SEKOLAH SULTAN ALAM SHAH, PUTRAJAYA

Additional MathematicsProject Work 2009

Luqman Hakim bin Abdul Wahid5 Jaya

CONTENTS

Part 1:

Introduction The history of pi (π)

Part 2:

Question 2(a) Question 2(b)

Part 3:

Question 3 Conjecture Conclusion

INTRODUCTION

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

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. Circles had been used in daily lives to help people in their living.

THE HISTORY OF PI(π)

Definition

Pi, π has the value of 3.14159265. 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.

History

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. π is one of the most important mathematical and physical constants: many formulae from mathematics, science, and engineering involve π.

π is an irrational number, which means that its value cannot be expressed exactly as a fraction m/n, where m and n are integers. Consequently, its decimal representation never ends or repeats. It is also a transcendental number, which means that no finite sequence of algebraic operations on integers (powers, roots, sums, etc.) can be equal to its value; proving this was a late achievement in mathematical history and a significant result of 19th century German mathematics. Throughout the history of mathematics, there has been much effort to determine π more accurately and to understand its nature; fascination with the number has even carried over into non-mathematical culture.

The Greek letter π, often spelled out pi in text, was adopted for the number from the Greek word for perimeter "περίμετρος", first by William Jones in 1707, and popularized by Leonhard Euler in 1737. The constant is occasionally also referred to as the circular constant, Archimedes' constant (not to be confused with an Archimedes number), or Ludolph's number (from a German mathematician whose efforts to calculate more of its digits became famous).

The name of the Greek letter π is pi, and this spelling is commonly used in typographical contexts when the Greek letter is not available, or its usage could be problematic. It is not normally capitalised (Π) even at the beginning of a sentence. 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/.

The constant is named "π" because "π" is the first letter of the Greek words περιφέρεια (periphery) and περίμετρος (perimeter), probably referring to its use in the formula to find the circumference, or perimeter, of a circle. π is Unicode character U+03C0 ("Greek small letter pi").

10 cm

Question 2(a)

Diagram shows 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 10 cm. Complete the table by using various values of d1 and d2 and the corresponding values of d2. Hence, determine the relation between the lengths of arcs PQR, PAB and BCR.

By using formula ½ πd to find the arc length of a semicircle;

d1 (cm) d2 (cm)Length of arc PQR in

terms of π (cm)Length of arc PAB in

terms of π (cm)Length of arc BCR in

terms of π (cm)

1 9 5π12

π92

π

2 8 5π π 4π

3 7 5π32

π72

π

4 6 5π 2π 3π

5 5 5π52

π52

π

6 4 5π 3π 2π

7 3 5π72

π32

π

8 2 5π 4π π

Q

A

C

P B Rd1 d2

9 1 5π92

π12

π

From the table we know that the length of arc PQR is not affected by the different in d1 and d2 in PAB and BCR respectively. We also can determine that the relation between the length of arc PQR, PAB and BCR is that the length of arc PQR is equal to the sum of the length of arcs PAB and BCR respectively. Therefore, we can get the equation;

Assuming that S is for the length of arc;

∴ SPQR = SPAB + SBCR

5π = 12

π d1 + 12

π d2

Let d1 = 6 cm, d2 = 4 cm, 5π = 12

π (6) + 12

π (4)

5π = 62

π + 42

π

5π = 102

π

5π = 5π

10 cm

Question 2(b)

Diagram 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.

(i) By using the same formula ½ πd as before to find the arc length of a semicircle;

d1 d2 d3 SPQR SPAB SBCD SDER

1 1 8 5π12

π12

π 4π

1 2 7 5π12

π π72

π

1 3 6 5π12

π32

π 3π

Q

EC

A

P B D Rd1 d2 d3

2 3 5 5π π32

π52

π

3 3 4 5π32

π32

π 2π

3 4 3 5π32

π 2π32

π

3 5 2 5π32

π52

π π

3 6 1 5π32

π 3π12

π

Therefore, the relation between the lengths of arcs PQR, PAB, BCD and DER based on the table is the sum of all the lengths of arcs of the inner semicircles which is PAB, BCD and DER respectively will equal to the length of arc of the outer semicircle, which is PQR.

Assuming that S is for the length of arc;

∴ SPQR = SPAB + SBCD + SDER

Let d1 = 1 cm, d2 = 2 cm and d3 = 7 cm,

5π = 12

π + π + 72

π

5π = 5π

(ii) Based on my findings in (a) and (b), I can conclude that the sum of the lengths of arcs of the inner semicircles for n inner semicircles where n = 2, 3, 4, … is equal to the length of arc of the outer semicircle.

