ADDITIONAL MATHEMATICS PROJECT WORK 2017jpnperak.moe.gov.my/jpn/attachments/article/5179/3.4 KPMT... · ADDITIONAL MATHEMATICS PROJECT WORK 2017 ... * A brief biography on Rene Descartes

PERAK STATE

ADDITIONAL

MATHEMATICS

PROJECT WORK

2017

TEACHER’S GUIDE

PERAK STATE EDUCATION DEPARTMENT

Additional Mathematics Project Work: TEACHER’S GUIDE YEAR 2017

PAGE | 1


Worksheet 1 : Let’s coordinate!

PART 1.1

* A brief biography on Rene Descartes.

* A simple anecdote on his contributions and achievements focusing on coordinate geometry.

PART 1.2

* A suitable name for the game.

* Involve basically the concept of coordinates. Other related mathematical concepts will be an extra

bonus for the student.

* Rules of the game.

* It can be an Indoor or outdoor game. Any modified board games like Snake and Ladder are acceptable

but preferably original.

Worksheet 2 : Let’s get into shape!

(a)

SET 1 SET 2

Triangle Coordinates of vertices Triangle Coordinates of vertices

A ( 0 , 1 ) ( 2 , 0 ) ( 2 , 2 ) P ( 5 , 0 ) ( 5 , 2 ) ( 7 , 1 )

B ( 0 , 3 ) ( 2 , 2 ) ( 2 , 4 ) Q ( 5 , 2 ) ( 5 , 4 ) ( 7 , 3 )

C ( 0 , 5 ) ( 2 , 4 ) ( 2 , 6 ) R ( 5 , 4 ) ( 5 , 6 ) ( 7 , 5 )

D ( 0 , 7 ) ( 2 , 6 ) ( 2 , 8 ) S ( 5 , 6 ) ( 5 , 8 ) ( 7 , 7 )

E ( 0 , 9 ) ( 2 , 8 ) ( 2 , 10 ) T ( 5 , 8 ) ( 5 , 10 ) ( 7 , 9 )

F ( 0 , 11 ) ( 2 , 10 ) ( 2 , 12 ) U ( 5 , 10 ) ( 5 , 12 ) ( 7 , 11 )

TABLE 1

(b) Refer page 2.

(c) Workmanship in building model : Skilful, neat and nice.

(d) Area of rectangular surfaces = 6 ( 6 4 ) = 144 cm2

Area of triangular surfaces = 12 (2

1 4 4 ) = 96 cm2

Total surface area of model = 240 cm2


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y

x 0

1 2 3 4 5 6 7

1

2

3

4

5

6

7

8

9

10

11

12


PAGE | 3


Worksheet 3 : Enjoy the journey!

Derive correctly the distance formula: PQ = √( 𝑥2 − 𝑥1)2 + ( 𝑦2 − 𝑦1)

2 .

1. BJ = 5 km , JA = 15 km , JC = 15 km

Total distance = 2 [ (15 + 5) + 15 ] = 70 km

2. Town C to Town A

CA = 18 km

Total distance = 18 + 15 + 5 + 20 = 58 km

Town C to Town B

CB = 17.09 km

Total distance = 15 + 5 + 17.09 + 15 + 15 = 67.09 km > 58 km

New road from Town C to Town A allows Ali to travel a shorter distance. Thus it is more preferable.

* Student must show that the total distance is the same irrespective of whether Ali goes to Town B or

Town C first.

Worksheet 4 : Coconut delight!

Derive correctly the formula: X = ( 𝑛𝑥1 + 𝑚𝑥2

𝑚+𝑛 ,

𝑛𝑦1 + 𝑚𝑦2

𝑚+𝑛 )

and midpoint of PQ = ( 𝑥1 + 𝑥2

2 ,

𝑦1 + 𝑦2

2 )

1. (𝑥+35

2 ,

𝑦+40

2) = (

5−10

2 ,

30+10

2) OR (

𝑥+5

2 ,

𝑦+30

2) = (

35−10

2 ,

40+10

2)

OR (𝑥−10

2 ,

𝑦+10

2) = (

35+5

2 ,

40+30

2)

Location of fourth coconut tree = ( – 40 , 0 ) OR ( 20 , 20 ) OR ( 50 , 60 )

* Student must show all the three possibilities

2. (a) (15+35

4 ,

40+90

4)

Location of coconut tree = ( 12.5 , 32.5 )

(b) AB = 25

(−30+10

5 ,

30+60

5) OR

5𝑥+107

= – 10 , 5𝑦+60

7= 10

Location of coconut tree = ( – 4 , 18 ) OR ( – 16 , 2 )

* Student must show both possibilities.


PAGE | 4


Worksheet 5 : Share and share alike!

Derive correctly the formula : Area of triangle ABC = 1

2 |𝑥1 𝑥2

𝑦1 𝑦2

𝑥3 𝑥1

𝑦3 𝑦1|.

