*additional Mathematics Project Work 2009*

PROJECT WORK FOR ADDITIONAL MATHEMATICS 2009

Circles In Our Daily Life Name :________________________ 

I.C. No. :______________________



Teacher :______________________

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Class :________________________

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 all that. 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, Sir _____ for guiding me and my friends throughout this project. 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.

Objectives The aims of carrying out this project work are: to apply and adapt a variety of problem-solving strategies

to solve problems;  to improve thinking skills;  to promote effective mathematical communication;  to develop mathematical knowledge through problem

solving in a way that increases students’ interest and confidence;  to use the language of mathematics to express

mathematical ideas precisely;  to provide learning environment that stimulates and

enhances effective learning;  to develop positive attitude towards mathematics.

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. 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. 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 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. The Greek letter π, often spelled out pi in text was adopted for the number from the Greek word for perimeter "περίµετρος", 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 efforts to calculate more of its digits became famous).

Contents

No. 1 2 3 4 5 6 7 8

Contents Acknowledgement Objectives Introduction Part 1 Part 2 Part 3 Conclusions References

Page 1 2 3-4 5-7 8-13 14-20 20 21

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(π), we accept 3.142 or 22/7 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 = ½(2πr). 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 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.

References  www.wikipedia.org  www.one-school.net 

Additional Mathematics Textbook Form 4 and Form 5.

Part 1(a) There are a lot of things around us related to circles or parts of a circle.

CDs

Coins

Round-mirror

Round-table

Round-clock

Smallest flower bed

Part 1(b) Pi or π is a mathematical constant related to circles. Definition of π :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: 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 of π :The ancient Babylonians calculated the area of a circle by taking 3 times the square of its radius, which gave a value of π=3. One Babylonian tablet (ca. 1900–1680 BC) indicates a value of 3.125 for π, 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 π. The ancient cultures mentioned above found their approximations by measurement. The first calculation of π 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 π but only an approximation within those limits. In this way, Archimedes showed that π 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 π, 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 π based on probability.

Part 2(a) Diagram 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. Q

C A

P d1

R

B

d2 10 cm

Diagram 1 Using formula: Arc of semicircle = ½ π d d1 d2 Length of arc PQR Length of arc PAB (cm) (cm) 1 2 3 4 5 6 7 8 9 10

9 8 7 6 5 4 3 2 1 0

in terms of π (cm) 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π

in terms of π (cm) 0.5 π 1.0 π 1.5 π 2.0 π 2.5 π 3.0 π 3.5 π 4.0 π 4.5 π 5.0 π Table 1

Length of arc BCR in terms of π (cm) 4.5 π 4.0 π 3.5 π 3.0 π 2.5 π 2.0 π 1.5 π 1.0 π 0.5 π 0.0 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 = S + S PAB

BCR

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

Let d1= 3, and d2 =7

SPQR = S + S PAB

BCR

5π

= ½ π(3) + ½ π(7)

5π

= 3/2 π + 7/2 π

5π

= 10/2 π

5π = 5 π

Part 2(b)

Diagram 2 shows a semicircle PQR of diameter 10 cm. Semicircles PAB, BCD and DER of diameter d1, d2 and d3 is equal to 10 cm.

Q

E C

A P d1

B

D

d2

R d3

10 cm

Diagram 2

2(b)(i) d1 (cm)

d2 (cm)

d3 (cm)

1

1

8

Length of arc PQR in terms of π (cm) 5π

Length of arc PAB in terms of π (cm) 0.5 π

Length of arc BCD in terms of π (cm) 0.5 π

Length of arc DER in terms of π (cm) 4.0 π

1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 5 5 5 5 6 6 6 7 7 8

2 3 4 5 6 7 8 1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 5 1 2 3 4 1 2 3 1 2 1

7 6 5 4 3 2 1 7 6 5 4 3 2 1 6 5 4 3 2 1 5 4 3 2 1 4 3 2 1 3 2 1 2 1 1

5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π

0.5 π 0.5 π 0.5 π 0.5 π 0.5 π 0.5 π 0.5 π 1.0 π 1.0 π 1.0 π 1.0 π 1.0 π 1.0 π 1.0 π 1.5 π 1.5 π 1.5 π 1.5 π 1.5 π 1.5 π 2.0 π 2.0 π 2.0 π 2.0 π 2.0 π 2.5 π 2.5 π 2.5 π 2.5 π 3.0 π 3.0 π 3.0 π 3.5 π 3.5 π 4.0 π

1.0 π 1.5 π 2.0 π 2.5 π 3.0 π 3.5 π 4.0 π 0.5 π 1.0 π 1.5 π 2.0 π 2.5 π 3.0 π 3.5 π 0.5 π 1.0 π 1.5 π 2.0 π 2.5 π 3.0 π 0.5 π 1.0 π 1.5 π 2.0 π 2.5 π 0.5 π 1.0 π 1.5 π 2.0 π 0.5 π 1.0 π 1.5 π 0.5 π 1.0 π 0.5 π

