PROGRAM SENTUHAN AKHIR MATEMATIK PMR • Tahun 2010
PN. ZURAIDA BINTI ADAM SMK SUNGAI PUSU GOMBAK, SELANGOR.
U SUCCESS WILL BE YOURS IF…. DREAM FOR IT. PLAN FOR IT. WORK FOR IT.
RACE FOR IT.
Objectives of the Seminar is to: Make sure, you understand the requirements of the PMR Mathematics Examination. Show the simplest steps to solve problems with accurate answer. Understand the marking scheme.
GIVE ATTENTION TO: 1. Interpretation of the answer:
(a) (-3)(-2) (b)
(c)
≠
6
−3 3 ≠ −2 2
3 − 3 ≠ −2 2
(d)
(e)
0 ≠0 5
3 ≠3 1
1. Interpretation of answer:
(f)
(g)
(h)
(i)
4 1 ≠ 8 2
Except in certain cases
7 1 = 3 = 3.5 2 2
Except in certain cases
2 ≠8 3
4x − 6 y 2x − 3y ≠ 2x x
THE CORRECT WAY TO WRITE THE FINAL ANSWER.
5 8
15 1. 24
=
2. 62
= 36
3. 4.
6 x y − 18 x 24 x y
5.
=
1x = x x =x 1
= 6.
2
6 x ( y −3 x) 24 x y y − 3x 4y
x =1 0
THE WRONG WAY TO WRITE THE FINAL ANSWER
7 ( x − 2)
1.
7 x − 14
2.
15 no need to change to 2 1 7 7
3.
−4 6
=
≠
− 4 6
≠
( −4 , 6 )
Things to take note of… Do NOT use liquid paper; a) Old words will appear b) Forget to write the correct answer
For subjective questions, show working systematically!
Don’t spend too much time on one question. Before starting, quickly glance through all the questions (Objective: 40, Subjective: 20)
Don’t wait till the last minute to shade the answers (objective)
PAPER 1
Questions se-tup • Objective questions with multiple choices (Four choices of answers) • Choose one correct answer • Darken/shade the prepared answer space in objective paper immediately.
How to get the answer? • Mental arithmatic • Do direct calculation • By using the given answer, work backwards
•The method of calculation need not be shown.
Objective question answering skills SUPERSTEP TO SUCCESS!!!
Simplify 1. ( – 4x + 5y ) – (8x – 3y) = A. 4x – 8y B. 8y – 12 x C. 8y + 12x D. 9x – 5y
SIMPLIFY BY OPENING THE BRACKET..
( − 4X + 5Y − 8X + 3Y) −12X + 8Y REARRANGE 8Y – 12 X ANSWER B.
What is the answer? 2.
( 4h – 7)2 =
A . 16h – 49 B . 16h2 + 49 C . 8h2 −12h + 49 D . 16h2 − 56h + 49
•
Simplify the multiplication
•
( 4h – 7) =
SQUARED THE BRACKET
2
2
1
( 4h – 7)(4h – 7) 3 4
16h2
-28h -28h +49
16h2 – 28h – 28h + 49 16h2 – 56h + 49 Answer: D
3. Simplify to the most simplest form. (2m( m – 5) + 7m) A. B. C. D.
2m2 2m2 2m2 2m2
+ − − +
2m − 5 12 m 3m 2m
Simplify the multiplication
2m( m – 5) + 7m Open the bracket …
2m2 – 10m + 7m 2m2 – 3m The answer is C.
Simplify this question.. 4).
( x – 3y)2 + 7xy = A. B. C. D.
x2 (x – 2) + xy x2 +7xy – 9y2 x2 + 7y2 + 8xy x2 + 9y2 + xy
Solution … 4) ( x – 3y)2 + 7xy = ( x – 3y) ( x – 3y) + 7xy x2 − 3xy −3xy + 9y2 + 7xy x2 + 9y2 − 6xy + 7xy = x2 + 9y2 + xy answer D.
5 . The difference between median and mode in the number row 2,4,5,4,3,2,2,5,7 is A. B. C. D.
1 2 3 4
Solution. Determine the median for 2, 2, 2, 3, 4 , 4, 5, 5, 7 The median is 4. And mode is 2 the difference between median and mode is 2 answer : B
1. 923 758 becomes 924 000 after it is rounded off to A B C D
one ten hundred thousand
923 758 3 000
Answer D
3 758
4 000
2. Diagram 2 shows some of the factors of 270. The possible value of y is 2 y
270 9
15 DIAGRAM 2
A B C D
4 6 8 12
3 10
270 6
= 45
3.
