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8/3/2019 Addmath Proj to Print http://slidepdf.com/reader/full/addmath-proj-to-print 1/21 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 is 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.

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Introduction

There are a lot of things around us related to circles or parts of a circle. Acircle 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. Thecommon 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 usedinterchangeably to refer to either the boundary of the figure (known as theperimeter) or to the whole figure including its interior. However, in strict technicalusage, "circle" refers to the perimeter while the interior of the circle is called adisk. The circumference of a circle is the perimeter of the circle (especially whenreferring to its length).

A circle is a special ellipse in which the two foci are coincident. Circles areconic sections attained when a right circular cone is intersected with a planeperpendicular 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, makesmuch of modern civilization possible. In mathematics, the study of the circle hashelped inspire the development of geometry and calculus. Circles had been usedin daily lives to help people in their living.

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 is constant, regardless of a circle's size. For example, if a circle has

twice the diameter of another circle it will also have twice the circumference, C,

preserving the ratio .

Alternatively π can be also defined as the ratio of a circle's area (A) tothe 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'scircumference 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 to3.14159 in the usual decimal notation. π is one of the most importantmathematical and physical constants: many formulae from mathematics, science,and engineering involve π.

π is an irrational number, which means that its value cannot be expressedexactly as a fractionm/n, wherem andn are integers. Consequently, its decimalrepresentation never ends or repeats. It is also a transcendental number, whichmeans 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 Germanmathematics. Throughout the history of mathematics, there has been much effortto determine π more accurately and to understand its nature; fascination with thenumber has even carried over into non-mathematical culture.

The Greek letter π, often spelled outpi in text, was adopted for the number from the Greek word forperimeter "περίμετρος", first by William Jones in 1707,and popularized by Leonhard Euler in 1737. The constant is occasionally alsoreferred to as the circular constant, Archimedes' constant (not to be confusedwith 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 intypographical contexts when the Greek letter is not available, or its usage couldbe problematic. It is not normally capitalised (Π) even at the beginning of asentence. When referring to this constant, the symbol π is always pronouncedlike "pie" in English, which is the conventional English pronunciation of the Greekletter. In Greek, the name of this letter is pronounced /pi/.

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

PART ICakes come in variety of forms and flavours and are among favourite dessertsserved during special occasions such as birthday parties, Hari Raya , weddingsand etc. Cakes are treasured not only because of their wonderful taste but also in

the art of cake baking and cake decorating.Find out how mathematics is used in cake baking and cake decorating andwrite about your findings.

Constructing the structure of a cakeThese cakes are made by using different sizes of circular pans, then stacking thebaked cake sections on top of each other.You are to plan for a cake that will serve between 200 and 250 people.

• The wedding cake must feed between 200 and 250 people.

• You have 4 different sizes of pans of you can use. ( All pans have the

same height )

r = 10 cm r = 15 cm r = 20 cm r = 25cm

• Each layer of cake must remain a cylinder

• You can stack layers . Each layer can then be separated and cutindividually.

• Each layer of cake will be cut into sectors that have a top area of exactly50 cm2

• You may have some left-over cake from a layer

Example of 50 cm2

top area of sector

• One sector feeds one person.

• Your final ingredients list must be proportional to the ingredients list

provided for you.

By using the theory of arithmetic and geometric progressions in Chapter 1 Form

5, the concept can be used to

• Decide on how many layers of each size of cake you will need for your

cake.• Show how you can cut the layers of the cake into equivalent sectors

having a top area of 50 cm2 each, in order to feed between 200 and 250people.

• Complete the ingredients list by identifying the quantities needed for eachingredient in the cake.

Work and calculations to determinethe ingredients of the cakeBaking a cake offers a tasty way to practice math skills, such as fractions andratios, in a real-world context. Many steps of baking a cake, such as countingingredients and setting the oven timer, provide basic math practice for youngchildren. Older children and teenagers can use more sophisticated math to solvebaking dilemmas, such as how to make a cake recipe larger or smaller or how todetermine what size slices you should cut. Practicing math while baking not onlyimproves your math skills, it helps you become a more flexible and resourcefulbaker.

• Calculate the proportions of different ingredients. For example, a frosting recipe thatcalls for 2 cups cream cheese, 2 cups confectioners' sugar and 1/2 cup butter has acream cheese, sugar and butter ratio of 4:4:1. Identifying ratios can also help youmake recipes larger or smaller.

• Use as few measuring cups as possible. For example, instead of using a 3/4 cup,use a 1/4 cup three times. This requires you to work with fractions.

• Determine what time it will be when the oven timer goes off. For example, if your

cake has to bake for 30 minutes and you set the timer at 3:40, the timer will go off at4:10.

