Sekolah Menengah Kebangsaan Derma

Additional Mathematics Project Work : Option 2

NAME : MOHAMAD MUKHRIZ BIN ZUBAIDI

CLASS : 5 EFISIEN

TEACHER’S NAME : MADAM KOH GAIK BOAY

CIRCULAR SCHOOL GARDEN

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, however

fractions such as 22/7 and other rational numbers such as 3.142 are commonly used to

approximate pi.

History of 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

treats π as (16/9)2 ≈ 3.1605.

THE RHIND PAPYRUS

In India around 600 BC, the Shulba Sutras (Sanskrit texts that are rich in

mathematical contents) treat π as (9785/5568)2 ≈ 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 = 1m

Radius of whole circular plot, R = 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

Common differenced, d = 9.6 π − 8.8π

= 0.8 π

S20 = (20÷2)[2(8π)+(20−1)(0.8π)]

= 312π

Therefore, the estimated number of tiles required = 312π tiles

(a) ii)

Area of the pavement covered by tile

= Area of the circular pavement − Area of the garden

= πR2 − πr2

= (3)2π − (1)

2π

= 8πm2

Area of one tile = length x breath

= 0.25 x 0.1

= 0.025m2

The estimated number of tiles required = 8πm2÷0.025m

2

= 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 = = 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

= 26.9409

T2 = (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 . = = 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+√ a

2

= 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

Number of bricks required = 22.627/0.025

= 905.08

= 905 tile

j)

Circular design Pros 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 Pros

Difficult 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.