Additional Mathematics Project Work 2017
Additional Mathematics Project Work 2017
Additional Mathematics Project Work 2017
2017
IC : 000130-08-1217
School : S.M.J.K. Sam Tet
Class : 5S5
Teacher : Mr. Lai
CONTENT
Ren Descartes
Biography of Ren Descartes
Philosopher Ren Descartes was born on March 31, 1596, in La Haye en Touraine, a small
town in central France. He was the youngest of three children. He started to study at the age of
8 at the Jesuit college of Henri IV in La Flche for seven years. The subjects he studied, such
as rhetoric and logic and the mathematical arts, which included music and astronomy, as well
as metaphysics, natural philosophy and ethics, equipped him well for his future as a
philosopher. So did he spending the next four years earning a baccalaureate in law at the
University of Poitiers. Descartes later added theology and medicine to his studies. But he
eschewed all this, resolving to seek no knowledge other than that of which could be found in
myself or else in the great book of the world, he wrote much later in Discourse on the Method
of Rightly Conducting the Reason and Seeking Truth in the Sciences, published in 1637.
Descartes is considered by many to be the father of modern philosophy, because his
ideas departed widely from current understanding in the early 17th century, which was more
feeling-based. While elements of his philosophy werent completely new, his approach to them
was. Descartes believed in basically clearing everything off the table, all preconceived and
inherited notions, and starting fresh, putting back one by one the things that were certain, which
for him began with the statement I exist. From this sprang his most famous quote: I think;
therefore I am. Although philosophy is largely where the 20th century deposited Descartes
each century has focused on different aspects of his workhis investigations in theoretical
physics led many scholars to consider him a mathematician first. He introduced Cartesian
geometry, which incorporates algebra; through his laws of refraction, he developed an empirical
understanding of rainbows; and he proposed a naturalistic account of the formation of the solar
system, although he felt he had to suppress much of that due to Galileos fate at the hands of
the Inquisition. His concern wasnt misplacedPope Alexander VII later added Descartes
works to the Index of Prohibited Books.
His revolutionary ideas made him a centre of controversy in his day, and he died in 1650 far
from home in Stockholm, Sweden. 13 years later, his works were placed on the Catholic
Church's "Index of Prohibited Books".
6) Players that step on the are allow to roll the dice once
again.
S
Dice Tokens
Set1
Triangle Coordinates of vertices
A ( 0,1 ) ( 2,0 ) ( 2,2 )
B ( 0,2 ) ( 2,2 ) ( 2,4 ) Set2
C ( 0,3 ) ( 2,4 ) ( 2,6 ) Triangle Coordinates of vertices
D ( 0,4 ) ( 2,6 ) ( 2,8 ) P ( 5,0 ) ( 5,2 ) ( 7,1 )
E ( 0,5 ) ( 2,8 ) ( 2,10 ) Q ( 5,2 ) ( 5,4 ) ( 7,3 )
F ( 0,6 ) ( 2,10 ) ( 2,12 ) R ( 5,4 ) ( 5,6 ) ( 7,5 )
S ( 5,6 ) ( 5,8 ) ( 7,7 )
TABLE 1 T ( 5,8 ) ( 5,10 ) ( 7,9 )
U ( 5,10 ) ( 5,12 ) ( 7,10 )
( b ) Six rectangles, drawn on a Cartesian
plane, are shown on page 3. On the Cartesian plane, draw all the twelve triangles shown in
table 1 to complete the layout for the model of a quartz crystal.
( REFER TO THE GRAPH PAPER ON NEXT PAGE )
( c ) Use the exact layout creatively to build the model.
( REFER TO THE MODEL BUILT )
( d ) Calculate the total surface area, in cm, of the model that you have built.
Area of a triangle Area of 12 triangles
= 2 12cm
= |
1 0 2 20
2 1 0 21 | cm
= 24 cm
1
= 2
|( 0 0 )+ (2 2 )+ ( 2 1 ) ( 2 1 )( 2 0 ) (0 2)| cm
1
= |0+ 4+2200| cm
2
1
= |4| cm
2
= 2cm
Area of 6 rectangles
= 6 (3 2)
=36cm
Since each block in the graph paper represent 2cm
Total surface area
= (36+24)x2
=120cm
Distance of QR= y2 - y1
Distance of PR= x 2 - x1
Use
Ali drives from Town A to visit his friends who stay in Town B and Town C. After the visit, he
returns to Town A.
