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1. Brief History of Geometry

Geometry began with a practical need to measure shapes. The word geometry means to

measure the earth and is the science of shape and size of things. It is believed that geometryfirst became important when an Egyptian pharaoh wanted to tax farmers who raised crops along

the Nile River. To compute the correct amount of tax the pharaohs agents had to be able tomeasure the amount of land being cultivated.

Around 2900 BC the first Egyptian pyramid was constructed. Knowledge of geometry was

essential for building pyramids, which consisted of a square base and triangular faces. The

earliest record of a formula for calculating the area of a triangle dates back to 2000 BC. TheEgyptians (5000500 BC) and the Babylonians (4000500 BC) developed practical geometry tosolve everyday problems, but there is no evidence that they logically deduced geometric facts

from basic principles.

It was the early Greeks (600 BC400 AD) that developed the principles of modern geometrybeginning with Thales of Miletus (624547 BC). Thales is credited with bringing the science ofgeometry from Egypt to Greece. Thales studied similar triangles and wrote the proof thatcorresponding sides of similar triangles are in proportion.

The next great Greek geometer was Pythagoras (569475 BC). Pythagoras is regarded as thefirst pure mathematician to logically deduce geometric facts from basic principles. Pythagoras

founded a brotherhood called the Pythagoreans, who pursued knowledge in mathematics, science,

and philosophy. Some people regard the Pythagorean School as the birthplace of reason andlogical thought. The most famous and useful contribution of the Pythagoreans was the

Pythagorean Theorem. The theory states that the sum of the squares of the legs of a right triangle

equals the square of the hypotenuse.

Euclid of Alexandria (325265 BC) was one of the greatest of all the Greek geometers and isconsidered by many to be the father of modern geometry. Euclid is best known for his 13-booktreatise The Elements. The Elements is one of the most important works in history and had a

profound impact on the development of Western civilization.

Euclid began The Elements with just a few basics, 23 definitions, 5 postulates, and 5 common

notions or general axioms. An axiom is a statement that is accepted as true. From these basics, he

proved his first proposition. Once proof was established for his first proposition, it could then be

used as part of the proof of a second proposition, then a third, and on it went. This process is

known as the axiomatic approach. Euclids Elements form the basis of the modern geometry thatis still taught in schools today.


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Archimedes of Syracuse (287212 BC) is regarded as the greatest of the Greek mathematiciansand was also the inventor of many mechanical devices including the screw, the pulley, and thelever. The Archimedean screwa device for raising water from a low level to a higher oneisan invention that is still in use today. Archimedes works include his treatiseMeasurement of aCircle, which was an analysis of circular area, and his masterpiece On the Sphere and the

Cylinderin which he determined the volumes and surface areas of spheres and cylinders.

There were no major developments in geometry until the appearance of Rene Descartes(15961650). In his famous treatiseDiscourse on the Method of Rightly Conducting the Reasonin the Search for Truth in the Sciences, Descartes combined algebra and geometry to create

analytic geometry. Analytic geometry, also known as coordinate geometry, involves placing ageometric figure into a coordinate system to illustrate proofs and to obtain information using

algebraic equations.

The next great development in geometry came with the development of non-Euclideangeometry. Carl Friedrich Gauss (17771855) who along with Archimedes and Newton is

considered to be one of the three greatest mathematicians of all time, invented non-Euclidiangeometry prior to the independent work of Janos Bolyai (18021860) and Nikolai Lobachevsky(1792-1856). Non-Euclidian geometry generally refers to any geometry not based on the

postulates of Euclid, including geometries for which the parallel postulate is not satisfied. The

parallel postulate states that through a given point not on a line, there is one and only one lineparallel to that line. Non-Euclidian geometry provides the mathematical foundation for EinsteinsTheory of Relativity.

The most recent development in geometry is fractal geometry. Fractal geometry was

developed and popularized by Benoit Mandelbrot in his 1982 bookThe Fractal Geometry ofNature. A fractal is a geometric shape, which is self-similar (invariance under a change of scale)

and has fractional (fractal) dimensions. Similar to chaos theory, which is the study of non-linearsystems; fractals are highly sensitive to initial conditions where a small change in the initial

conditions of a system can lead to dramatically different outputs for that system.


