Circle

Ad dit io nal Mathe matics 2009 Pr oj ect W ork Circles Around Us Name: Class: 5 Science 1 IC number: Teacher: School: SMK

Appreciation

wira penrisen

First and foremost, I would like to thank God that finally, I had succeeded in finishing this project work. I would like to thank my beloved Additional Mathematic Teacher for all the assistance he has provided me during my job search. I appreciate the information and advice he have given, as well as the connections he have shared with me. His expertise and help have been very precious during this process. Also, thanks to my parents for giving me fully moral and finance support in completing this project work and permission to use their notebook for further research in completing this project work. I thank them very much for sacrificing their time and money in helping me to complete this project work. I would like to give my special thank to my fellow friends who had given me extra information on the project work and study group that we had done. Thank you for spending time with me to discuss about the coursework. I would also like to thank the Ministry Of Education Malaysia for giving me a chance to apply Additional Mathematics skills in daily life through this valuable coursework. Without their effort, I would not have a chance to sharpen my Additional Mathematics skills. Last but not least, I would like to express my highest gratitude to all those who gave me the possibility to complete this coursework. I really appreciate all your help. Again, thank you so much.

Introduction 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.

Circle 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. However, 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 diameter of a circle is the length of a line segment whose endpoints lie on the circle and which passes through the centre of the circle. This is the largest distance between any two points on the circle. The diameter of a circle is twice its radius. The term "radius" can also refer to a line segment from the centre a circle to its perimeter, and similarly the term "diameter" can refer to a line segment between two points on the perimeter which passes through the centre. In this sense, the midpoint of a diameter is the centre and so it is composed of two radii. A chord of a circle is a line segment whose two endpoints lie on the circle. The diameter, passing through the circle's centre, is the largest chord in a circle. A tangent to a circle is a straight line that touches the circle at a single point. A secant is an extended chord: a straight line cutting the circle at two points. An arc of a circle is any connected part of the circle's circumference. A sector is a region bounded by two radii and an arc lying between the radii, and a segment is a region bounded by a chord and an arc lying between the chord's endpoints. 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 mostmedieval scholars, and many believed

that there was something intrinsically "divine" or "perfect" that could be found in circles.

Part 1

There are a lot of things around us related to circles or part of a circle. (a)Collect pictures of 5 such objects. You may use a camera to take pictures around your school compound or get pictures from magazines, newspaper, the internet or any other sources. (b) Pi or π is a mathematical constant related to circles. Define π and write a brief history of π.

Part 1(b)

Pi

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 (see the table for its representation in some other bases). 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, itsdecimal 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). The letter π

The name of the Greek letter π 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 (Π) even at the beginning of a sentence. When referring to this constant, the symbol π 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 "π" because "π" is the first letter of the Greek words περιφέρεια (periphery) and περίμετρος (perimeter), probably referring to its use in the formula to find the circumference, or perimeter, of a circle. π isUnicode character U+03C0 ("Greek small letter pi").

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. 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. Irrationality and transcendence Being an irrational number, π cannot be written as the ratio of two integers. This was proved in 1768 byJohann Heinrich Lambert. In the

20th century, proofs were found that require no prerequisite knowledge beyond integral calculus. One of those, due to Ivan Niven, is widely known. A somewhat earlier similar proof is by Mary Cartwright. Furthermore, π is also transcendental, as was proved by Ferdinand von Lindemann in 1882. This means that there is no polynomial with rational coefficients of which π is a root. An important consequence of the transcendence of π is the fact that it is not constructible. Because the coordinates of all points that can be constructed with compass and straightedge are constructible numbers, it is impossible to square the circle: that is, it is impossible to construct, using compass and straightedge alone, a square whose area is equal to the area of a given circle. This is historically significant, for squaring a circle is one of the easily understood elementary geometry problems left to us from antiquity; many amateurs in modern times have attempted to solve each of these problems, and their efforts are sometimes ingenious, but in this case, doomed to failure: a fact not always understood by the amateur involved. Numerical value The numerical value of π truncated to 50 decimal places is: 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 While the value of π has been computed to more than a trillion (1012) digits, elementary applications, such as calculating the circumference of a circle, will rarely require more than a dozen decimal places. For example, a value truncated to 11 decimal places is accurate enough to calculate the circumference of a circle the size of the earth with a precision of a millimeter, and one truncated to 39 decimal places is sufficient to compute the circumference of any circle that fits in the observable universe to a precision comparable to the size of a hydrogen atom. Because π is an irrational number, its decimal expansion never ends and does not repeat. This infinite sequence of digits has fascinated mathematicians and laymen alike, and much effort over the last few centuries has been put into computing more digits and investigating the number's properties. Despite much analytical work, and supercomputer calculations that have determined over 1 trillion digits of π, no simplebase-10 pattern in the digits has ever been found. Digits of π are available on many web pages, and there is software for calculating π to billions of digits on any personal computer.