1) Calculate the total distance of the journey.
Distance of AJ Distance of JB Distance of JC
= (72 ) +(84 )
2 2
= (52 ) +( 84 )
2 2
= (112 ) + (84 )
2 2
= (9 ) +(12 )
2 2
= (3) + (4 )
2 2
= ( 9 ) +(12 )
2 2
= 225 = 25 = 225
=15 =5 =15
= (18 ) 2
= ( 6 ) +(16 )
2 2
= 18 = 292
= 17.088
If AC is built, If BC is built,
AJ + JB + BJ + JC + CA AJ + JB + BC + CJ + JA
= 15 + 5 + 5 + 15 + 18 =15 + 5 + 17.088 + 15 +15
= 58km = 67.088km
n( OX - OP )=m(
OQ - OX )
n OX -n OP =m
OQ -m OX
m OX + n OX = m
OQ + n OP
(m+n) OX = m OQ
+ n OP
1
OX = m+n (m
OQ + n OP )
1
OX = m+n [m
() ()
x2
y2 +n
x1
y1
m x 2 +n x 1
OX =
( ) m+n
m y 2 +n y 1
m+n
X=
( m xm++ nn x , m ym+n+n y )
2 1 2 1
Use
1 )Salleh, the owner of the land, would like to plant another coconut tree in such a way the four
coconut trees are at the vertices of a parallelogram. Where should Salleh plant the fourth
coconut tree? Show your calculation.
Midpoint of Midpoint of
AC=Midpoint of BD CD=Midpoint of AB
10+ 40 x+35 y + 40 510 20+10
( 10+35
2
,
2 ) ( 2
,
2 )=(
2
,
2 )
x+5 y +30
= ( 2 , 2 ) ( x+35
2
,
y + 40
2 )=(
5 30
2 2)
,
x +35 5
( 252 , 502 ) = 2
=
2
x+35=-5
x+5 y +30
( 2
,
2 ) x=-40
The fourth coconut
tree should be plant
25 x +5
at ( -40,-10 ).
2 = 2
25=x+5
x=20
The fourth coconut
tree should be plant
at ( 20,20 ).
2 ) Puteri, Sallehs wife, would like to plant two more coconut trees.
( a ) One tree is to be planted between the coconut trees at location B and C such that the three
coconut trees form a straight line. Further, the location of the tree is such that its distance from
C is three times its distance form B. Where should Puteri plant the coconut tree?
Location of the coconut tree,
= ( 3(5)+3+1(35)
1
,
3(30)+1(40)
3+1 )
= ( 504 , 1304 )
= (12 12 ,32 12 )
Puteri should plant the coconut tree at (12 12 ,32 12 ) .
( b ) The other tree is to be planted 10m away from the coconut at location A such that this
coconut tree and the two coconut trees at location A and B form a straight line. Where should
Puteri plant the coconut tree?
Distance of AB,
= (105 ) ( 1030 )
2 2
= 152202
= 625
=25
Let the location of the coconut tree that is going to be plant = S
AS:SB
10:15
2:3
S
= ( 3(10)+2(5)
3+ 2
,
3(10)+2(30)
3+2 )
= ( 205 , 905 )
=( -4,18 )
Puteri should plant the tree at ( -4,18 ).
=
1
[ x y + x y x y x y + x y + x y x y x y x y x y + x y + x
2 2 1 2 2 1 1 1 2 3 2 3 3 2 2 2 3 3 1 3 3 1 1 1
1
= 2 [ x 1 y2 x2 y 3x 3 y 1+ x 2 y 1+ x 3 y 2+ x 1 y 3 ]
= 2 y
1
1 x1
| x2
y2
x3 x1
y3 y1 |
Area of triangle ABC= 2 y
1
1 x1
| x2
y2 |
x3 x1
y3 y1
Use
The four brothers agreed to share this piece of land equally among themselves. They decided to
divide it into four triangular plots of equal areas as shown in Diagram 3.
