Additional Mathematics Project Work 2009

ADDITIONAL MATHEMATICS PROJECT WORK 2009

NAME

: WAN MUHAMMAD SHAFIQ BIN WAN ADNAN

FORM

: 5 IBNU SINA

I/C NO. : 920322-14-6187 TEACHER’S NAME : CIK KAMALIAH WAHIDA BINTI OMAR

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TITLE: PAGE: APPRECIATION…………………………………… …. 1 INTRODUCTION. ……………………………………. 2 PART 1 ~ TASK 1…………………………………………………. 4 PART 2 ~ TASK 2…………………………………………………. 10

After weeks of struggle and hard work to complete assignment given to us by our teacher, Cik Kamaliah Wahida binti Omar. I finally did it within 2 weeks with satisfaction and senses of success because I

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have understood more deeply about the interest and investment more than before. I have to be grateful and thankful to all parties who have helped me in the process of completing my assignment. It was a great experience for me as I have learnt to be more independent and to work as group. For this, I would like to take this opportunity to express my thankfulness once again to all parties concerned. Firstly, I would like to thanks my Additional Mathematics’ teacher, Cik Kamaliah Wahida binti Omar for patiently explained to us the proper and precise way to complete this assignment. With her help and guidance, many problems I have encountered had been solved. Beside that, I would like to thanks my parents for all their support and encouragement they have given to me. In addition, my parents had given me guidance on the methods to account for investment which have greatly enhanced my knowledge on particular area. Last but not least, I would like to express my thankfulness to my cousin and friends, who have patiently explained to me and did this project with me in group.

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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. Circle is also the set of all points equidistant from a given point. The point from which all the points on a circle are equidistant is called the center of the circle, and the distance from that point to the circle is called the radius of the circle. A circle is named with a single letter, its center. See the diagram below.

Figure 1.1: A circle The circle above has its center at point C and a radius of length r. By definition, all radii of a circle are congruent, since all the points on a circle are the same distance from the center, and the radii of a circle have one endpoint on the circle and one at the center. All circles have a diameter, too. The diameter of a circle is the segment that contains the center and whose endpoints are both on the circle. The length of the diameter is twice that of the radius. Therefore, all diameters of a circle are congruent, too.

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Figure 1.2: The diameter of a circle Keep in mind that an infinite number of radii and diameters can be drawn in a circle. Although they are all congruent, they are not the same. Sometimes a strategically placed radius will help make a problem much clearer. Likewise, diameters can be drawn into a circle to strategically divide the area within the circle. Each of these techniques is prevalent in geometric proofs, and each is based on the facts that all radii are congruent, and all diameters are congruent. However, their position when drawn makes each one different. 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.

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TASK 1 There are a lot of things around us related to circles or parts of a circle.

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(a) Collect pictures of 5 such objects. You may use camera to take pictures around your school compound or get pictures from magazines, newspapers, the internet or any other resources.

BALL

CYLINDRICAL CYLINDRICAL CAN PIPE

CYLINDRICAL ROLLER BEARINGS

NECKLACE CIRCLE

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(b)Pi or π is a mathematical constant related to circles. Define π and write a brief story of π .

Definition of π : 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. In Euclidean plane geometry, π is defined

as

the

ratio

of

a

circle's

circumference to its diameter:

The ratio

/d is constant, regardless of a circle's size. For

C

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]

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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 of π : 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.

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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. And next, we find progress in India, where the mathematician Madhava worked out a power series definition of π, which allowed him

to

compute

π

to 13 decimal

places. 13 decimal

places,

computing a power series completely by hand Astounding! Even better, during the same century, when this work made its way to the great Persian Arabic mathematicians, they worked it out to 9 digits in base-60 (base-60 was in inheritance from the Babylonians). 9 digits in base 60 is roughly 16 digits in decimal! And finally, we get back to Europe; in the 17th century, van Ceulen used the power series to work out 35 decimal places of π. Alas, the publication of it was on his tombstone. Then we get to the 19th century, when William Rutherford calculated 208 decimal places of π. The real pity of that is that he

