Addmath Project 2009

PROJECT WORK FOR ADDITIONAL MATHEMATICS 2009

Nama: Muhammad Yusuf Bin Nor Hashim No i/c: 920329-10-5353 Sekolah Menengah Kebangsaan Bandar Baru Bangi

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-ContentNo. 1 2 3 4

Contents Introduction Part 1 Part 2 Part 3

Page 3 4-5 6 - 9 10 - 16

-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

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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.[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]

PART 1 a) Collect pictures of 5 such objects. You may use camera to take picture around your school compound or get from magazines, newspaper, the internet or any other resources.

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Wheel of a bicycle

Fish pond

Circles on water surface

School park

Round table at school compound

b) Pi or π is a mathematical constant related to circles. Define π and write a brief history of π What is Pi (π)? Who first used Pi (π)? How do you find its value? What is it for? How many digits is it?

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By definition, Pi is the ratio of the circumference of a circle to its diameter. Pi is always the same number, no matter which circle you use to compute it. For the sake of usefulness people often need to approximate Pi. For many purposes you can use 3.14159, which is really pretty good, but if you want better approximation you can use a computer to get it. Here’s Pi to many more digits 3.14159265358939323846. The area of a circle is Pi times the square of the length of the radius or = πr A very briefly history about Pi (π) Pi is very old number. We know that the Egyptians and the Babylonians knew about the existence of the constants ratio Pi, although they didn’t know its value nearly as well as we do today. The had figured out that it was a little bigger than 3; the Babylonians had an approximation of 3 1/8(3.125, and the Egyptians had a somewhat worse approximation of 4(8/9)2 (about 3.160484) , which is slightly less accurate and much harder to work with. With modern symbol for π was first used in our modern sense in 1706 by William, Jones, who wrote: There are various other ways of finding the lengths or areas of particular curve lines, or planes, which may very much facilitate the practice, as for instances, in the circle, the diameter is to the circumference as 1 to 3.14159. Pi is an infinite decimal. Unlike numbers such as 3, 9.876 and 4.5 which has finitely many nonzero numbers to the right of the decimal place, Pi has infinitely many numbers to the right decimal point If you write Pi down in the decimal form, the numbers to the right of the 0 never repeat in a pattern. Some infinite decimals do have patterns – for instance, the infinite decimal .333333…. has all 3’s to the right of the decimal point, and the numbers .123456789123456789… the sequence 123456789 is repeated. However, although many mathematicians have tried to find it, no repeating pattern for Pi has been discovered – in fact, in 1768 Johann Lambert proved that there cannot be any such repeating pattern. Pi shows up in some unexpected places like probability and the ‘famous five’ equation connecting the five most important numbers in mathematics, 0, 1, e, Pi and i: e^ (iπ) + 1 = 0.

PART 2 a)

Diagram 1 shows a semicircle PQR of diameter 10 cm. semicircles PAB and BCR of diameter d1 and d2 respectively are inscribed in the semicircle PQR such that the sum of d1 and d2 is equal to 10cm. 5

Complete Table 1 by using various values of d1 and the corresponding values of d2. Hence, determine the relation between the lengths of arcs PQR, PAB and BCR. The length of arc (s) of circle can be found by using the formula

s = (2πr) = πr2

d1

d2

Length of arc PQR in

Length of arc PAB in

Length of arc BCR in

(cm) 1 2

(cm) 9 8

terms of π (cm) 5π 5π

terms of π (cm) ½π π

terms of π (cm) 9/2 π 4π 6

3 4 5 6 7 8 9

7 6 5 4 3 2 1

5π 5π 5π 5π 5π 5π 5π

3/2 π 2π 5/2π 3π 7/2 π 4π 9/2 π

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 are we can get the equation:

SPQR = S + S PAB

Let d1= 3, and d2 =7

BCR

SPQR = S + S PAB

BCR

5π

= ½ π (3) + ½ π (7)

5π

= 3/2 π + 7/2 π

5π

= 10/2 π

5π = 5 π

b) Diagram 2 shows a semicircle PQR of diameter 10cm. semicircle 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 10cm

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b) i) Using various values of d1, d2 and the corresponding values of d3, determine the relation between the lengths of arcs PQR, PAB, BCD and DER

d1 d2 d3 SPQR

SPAB

SBCD

SDER

1

2

7

5π

1/2 π

π

7/2 π

2

2

6

5π

π

π

3π

2

3

5

5π

π

3/2 π

5/2 π

2

4

4

5π

π

2π

2π

2

5

3

5π

π

5/2 π

3/2 π

SPQR = SPAB + SBCD + SDER

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Let d1 = 2, d2 = 5, d3 = 3 SPQR = SPAB + SBCD + SDER 5 π = π + 5/2 π + 3/2 π 5π= 5π

b) ii) Based on your findings in (a) and (b), make generalizations about the length of the arc of the outer semicircles and the lengths of arcs of the inner semicircles for n inner semicircles where n = 2, 3, 4… 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) For different values of diameters of the outer semicircle, shows\ that the generalisations stated in b (ii) is still true. 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

d2

d3

d4

SABC

SAPQ

SQRS

SSTU

SUVC

10

8

6

6

15 π

5π

4π

3π

3π

12

3

5

10

15 π

6π

3/2 π

5/2 π

5π

14

8

4

4

15 π

7π

4π

2π

2π

15

5

3

7

15 π

15/2 π

5/2 π

3/2 π

7/2 π

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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

PART 3 The Mathematics Society is given a task to design a garden to beautify the school by using the design as shown in the 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 fish ponds is x m. express y in terms of π and x. Area of flower plot = y m2 y = (25/2) π - (1/2(x/2)2 π + 1/2((10-x)/2)2 π)

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= (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

y=

((10x – x2)/4) π

b) Find the diameter of the two fish ponds if the area of the flower plot is 16.5 m2. (Use π = 22/7) y = 16.5 m2 ((10x – x2)/4) π

16.5 = 66 =

(10x - x2) 22/7

66(7/22) = 10x – x2 0 = x2 - 10x + 21 0 = (x-7) (x – 3) x=7 , x=3

c) Reduce the non-linear equation obtained in (a) to simple linear form 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=

((10x – x2)/4) π

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 11

Y/x

8.0

7.0

6.0

5.0

4.0

3.0

2.0 0

1

2

3

4

5

12

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) 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 the cost of constructing the garden is minimum. 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

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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 cm. n = 12, a = 30cm, S 12 = 1000cm

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.

S12 = n/2 (2a + (n – 1) d 1000 = 12/2 (2(30) + (12 – 1) d) 1000 = 6 (60 + 11d)

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1000 = 360 + 66d 1000 – 360 = 66d 640 = 66d d = 9.697

Tn (flower

Diameter

bed) T1 T2 T3 T4 T5 T6

(cm) 30 39.697 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

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