1 )Calculate the area, in km, of each plot of triangular land.
Area of whole land Area of each triangular land
=56 4
=14 km
= 2 2
1 0
| 2 8 10 0
10 6 0 2 |
1
= 2 |0+12+0+20480600|
1
= 2 |112|
1
= 2 ( 112 )
=56 km
|
1 0 10 a 0
2 2 0 b2 | =14 |
1 10 8 a 10
2 0 6 b0 | =14
|0 10 a 0
2 0 b2 | =28 |10 8 a 10
0 6 b0 | =28
10b+2a-20=28 60+8b-6a-10b=28
10b+2a-20=28 10b+2a-20= -28
Whole equation is multiplied by 3, so
6a=-30b+144 (1) 6a=-30b-24(2) 6a=-2b+32 (3)6a=-2b+88 (4)
1 ) A straight line passes through point A ( x1 , y1 ) and point B ( x2 , y2 ) , State the formula for
the gradient of the straight line.
y2 y1
Gradient = x 2x
1
2 ) The gradients of two straight lines are m1 and m 2 respectively. State the condition for
the two straight lines to be:
( a ) parallel
The gradient of the two lines must be the same, which means m1 = m2 .
( b )perpendicular
The product of the gradient of the two lines must be -1, which means m1 m 2 = -1
Use
a) Rita begins from location (12,18) and drives along a straight road which allows her to be
always equidistant from the police station and the fire station at any time. By considering
gradients only, determine the location on the road when Rita is nearest to the petrol station.
Let, Police station = A, Petrol station = B, Fire station = C, Position of Rita = D,
Position nearest to the petrol station = M
Gradient of AC Gradient of DM
5+ 1 3
M DM )=-1
= 44 4 (
6 4
= 8 M DM =
3
3
= 4
Equation of BM Equation of DM
b2 3 b18 4
= =
a+3 4 a12 3
4b-8=-3a-9 3b-54=4a-48
4b+3a=-1 (1) 3b-4a=6 2
( 1 )multiple by 3, ( 2 )multiple by 4
12b+ 9a=-3 27
12b-16a=24 When a= 25
25a=-27
27 27
3b=4( +6
a= 25 25
42
3b= 25
14
b= 25
27 14
The position on the road when Rita is nearest to the petrol station is ( , .
25 25
1
b ) If Rita drives with an average speed of 80km h , calculate the time, in minutes, she
takes to reach this nearest location.
Distant of DM
= 11881
25
109
= 5
=21.8
21.8
Time = 80 x 60
=0.2725 x 60
=16.35minutes
Worksheet 7: Nursery Fun!
State
State the formula for finding the equation of a straight line given in the following:
1 ) Gradient and the y-intercept
2) Gradient and one point on the line
3 ) Two points on the line
4) The x-intercept and y-intercept
y y1 = m(x x1)
Use
1
1 ) The gradients of boundaries AB and BC are 2 and -1 respectively. Find all the equations
of all the five boundaries of the nursery bed.
5 2 =2
= 3 = 3
Equation of CD
y 4
Equation of AE Equation of DE x7 =2
y 5 5 y 0 2
x0 = 3 x3 = 3
y-4=2x-14
5 2
y= x +5 y= 3 x -2
3
Equation of AB Equation of BC
y 5 1 y 7
= =1
x0 2 x4
2y-10=x y-7=-x+4
1 y=-x+11
y= 2 x+5
2 ) Calculate the area of the nursery bed by using two methods, one of which must be
integration.
