Additional Mathematics project work for SPM 2014 ( PERLIS ) - OPTION 2Full description...

Author:
Mukhriz Zubaidi

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NAME CLASS

TEACHER’S NAME

: MOHAMAD MUKHRIZ BIN ZUBAIDI : 5 EFISIEN : MADAM KOH GAIK BOAY

Part 1 (a)

Example of round

(b)

Pi (π) The number π is a mathematical constant, the ratio of a circle's circumference to its diameter, approximately equal to 3.141592654. It has been represented by the Greek letter "π" since the mid-18th century though it is also sometimes spelled out as "pi". Pi is an irrational number, it cannot be expressed exactly as a common fraction, ho wever fractions such as 22/7 and other rational numb ers such as 3.142 are commonly used to approximate pi.

The earliest written approximations of π are found in Egypt and Babylon, both within 1 percent of the true value. In Babylon, a clay tablet dated 1900 – 1600 BC has a geometrical statement that, by implication, treats πas 25/8 = 3.1250. In Egypt, the Rhind Papyrus, dated around 1650 BC, but copied from a document dated to 1850 BC has a formula for the area of a circle that 2 treats π as (16/9) ≈ 3.1605.

THE RHIND PAPYRUS

In India around 600 BC, the Shulba Sutras (Sanskrit texts that are rich in 2 mathematical contents) treat π as (9785/5568) ≈ 3.088. In 150 BC, or perhaps earlier, Indian sources treat π as :

√10 ≈ 3.1622

The first recorded algorithm for rigorously calculating the value of π was a geometrical approach using polygons, devised around 250 BC by the Greek mathematician Archimedes. Archimedes computed upper and lower bounds of π by drawing a regular hexagon inside and outside a circle, and successively 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 < π < 22/7(3.1408 < π < 3.1429).

The way Archimedes find the limit of π. he estimated π by computing the perimeters of circumscribed and inscribed polygons

Around 150 AD, Greek-Roman scientist Ptolemy, in his Almagest, gave a value for π of 3.1416, which he may have obtained from Archimedes or from Apollonius of Perga. Mathematicians using polygonal algorithms reached 39 digits of π in 1630. The development of computers in the mid-20th century again revolutionized the hunt for digits of π. American mathematicians John Wrench and Levi Smith reached 1,120 digits in 1949 using a desk calculator. Using an inverse tangent (arctan) infinite series, a team led by George Reitwiesner and John von Neumann that same year achieved 2,037 digits with a calculation that took 70 hours of computer time on the ENIAC computer. The record, always relying on an arctan series, was broken repeatedly from7,480 digits in 1957 to10,000 digits in 1958 to 100,000 digits in 1961 until 1 million digits was reached in 1973.

Part 2 (a) i)

Radius of circular plot of flower, r Radius of whole circular plot, R

= 1m = 1m + 2m = 3m Width of the tile, L = 10cm = 0.1m Length of the tile, k = 25cm = 0.25m Number of the row that can be made arranged by the tiles

= (R−r) ÷ L = (3-1)÷0.1 = 20 The number of tiles that cover up the circumference of the circular plot of flower = 2πr ÷k = 2π (1)÷0.25 = 8π The number of bricks that cover up the circumference of the circular arrangement of tiles forms an arithmetic progression,where T1 = 8π T2 = (2πr)÷k = [2(1+0.1)π]÷0.25 = (2.2π)÷0.25 = 8.8π T3 = (2πr)÷0.25 = [2(1+0.2) π]÷0.25 = (2.4π)÷0.25 = 9.6π

= 9.6 π − 8.8π = 0.8 π = (20÷2)[2(8π)+(20−1)(0.8π)] = 312π

Common differenced, d S20

= 312π tiles

Therefore, the estimated number of tiles required

(a) ii)

Area of the pavement covered by tile

= Area of the circular pavement − Area of the garden 2 = πR − πr2 2 2 = (3) π − (1) π 2 = 8πm Area of one tile

= length x breath = 0.25 x 0.1 2 = 0.025m

The estimated number of tiles required

2

2

= 8πm ÷0.025m = 320π

(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 circular pavement 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.

(b)

Half of the diagonal of the inner octagon, r = 2÷2 = 1m Half of the diagonal of the outer octagon, R = 6÷2 = 3m Length 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

= (R−r)/L

= (3−1)/0.1 = 20

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

a

= = 0.5858 = √ = 0.7654

The number of tiles that cover up the perimeter of the octagonal arrangement of bricks form an arithmetic progression,where T1

T2

T2

= (8a)÷0.25 = [8(0.7654)]÷0.25 = 24.4928 = (8a)÷0.25 = [8 ]]÷0.25 = [8(0.8419)]÷0.25 = 26.9409 = (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

We assume that d = 2.44 S20

= (20÷2)[2(24.49) + (20−1) (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

a

.

= = 5.2720 = √ = 2.2961m 2,

Given the formula of octagon , A=2(1+√2)a , where a= side length Area of the pavement = Area of outside octagon area of inner octagon 2 2 = 2(1 + √ a 2(1+√ a 2 2 = 2(1+√ ) (2.2961) 2(1+√ ) (0.7654) 2 = 22.627m Area of one tile

= length x breath = 0.25 x 0.1 2 = 0.025m

Number of bricks required

= 22.627/0.025 = 905.08 = 905 tile

j)

Circular design Pros Easy to design on a piece of paper Easy to make a circle. A mason needs only a piece of rope, stick and a pencil.

Cons Difficult to fill rectangle tile in circle Difficult to calculate amount of tile needed to fill in circle.

Octagonal Design Cons Difficult to design on a piece of paper

Difficult to make an octagon. It requires man y steps to make an octagon.

Pros Rectangle tile can fit in octagon better than circle. Easy to calculate the number of tiles to fill in octagon.

Further Exploration A) Based on the result in b) i) where a = 0.7654m. The area of inner octagon can be calculated by 2 using the formula A=2[2(1+√2)a ]

The area of inner octagon

2

= 2[ 2(1+√2)(0.7654) ] 2 = 5.627 m

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 2 eight triangle) can be calculated by using the formula A=2(1+√2)a Area of outside octagon

2

= 2(1+√2)(2.2961) 2 = 25.4558 m

From the diagram, this 4 triangle can be obtained

Based on the result in b) i) where, a = 2.2961m. 2 The area for 4 triangle = ½ [2(1+√2)x2.2961 ] 2 = 12.7279 m From diagram this can be obtained 4 of this triangle can be obtained

sin 45° = x÷3 x = sin 45° (3) = 2.1213m cos 45°= y÷3 y = cos 45° (3) = 2.1213m

area for 4 of this triangle

= 4 [1/2 (2.1213) (2.1213) ] 2 =9m

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 = 5m – 2.1213m – 2.1213m = 0.7574m

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 rectangle – area of 2 triangle = 0.7574 (4.2426) – 0.7574 (0.9143) 2 = 2.5209 m

2

2

2

2

2

Therefore, area of pavement = (2.5209 m + 12.7279 m +9 m + 25.4558 m ) – 5.627 m 2 = 44.0776 m Number of tiles

2

2

= 44.0776m ÷0.025m = 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.

Reflection

“With brilliant mind , comes great responsibility” – Mukhriz While I’m conducting my additional project, I learn that there are many undiscovered things abundant 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 !

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