History of Pi 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 Bibleappears 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.

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 inregular 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, as follows: π ≈ A3072 = 3 ⋅ 28 ⋅ √(2 - √(2 + √(2 + √(2 + √(2 + √(2 + √(2 + √(2 + √(2 + 1) ≈ 3.14159

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 Until the second millennium, π was known to fewer than 10 decimal digits. The next major advance in π studies came with the development ofcalculus, 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[28][29] 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.[30] 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."[31] In 1706 John Machin was the first to compute 100 decimals of π, using the formula

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 Eulerin 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}}

Part 2

(a)

Diagram 1 shows a semicircle PQR of diameter 10cm. Semicircles PAB and BCR of diameter d1 and d2 are respectively inscribed in the semicircle PQR such that the sum of d1 and d2 is equal to 10cm.

d1 cm Q 10 d2

Diagram 1

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.

(b)

E321 cm D Q 10 d

(i) (ii)

Diagram 2 shows a semicircle PQR of diameter 10cm. Semicircles PAB, BCD and DER of diameter d1, d2 and d3 respectively inscribed in the semicircle PQR such that the sum of d1, d2 and d3 is equal to 10cm.

Using various values of d1 and d2 and the corresponding values of d3, determine the relation between the lengths of arc PQR, PAB, BCD and DER. Based on your findings in (a) and (b), make generalisations 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 ......

(c)

d1 cm Q 10 d2

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

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. By using various values of d1 and corresponding values of d2, I determine the relation between length of arc PQR, PAB, and BCR. Using formula: Arc of semicircle = ½πd d1

d2

Length of arc PQR in

Length of arc PAB in

Length of arc BCR in

(cm) 1 2 3 4 5 6 7 8 9

(cm) 9 8 7 6 5 4 3 2 1

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

terms of π (cm) ½π π 3/2 π 2π 5/2π 3π 7/2 π 4π 9/2 π

terms of π (cm) 9/2 π 4π 7/2 π 3π 5/2 π 2π 3/2 π π ½π

Table 1 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

Let d1= 3, and d2=7

BCR

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

E321 cm D Q 10 d

b (i)

d1 1 2 2 2 2

d2 2 2 3 4 5

d3 7 6 5 4 3

SPQR 5π 5π 5π 5π 5π

SPAB 1/2 π π π π π

SBCD Π Π 3/2 π 2π 5/2 π

SDER 7/2 π 3π 5/2 π 2π 3/2 π

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

3/2 π 5π = 5π

b (ii) The length of arc of outer semicircle is equal to the sum of the length of arc of inner semicircle for n = 1,2,3,4,….

Souter = S1 + S2 + S3 + S4 + S5 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

d2 8 3 8 5

d3 6 5 4 3

d4 6 10 4 7

SABC 15 π 15 π 15 π 15 π

let d1=10, 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

Part 3 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 semicircles are fish ponds.

(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. (b) Find the diameters of two fish ponds if the area of the flower pot is 16.5 m2. (Use = 22/7) (c) Reduce the non-linear equatin 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 pot if the diameter of one of the fish ponds is 4.5 m.

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

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

Part 3

(a)

The area of the flower plot is y m2 and the diameter of one of the fish pond is x cm.