Method one
Coordinate of Point B Area
1
y= 2 x+5 (1) = 2 5 |
1 0 4 76 3 0
7 42 0 5 |
y=-x+11 (2) 1
sub (1) to (2) = 2 |16+14+ 152049246|
1
x+5=x+11
2 1
= 2 |54|
3
x
2 =6 =27unit
X=4
When x=4
1
y= 2 ( 4 )+ 5
y=7
coordinate of B is B( 4,7 )
Method two
Area
4 7 3 6 7
= AB+ BC AE DE CD
0 4 0 3 6
4 7 3 6 7
1 5 2
= 2 x+ 5 dx+ x +11 dx 3 x+5 dx 3 x2 dx 2 x 10 dx
0 4 0 3 6
4 7 3 6
= [ 1 2
4
x +5 x ] [
0
+
x2
2
+11 x ] [
4
-
5 2
6
x +5x ] [
0
-
2 2
6
x 2 x ]
3
-
7
[ x 210 x ]6
1 ( 7 )2 ( 4 )2 5 2
=[( 4 (4) +5(4))-0] + [( +11(7)( + 11(4 ))[( ( 3 ) +5(3))0]
2 2 6
2 2 2 2
( ( 6 ) 2(6))( (3 ) 2(3)) 2 2
6 6 ] - [( ( 7 ) 10( 7)( ( 6 ) 10(6))
33 15
=24+ 2 - 2 -3-3
=27unit
Worksheet 8:Money-minded!
1 ) A road junction is to be built along the highway connecting village B and Village Q to the
highway. The equation of the highway is y=2x+1. Determine the location of the road junction so
as to minimize the construction cost. Use two methods.
Method one
Let, the junction to village Q=M
the junction to village P=N
Village Q= Q
Village P=P
Method two
Let, the junction to village Q=M
the junction to village P=N
Village Q= Q
Village P=P
Equation of NP
(1) 1
y 2 x +1
Equation of MQ,
1 1
y 2 x + 11 gradient of equation NP
2
1 gradient of equation NP in vector form=
gradient of equation MQ
2
gradient of equation MQ in vector form=
(12 )
(12 ) point = (4,-1 )
point = ( 8,7 )
Point in vector form= (14 )
()8
Point in vector form = 7
2
Equation of vector NP , r= 1 s + ( ) (14 )
Equation of vector
MQ , r= (12 ) s+
(2)
(87) (2) ( 1 ) t+
2 (13)=(12 ) s (14 )
+
() () ( )
1
t+
1
=
2
s+ () 8 (2t+t+13)=(s 1)
2 s+ 4
2 3 1 7
t+1=2s+4 2t+3=-s-1
1
t=2s+3 (1) t=- 2 s2 (2)
(2t+t+13) = (s2 s ++87) sub(1) to (2)
t+1=2s+8 2t+3=-s+7 1
2s+3=- 2 s2
1 s=-2
t=2s+7 (1) t= s+2 (2)
2 when s=-2
((2)+ 7 )
2(2)+ 8
( 49)
=
2) A school is to be built as far away from the highway as possible and it must be 5km from each
of the two villages P and Q. Determine the location of the school by solving simultaneously.
Linear equation Non- linear equation (1) Non- linear equation (2)
y 3 1 Let the location of school=S Let the location of school=S
=
x6 2 QS=5 PS=5
2y-6=-x+6 ( x 8 ) +( y 7)
2
=5 ( x 4 ) +( y +1)
2
=5
1 ( x8 )2 + ( y7 )2=25 ( x4 )2 + ( y +1 )2=25
y= x +6
2
x-16x+64+y-14y+49=25 x-8x+16+y+2y+1=25
x=-2y+12 (1) x+y-16x-14y+88=0 --(2) x+y-8x+2y-8=0 (3)
P = ( xy ) =
x
i+
y
j
Question
The position vectors of Johan, Kassim and Latif relative to thr origin O are OJ ,OK and OL
(
4t
respectively. It is given that OJ = 63 t ) ,
OK = (4t
92 t
) and OL
= ( 424
t12)
t
(
OJ = 4t
63 t )
OK = (4t
92 t
)
OL = ( 424
t12)
t
OJ = 3
3 () (49 ) (4(1)
92(1)
) (122 ) ( ( ) ( ))
24 1
4 1 12
Gradient,
OK =
( )