The water level in a container is 2.0 m on Sunday. The water level drops by 25 % on Monday. It rises by 40% on Tuesday as compared to Monday’s level. What is the height, in m, of the water level on Tuesday? A 0.9 B 1.2 C 1.9 D 2.1
Sunday = 2.0m Monday = 2.0 – (25% × 2.0) = 2.0 – 0.5
= 1.5
Tuesday = 1.5 + (40% × 1.5) = 1.5 + 0.6 = 2.1
4.
A group of students consists of 42 boys and 70 girls.
2 of the boys and 1 of the girls attended a youth camp. 3 2 Calculate the percentage of students who attended the youth camp.
A B C D
43.75 56.25 60.00 62.50
Total number of the student = 112
Number of students attended the youth camp
1 2 = ( x 42 boys) + ( x 70 girls) 3 2
= 28 boys + 35 girls = 63 students Therefore % of students attended the youth camp 63 x 100 = 56.25 % = 112
5. Diagram 5 shows four rectangles drawn on a square grid. B
C
A
D DIAGRAM 5
Among A, B , C and D , choose the rectangle with the smallest fraction shaded.
A. 4 12 B. C. D.
=
1 3
1 4 = 4 16 4 = 1 20 5 8 1 = 3 24
6. Diagram 6 shows the change in the reading of a weighing machine when two cakes of equal mass are removed.
0.35 kg 0.35 kg
0.18 kg
0.18 kg
2 cakes are removed DIAGRAM 6
Find the mass, in g, of each cake that is removed A 42 B 85 0.35kg – 0.18kg = 0.17kg C 425 0.17kg ÷ 2 = 0.085kg D 850 0.085kg × 1000 = 85 g
7. In Diagram 7, PQST and RSUV are rectangles. R
and U are the midpoints of QS and ST respectively. P
8 cm
Q
11 cm
VV
R
T
U
DIAGRAM 7
S
Find the area of the shaded region, A 44 cm2 B 55 cm2 C 66 cm2 D 88 cm2
Area of shaded region = area of PQT + RSUV Area of PQT = ½ × 11cm × 8cm = 44cm2 Area of RSUV = 4cm × 5.5cm = 22cm Area of shaded region = 44cm2 + 22cm2 = 66cm2
8. In Diagram 8, PQUV is a square and RSTU is a
rectangle.
T 7 cm
V
U
S
5 cm
P
Q
R
DIAGRAM 8
Given that PR = RS, perimeter of the whole diagram is A 37 cm B 47 cm C 57 cm D 67 cm
T 7 cm
10cm V` 5cm P
5 cm U
S 10cm
5cm
Q
5cm
R
Perimeter = 5cm + 5cm + 5cm + 5cm + 10 + 7 + 10 = 47cm
9. Given that the mean of 5, 7, 6, 3, p, 8 and 7 is 6, find the value of p is A. B. C. D.
5 6 7 8
5+7+6+3+p+8+7 7
=6
36 + p = 6 x 7 36 + p = 42 p = 42 – 36 p=6
10. Diagram 10 is a circle with center O. PQR is a straight line. R 32o 53o
Q y
o
P The value of y is A B C D
53o 69o 74o 85o
O
DIAGRAM 10
90o – 32o = 58o yo = 180o – (58o + 53o) = 69o
11
Diagram 11 shows a set of numbers. 7, 3, 6, 9, 8, 4, 2, 3 DIAGRAM 11 The difference between mode and median of the above numbers is A B C D
0.25 0.50 1.00 2.00
mode = 3 median = 2,3,3,4,6,7,8,9 4+6 =5 2 median – mode = 5 – 3 =2
If you are smart enough, nothing can be a problem.
CCT Question Table shows the time allocation for a test. Test Paper 1
Time Allocation 1 Âź hours
Break 20 minutes Paper 2 1 ½ hours All candidates must be in the examination hall 10 minutes before Paper 1 starts. Paper 2 ends at 1.05 p.m. At what time must the candidates be in the hall before paper 1 starts? A 9.35 a.m.
C 10.00 a.m.
B 9.50 a.m.
D 10.10 a.m.
SOLUTION To solve, work backwards. Convert 1.05 p.m. 1.05 + 1200 = in 24-hour system. 0105 + 1200 = 1305 Find the duration of the papers including break.
1¼ hours + 20 mints + 1½ hours + 10 mints = 3hrs 15mins = 0315
Subtract it.
1305 – 0315 = 0950 That is,
Write the answer.
9.50 a.m.