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The Formulas and Application in Geometry

Right Triangle and Pythagora's theorem

Pythagora's theorem: The two sides a and b of a right triangle and the hypotenuse c are related

by

a2

+ b2

= c2

Area and Perimeter of Triangle

Perimeter = a + b + c

There are several formulas for the area.

If the base b and the corresponding height h are known, we use the formula

Area = (1 / 2) * b * h.

If two sides and the angle between them are known, we use one of the formulas, depending on

which side and which angle are known

Area = (1 / 2)* b * c sin A

Area = (1 / 2)* a * c sin B

Area = (1 / 2)* a * b sin C .

If all three sides are known, we may use Heron's formula for the area.


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Area = sqrt [ s(s - a)(s - b)(s - c) ] , where s = (a + b + c)/2.

Area and Perimeter of Rectangle

Perimeter = 2L + 2W

Area = L * W

Area of Parallelogram

Area = b * h

Area of Trapezoid

Area = (1 / 2)(a + b) * h


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Circumference of a Circle and Area of a Circular Region

Circumference = 2*Pi*r

Area = Pi*r2

Arc length and Area of a Circular Sector

Arclength: s = r*t

Area = (1/2) *r2

* t

where t is the central angle in RADIANS.


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Volume and Surface Area of a Rectangular Solid

Volume = L*W*H

Surface Area = 2(L*W + H*W + H*L)

Volume and Surface Area of a Sphere

Volume = (4/3)* Pi * r3

Surface Area = 4 * Pi * r2

Volume and Surface Area of a Right Circular Cylinder

Volume = Pi * r2

* h

Surface Area = 2 * Pi * r * h


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Volume and Surface Area of a Right Circular Cone

Volume = (1/3)* Pi * r2

* h

Surface Area = Pi * r * sqrt (r2

+ h2)

2.A Brief History of Geometry

Sir Isaac Newton Gottfried Wilhelm von Leibniz

The discovery of calculus is often attributed to two men, Isaac Newton and Gottfried Leibniz,

who independently developed its foundations. Although they both were instrumental in its

creation, they thought of the fundamental concepts in very different ways. While Newton

considered variables changing with time, Leibniz thought of the variables x and y as ranging

over sequences of infinitely close values. He introduced dx and dy as differences betweensuccessive values of these sequences. Leibniz knew that dy/dx gives the tangent but he did not

use it as a defining property. On the other hand, Newton used quantities x' and y', which were

finite velocities, to compute the tangent. Of course neither Leibniz nor Newton thought in termsof functions, but both always thought in terms of graphs. For Newton the calculus was

geometrical while Leibniz took it towards analysis.
http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Newton.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Leibniz.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Leibniz.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Newton.html


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It is interesting to note that Leibniz was very conscious of the importance of good notation

and put a lot of thought into the symbols he used. Newton, on the other hand, wrote more forhimself than anyone else. Consequently, he tended to use whatever notation he thought of on that

day. This turned out to be important in later developments. Leibniz's notation was better suited to

generalizing calculus to multiple variables and in addition it highlighted the operator aspect of

the derivative and integral. As a result, much of the notation that is used in Calculus today is dueto Leibniz.

The development of Calculus can roughly be described along a timeline which goes through

three periods: Anticipation, Development, and Rigorization. In the Anticipation stage techniques

were being used by mathematicians that involved infinite processes to find areas under curves ormaximaize certain quantities. In the Development stage Newton and Leibniz created the

foundations of Calculus and brought all of these techniques together under the umbrella of the

derivative and integral. However, their methods were not always logically sound, and it took

mathematicians a long time during the Rigorization stage to justify them and put Calculus on asound mathematical foundation.

In their development of the calculus both Newton and Leibniz used "infinitesimals",quantities that are infinitely small and yet nonzero. Of course, such infinitesimals do not really

exist, but Newton and Leibniz found it convenient to use these quantities in their computations

and their derivations of results. Although one could not argue with the success of calculus, thisconcept of infinitesimals bothered mathematicians. Lord Bishop Berkeley made serious

criticisms of the calculus referring to infinitesimals as "the ghosts of departed quantities".

