Project Work Addmath Form 5

CONTENT 1. ACKNOWLEDGEMENT 2. INTRODUCTION 3. PART 1 4. PART 2 5. PART 3

ACKNOWLEDGEMENT 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, Mrs. Wan Nurhalina bt. __________for all the assistance she has provided me during my job search. I appreciate the information and advice she have given, as well as the connections she have shared with me. Her expertise and help have been invaluable during this process. Also, thanks to my mom and my dad for giving me fully support in completing this project work and permission to use their notebook for further research in completing this project work. I sincerely appreciate their generosity. 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. 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 helps. Again, thank you so much. Best Regards,

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 centre 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, 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.

PART 1 There are a lot of things are a lot of things around us related to circles or parts of a circle. Circle exists in our everyday lives and without circles, we could not imagine what it would cause to this world as the most important thing, the Earth itself is a circle. In this project, I will use the principle of circle that I had studied to design a garden to beautify the school.

Before I further my task, first, we have to know what do pi (π) related to a circle. 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/. Pi or π is a mathematical constant whose value is the ratio of any circle's circumference to its diameter. 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 early history of pi is believed to be built during the Fourth Dynasty of Egypt's Old Kingdom, the Great Pyramid was constructed with an approximate ratio of height to circumference of the base of 2π. Each side is 440 cubits long, and the height is believed to have been 280 cubits tall at the time of its construction. This puts the value at approximately 3.142, or 0.04% above the exact value. An Egyptian scribe named Ahmes wrote the oldest known text to give an approximate value for π. The Rhind Mathematical Papyrus dates from the Egyptian Second Intermediate Period—though Ahmes stated that he copied a Middle Kingdom papyrus (i.e. from before 1650 BC)—and describes the value in such a way that the result obtained comes out to 256⁄81, which is approximately 3.16, or 0.6% above the exact value. As early as the 19th century BC, Babylonian mathematicians were using π ≈ 25⁄8, which is about 0.5% below the exact value. The Indian astronomer Yajnavalkya gave astronomical calculations in the Shatapatha Brahmana (c. 9th century BC) that led to a fractional approximation of π ≈ 339⁄108 (which equals 3.13888..., which is correct to two decimal places when rounded, or 0.09% below the exact value). In the third century BC, Archimedes proved the sharp inequalities 223⁄71 < π < 22⁄7, by means of regular 96-gons; these values are 0.02% and 0.04% off, respectively. (Differentiating the arctangent function leads to a simple modern proof that indeed 3+1⁄7 exceeds π.) Later, in the second century AD, Ptolemy, using a regular 360-gon, obtained a value of 3.141666...., which is correct to three decimal places.[1] The Chinese mathematician Liu Hui in 263 AD computed π with to between 3.141024 and 3.142708 with inscribe 96-gon and 192-gon; the average of these two values is 3.141864, an error of less than 0.01%. However, he suggested that 3.14 was a good enough approximation for practical purpose. Later he obtained a more accurate result π ≈ 3927⁄1250 = 3.1416.

PART 2 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.

Diagram 1 (a) I had completed the Table 1 by using various values of d1 and the corresponding values of d2, in which d1 + d2 = 10cm, to determine the relation between the lengths of arcs PQR, PAB, and BCR. To find the length of arc, I had used the formula: Arc of semicircle = ½πd d1 (cm)

d2 (cm)

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

Length of arc PAB in Length of arc BCR in terms of π terms of π (cm) (cm) 1 9 ½π 9/2 π 2 8 π 4π 3 7 3/2 π 7/2 π 4 6 2π 3π 5 5 5/2π 5/2 π 6 4 3π 2π 7 3 7/2 π 3/2 π 8 2 4π π 9 1 9/2 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 the equation is: SPQR = SPAB + SBCR Let d1= 3, and d2 =7 SPQR = SPAB + SBCR 5π = ½ π (3) + ½ π (7) 5π = 3/2 π + 7/2 π 5π = 10/2 π 5π = 5 π

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

Diagram 2 (b)(i) By using various values of d1, d2, and the corresponding values of d3, in which d1 + d2 + d3 = 10cm, I had make a table to determine the relation between the lengths of arcs PQR, PAB, BCD, and DER. All the results are tabulated in Table 2 below: d1 (cm)

d2 (cm)

d3 (cm)

1 2 2 2 2

2 2 3 4 5

7 6 5 4 3

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

Length of arc PAB Length of arc in terms of π BCD in terms of π (cm) (cm) 1/2 π π π π π 3/2 π π 2π π 5/2 π Table 2

Length of arc DER in terms of π (cm) 7/2 π 3π 5/2 π 2π 3/2 π

From the Table 2, we know that the length of arc PQR is not affected by the different in d1, d2, and d3 in PAB, BCD, and DER respectively. The relation between the length of arcs PQR, PAB, BCD, and DER is that the length of arc PQR is equal to the sum of the length of arcs PAB, BCD, and DER, which the equation is: 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) Therefore, we can generalise that the length of arc of outer semicircle is equal to the sum of the length of arc of inner semicircle for x = 2, 3, 4, 5….. Souter = S1 + S2 + S3 + S4 + S5 (c) Assume that the diameter of the outer semicircle is 30cm and 4 semicircles are inscribed in the outer semicircle such that the sum of d1 (arc1), d2 (arc2), d3 (arc3), d4 (arc4) is equal to 30cm. Again, by using various values of d1, d2, d3, and the corresponding values of d4, in which d1 + d2 + d3 + d4 = 30cm, I had make a table to determine the relation between the lengths of arc1, arc2, arc3, and arc4. All the results are tabulated in Table 3 below: d1 (cm)

d2 (cm)

d3 (cm)

d4 (cm)

10 12 14 15

8 3 8 5

6 5 4 3

6 10 4 7

Length of arc of outer semicircle in terms of π (cm) 15 π 15 π 15 π 15 π

Length of arc 1 in terms of π (cm) 5π 6π 7π 15/2 π

Length of arc 2 in terms of π (cm)

Length of arc 3 in terms of π (cm)

4π 3π 3/2 π 5/2 π 4π 2π 5/2 π 3/2 π Table 3

Length of arc 4 in terms of π (cm) 3π 5π 2π 7/2 π

From the Table 3, we know that the length of arc PQR is not affected by the different in d1, d2, d3, and d4 in arc1, arc2, arc3, and arc4 respectively. The relation between the length of arc of outer semicircle, arc1, arc2, arc3, and arc4 is that the length of arc of outer semicircle is equal to the sum of the length of arc1, arc2, arc3, and arc4 which the equation is: Souter semicircle = Sarc1 + Sarc2 + Sarc3 + Sarc4 Let d1=10, d2=8, d3=6, d4=6 Souter semicircle = Sarc1 + Sarc2 + Sarc3 + Sarc4 15 π = 5 π + 4 π + 3 π + 3 π 15 π = 15 π Therefore, we can conclude that length of arc of outer semicircle is equal to the sum of the length of arc of inner semicircle for x = 2, 3, 4, 5….. Souter = S1 + S2 + S3 + S4 + S5......

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 flower and the two inner semicircles are fish ponds.

Diagram 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)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 (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 3 m while the diameter of fish pond F is 7 m. (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

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

8.0

7.0

6.0

5.0

4.0

3.0

2.0 0

1

2

3

4

5

6

7

X

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

fore, 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 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.

Diagram 4 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) 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

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.