Deduce correctly the formula : Area of quadrilateral ABCD = 1

2 |𝑥1 𝑥2

𝑦1 𝑦2

𝑥3 𝑥4

𝑦3 𝑦4

𝑥1

𝑦1|.

Condition: The four vertices must be arranged clockwise or anticlockwise.

1. 1

2 |0 102 0

8 26 10

02| 4 = 14 km2

2. 8x – 2y = 24 and 2x + 10y = 48 or other equivalent simultaneous linear equations. X ( 4 , 4 )

Worksheet 6 : Close encounter!

1. 𝑦2− 𝑦1

𝑥2 − 𝑥1

2. (a) m1 = m2

(b) m1 m2 = – 1

Parallel : 𝑦−2

𝑥+3 =

5+1

−4−4

Police station, Fire station : Midpoint = ( 0 , 2 )

Perpendicular : 𝑦−2

𝑥−0 =

4

3

Location Rita is nearest the petrol station = ( – 1.08 , 0.56 )

Distance travelled = 21.8 km

Time taken = 16.35 min


PAGE | 5


Worksheet 7 : Nursery fun!

1. y = mx + c

2. y – y1 = m( x – x1 )

3. 𝑦−𝑦1

𝑥− 𝑥1 =

𝑦2− 𝑦1

𝑥2 − 𝑥1

4. 𝑥

𝑎 +

𝑦

𝑏 = 1

1. AB : y = 1

2 x + 5

BC : y + x = 11

CD : y = 2x – 10

DE : 3y = 2x – 6

EA : 𝑥

3 +

𝑦

5 = 1

* Accept equivalent form of the equation for all boundaries

2. At vertex B, y = 1

2 x + 5 and y + x = 11

B( 4 , 7 )

Method 1

Area of nursery plot = 1

2 |3 60 2

7 44 7

05

30 | or equivalent

= 27 m2

Method 2 [Integration]

Area under AB = ∫1

2

4

0 𝑥 + 5 𝑑𝑥 = 24

Area under BC = ∫ −𝑥 + 11 𝑑𝑥7

4 = 16.5

Area under AE = ∫ −5

3

3

0𝑥 + 3 𝑑𝑥 = 7.5

Area under ED = ∫ 2

3

6

3𝑥 − 2 𝑑𝑥 = 3

Area under DC = ∫ 2𝑥 − 10 𝑑𝑥7

6 = 3

Area of nursery plot = 24 + 16.5 – 7.5 – 3 – 3

= 27 m2


PAGE | 6


Worksheet 8 : Money-minded!

1. Method 1

PC + CQ = BC + CQ

BCQ straight line, distance shortest

Method 2

Create the following table. Excel recommended.

C , location on highway y = 2x + 1 Distance CP Distance CQ Distance CP + CQ

Student observed C(2 , 5) gives smallest value for CP + CQ.

2. (a) Non-linear equations : ( x – 8 ) 2 + ( y – 7 )2 = 52 and ( x – 4 )2 + ( y + 1 )2 = 52

Possible locations : (4 , 4) , ( 8 , 2)

Furthest location : ( 8 , 2)

(b) PQ : midpoint = ( 6 , 3 ), gradient = 2

Perpendicular : 𝑦−3

𝑥−6 = –

1

2

Linear equation: x + 2y = 12

Non-linear equation : Either ( x – 8 ) 2 + ( y – 7 )2 = 52 or ( x – 4 )2 + ( y + 1 )2 = 52

Possible locations : (4 , 4) , ( 8 , 2)

Furthest location : ( 8 , 2)

P ( 4 , – 1 )

Q ( 8 , 7 )

A

B

C

Location for

shortest distance and

hence minimum

construction cost

A( x , 2x + 1), P ( 4 , – 1 )

Perpendicular : 2𝑥+2

𝑥−4 = –

1

2

x = 0

A( 0 , 1 )

Midpoint : B(– 4 , 3 )

C( x , 2x + 1), B(– 4 , 3 ) , Q( 8 , 7)

Collinear : 2𝑥−6

𝑥−8 =

7−3

8+4

x = 2

C( 2 , 5 )

* Accept other correct methods except drawing.


PAGE | 7


Worksheet 9.1 : Hello! Nice meeting you!

Define correctly : 𝑂𝑃⃗⃗⃗⃗ ⃗ = (𝑥𝑦) or 𝑂𝑃⃗⃗⃗⃗ ⃗ = xi + yj

(a) Initial positions : 𝑂𝐽⃗⃗⃗⃗ = (46) , 𝑂𝐾⃗⃗⃗⃗⃗⃗ = (

9−4

) , 𝑂𝐿⃗⃗⃗⃗ ⃗ = (2

−12)

Student A : Johan, Student B : Latif , Student C : Kassim

(b) Student A (Johan) : x = 4 – t , y = 6 – 3t , equation of path : y = 3x – 6

Student B (Latif) : x = 2 – 4t , y = 4t – 12 , equation of path : 2y = x – 17

Student C (Kassim) : x = 9 – 2t , y = – 4 – t , equation of path : x + y = – 10

(c) If Johan meets Kassim, 𝑂𝐽⃗⃗⃗⃗ = 𝑂𝐾⃗⃗⃗⃗⃗⃗ giving t = 5

If Kassim meets Latif, 𝑂𝐾⃗⃗⃗⃗⃗⃗ = 𝑂𝐿⃗⃗⃗⃗ ⃗ giving t < 0

Conclusion: Only Johan and Kassim meet one another at a common location.

