cm until the last flower bed along the fish ponds of 10m. Overall, we can look

that the flower beds are increasing in size.


Part 4CONCLUSION

Based on the project, it is shown that our life is related to circles. In addition circles must

not be apart from pi. All the calculation in this project involve pi as it is about the circles or parts

of it. Mathematics knowledge for the calculation is indeed needed as it is to help to understand

the concept of circle and pi.

This project is revolved between circle and pi and also the application of them in our

daily life. Having the concept of circles is really helpful. In the project of constructing the garden,

the application of circles is shown as it is needed in order to get the accurate size and area of it.


There are some conclusion can be concluded from the project. The length of arc can be

calculated by dividing the diameter into two if it is in terms of pi. Other than that, it is shown that

the conjectures proposed cannot be accepted as the relationship of the arcs is proven such that

the sum of the total lengths of arcs of the inner semicircles is equivalent to the length of arc of the

outer semicircle. Moreover, the generalization had been proven by having different diameter of

the outer semicircle. The result came out exactly the same.

From this project, I have learnt a lot. This project had made me understand more about

the circles in our life. Furthermore, this project really made me give out the positive attitude

towards additional mathematics. It is also improve my thinking skill in solving the problems.

Nevertheless, it had taught me of applying the formulae and concept that I have learnt before.

From this project, I have learnt how to relate the concept I have learnt.

In this conclusion I would like to say my greatest gratitude to the Almighty ALLAH in

giving me the thoughts and life until today, my parents who keep on giving me support in my

study, my Additional Mathematics teacher, Pn. Saadiah bt Abu Bakar who had taught me this

subject and not to forget all my friends either my classmates or my dorm mates. You have helped

a lot in finishing this project. Thank you a lot. May God bless you all.

19

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