Area of flower plot = y m2 y y y y y y y y

= = = = = = = =

(25/2) π - (1/2(x/2)2π + 1/2((10-x )/2)2 π) (25/2) π - (1/2(x/2)2π + 1/2((100-20x+x2)/4) π) (25/2) π - (x2/8 π + ((100 - 20x + x2)/8) π) (25/2) π - (x2π + 100π – 20x π + x2π )/8 (25/2) π - ( 2x2– 20x + 100)/8) π (25/2) π - (( x2 – 10x + 50)/4) (25/2 - (x2 - 10x + 50)/4) π ((10x – x2)/4) π

Therefore, the area of flower plot is equal to ((10x – x2)/4) π. (b) By using π = 22/7, we can find the diameters of the two fish ponds if the area of the flower plot is 16.5 m2. Area of flower plot = 16.5 m2 = ((10x – x2)/4) π 16.5 66 66(7/22) 0 0 x=7

= = = = = ,

((10x – x2)/4) π (10x - x2) 22/7 10x – x2 x2 - 10x + 21 (x-7) (x – 3) x=3

Therefore, the diameter of fish pond E is 3m while the diameter of fish pond F is 7m.

(c)

We can reduce the non-linear equation obtained in (a) to simpler linear form. y = ((10x – x2)/4) π y/x = ((10x – x2)/4x) π y/x = (10/4 - x/4) π

To determine the area of the flower plot, we have to plot a straight line graph by using the equation: y/x = (10/4 - x/4) π x y/ x

1 7.1

2 6.3

3 5.5

4 4.7

5 3.9

6 3.1

7 2.4

From the graph, we can determine the area of the flower pot if the diameter of one of the fish pond is 4.5 m. Area of flower plot = y When x = 4.5, y/x = 4.3 y = y/x * x y = 4.3 * 4.5 y = 19.35m2 Therefore, the area of the flower pot if the diameter of one of the fish pond is 4.5 m is equal to 19.35m2.

(d) The cost of constructing the fish ponds is higher than that of the flower plot. There are two methods to determine the area of the flower plot such that the cost of constructing the garden is minimum. The area of the flower plot can be determined by using differentiation method and completing square method. (i)

Differentiation method: dy/dx = ((10x-x2)/4) π = ( 10/4 – 2x/4) π = 5/2 π – x/2 π

Since the cost of constructing the garden is minimum, thus dy/dx is equal to 0. 0 = 5/2 π – x/2 π

5/2 π = x/2 π x = 5 (ii)

Completing Square method: y = ((10x – x2)/4) π = 10x/4 π - x2/4 π = -1/4 π (x2– 10x) = -1/4 π (x – 5)2 - 52 = -1/4 π (x - 5)2 – 25

Since the cost of constructing the garden is minimum, thus (x-5)2 is equal to 0. x–5=0 x=5 Therefore, the area of the flower pot is 5 m2.

(e) The diameter of the smallest flower bed is 30 cm and the diameter of the flower beds are increased by a constant value successively. We can determine the diameter of the remaining flower beds by using the formula : Sn = n/2 (2a + (n – 1) d n = 12, a = 30cm, S12 = 1000cm 1000 = 12/2 (2(30) + (12 – 1) d) 1000 = 6 (60 + 11d) 1000 = 360 + 66d 1000 – 360 = 66d 640 = 66d d = 9.697 Since d is equal to 9.697, thus, we can find the diameter of the following flower beds

Tn(flower bed) Diameter

T1

T2

T3

T4

T5

T6

T7

T8

T9

T10

30

39.697

49.394

59.091

68.788

78.485

88.182

97.879

107.576

117.273

T11

T12

126.97 136.667

(cm)

The diameter of the remaining flower beds = T5 + T6 + T7 + T8 + T9 + T10 + T11 +T12 = 68.788 + 78.485 + 88.182 + 97.879 + 107.576 + 117.273 + 126.97 + 136.667 =821.82 cm Therefore, the diameter of the remaining flower beds is equal to 821.82 cm.

CONCLUSION Pi( ) 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 works 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 give thanks to my additional mathematics teacher as without his help, I would not be able to accomplish this project.

Reference

Websites * http://images.google.com/ * http://en.wikipedia.org/wiki/Circle * http://www.gap-system.org/~history/HistTopics/Pi_through_the_ages.html www.pdfcoke.com www.4shared.com www.dogpile.com