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.