• Calculate the surface area of the part of the cake that needs frosting. For example, asheet cake in a pan only needs the top frosted, while a sheet cake on a tray needsthe top and four sides frosted. A round layer cake requires frosting on the top, oneach layer and on the sides.

• Determine how large each slice should be if you want to serve a certain amount of people. For example, an 18 by 13 inch sheet cake designed to serve 25 peopleshould be cut into slices that measure approximately 3 by 3 inches.

Add up the cost of your ingredients to find the cost of your cake. Estimate the cost of partially used ingredients, such as flour, by determining the fraction of the container used and multiplying that by the cost of the entire container

Initial draft of the cake

r = 10 cm

r = 15 cmr = 20 cm h

= 20 cm

r = 25 cm

PART IIBest Bakery shop received an order from your school to bake a 5 kg of round cake asshown in Diagram 1 for the Teacher’s Day celebration.

h cm

d cm

1) If a kilogram of cake is has a volume of 3800 cm2, and the height

of the cake is to be 7.0cm, calculate the diameter of the baking

tray to be used to fit the 5 kg cake ordered by your school. [ use

]

Volume of 5kg cake = Base area of cake x Height of cake

3800 x 5 = (3.142)( )² x 7

(3.142) = ( )²

863.872 = ( )²

= 29.392

d = 58.784 cm

i) State the range of heights that is NOT suitable for

h < 7cm is NOT suitable, because the resulting diameter

produced is too large to fit into the oven. Furthermore,the cake would be too short and too wide, making it less

attractive.

ii) Suggest the dimensions that you think most

suitable for the cake. Give the reasons for your

h = 8cm, d = 54.99cm, because it can fit into the oven, and the

size is suitable for easy handling.

c) i) Form an equation to represent the linear relation

between h and d . Hence, plot a suitable graph based on

the equation that you have formed.

[You may draw your graph with the air of computer

software]

19000 = (3.142)( )²h

19000/(3.142)h =

= d²

d =

d =

log d =

log d = log h + log 155.53

Log h 0 1 2 3 4Log d 2.19 1.69 1.19 0.69 0.19

ii)

(a) If Best Bakery received an order to bake a cake where

the height of the cake is 10.5 cm, use your graph based onthe equation that you have formed.

h = 10.5cm, log h = 1.021, log d = 1.680, d = 47.86cm

(b) If Best Bakery used a 42 cm diameter round cake tray,

use your graph to estimate the height of the cake obtained.

Answer:d = 42cm, log d = 1.623, log h = 1.140, h = 13.80cm

1) Best bakery has been requested to decorate the cake with freshcream. The thickness of the cream is normally set to a uniform

(a) Estimate the amount of fresh cream required to decorate

the cake using the dimensions that you have suggested in 2 (b)

(ii)

h = 8cm, d = 54.99cm

Amount of fresh cream = VOLUME of fresh cream needed (area x height)

Amount of fresh cream = Vol. of cream at the top surface + Vol. of cream at the side

surface

Vol. of cream at the top surface

= Area of top surface x Height of cream

= (3.142)( )² x 1

= 2375 cm³

Vol. of cream at the side surface

= Area of side surface x Height of cream

= (Circumference of cake x Height of cake) x Height of cream

= 2(3.142)(54.99/2)(8) x 1

= 1382.23 cm³

Therefore, amount of fresh cream = 2375 + 1382.23 = 3757.23 cm

(b) Suggest three other shapes for cake, that will have the

same height and volume as those suggested in 2(b)(ii).

Estimate the amount of fresh cream to be used on each of

the cakes.

1. Rectangle shaped-base (cuboid)

19000 = base area x height

base area =

length x width = 2375By trial and improvement, 2375 = 50 x 47.5 (length = 50, width = 47.5, height = 8)

Therefore, volume of cream= 2(Area of left/right side surface)(Height of cream) + 2(Area of front/back sidesurface)(Height of cream) + Vol. of top surface= 2(8 x 50)(1) + 2(8 x 47.5)(1) + 2375 = 3935 cm³

2. Triangle-shaped base

19000 = base area x heightbase area = 2375

x length x width = 2375

length x width = 4750By trial and improvement, 4750 = 95 x 50 (length = 95, width = 50)Slant length of triangle = √(95² + 25²)= 98.23Therefore, amount of cream= Area of rectangular front side surface(Height of cream) + 2(Area of slantrectangular left/right side surface)(Height of cream) + Vol. of top surface

= (50 x 8)(1) + 2(98.23 x 8)(1) + 2375 = 4346.68 cm³

3. Pentagon-shaped base

19000 = base area x heightbase area = 2375 = area of 5 similar isosceles triangles in a pentagontherefore:

2375 = 5(length x width)475 = length x widthBy trial and improvement, 475 = 25 x 19 (length = 25, width = 19)

Therefore, amount of cream= 5(area of one rectangular side surface)(height of cream) + vol. of top surface= 5(8 x 19) + 2375 = 3135 cm³

(c) Based on the values that you have found which shape

requires the least amount of fresh cream to be used?

