Folio Add Math 4

♥ AIMS ♥

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To apply and adapt variety of problem-solving strategies to solve problems.

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To improve thinking skills.

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To promote effective mathematical communication.

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To use the language of mathematics to express mathematical ideas precisely.

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To develop positive attitude towards mathematics.

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To provide learning environment that stimulates and enhances effective learning.

♥CONTENT♥ AIMS

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INTRODUCTION OF CIRCLE

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1

PICTURES OF CIRCLES DEFINITION OF PI HISTORY OF PI PART TWO PART THREE CONCLUSION REFERENCES

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3 6 7 9 13 21 22

INTRODUCTION A circle is a simple shape of Euclidean geometry consisting of those points in a plane which is the same distance from a given point called the centre. The common distance of the points of a circle from its centre is called the radius. The circle has been known since before the beginning of recorded history. It is the basis for a 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 since, particularly geometry and astrology and also 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 ∏.

♥PICTURES OF CIRCLES♥ There are a lot of things around us related to circles.

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♥DEFINITION OF PI, π In Euclidean plane geometry, π is defined as the ratio of a circle’s circumference to its diameter. π=Cd

The ratio Cd 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. Area of the circle = π × area of the shaded square. Alternatively, π can also be defined as the ratio of a circle’s area, A, to the area of square whose side is equal to the radius = π = Ar2 These definitions depend on results of Euclidean geometry, such as the fact that all circles are similar. This can be considered a problem when 5

pi occurs in areas mathematics that otherwise do not involve geometry. For this reason, mathematician often prefer to define pi without reference to geometry, instead selecting one of its analytic properties as a definition. A common choice is to define pi as twice the smallest positive x for which cos x.

♥ HISTORY OF PI ♥

The ancient Babylonians calculated the area of a circle by taking 3 times the square of its radius, which gave a value of pi equal to three. One Babylonians indicates value of 3.125 for pi, which is closer approximation. In the Egyptian Rhind Papyrus, there is evidence that the Egyptian calculated the area of circle by using 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 for pi was done by Archimedes of Syracuse, 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 the circle was circumscribed. Since the actual area of the circle was lying between areas of the inscribed and circumscribed polygons, the areas of the polygons have upper and lower bounds for the area of the circle. Archimedes knows that

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he had not found the value of pi but only an approximation within those limits. A similar approach was used by Zu Chongzhi, a brilliant Chinese Mathematician and an astronomer. He 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 is 355113. To compute this accuracy for pi, he must have started with an inscribed regular 24576-gon and performed lengthy calculations involving hundreds of square roots carried out to nine decimal places. Mathematicians began using the Greek letter, n , in the 1700s. Introduced by William Jones in 1706, use of the symbol was popularized by Euler, who adapted it in 1737. An 18th century, French Mathematicians named George Buffon devised a way to calculate pi based on probability. Pi or π is a mathematical constant related to circles. Thus, pi = π = 227 or 3.142

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

D E Q321 cm 10 d

Table 1 is completed by using various values of d1 and the corresponding value of d2. Then, the relation between the lengths of arcs PQR, PAB and BCR are determined.

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d1

d2

(cm) 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5

(cm) 9.0 8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5

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

(15.71cm) (15.71cm) (15.71cm) (15.71cm) (15.71cm) (15.71cm) (15.71cm) (15.71cm) (15.71cm) (15.71cm)

Length of PAB in terms of π (cm)

Length of BCR in terms of π (cm)

π2 34π π 54π 32π 74 π 2π 94 π 52 π 114 π

92π 174π 4π 154π 72 π 134 π 3π 114 π 52 π 94 π

TABLE 1

The tabulated data below show various values of d1 and d2. Therefore, from the obtained result, the relation between the lengths of arc PQR, PAB, BCR is Length of arc PQR = length arc PAB + length arc BCR SPQR = SPAB + SBCR Let d1 = 2, and d2 = 8 5π=π+4π 5π=5π

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

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D E Q321 cm 10 d

i) By using various values of d1 and d2 and the corresponding values of d3, the relation between the length of arc PQR, PAB, BCD and DER are determined. The findings are tabulated.

D1

D2

( ( cm) cm) 0 1 .25 .00 0 1 .50 .50 1 2 .00 .00 1 2 .50 .50 2 3 .00 .00

D3 ( cm) 8 .75 8 .00 7 .00 6 .00 5 .00

Length of arc PQR in π

Length of arc PAB in π

Length of arc BCD in π

Length of arc DER in π

5 π (15.71 cm) 5 π (15.71 cm) 5 π (15.71 cm) 5 π (15.71 cm) 5 π (15.71 cm)

π8

π2

358

π4

34

4π

π2

π

72

34

54

3π

π

32

52

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2 .50 3 .00 3 .50 4 .00 4 .25

3 .50 4 .00 4 .50 5 .00 5 .25

4 .00 3 .00 2 .00 1 .00 0 .50

5 π (15.71 cm) 5 π (15.71 cm) 5 π (15.71 cm) 5 π (15.71 cm) 5 π (15.71 cm)

54

74

2π

32

2π

32

74

94

π

2π

52

π2

178

218

π4

TABLE 2

Therefore, from the result in table 2: Length of arc PQR = length arc PAB + length arc BCD + length of arc DER

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

5π=5π

(ii) Based on the findings in (a) and (b), the generalisations about the length of the arc of the outer semicircle and the length of arcs of the inner semicircle are made n = 2, 3, 4... The length of arc PQR equals to the sum of the length of arcs of all the inscribed semicircle. 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 + ...

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(c) For different values of diameters of the outer semicircle, the generalisations showed stated in b (ii) is still true. The length of 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.

In other words, for different values of diameters of the outer semicircle, show that the generalisations in b (ii) is still true.

PART THREE 12

The Mathematics society is given a task to design a garden to beautify the school by using the design as shown below. The shaded area region will be planted with flowers and the two inner semicircles are fish ponds.

E321 cm D Q 10 d

(a) The area of the flower plot is y m2 and the diameter of one of the fish ponds is x m.y is expressed in terms of π and x.

Area of flower plot = y m2 13

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

y=

((10x – x2)/4) π

(b) If the area of the flower plot is 16.5 m2, the diameters of the two fish ponds are found by using π = 227 .

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

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(c) The non-linear equation obtained in (a) is reduced to simple linear form and the, a straight line graph is plotted. By using the area of the flower plot is determined if the diameter of one of the fish ponds is 4.5 m. .

y=

((10x – x2)/4) π

y/x = (10/4 - x/4) π

x y/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

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

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

(e) 16

The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society.

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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

30 39.697 17

T3 T4 T5 T6

49.394 59.091 68.788 78.485

T7 T8 T9 T10 T11 T12

88.182 97.879 107.576 117.273 126.97 136.667

The diameter of each of the flower beds

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CONCLUSION♥ Based on this project, I learned lots of things then I realized that circles do related to our daily lives. From this project, I also have learnt to work in group and get new knowledge that can be used in the future. Moreover, I am able to apply and use the functions of circles in my daily life. I also learned to solve questions that I have never tried to solve before.

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REFERENCES

♥SOURCES ♥ friends; students in 5 Berlian, SMK Perempuan Likas ♥ Additional Mathematics teacher; Sir Mohd. Anuar Ali ♥ Internet; www.pdfcoke.com

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