At common location: Method 1

𝑂𝐽⃗⃗⃗⃗ = 𝑂𝐾⃗⃗⃗⃗⃗⃗ = (−1−9

). Common location = ( – 1 , – 9 )

Method 2

y = 3x – 6 and x + y = – 10. Common location = ( – 1 , – 9 )

(d) Vx = 𝑑𝑥

𝑑𝑡 , Vy =

𝑑𝑦

𝑑𝑡

Johan : Vx = – i , Vy = – 3j , V = – i – 3j , V = √10

Kassim : Vx = – 2i , Vy = – j , V = –2i – j , V = √5

Latif : Vx = – 4i , Vy = 4j , V = – 4i + 4j , V = √32

Jogs fastest : Latif


PAGE | 8


Worksheet 9.2 : The grand line-up!

(a) 𝑂𝑆⃗⃗⃗⃗ ⃗ = (03) , t = 1

Paul : ( 4 , – 3 ) , Martin : ( 5 , – 4 1

3 )

(b) Student draws on the same axes the following straight lines.

Sam : 3x + y = 3 , Paul : y – x = – 7 , Martin : x + 3y = – 8

(c) (i) Method 1

𝑆𝑃⃗⃗ ⃗⃗ = k 𝑃𝑀⃗⃗⃗⃗ ⃗⃗ or equivalent

t = 3

Method 2

Gradient : 5𝑡−11

7−3𝑡 =

5𝑡−9

6−3𝑡 or equivalent

t = 3

(ii) Collinear : S(4 , – 9) , P(2 , – 5) , M(1 , – 3)

Equation of straight line : 2x + y = – 1 .

(iii) (4 – 2) : (4 – 1) OR (– 9 + 5 ) : (– 9 + 3)

Ratio = 2 : 3


PAGE | 9


Worksheet 10 : Loving it, HOTS!

1.

Area = 2 1

2 |0 12 30 5 4

00| = 33 unit2 Area = 2

1

2 |0 −4 −120 3 −5

00| = 56 unit2

2. (a) (0 , 0) , (2 , 2m1) , (2 , 2m2)

Pythagoras Theorem : (2 – 0)2 + (2m1 – 0 )2 + (2 – 0)2 + (2m2 – 0 )2 = (2m1 – 2m2 )2

m1 m2 = – 1

(b) Parallel : 2

𝑚+1 =

1

3𝑚 , m =

1

5 , solution exists , possible for lines to be parallel

Perpendicular : 2

𝑚+1

1

3𝑚 = – 1

3m2 + 3m + 2 = 0

b2 – 4ac < 0 , no solution , lines impossible to be perpendicular

3. Perpendicular : : 2𝑥−6

𝑥−8 = –

1

2

x = 4 , y = 5

Perpendicular distance = √20 = 4.472

4. (a) Locus of P : Circle with centre ( 3 , 10 ) and radius 5 units

Equation of locus : (x – 3 )2+ ( y – 10 )2 = 52

x2 + y2 – 6x – 20y + 84 = 0.

(b) x = – 2 , y2 – 20y + 100 = 0

y = 10

Method 1

1

2 |−2 7 −110 13 7

−210

| = 15 unit2

Method 2

1

2 √90 × √10 = 15 unit2

O

A B

5 units

13 units

5 units

13 units

x

y

(12 , 5) (3 , 4)

y

x O

(– 4 , 3 )

(– 12 , – 5 )


PAGE | 10


5. (a) Equation of locus : x2 + ( y – 12 )2 = (0 – 4)2 + (12 – 4)2 or equivalent

x2 + y2 – 24y + 64 = 0

(b) mAC mBC = – 1

ACB = 90o

ACB is a quadrant of a circle

Area = 1

2(8 + 12)(8) +

1

2(12 + 4)(4) –

1

4𝜋 (√80 )2

= 112 – 20𝜋 unit2

6. (a) Constant rate = 1.2−1.5

2 = – 0.15 m per week

Thickness y = 1.5 – 0.15x or equivalent

(b)

(c) 10 weeks

Worksheet 11 : Reflection

* Mention other topics like Linear Law, Linear Programming, …….

* Stresses and marvels creatively at the usefulness of coordinates in solving various problems.

O x

y

1.5

10

Documents

ADDITIONAL MATHEMATICS PROJECT WORK 2017jpnperak.moe.gov.my/jpn/attachments/article/5179/3.4 KPMT... · ADDITIONAL MATHEMATICS PROJECT WORK 2017 ... * A brief biography on Rene Descartes