additional mathematics - circular measure ( spm project work )

8/12/2019 Additional Mathematics - Circular Measure ( SPM Project Work )

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Sekolah Menengah Kebangsaan erma

Additional Mathematics Project Work : Option

NAME : MOHAMAD MUKHRIZ BIN ZUBAIDI

CLASS : 5 EFISIEN

TEACHERS NAME : MADAM KOH GAIK BOAY

CIRCULAR SCHOOL GARDEN


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History of Pi

The earliest written approximations of are found inEgypt andBabylon,both within 1 percent of the true value. In Babylon, aclay tablet dated 19001600 BC has a geometrical statement that, by implication, treats as25/8 = 3.1250.

In Egypt, theRhind Papyrus, dated around 1650 BC, but copied from a

document dated to 1850 BC has a formula for the area of a circle that

treats as (16/9)23.1605.

THE RHIND PAPYRUS

InIndia around 600 BC, theShulba Sutras (Sanskrit texts that are rich in

mathematical contents) treat as (9785/5568)23.088. In 150 BC, orperhaps earlier, Indian sources treat as :

10 3.1622
http://en.wikipedia.org/wiki/Egypthttp://en.wikipedia.org/wiki/Babylonhttp://en.wikipedia.org/wiki/Clay_tablethttp://en.wikipedia.org/wiki/Rhind_Papyrushttp://en.wikipedia.org/wiki/Indiahttp://en.wikipedia.org/wiki/Shulba_Sutrashttp://en.wikipedia.org/wiki/Sanskrithttp://en.wikipedia.org/wiki/Sanskrithttp://en.wikipedia.org/wiki/Shulba_Sutrashttp://en.wikipedia.org/wiki/Indiahttp://en.wikipedia.org/wiki/Rhind_Papyrushttp://en.wikipedia.org/wiki/Clay_tablethttp://en.wikipedia.org/wiki/Babylonhttp://en.wikipedia.org/wiki/Egypt


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The first recorded algorithm for rigorously calculating the value of was ageometrical approach using polygons, devised around 250 BC by the Greek

mathematicianArchimedes. Archimedes computed upper and lower bounds

of by drawing a regular hexagon inside and outside a circle, andsuccessively doubling the number of sides until he reached a 96-sided regular

polygon. By calculating the perimeters of these polygons, he proved that

223/71 < < 22/7(3.1408 < < 3.1429).

The way Archimedes find the limit of . he estimated by computing the perimeters of

circumscribed and inscribed polygons

Around 150 AD, Greek-Roman scientistPtolemy,in hisAlmagest,gave a

value for of 3.1416, which he may have obtained from Archimedes orfromApollonius of Perga.Mathematicians using polygonal algorithms

reached 39 digits of in 1630.

The development of computers in the mid-20th century again revolutionized

the hunt for digits of . American mathematiciansJohn Wrench and LeviSmith reached 1,120 digits in 1949 using a desk calculator. Using aninverse

tangent (arctan) infinite series, a team led by George Reitwiesner andJohn

von Neumann that same year achieved 2,037 digits with a calculation that

took 70 hours of computer time on theENIAC computer. The record, always

relying on an arctan series, was broken repeatedly from7,480 digits in 1957to10,000 digits in 1958 to 100,000 digits in 1961 until 1 million digits was

reached in 1973.
http://en.wikipedia.org/wiki/Archimedeshttp://en.wikipedia.org/wiki/Ptolemyhttp://en.wikipedia.org/wiki/Almagesthttp://en.wikipedia.org/wiki/Apollonius_of_Pergahttp://en.wikipedia.org/wiki/John_Wrenchhttp://en.wikipedia.org/wiki/Inverse_tangenthttp://en.wikipedia.org/wiki/Inverse_tangenthttp://en.wikipedia.org/wiki/John_von_Neumannhttp://en.wikipedia.org/wiki/John_von_Neumannhttp://en.wikipedia.org/wiki/ENIAChttp://en.wikipedia.org/wiki/ENIAChttp://en.wikipedia.org/wiki/John_von_Neumannhttp://en.wikipedia.org/wiki/John_von_Neumannhttp://en.wikipedia.org/wiki/Inverse_tangenthttp://en.wikipedia.org/wiki/Inverse_tangenthttp://en.wikipedia.org/wiki/John_Wrenchhttp://en.wikipedia.org/wiki/Apollonius_of_Pergahttp://en.wikipedia.org/wiki/Almagesthttp://en.wikipedia.org/wiki/Ptolemyhttp://en.wikipedia.org/wiki/Archimedes


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Part 2(a) i)

Radius of circular plot of flower, r = 1m

Radius of whole circular plot, R = 1m + 2m= 3m

Width of the tile, L = 10cm

= 0.1mLength of the tile, k = 25cm

= 0.25m

Number of the row that can be made arranged by the tiles = (Rr) L= (3-1)0.1= 20

The number of tiles that cover up the circumference of the circular plot of flower= 2rk= 2 (1)0.25= 8

The number of bricks that cover up the circumference of the circular arrangement of tiles forms

an arithmetic progression,where

T1 = 8T2 = (2r)k

= [2(1+0.1)]0.25= (2.2)0.25= 8.8

T3 = (2r)0.25= [2(1+0.2) ]0.25= (2.4)0.25= 9.6


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Common differenced, d = 9.6 8.8= 0.8

S20 = (202)[2(8)+(201)(0.8)]= 312

Therefore, the estimated number of tiles required = 312tiles

(a) ii)

Area of the pavement covered by tile= Area of the circular pavement Area of the garden= R2 r2= (3)

2 (1)2= 8m2

Area of one tile = length x breath= 0.25 x 0.1

= 0.025m2

The estimated number of tiles required = 8m20.025m2

= 320

(a) iii) The method in 2(a) i) uses arithmetic progression for finding the number of tiles.