OL = 2
8
36
34
=3 (57 ) Gradient,
8+12 4
=
Equation of Johan, Gradient, 22 4
y 6 5+ 4 1 = -1
=3 =
x4 79 2 Equation of Latif
y-6=3x-12 1 y+8
=1
y=3x-6 = 2 x +2
(63 t ) (4s )
4t = 92 s
(63 t ) ( 4 m12)
4t = 24 m
4-t=9-2s 4-t=2-4m
2s-t=5(1) 4m=t-2(1)
6-3t=-4-s 6-3t=4m-12
s-3t=-10(2) 4m=18-3t(2)
when (2) is multiple by 2 Sub (1) to (2)
2s=6t-20(3) t-2=18-3t
Sub (1) to (2) 4t=20
6t-20-t=5 t=5
5t=25 when t=5
t=5 4m=18-3(5)
when t=5 3
2s=6(5)-20 m= 4
s=5
When s=5 3
When m= 4
OK = (92(5)
45 ) (9 )
1
=
( )
3
24( )
OL = 3
4
4 ( )12
= (1
9 )
4
Answer: Johan, Kassim and Latif will meet one another, they will meet at ( -1,-9 )
Method Two
y=3x-6(1) Sub (1) to (2) Sub (1) to (3) Sub (2) to (3)
3x-6= 3x-6=-x-10 1 17
x
1 17 1 17 4x=-4 2 2 =-x-10
y= 2 x 2 2
x
2 x=-1
when x=-1
(2) 5 5 y=3(-1)-6 3 3
x= x=
2 2 y=-9 2 2
y=-x-10(3) ( -1,-9 )
x=-1 x= -1
when x=-1 when x=-1
y=3(-1)-6 y=-(-1)-10
y=-9 y=-9
( -1,-9 ) ( -1,-9 )
Answer: Johan, Kassim and Latif will meet one another, they will meet at ( -1,-9 )
i j
express, for each student V x and V y in terms of and by using differentiation.
~ ~
Hence, find the resultant velocity V of each student in the form x i +y j to determine who
jogs the fastest.
(
OJ = 4tOK
63 t
=
) =
OL
x=4-t
dx
(4t
92 t
) ( 424
t12)
t
=1
dt
x=9-2t x=2-4t
V x =i dx
=2
dx
=4
dt dt
V x =2 i V x =4 i
Resulta
nt Resulta Resulta
velocity nt nt
= velocity velocity
2
=
(1 ) +(3) =
= (2 ) +(1)
2
(4 ) +(4)
2
10 = 5 =
32
Answer: Latif (student B) jogs the fastest because the resultant velocity is the highest which is
32 .
( )
72t
OM
(
2 t2
respectively. It is given that OS = 6 t+9 ) (
t +5
, OP = t2 ) and OM = 2 t5
3
where t is the time in minutes.
( a ) Find the coordinates of Paul and Martin when they are in the location shown in Diagram 8.
In the diagram the coordinate of Sam is ( 0,3 ).
(62 t2
t+ 9) (03)
=
2t-2=0
2t=2
t=1
when t=1
= ((1)2
(1)+5
) =
( )
72(1)
2
(1)5
3
= (34 )
( )
5
= 13
The coordinate of Paul = ( 4,-3 )
3
13
The coordinate of Martin = ( 5, 3 )
(
OS = 2 t2 ) ( )
OS = 2 t2
6 t +9 ( )
OS = 2 t2
6 t+9 ( )
OS = 2 t2
6 t+9
6 t+9
OS =(2) OS =(0 ) OS =( 2 )
9 3 3
Coordinate = ( -2,9 ) Coordinate = ( 0,3 ) Coordinate = ( 2,-3 )
Paul
(
OP = t +5 ) ( )
OP = t +5
t2 ( )
OP = t +5
t2 ( )
OP = t +5
t2
t2
OP =( 5 ) OP =( 4 ) OP =( 3 )
2 3 4
Coordinate = (5,-2 ) Coordinate = ( 4,-3 ) Coordinate = ( 3,-4 )
Martin
( ) ( ) ( )
72t 72t 72t
( )
72t OM = 2 OM = 2 OM = 2
t5 t5 t5
OM = 2 3 3 3
t5
3
OM = 7
5 ( ) 5
OM = 13
( )
3
OM = 11
( )
3 3
Coordinate = ( 7,-5 )
Coordinate = ( 5, Coordinate = ( 3,
13 11
)
3 3
( The loci is draw on the graph paper on next page )
( c ) At a certain time, the position of Sam, Paul and Martin lie on a straight line.