Pentagon-shaped cake, since it requires only 3135 cm³ of cream to beused.

PART IIIFind the dimension of a 5 kg round cake that requires the minimum amount freshcream to decorate. Use at least two different methods including calculus. Statewhether you would choose to bake a cake of such dimensions. Give reasons for

Method 1: DifferentiationUse two equations for this method: the formula for volume of cake (as in Q2/a),and the formula for amount (volume) of cream to be used for the round cake (asin Q3/a).19000 = (3.142)r²h → (1)V = (3.142)r² + 2(3.142)rh → (2)

From (1): h = → (3)

Sub. (3) into (2):

V = (3.142)r² + 2(3.142)r( )

V = (3.142)r² + ( )

V = (3.142)r² + 38000r -1

( ) = 2(3.142)r – ( )

0 = 2(3.142)r – ( ) -->> minimum value, therefore = 0

= 2(3.142)r

= r³

6047.104 = r³r = 18.22

Sub. r = 18.22 into (3):h =

h = 18.22therefore, h = 18.22cm, d = 2r = 2(18.22) = 36.44cm

Method 2: Quadratic FunctionsUse the two same equations as in Method 1, but only the formula for amount of cream is the main equation used as the quadratic function.Let f(r) = volume of cream, r = radius of round cake:19000 = (3.142)r²h → (1)

f(r) = (3.142)r² + 2(3.142)hr → (2)From (2):f(r) = (3.142)(r² + 2hr) -->> factorize (3.142)

= (3.142)[ (r + )² – ( )² ] -->> completing square, with a = (3.142), b = 2h and

c = 0= (3.142)[ (r + h)² – h² ]= (3.142)(r + h)² – (3.142)h²(a = (3.142) (positive indicates min. value), min. value = f(r) = –(3.142)h²,corresponding value of x = r = --h)

Sub. r = --h into (1):19000 = (3.142)(--h)²hh³ = 6047.104h = 18.22

Sub. h = 18.22 into (1):19000 = (3.142)r²(18.22)r² = 331.894r = 18.22therefore, h = 18.22 cm, d = 2r = 2(18.22) = 36.44 cm

I would choose not to bake a cake with such dimensions because itsdimensions are not suitable (the height is too high) and therefore less

attractive. Furthermore, such cakes are difficult to handle easily.

FURTHER EXPLORATION

Best Bakery received an order to bake a multi-storey cake for Merdeka Daycelebration, as shown in Diagram 2.

Diagram 2

The height of each cake is 6.0 cm and the radius of the largest cake is 31.0 cm.The radius of the second cake is 10% less than the radius of the first cake, theradius of the third cake is 10% less than the radius of the second cake and so

one.(a) Find the volume of the first, the second, the third and the fourth cakes. By

comparing all these values, determine whether the volumes of the cakesform a number patterns? Explain and elaborate on the number patterns.

height, h of each cake = 6cm

radius of largest cake = 31cmradius of 2nd cake = 10% smaller than 1st cakeradius of 3rd cake = 10% smaller than 2nd cake

31, 27.9, 25.11, 22.599…

a = 31, r =

V = (3.142)r²h

Radius of 1st cake = 31, volume of 1st cake = (3.142)(31)²(6) = 18116.772

Radius of 2nd cake = 27.9, vol. of 2nd cake = 14674.585

Radius of 3rd cake = 25.11, vol. of 3rd cake = 11886.414

Radius of 4th cake = 22.599, vol. of 4th cake = 9627.995

18116.772, 14674.585, 11886.414, 9627.995, …

a = 18116.772, ratio, r = T2/T1 = T3 /T2 = … = 0.81

(b)If the total mass of all the cakes should not exceed 1.5 kg,

calculate the maximum number of cakes that the bakery needs to

Sn =

Sn = 57000, a = 18116.772 and r = 0.81

57000 =

1 – 0.81n = 0.59779

0.40221 = 0.81n

og0.81 0.40221 = n

n =

n = 4.322

therefore, n ≈ 4

REFLECTION

I have done many researches throughout the internet and discussing witha friend who have helped me a lot in completing this project. Through thecompletion of this project, I have learned many skills and techniques. This projectreally helps me to understand more about the uses of progressions in our dailylife.

This project also helped expose the techniques of application of additional mathematics in real life situations. I have learnt how to bake a weddingtiered cake stands with good quality and proper height.

Apart from that, this project encourages the student to work together andshare their knowledge. It is also encourage student to gather information from theinternet, improve thinking skills and promote effective mathematicalcommunication.

Last but not least, I proposed this project should be continue because itbrings a lot of moral values to the student and also test the studentsunderstanding in Additional Mathematics.