Berkeley's criticisms were well founded and important in that they focused the attention of

mathematicians on a logical clarification of the calculus. It was to be over 100 years, however,

before Calculus was to be made rigorous. Ultimately, Cauchy, Weierstrass, and Riemann

reformulated Calculus in terms of limits rather than infinitesimals. Thus the need for theseinfinitely small (and nonexistent) quantities was removed, and replaced by a notion of quantities

being "close" to others. The derivative and the integral were both reformulated in terms of limits.

While it may seem like a lot of work to create rigorous justifications of computations thatseemed to work fine in the first place, this is an important development. By putting Calculus on a

logical footing, mathematicians were better able to understand and extend its results, as well as

to come to terms with some of the more subtle aspects of the theory.

When we first study Calculus we often learn its concepts in an order that is somewhat

backwards to its development. We wish to take advantage of the hundreds of years of thought

that have gone into it. As a result, we often begin by learning about limits. Afterward we definethe derivative and integral developed by Newton and Leibniz. But unlike Newton and Leibniz

we define them in the modern way - in terms of limits. Afterward we see how the derivative and

integral can be used to solve many of the problems that precipitated the development of Calculus.
http://www.uiowa.edu/~c22m025c/timeline.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Berkeley.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Cauchy.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Weierstrass.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Riemann.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Riemann.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Weierstrass.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Cauchy.htmlhttp://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Berkeley.htmlhttp://www.uiowa.edu/~c22m025c/timeline.html


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The Formula of Calculus

Derivatives

Integrals


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3.Brief History of Progressions

Arithmetic Progressions (AP)

The prehistory ofarithmetic is limited to a very small number of small artifacts which mayindicate conception of addition and subtraction, the best-known being the Ishango bone from

central Africa, dating from somewhere between 20 000 and 18 000 BC although its interpretation

is disputed.

The earliest written records indicate the Egyptians and Babylonians used all the elementary

arithmetic operations as early as 2 000 BC. These artifacts do not always reveal the specific

process used for solving problems, but the characteristics of the particular numeral system

strongly influence the complexity of the methods. The hieroglyphic system for Egyptiannumerals, like the later Roman numerals, descended from tally marks used for counting. In both

cases, this origin resulted in values that used a decimal base but did not include positional

notation. Although addition was generally straightforward, multiplication in Roman arithmeticrequired the assistance of a counting board to obtain the results.

In the Middle Ages, arithmetic was one of the seven liberal arts taught in universities. The

flourishing of algebra in the medieval Islamic world and in Renaissance Europe was anoutgrowth of the enormous simplification of computation through decimal notation. Various

types of tools exist to assist in numeric calculations. Examples include slide rules (for

multiplication, division, and trigonometry) and homographs in addition to the electrical

calculator.

Geometric Progressions (GP)

Geometric progression is also known as a geometric sequence, is a sequence of numberswhere each term after the first is found by multiplying the previous one by a fixed non-zero

number called the common ratio.

Books VIII and IX of Euclid'sElements analyze geometric progressions and give several of

their properties. Book IX, Proposition 35, proves that in a geometric series if the first term issubtracted from the second and last term in the sequence then as the excess of the second is to

the first, so will the excess of the last be to all of those before it.

Applying this to the geometric progression 31, 62, 124, 248, 496 (which results from 1, 2, 4,8, 16 by multiplying all terms by 31), we see that 62 minus 31 is to 31 as 496 minus 31 is to the

sum of 31, 62, 124, 248. Therefore the numbers 1, 2, 4, 8, 16, 31, 62, 124 and 248 add up to 496

and further these are all the numbers which divide 496. For suppose that p divides 496 and it isnot amongst these numbers. Assumepq equals 16 31, or 31 is to q asp is to 16. Nowp cannotdivide 16 or it would be amongst the numbers 1, 2, 4, 8 or 16. Therefore 31 cannot divide q. And

since 31 does not divide q and q measures 496, the fundamental theorem of arithmetic implies

that q must divide 16 and be amongst the numbers 1, 2, 4, 8 or 16. Let q be 4, thenp must be 124,which is impossible since by hypothesisp is not amongst the numbers 1, 2, 4, 8, 16, 31, 62, 124

or 248.