( i ) Determine this using two methods.
Method One
Let, Sam=S
Paul=P
Martin=M
SP =OP OS
= (t2) (6 t +9) ( 5 t11 )
t+5 2t2 = 3t +7
( )( ( )
72 t 2t
PM =OM OP =
2
3
t15
t+5 =
t2 )5
3
t3
Gradient SP =Gradient PM
5
t3
5t11 3
=
3 t+7 2t
35
10t-22-5t+11t=-5t+ t +9t-21
3
1
t
3 =1
t=3
Answer: At t=3, Sam Paul and Martin are on the straight line.
y+9=-2x+8
y=-2x-1
( iii )Find the ratio of the distance between Sam and Paul to the distance between Sam and
Martin at that time.
When t=3
Coordinate of Sam is ( 4,-9 ), Coordinate of Paul is ( 2,-5 ), Coordinate of Martin is ( 1,-3 )
Distance between Sam and Paul Distance between Sam and Martin
= ( 42 ) +(9+5)
2
= ( 41 ) +(9+3 )
2 2
= 4+ 16 = 9+36
= 20 = 45
SamPaul 20 2
= =
SamMartin 45 3
2
Answer: The ratio between Sam and Paul to the distance between Sam and martin is 3 .
( c ) Hence by considering integer coordinates only, calculate a possible area for each
parallelogram.
= 2 4 |
1 2 14 17 5 2
9 13 8 4 | |
1 5 17 13 1 5
= 2 2 7 10 5 2 |
1 1
= 2 |18+182+136+20561536516| = 2 |35+170+65+234911025|
1 1
= 2 |66| = 2 |112|
2 )( a ) Diagram 10 shows two straight lines, PQ and RS, intersecting at the origin O, and the
line x=2 drawn on a Cartesian plane. The gradients of PQ and RS are m1 and m2
Let the y-coordinate of the point that intersect When the point ( 2,4 ) if rotate 90 clockwise,
the line x=2 be 4 the new point is ( 4,-2 ).
Point = ( 2,4 )
1
m1 m2=(2)( )
2
m1 m2=1 (shown)
( b ) Show that it is possible for the two lines 2x-(m+1)y=5 and 3my=x-1 to be parallel but
impossible to be perpendicular.
2x-(m+1)y=5 2x-(m+1)y=5
(m+1)y=2x-5 (m+1)y=2x-5
2 2 5
y= m+1 x - y= m+1 x - m+1
5
m+1
gradient of this equation is
gradient of this
2 m1 m2=1
equation is m+1
2
3 ) Find the perpendicular distance of the point ( 8,3 )from the line y=2x-3
4 ) A( -1,7 ) and B( 7,13 ) are two points on the Cartesian plane. Point P moves such that the
product of the gradient of PA and PB is always -1
( a ) Find the equation of locus P
Let coordinate of P be P( x,y )
(gradient PA)(gradient PB)=-1
y 7 y 13
( ( =1
x +1 x7
y-13y-7y+91=-x+7x-x+7
x+y-6x-20y+84=0
( b ) The x-coordinate of point Q is -2 and it lies on the locus of P. Calculate the area of triangle
AQB in two ways
= 90
= 15 unit
=15 unit
5 ) Point P moves on a Cartesian plane such that it is always at a constant distance from point
C( 0,12 ). Diagram 11 shows part of the locus of point P which passes through point A( -8,8 )
and point B( 4,4 )
( a ) Find the equation of locus P.
AC PC = 80
= (80 ) +(812)
2
( x 0 ) +( y 12)
2
= 80
= 80
x+y-24y+144= 80
BC
x+y-24y+64=0
= ( 40 ) +(412)
2
= 80
1
( 2 (-2)=-1
since line AC and line BC are perpendicular and have the same length with a locus P passing
through, it is a sector of a circle with 90.
Area of trapezium ACO Area of trapezium CBO Area of the sector of circle
1 1 22 1
= ( 8+12 ) ( 8 ) = ( 4+ 12 )( 4 ) =( ( 80 ) ( 4
2 2 7
=80 unit =32 unit =62.857 unit
Thank you