Table 2

From the table, we can conclude that :-

SPQR

=

SPAB

+

SBCD

+

SDER

3.5 π 3.0 π 2.5 π 2.0 π 1.5 π 1.0 π 0.5 π 3.5 π 3.0 π 2.5 π 2.0 π 1.5 π 1.0 π 0.5 π 3.0 π 2.5 π 2.0 π 1.5 π 1.0 π 0.5 π 2.5 π 2.0 π 1.5 π 1.0 π 0.5 π 2.0 π 1.5 π 1.0 π 0.5 π 1.5 π 1.0 π 0.5 π 1.0 π 0.5 π 0.5 π

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

Let d1 = 2, d2 = 5, d3 = 3

SPQR

= SPAB + SBCD + 5 π = π + 5/2 π + 5π = 5π

SDER 3/2 π

2(b)(ii) Based 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

Souter = S1 + S2 + S3 + S4 + S5

Part 2(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 10 12 14 15 Let d1=10,

d2 d3 d4 SABC 8 6 6 15 π 3 5 10 15 π 8 4 4 15 π 5 3 7 15 π d2=8, d3=6, d4=6,

SAPQ 5π 6π 7π 15/2 π

SQRS 4π 3/2 π 4π 5/2 π

SSTU 3π 5/2 π 2π 3/2 π

SUVC 3π 5π 2π 7/2 π

SABC = SAPQ + SQRS + SSTU + SUVC 15 π = 5 π + 4 π + 3 π + 3 π 15 π = 15 π 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  0.5 π + 0.5 π + 4.0 π = 5 π The sum of the length of arcs of the inner semicircles  Factorise π/2 (1cm + 1cm + 8cm) =5 π 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.

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. The shaded region will be planted with flowers and the two inner semicircleas are fish ponds. Flower Plot

D

F E

Fish ponds

A xm

C

B 10 cm

Diagram 3 The area of the flower plot is ym2 and the diameter of one of the fish ponds is x m.

Area of ADC = ½ π (10/2) 2 = 25/2 π Area of AEB

= ½ π (x/2) 2 = ½ π (x2 /4) = x2 /8 π Area of BFC = ½ π (5 - x/2) 2 = ½ π (25–5x+x2/4) = (25/2) π – (5x/2) π + (x2/8) π Area of shaded region = = = = =

Area of ADC – (Area of AEB + Area of BFC) 25/2 π – [x2 /8 π + ((25/2) π – (5x/2) π + (x2/8) π )] 25/2 π – [x2 /8 π + (25/2) π – (5x/2) π + (x2/8) π ] 25/2 π – x2 /8 π - (25/2) π + (5x/2) π - (x2/8) π -x2/4 π + 5x/2 π

y = ((10x – x2)/4) π

Part 3(b) The area of the flower plot is 16.5 m2.

16.5 = -x2/4 π + 5x/2 π 16.5 = -x2/4 (22/7) + 5x/2 (22/7) To eliminate the π ,divide all the terms by π . 16.5/(22/7) = -x2/4 + 5x/2 5.25 = 5x/2 - x2/4 21 = 10x - x2 x2 - 10x + 21 = 0 Factorize the equation to get the value of x. (x-7)(x-3) = 0 X=7 or x=3  x = 7cm or x = 3cm.

Part 3(c) Linear Law y = -x2/4 π + 5x/2 π

Change it to linear form of Y = mX + C y/x = -x/4 π + 5/2 π Y = y/x m = - π/4 X=x C = 5/2 π A graph of y/x against x was plotted and the line of best fit is drawn. x 1 y/x 7.1

2 6.3

3 5.5

4 4.7

5 3.9

6 3.1

7 2.4

From the graph, 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 4.5m is  4.3 m ( 4.5 m) = 19.35 m2

Part 3(d) The cost of constructing the fish ponds is higher than that of the flower plot. Two methods are used to determine the area of the flower plot such that the cost of constructing the garden is minimum.

Method 1 : Differentiation y = -x2/4 π + 5x/2 π dy/dx = - πx/2 + 5/2 π d2y/dx2 = - π/2  y has maximum value. At maximum point, d2y/dx2 = 0 - πx/2 + 5/2 π = 0 πx/2 = 5/2 π x = 5m  Maximum value of y = -(5)2/4 π + 5(5)/2 π = 6.25 π m2. Method 2 : Completing The Square y = -x2/4 π + 5x/2 π = - π/4 (x2 – 10x) = - π/4 (x2 – 10x + 25 - 25) = - π/4 [(x-5)2 - 25] = - π/4 (x-5)2 + 25 π/4  y is a shape graph as, a = - π/4 It has a maximum value.  When x = 5m, maximum value of the graph is 6.25 π m2.

Part 3(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 10m. The diameter of the smallest flower bed is 30cm and the diameter of the A by a constant value successively. B flower beds are increased 10 cm

Diagram 4  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

Tn (flower bed)

Diameter (cm)

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T12

30.000 39.697 49.394 59.091 68.788 78.485 88.182 97.879 107.576 117.273 126.970 136.667