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The Formula of Progressions

Arithmetic Progressions (AP)

a, a + d, a + 2d, Tn= a + (n 1)d nth termSn= n/2 [2a + (n 1)d] sum of the first ntermsSn= n/2 (a + 1) sum of the first ntermsGeometric Progressions (GP)

a, ar, ar, Tn = ar^n1 nth term Sn = a(1r^n)/(1r), r < 1 sum of the first n terms Sn = a(r^n1)/(r1), r > 1 sum of the first n terms

S = a/(1r) sum to infinity


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4. Geometry, Calculus and Progressions Today

After a few centuries, its obvious that geometry, calculus and progressions are still used today inour daily life. Geometry is widely used in engineering, geology, mapping, aeronautics, fashion

and design, carpentry, welding, milling, optical illusions and etc. Calculus is used in everybranch of the physical sciences, in computer science, statistics, engineering, economics, business,medicine, demography, and in other fields wherever a problem can be mathematically modeled

and an optimal solution is desired. Progressions are being applied in distribution of proportion,

mathematical designs and patterns, engineering and etc. AP techniques can be applied in

engineering which helps this field to a large extent.

Geometry

Geometry is needed to determine suitable dimensions for the cake, to assist in designing anddecorating cakes that comes in many attractive shapes and designs, and also to estimate volumeof cake to be produced.

When making a batch of cake batter, you end up with a certain volume, determined by the

recipe. The baker must then choose the appropriate size and shape of pan to achieve the desiredresult. If the pan is too big, the cake becomes too short. If the pan is too small, the cake becomes

too tall. This leads into the next situation.

The ratio of the surface area to the volume determines how much crust a baked good will

have. The more surface area there is, compared to the volume, the faster the item will bake, and

the less "inside" there will be. For a very large, thick item, it will take a long time for the heat topenetrate to the center. To avoid having a rock-hard outside in this case, the baker will have tolower the temperature a little bit and bake for a longer time.

We mix ingredients in round bowls because cubes would have corners where unmixed

ingredients would accumulate, and we would have a hard time scraping them into the batter.

Calculus (Differentiation)

Calculus is used to determine minimum or maximum amount of ingredients for cake baking,

to estimate minimum or maximum amount of cream needed for decorating, and to estimate

minimum or maximum size of cake produced.


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Progressions

Progressions are functioned to determine total weight or volume of multi-storey cakes with

proportional dimensions, to estimate total ingredients needed for cake-baking, and to estimate

total amount of cream for decoration.

For example when we make a cake with many layers, we must fix the difference of diameter

of the two layers. So we can say that it used arithmetic progression. When the diameter of the

first layer of the cake is 8 and the diameter of second layer of the cake is 6, then the diameter ofthe third layer should be 4.

In this case, we use arithmetic progression where the difference of the diameter is constantthat is2. When the diameter decreases, the weight also decreases. That is the way how the cake is

balance to prevent it from smooch. We can also use ratio, because when we prepare the

ingredient for each layer of the cake, we need to decrease its ratio from lower layer to upper

layer. When we cut the cake, we can use fraction to divide the cake according to the total people

that will eat the cak

Question 1

Given:

- volume of a cake, v = 3 800 cm- height of a cake, h = 7.0 cm

- = 3.142

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

5v = rhr = 19 000/(h)r = 19 000/(3.142)(7)r = 19 000/21.994

r = 863.87

r = 29.39Therefore, 29.39 2 = 58.78

d = 58.78 cm


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Question 2

Given an oven with inner dimensions:- the length = 80.0 cm

- the width = 60.0 cm

- the height, h = 45.0 cm

(a) By using the above formula to forms a formula for d in terms of h.