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made an error in the 153rd digit, and so only the first 152 digits were correct. That was pretty much it until the first computers came along, and once that happened, the fun went out of trying to calculate it, since any bozo could write a program to do it. There's a website that will let you look at its computation of the first 2 hundred million digits of π. The name of π came from Euler (he of the great equation, eiπ + 1 = 0). It's an abbreviation for perimeter in Greek. There's also one bit of urban myth about π that is, alas, not true. The story goes that some state in the American Midwest (Indiana, Iowa, Ohio, Illinois in various versions) passed a law that π=3.

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TASK 2 (a) Diagram 1 shows a semicircle PQR of diameter 10 cm. Semicircles PAB and BCR of diameter d 1 and d 2 respectively are inscribed in the semicircle PQR such that the sum of d 1 and d 2 is equal to 10 cm.

Complete Table 1 by using various values of d 1 and the corresponding values of d 2 . Hence, determine the relation between the lengths of arcs PQR, PAB and BCR. Using formula: Arc of semicircle = ½πd d 1 (cm)

d 2 (cm)

Length of

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Length of

Length of

1.0

arc PQR (cm) 5π

9.0

1.1

8.9

5π

1.5

8.5

5π

2.3

7.7

5π

2.5

7.5

5π

3.0

7.0

5π

3.5

6.5

5π

3.7

6.3

5π

4.0 4.5

6.0 5.5

5π 5π

arc PAB (cm)

arc BCR (cm)

π

7 π 2 89 π 20 17 π 4 77 π 20 15 π 4 7 π 2 13 π 4 63 π 20 3π 11 π 4

2 11 π 20 3 π 4 23 π 20 5 π 4 3 π 2 7 π 4 37 π 20 2π 9 π 4

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 Let d1= 3, and d2 =7

PAB

+ S

BCR

SPQR = S

PAB

+ S

BCR

5π

= ½ π(3) + ½ π(7)

5π

= 3/2 π + 7/2 π

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5π

= 10/2 π

5π = 5 π Therefore, the length of arc PQR equals to the sum of the lengths of arcs of the two inscribed semicircles.

(b)Diagram 2 shows a semicircle PQR of diameter 10 cm. Semicircles PAB, BCD and DER of diameter d 1 , d 2 and d 3 respectively are inscribed in the semicircle PQR such that the sum of d 1 , d 2 and d 3 is equal to 10 cm.

(i)

Using various values of d 1 and d 2 , and the corresponding values of d 3 , determine the relation between the lengths of arcs PQR, PAB, BCD and DER. Tabulate your findings.

d 1 (c

d 2 (cm

d 3 (c

m)

)

m)

1.0

2.0

7.0

Length of arc PQR (cm) 5π

Length of arc PAB (cm)

π 2

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Length of arc BCD (cm)

π

Length of arc DER (cm) 7 π 2

1.2

2.8

6.0

5π

1.5

2.5

6.0

5π

1.5

4.0

3.5

5π

1.7

3.0

5.3

5π

1.8

2.5

5.7

5π

1.9

3.6

5.5

5π

2.0

3.5

4.5

5π

(ii)

3 π 5 3 π 4 3 π 4 17 π 20 9 π 10 19 π 20

π

7 π 5 5 π 4 2π 3 π 2 5 π 4 9 π 5 7 π 4

3π 3π 7 π 4 53 π 20 57 π 20 11 π 4 9 π 4

Based on your findings in (a) and (b), make generalizations 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,… The length of arc PQR equals to the sum of the lengths of arcs of all inscribed semicircles.

