I have done many researches throughout the internet and discussing with a friend who have helped me a lot in completing this project. Through the completion of this project, I have learned many skills and techniques. This project really helps me to understand more about the uses of circles in our daily life. This project also helped expose the techniques of application of additional mathematics in real life situations.
First of all, I want to express my utmost gratitude to everyone that has contributed to the success of this project. Though it’s kind of hard for me to accomplish the project alone but with the help and never-ending supports from these people; •
my parents, my additional mathematics teacher, Puan Intan Roselyza, my friends and those that have contributed in this project,
I managed to get this project done in just a few days. Thank you.
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Contents Introduction Part 1 Part 2a Part 2b Part 3
Page 3 - 4 5 - 8 9 - 10 11 - 12 13 - 16
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.
PART 1 a) The task is carried out by collecting 5 such objects related to circles or parts of a circle. Some of the examples are clock, wheel, donut, CD-ROM and plate. The aim is to create awareness among students that mathematics is applicable in our daily lives. b) For this part, I used the internet resources as a reference to get the definition of pi and brief history of pi (π).
a) Semicircle PAB and BCR of diameter d1 and d2 are inscribed in the semicircle PQR such that the sum of d1 and d2 is equal to 10 cm. Table 1 can be completed by using various values of d1, and the corresponding value of d2. For this part, I used the formulae to find length of arcs PQR, PAB and BCR is s = π r. The relation between the lengths of arcs PQR, PAB and BCR is determined. Thus, d1 + d2= 10. b) (i) Semicircle PAB, BCD and DER of diameter d1, d2 and d3 are inscribed in the semicircle PQR such that the sum of d1, d2 and d3 is equal to 10 cm. The task is carried out by using the various values of d1,and d2 ¬and the corresponding values d3. The relation between the lengths of arcs PQR, PAB, BCD and DER is determined and the findings are tabulated. Thus, d1 + d2 + d3 = 10. (ii) Based on the findings in (a) and (b), generalizations are made 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) For different values of diameters of the outer semicircle, generalizations stated in b (ii) is still true that the length of arc of the outer semicircle is equal to the sum of the lengths of arc of the inner semicircles for n semicircles where n = 1,2,3,4...
PART 3 a) The Mathematics Society is given a task to design a garden to beautify the school. The shaded region in Diagram 3 will be planted with flowers and the two inner semicircles are fish ponds. The area of the flower plot is y m2 and the diameter of one of the fish ponds is x m. To find the area of semicircles, the formula, π r2 is applied. Thus, y is expressed in terms of π and x. b) The diameters of the two fish ponds are calculated when the area of the area of the flower plot is 16.5 m2. Quadratic equation is used to get the diameters of the two fish ponds.
c) The non-linear equation obtained in (a) is reduced to simpler linear form that is values for the vertical axis and x values for the horizontal axis. A straight line graph is plotted using Microsoft Excel. Using the straight line graph, the area of the flower plot is determined when the diameter of one of the fish ponds is 4.5 m. d) The cost of the fish ponds is higher than that of the flower plot. Thus, differentiation and completing the square methods are used to determine the area of the flower plot such that the cost of constructing the garden is minimum. e) The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society. The sum of the diameters of the semicircular flower beds is 10 m. The diameter of the smallest flower bed is 30 cm and the diameter of the flower beds are increased by a constant value successively. Arithmetic progression is used to determine the common difference in order to determine the diameters of the remaining flower beds.
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 design a garden to beautify the school.
Circle-shaped bottle cover
Circle design on jewelry
The CD of an album
Circular wave of water
Before I continue the task, first, we do have to know what do pi(π) related to a circle.
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.
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.
10 d2 d1 Q cm
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 (cm) 1 2 3 4 5 6 7 8 9 d2 (cm) 9 8 7 6 5 4 3 2 1 Length of arc PQR in terms of π (cm) 5π 5π 5π 5π 5π 5π 5π 5π 5π Length of arc PAB in terms of π (cm) ½π π 3/2 π 2π 5/2π 3π 7/2 π 4π 9/2 π Table 1 Length of arc BCR in terms of π (cm) 9/2 π 4π 7/2 π 3π 5/2 π 2π 3/2 π π ½π
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
Let d1= 3, and d2=7
SPQR = S + S
5π 5π 5π
= ½ π(3) + ½ π(7) = 3/2 π + 7/2 π = 10/2 π
5π = 5 π
E3 D1 cm 10 d2 Q
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
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
let d1=10, d2=8, d3=6, d4=6, SABC = SAPQ + SQRS + SSTU + SUVC 15 π = 5 π + 4 π + 3 π + 3 π 15 π = 15 π
a) Area of flower plot = y m2
y = (25/2) π - (1/2(x/2)2 π + 1/2((10-x )/2)2 π) = (25/2) π - (1/2(x/2)2 π + 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 = = y= (25/2) π - (( x2 – 10x + 50)/4) (25/2 - (x2 - 10x + 50)/4) π ((10x – x2)/4) π
y = 16.5 m2 16.5 = 66 = ((10x – x2)/4) π (10x - x2) 22/7
66(7/22) = 10x – x2 0 = x2 - 10x + 21 0 = (x-7)(x – 3) x=7 , x=3
((10x – x2)/4) π
y/x = (10/4 - x/4) π
x 1 y/x 7.1
2.0 0 1 2 3 4 5 6 7
When x = 4.5 , y/x = 4.3 Area of flower plot = y/x * x = 4.3 * 4.5 = 19.35m2
d) Differentiation method dy/dx = ((10x-x2)/4) π = ( 10/4 – 2x/4) π
0 = 5/2 π – x/2 π 5/2 π = x/2 π x = 5
Completing square method y= = = ((10x – x2)/4) π 5/2 π - x2/4 π -1/4 π (x2 – 10x)
y+ 52 = -1/4 π (x – 5)2 y = -1/4 π (x - 5)2 - 25 x–5=0 x=5
Tn (flower bed) T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T12
Diameter (cm) 30 39.697 49.394 59.091 68.788 78.485 88.182 97.879 107.576 117.273 126.97 136.667
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
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 circunference, 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 = 1/2(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.