5v = rh19 000 = (3.142)(d/2) h

19 000/(3.142)h = d/4

24 188.415/h = dd = 155.53/h

(b) Based on the table above,

(i) The range of heights that is not suitable for the cakes is 1 cm h 7 cm, because theresulting diameter produced is too large to fit into the oven, as the inner width of the

oven is 60 cm. Furthermore, the cake would be too short and too wide, making it less

attractive

(ii) The dimensions that I think most suitable for the cake is cake with 8.0 cm in height

and 54.99 cm in diameter, because it can fit into the oven, and the size is suitable foreasy handling. Moreover, this dimension is more symmetrical and easier to decorate.

h/cm d/cm

1 155.52625192 109.9736674

3 89.79312339

4 77.76312594

5 69.5534543

6 63.49332645

7 58.78339783

8 54.98683368

9 51.84208396

10 49.18171919


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(c) (i) The same formula in 2(a) is used, that is 19 000 = (3.142)(d/2)h. The same process is also

used, that is, make d the subject. An equation which is suitable and relevant for the graph:

19 000 = (3.142)(d/2)h

19000/(3.142)h = d/4

24188.415/h = dd = 155.53/hd = 155.53h -1/2

log d = log 155.53h -1/2log d = -1/2 log h + log 155.53

Graph of log10d against log10 h

log10 h log10 d

0 2.19

1 1.69

2 1.19

3 0.69

4 0.19


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( ii)

a) When h = 10.5 cm, log10 h = 1.0212According to the graph, log10 d = 1.7 when log10 h = 1.680

Therefore, d = 47.86 cm


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b) When d = 42 cm, log10 d = 1.6232

According to the graph, log10 h = 1.140 when log10 d = 1.6232

Therefore, h = 13.80 cm

Question 3

(a) 27.49cm

8cm

Diagram 1: Cake without Cream

h = 8cm, d = 54.99cm ,r=27.49cm

1cm

28.49cm

1cm

Diagram 2: Cake with Cream

To calculate volume of cream used, the cream is symbolised as the larger cylinder and the cake is

symbolised as the smaller cylinder.

9 cm


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Therefore, amount of fresh cream= Volume adding fresh creamVolume before=(3.142 x 28.49

2x 9) cm

3-19000 cm

3

=22952.7cm3-19000cm

3

=3952.7cm

3

(b) The three shapes that I suggested for cake,that will have the same height and volume as those

suggested in 2(b)(ii),with h=8cm and 54.99cm are:1. Rectangle-shaped base (cuboids)

2. Triangle-shaped base

3. Pentagon-base

1. Rectangle-shaped base (cuboids)

Volume = base area x heightBase area = 19000/8

Length x width = 2375

By 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 side surface)(Height

of cream)+Volume of top surface

= 2(8 x 50)(1) + 2(8 x 47.5)(1) + 2375

=3935 cm3


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2. Triangle-shaped base

19000 = base area x heightbase area = 2375

x length x width = 2375

length x width = 4750

By trial and improvement, 4750 = 95 x 50 (length = 95, width = 50)

Slant length of triangle = (95 + 25)= 98.23

Therefore, amount of cream

= Area of rectangular front side surface(Height of cream) + 2(Area of slant rectangular left/right

side surface)(Height of cream) + Vol. of top surface= (50 x 8)(1) + 2(98.23 x 8)(1) + 2375 = 4346.68 cm


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3. Pentagon-base

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

therefore:2375 = 5(length x width)475 = length x width

By 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 cm

(c)

Based on the values above,the shape that requires the least amount of fresh cream to be used is:Pentagon-shaped cake, since it requires only 3135 cm of cream to be used.

When there's minimum or maximum, well, there's differentiation and quadratic functions.Use both to find the minimum height, h and its corresponding minimum diameter, d is calculated

by using the differentation and quadratic functions.


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Method 1: Differentiation

Use two equations for this method: the formula for volume of cake (as in Q2/a), and the formulafor amount (volume) of cream to be used for the round cake (as in Q3/a).