Therefore, this method implies that there will be spaces in the circular pavement not

covered by between the tiles. Meanwhile the method in 2 (a) ii) uses area of the circularpavement and the area of tiles to find the number of brick. This method apply that all the

area of circular pavement is covered by the tiles. However, the calculation will not be

accurate as there will always be space in the arrangement between the tiles. This can be

proved by the difference of the result in both calculations that shows that the result in 2(a)

ii) gives a greater value than 2 a) i). The most accurate method is the circumference

methods because a rectangle tile can never fit perfectly in a circle. There will always be

spaces between the rectangle tiles in the circle.

(a) iv) A mason will use the method in (a) iii). If the mason uses the method in (a) iii), the

number of tiles needed to cover the circular pavement will be more than the number of

brick needed. This extra bricks will result in a waste of budget and will further increase

the cost to build the circular pavement.


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(b)

Half of the diagonal of the inner octagon, r = 22

= 1m

Half of the diagonal of the outer octagon, R = 62

= 3mLength of the brick, k = 25cm

= 0.25m

Width of the brick, L = 10cm

= 0.1m

Number of the row that can be made arranged by the tiles = (Rr)/L= (31)/0.1= 20


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Based on the triangle, a can be find by using the equation

= = 0.5858

a = = 0.7654

The number of tiles that cover up the perimeter of the octagonal arrangement of bricks form an

arithmetic progression,where

T1 = (8a)0.25= [8(0.7654)]0.25

= 24.4928

T2 = (8a)0.25

= [8 ]]0.25= [8(0.8419)]0.25

= 26.9409T2 = (8a)0.25

= [8 ]]0.25= [8(0.9184)]0.25

= 29.3901

Common difference, d = 26.9409 24.4928= 2.4481

= 29.3901 26.9409= 2.4492


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We assume that d = 2.44

S20 = (202)[2(24.49) + (201)(2.44)]= 953.4

Therefore the number of tiles = 953 tile

Method 2

Based on the triangle, a can be find by using the equation

.

= = 5.2720

a = = 2.2961m

Given the formula of octagon , A=2(1+2)a2,, where a= side length

Area of the pavement = Area of outside octagon area of inner octagon

= 2(1 + a2

2(1+a2

= 2(1+) (2.2961)2 2(1+) (0.7654)2

= 22.627m2

Area of one tile = length x breath= 0.25 x 0.1

= 0.025m2


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Number of bricks required = 22.627/0.025

= 905.08

= 905 tile

j)

Circular designPros Cons

Easy to design on a piece of paper Difficult to fill rectangle tile in circle

Easy to make a circle. A mason needs only a

piece of rope, stick and a pencil.

Difficult to calculate amount of tile needed to

fill in circle.

Octagonal Design

Cons ProsDifficult to design on a piece of paper Rectangle tile can fit in octagon better than

circle.

Difficult to make an octagon. It requires many

steps to make an octagon.

Easy to calculate the number of tiles to fill in

octagon.


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Further Exploration

A)

Based on the result in b) i) where a = 0.7654m. The area of inner octagon can be calculated by

using the formula A=2[2(1+2)a

2

]

The area of inner octagon = 2[ 2(1+2)(0.7654)2]= 5.627 m

2

From the diagram, this octagon can be obtained

Based on the result in b) i) where a = 2.2961m. The area of 8 triangle or an octagon(octagon has

eight triangle) can be calculated by using the formula A=2(1+2)a2

Area of outside octagon = 2(1+2)(2.2961)2= 25.4558 m

2


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From the diagram, this 4 triangle can be obtained

Based on the result in b) i) where, a = 2.2961m.

The area for 4 triangle = [2(1+2)x2.29612 ]= 12.7279 m

2

From diagram this can be obtained 4 of this triangle can be obtained

sin 45 = x3

x = sin 45 (3)

= 2.1213m

cos 45 = y3

y = cos 45 (3)

= 2.1213m


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area for 4 of this triangle = 4 [1/2 (2.1213) (2.1213) ]

= 9 m2

From the diagram, this rectangle can be obtained

Interior angle of octagon = (8-2) (180)

= 1080

angle for each side in octagon = 1080/8

= 135

m = (135 = 22.5

n = 90-22.5= 67.5

z = 5m2.1213m2.1213m= 0.7574m


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tan n = v/ (z/2)

tan 67.5 = v/ 0.3787

v = tan 67.5 (0.3787)

= 0.9143m

area of shaded region = area of rectanglearea of 2 triangle= 0.7574 (4.2426)0.7574 (0.9143)= 2.5209 m

2

Therefore, area of pavement = (2.5209 m2+ 12.7279 m

2+9 m

2+ 25.4558 m

2)5.627 m2

= 44.0776 m2

Number of tiles = 44.0776m2

0.025m2

= 1763.104

= 1763 tiles

(b)An altenative design for the pavement :

Reasons :

1. The design is very easy to make by mason.

2. The rectangle tiles fit very easily in the rectangle design.

3. The number of tile can be easily calculated.

4. The tile lay can be easily laid into the design. No big spaces will between tiles.


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Reflection

With brilliant mind , comes great responsibility Mukhriz

While Im conducting my additional project, I learn that there are many undiscovered thingsabundant around the the globe . And with knowledge, we can do anything to explore the

undiscovered amazing things and create a better world for everyone although we have to face

many challenges. But it is not the problem . Life is always have it ups and down . That is the

cycle of LIFE !

additional mathematics - circular measure ( spm project work )

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