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PROJECT WORK FOR ADDITIONAL MATHEMATHICS 2009 Circles In Our Daily Life Ahli Kumpulan : 1. Muhammad Muhsan b. Mohd Mashor 2. Teoh Kok Siang Guru Pembimbing : Pn. Chan Siew Thoe Sekolah Menengah Sains Tun Syed Shahabudin 2 -ContentNo. 1 2 3 4 5 Contents Introduction Part 1 Part 2a Part 2b Part 3 Page 3 - 4 5 - 8 9 - 10 11 - 12 13 - 16 2 -IntroductionA 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: 2 • 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.[1] • • 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.[2] 2 Part 1 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. Wheel of a bicycle Circles on water surface School park Fish pond Round table at school compound 2 Before I continue the task, first, we do have to know what do pi(π) related to a circle. 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. 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:[3][5] 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.[6] The formulas below illustrate other (equivalent) definitions. History 2 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. 2 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. 2 Part 2 (a) Q C A P d1 B 10 cm d2 R 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 2 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 Let d1= 3, and d2 =7 SPQR = S + S PAB BCR 5π 5π 5π = ½ π(3) + ½ π(7) = 3/2 π + 7/2 π = 10/2 π 5π = 5 π Q E A P d1 B d2 10 D d3 C (b) R 2 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 bii) 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 d2 8 d3 6 d4 6 SABC 15 π SAPQ 5π SQRS 4π SSTU 3π SUVC 3π 2 12 14 15 3 8 5 5 4 3 10 4 7 15 π 15 π 15 π 6π 7π 15/2 π 3/2 π 4π 5/2 π 5/2 π 2π 3/2 π 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 Part 3 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 2 = = y= (25/2) π - (( x2 – 10x + 50)/4) (25/2 - (x2 - 10x + 50)/4) π ((10x – x2)/4) π b. 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 Y/x 8.0 7.0 c. y = ((10x – x2)/4) π 6.0 y/x = (10/4 - x/4) π x y/x 5.0 1 7.1 4.0 2 6.3 3 5.5 4 4.7 5 3.9 6 3.1 7 2.4 3.0 2.0 0 1 2 2 3 4 5 6 7 X 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 2 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 e. n = 12, a = 30cm, S12 = 1000cm Tn (flower bed) T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T12 Diameter (cm) 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 30 39.697 49.394 59.091 68.788 78.485 88.182 97.879 107.576 117.273 126.97 136.667 2 d = 9.697 2