19000 = (3.142)rh (1)V = (3.142)r + 2(3.142)rh (2)From (1): h = 19000/(3.142)r (3)Sub. (3) into (2):

V = (3.142)r + 2(3.142)r(19000/(3.142)r)V = (3.142)r + (38000/r)

V = (3.142)r + 38000r-1

dV/dr = 2(3.142)r(38000/r)

0 = 2(3.142)r(38000/r) -->> minimum value, therefore dV/dr = 038000/r = 2(3.142)r38000/2(3.142) = r

6047.104 = rr = 18.22

Sub. r = 18.22 into (3):h = 19000/(3.142)(18.22)

h = 18.22

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


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Method 2: Quadratic Functions

Use 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)rh (1)f(r) = (3.142)r + 2(3.142)hr (2)

From (2):

f(r) = (3.142)(r + 2hr) -->> factorize (3.142)

= (3.142)[ (r + 2h/2)(2h/2) ] -->> completing square, with a = (3.142), b = 2h and c =0

= (3.142)[ (r + h)h ]= (3.142)(r + h)(3.142)h(a = (3.142) (positive indicates min. value), min. value = f(r) =(3.142)h, corresponding

value of x = r = --h)

Sub. r = --h into (1):19000 = (3.142)(--h)h

h = 6047.104

h = 18.22

Sub. h = 18.22 into (1):

19000 = (3.142)r(18.22)

r = 331.894r = 18.22

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

I would choose not to bake a cake with such dimensions because its dimensions are not suitable(the height is too high) and therefore less attractive. Furthermore, such cakes are difficult to

handle easily.


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Best Bakery received an order to bake a multi-storey cake for Merdeka Day celebration, as

shown in the Diagram 2.

Given:

-height, h of each cake = 6cm

-radius of largest cake = 31cm-radius of 2nd cake = 10% smaller than 1st cake

-radius of 3rd cake = 10% smaller than 2nd cake

- =3.142

From the given information,Ive found out the geometric progression(GP)with a=31, r=9/10,27.9, 25.11 and 22.599.

a = 31, r =

V = (3.142)rh

(a)

Volume of cake 1 Volume of cake 2

= r2h = r2h= 3.142 x 31 x 31 x 6 = 3.142 x (0.9 x 31)2 x 6

= 18116.772 cm3

= 3.142 x (27.9)2x 6

= 14676.585 cm3

Volume of cake 3 Volume of cake 4

= r2h = r2h= 3.142 x (0.9 x 0.9 x 31)

2x 6 = 3.142 x (0.9 x 0.9 x 0.9 x 31)

2x 6

= 3.142 x (25.11)2 x 6 = 3.142 x (22.599)2 x 6

= 11886.414 cm3

= 9627.995 cm3


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The values 118116.772, 14676.585, 11886.414, 9627.995 form a number pattern.The pattern formed is a geometrical progression.

This is proven by the fact that there is a common ratio between subsequent numbers, r = 0.81.

14676.585 = 0.81 11886.414 = 0.8118116.772 14676.585

. 9627.995 = 0.8111886.414

(b) The total mass of all cakes should not exceed 15 kg, therefore:

By using the formula of the sum of the first n terms, Sn, of a geometric progression:

Sn = (a(1r n)) / (1r), with Sn = 57 000, a = 18 116.772 and r = 0.81 to find n:57 000 = (18 116.772(1(0.81)n)) / (10.81)10.81n = 0.597790.40221 = 0.81n log 0.81

0.40221 = nn = log 0.40221 / log 0.81n = 4.322Therefore, n 4

Verifying the answer:

When n = 5:S5 = (18 116.772(1(0.81)5)) / (10.81) = 62 104.443 > 57 000(Sn > 57 000, n = 5 is not suitable)

When n = 4:S4 = (18 116.772(1(0.81)4)) / (10.81) = 54 305.767 < 57 000(Sn < 57 000, n = 4 is suitable)


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After doing some research,answering questions,drawing graphs,and some problem solving,I saw

that the usage of geometry,calculus and progressions are important in daily life.

-Geometry is the study of angles and triangles,perimeter,area and volume.It differs from

algebrainthat one develops a logical structure where mathematical relationships areproved and

applied.

-An arithmetic progression(AP) is a sequence of numbers such that the difference of any two

successive members of the sequence is a constant.

-Differantation essentially the process of finding an equation which will give you gradient (slope)

at any point along the curve.Say you have y=x^2.The equation y=2x will give you the gradientof y at any point along that curve.

As the conclusion,geometry,calculus and progressions are partof our necessities.Thus,we should

be thankful of the people who contribute inthe ide of geometry,calculus and progressions

because without them,we cdant done the multi-storey cake,and its hard to find out the volumeof ingredients needed for the cake.

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