Assuming that S is for the length of arc;

∴ SOUTER = ∑ SINNER

Question 2(c)

Assume that the diameter of the outer semicircle is 20 cm and 4 semicircles are inscribed in the semicircle such that the sum of d1, d2 , d3 , d4 is equal to 20 cm. Again, by using various values of d1 (arc 1), d2 (arc 2), d3 (arc 3), and the corresponding values of d4 (arc 4), in which d1 + d2 + d3 + d4 = 20 cm, I had make a table to determine the relation between the lengths of the arcs of the inner semicircles with the length of arc of the outer semicircle. All the results are tabulated in table below:

By using the same formula ½ πd as before to find the arc length of a semicircle;

d1 (cm) d2 (cm) d3 (cm) d4 (cm)

Length of arc of the

outer semicircle in terms of π (cm)

Length of arc 1 in

terms of π (cm)

Length of arc 2 in

terms of π (cm)

Length of arc 3 in

terms of π (cm)

Length of arc 4 in

terms of π (cm)

1 2 3 14 10π12

π π32

π 7π

2 3 4 11 10π π32

π 2π112

π

3 4 5 8 10π32

π 2π52

π 4π

4 5 6 5 10π 2π52

π 3π52

π

5 6 7 2 10π52

π 3π72

π π

From the table above, we know that the length of arc of the outer semicircle is not affected by the different in d1, d2, d3, and d4 in arc 1, arc 2, arc 3, and arc 4 respectively. The relation between the length of arc of outer semicircle and the lengths of arcs of the inner semicircle which are arc 1, arc 2, arc 3, and arc 4 is that the length of arc of outer semicircle is equal to the sum of the length of arc 1, arc 2, arc 3, and arc 4 respectively and the equation is:

SOUTER SEMICIRCLE= Sarc 1 + Sarc 2 + Sarc 3 + Sarc 4

Let d1 = 1 cm, d2 = 2 cm, d3 = 3 cm, and d4 = 14 cm,

10π = 12

π + π + 32

π + 7π

10π = 10π

Therefore, I can conclude that the length of arc of the outer semicircle is equal to the lengths of arcs of the inner semicircles where n = 2, 3, 4, … .

10 m

Question 3

The Mathematics Society is given a task to design a garden to beautify the school by using the design as shown in diagram above. The shaded region will be planted with flower and the two inner semicircles are fish ponds.

D

E

F

A Cx m

B

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

Area of flower plot = y m²

Using the formula 12

πr² to find the area of the semicircle;

y = 12

π (5)² – (12

π (x2

)² + 12

π (1 0 - x2

)²)

y = 252

π – (x ²8

π + (1 0 0 – 20 x + x ² )8

π)

y = 252

π – 2x² – 20x + 1008

π

y = 252

π – x² – 10x + 504

π

y = 10x – x ²4

π Therefore, the area of the flower plot is; y = 10x – x ²4

π.

(b) Find the diameters of the two fish ponds if the area of the flower plot is 16.5 m².

(Use π = 227 )

Area of flower plot = 16.5 m²

∴ 16.5 = 10x – x²4

π

66 = 10x – x² (227

)

66 (722

) = 10x – x²

0 = x² – 10x + 21

0 = (x – 7) (x – 3)

∴ x = 7 , x = 3

The diameter of the fish ponds is 7 m and 3 m respectively.

(c) Reduce the non linear equation to simple linear form. 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.

Since y = 10x – x ²4

π;

∴ yx

= 10x – x ²4 x

π

yx

= 10 – x4

π

yx

= (52

– 14x) π

x 1 2 3 4 5 6 7

yx

7. 07 6. 28 5. 50 4. 71 3. 93 3. 14 2. 36

(See graph paper.)

From the graph, we can determine the area of the flower plot if the diameter of one of the fish pond is 4.5 m.

Area of flower plot = y

When x = 4.5, yx

=

y = yx

(x)

y =

y =

Therefore, the area of the flower plot if the diameter of one of the fish pond is 4.5 m is equal to __________ m².

(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.

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 the square method.

I. Differentiation method:

y = 10x – x ²4

π

dydx

= 10x – x ²4

π

= 52

π – x2

π

Since the cost of constructing the garden is minimum, thus dydx

is equal to 0.

0 = 52

π – x2

π

52

π = x2

π

∴ x = 5 m²

II. Completing the square method:

y = 10x – x ²4

π

= 10x4

π – x ²4

π

= – 14

π (x² – 10x)

= – 14

π (x – 5)² – 5²

= – 14

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 m²

Therefore, the area of the flower plot is 5 m².

(e) The principle suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society. 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.

By using the formula Sn = n2

[2a + (n – 1)] d;

n (no. of 12 semicircular flowers) = 12

a (the 1st flower bed) = 30 cm

S₁₂ (sum of all the flower beds) = 1000 cm / 10 m

S₁₂ = 122

[2(30) + (12 – 1)] d

1000 = 6 (60 + 11d)

1000 = 360 + 66d

1000 – 360 = 66d

640 = 66d

d = 9.697

Therefore, I can now find all the diameter of the remaining flower beds by adding the constant value of the arithmetic progressions, which is d, and the result is tabulated in the table below.

Tn (flower bed) Diameter (cm) Tn (flower bed) Diameter (cm)

T1 30 T7 88.182

T2 39.697 T8 97.879

T3 49.394 T9 107.576

T4 59.091 T10 117.273

T5 68.788 T11 126.97

T6 78.485 T12 136.667

CONCLUSION

So, when I finished this project work for additional mathematics 2009, I had draw a general conclusion about the length of the 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. This can be simplified by; SOUTER = ∑ SINNER, where S represents the length of arc. Furthermore, I also can conclude that we can use the concept of additional mathematics in our daily life such as when we want to built a circle fish pond, we need to use the concept of circular measure to measure the pond’s circumference and area. Therefore, I had learn many things while making this project and realize that additional mathematics is important and it make our life easier.