Souter = S1 + S2 + S3 + S4 + S5

(c) For different values of diameters of the outer semicircle, show that the generalizations stated in (b)(ii) is still true. Prove: D = d1 + d 2 + d 3 + ... + d n , where D is the diameter of the outer

semicircle and d 1 is the diameter of the smallest inner semicircle. D π = D 2 2

Arc length of PQR = π 

Sum of the lengths of the arcs

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d  d  d  = π  1  + π  2  + ... + π  n   2  2   2  =

=

π 2

π 2

[ d1 + d 2 + d 3 +... + d n ] D

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TASK 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 plot is y m 2 and the diameter of one of the ponds is x m. Express y in terms of π and x. Area of semicircle ACD = =

1 2

π

2

10    2 

25 π 2

Total area of two fish ponds 2

=

1 x 1 10 − x  π  + π   2 2 2  2 

2

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[

1 8

2 = π x 2 + (10 − x )

]

[

=

1 π x 2 + 100 − 20 x + x 2 8

=

1 π 2 x 2 − 20 x +100 8

=

1 π x 2 10 x + 50 4

[

[

]

]

]

‘ y = area of the flower plot y=

25 π π − x 2 −10 x + 50 2 4

y=-

[

π

5 x2 + π 4 2

]

x

(b) Find the diameter of the two fish ponds if area of the flower (Use π =

plot is 16.5 m 2

22 ) 7

Given y = 16.5, 16.5 = -

π 2 5 x + π 4 2

x (use π = 22 ) 7

x 2 − 10 x + 21 = 0 x = 3,7

(c) Reduce the non-linear equation obtained in (a) to simple linear from and hence, plot a straight line graph. Using the straight line graph, determine the area of the flower plot if the diameter of one of the fish ponds is 4.5 m . y=-

π

5 x2 + π 4 2

x

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y π 5 =- x + x 4 2π

x Y y x

1 7.1 7.1

2 12.6 6.3

3 16.5 5.5

4 18.9 4.7

Plot graph manually:

When x=4.5, Y/x = 4.3 Area of flower pot =

= y/x(x) 4.3 x 4.5 = 19.35 m2

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5 19.6 3.9

6 18.9 3.1

7 16.8 2.4

(d) The cost of constructing the fish ponds is higher than that of the flower plot. Use two methods to determine the area of the flower plot such that that the cost of constructing the garden is minimum. In order that the cost of the garden to be minimum, find the maximum area of the flower plot. Note that the cost of constructing the ponds is higher than that of the flower plot.

(Method 1) Tabulation y=-

x 1 2 3 4 5 6 7 8 9

π

5 x2 + π 4 2

x

7.07 12.57 16.50 18.85 19.63 18.85 16.50 12.57 7.07

When x = 5, y =19.63. Therefore, for minimum cost of constructing the garden, the area of the flower plot should be 19.63m 2 .

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(Method 2) Differentiation y=

5 π π x − x2 2 4

dy 5 π = π− x dx 2 2

When

dy =0, dx

5 π π − x =0 2 2

x =5

d2y π =− <0 2 2 dx ⇒ y is maximum when y max =

x =5

5 π π ( 5) − ( 5) 2 2 4

= 19.63 m 2 For minimum cost of constructing the garden, the area of the flower plot should be 19.63m 2 .

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(e) The principal 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 30 cm and the diameter of the flower beds are increased by a constant value successively . Determine the diameter of the remaining flower beds . Diameters of the semicircular flower beds: 30, 30 + d, 30 + 2d,…which is an Arithmetic Progression whereas a = 30, n = 12 and d is the common difference. S n = AB = 1000 cm

S12 =

n [ 2a + (n −1)d ] = 1000 2

S12 =

12 [ 2( 30 ) + (12 −1) d ] = 1000 2 6( 60 +11 d ) = 1000

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360 + 66 d = 1000 1000 − 360 = 66 d

640 = 66 d d = 9.697 = 9.7

∴ Common difference, d = 9.7 cm

Hence, the diameters of the flower beds are: Tn (Flower Bed) 1 2 3 4 5 6 7 8 9 10 11 12

Diameter of flower beds (cm) 30.0 39.7 49.4 59.4 68.8 78.5 88.2 97.9 107.6 117.3 127.0 136.7

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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 circumference, 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(2_r). 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

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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.

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