Answer: B
Q
x째
R
O
100째 85 째 P
S
T
In diagram 16, PQRS is a circle with centre O and PST is a straight line. Find the value of x. A 15o C 45o B 35o D 50o
Q
SOLUTION
x°
R
O
100 ° 85 ° P
S
T
Knowing that angles subtended at the circumference by an arc are half the one at centre, ∠ PQS = ½ ∠ POS. Therefore ∠ PQS = 50o. Since PQRS is cyclic quadrilateral, ∠ PQR = ∠ RST = 85o
Therefore, ∠ SQR = ∠ PQR − ∠ PQS = 85o − 50o = 35o
Answer: B
PAPER 2
I swear! I didn't use the calculator.
Mathematics
QUESTION FORMAT • Answers must be written in the space provided in question paper. • Marks given based on steps made during calculation and accuracy of answer.
If a question needs candidates to show the proper solving steps ..,
BUT
Only the final answer written in the space provided...
NO Marks will be given, because it is assumed candidates did not Do calculation.
• A. Calculate the value of
6(57 ÷ 3 − 4) + 51 Solution: 6(19 − 4) + 51 6(15) + 51 90 + 51 141
K1 N1
Solve each of the following linear equation a) 2q = −5q −14 b) 3(4m − 1) = 2m + 5 Solution: a) 2q + 5q = −14 7q = −14 q = −2
P1
• Factorise completely • b ) p(5 – x) + 2q(x – 5) 5p – px + 2qx – 10q 5p – 10q – px + 2qx 5( p – 2q)− x( p −2q) (p – 2q)(5 – x)
K1 K1
Given p – 2 = q2 + 3, express q in the term of p
• Rearrange the expression q2 + 3 = p – 2 q2 = p – 2 – 3 q = p − 2 −3 q=
p −5
K1 Not complete yet
N1
Paper 2 topics with ‘BIG’ marks !!!! TOPIC 1. Algebraic Expressions (Form1,2 & 3) i. Expansion ii. Factorization iii. Algebraic Fractions iv. Algebraic formulae 2. Loci in Two Dimensions (Form 2) 3. Transformations 1 or 2 questions
MARKS
2 marks 3 marks 3 marks 3 marks 5 marks 4 to 6 marks
4.
Statistics (Form 2 & 3) ~ pictograph (2004), bar chart (2004), pie chart (2005), line graph (2006),bar chart (2007)
4 marks
5.
Geometrical Constructions (Form 2)
6 marks
6.
Graph of Functions (Form 3)
4 marks
TOTAL (approximately) :
35 marks
FRACTIONS 4 basic operations Simplifying Involving LCM Transferring the answer
1. Calculate the value of 3 1 2 2 ÷ 1 + 5 3 5
1. Calculate the value of 3 1 2 2 ÷ 1 + 5 3 5
13 4 2 = ÷ + 5 3 5 13 26 = ÷ 5 15
K1
13 15 195 = × = 5 26 130 1 3/ 2 = 2
1
N1
2. Calculate the value of
2 1 3 −1 × − 3 5 4 and express the answer as a fraction in its lowest term. [2 marks]
5 11 − × − 3 20
11 12
K1 N1
Calculate the value of
3 1 2 2 − ÷ 2 5 3 5 and express the answer as a fraction in its lowest term.
STEPS TO BE SHOWN 7 2 13 3 1 2 = − ÷ 2 − ÷ 2 3 5 5 5 3 5
=
35 6 5 − × 15 15 13
29 5 1 = × 3 15 13 =
29 39
N1
K1
1 1.Calculate 15 ÷ 1 + ( −16) 4
5 15 ÷ − 16 4
4 15 × − 16 5 12 − 16
−4
K1
N1
DECIMALS Place value 4 basic operations Change decimal to fractions and vice versa Bracket operations
1. Calculate
3 − 6.04 ÷ 0.2 − (−1 ) 4
− 6.04 + 1.75 0.2
K1
60.4 − + 1.75 2
− 30.2 + 1.75 − 28.45
N1
3. Calculate
2 5.43 ÷ 0.3 − (−1 ) 5
(5.43 ÷ 0.3) + 1.4
K1
18.1 + 1.4 19.5
N1
Calculate the value and express the answer correct to two decimal places 12 − −
1 × 0.456 6
= 12 - ( - 0.076 ) = 12 + 0.076
K1
= 12.076 = 12.08
N1
Calculate the value and express the answer correct to two decimal places
2 2 − ( − 3.12 ) ÷ 0.3 5
= 2.4 + 3.12 ÷ 0.3
K1
= 12.80
N1
= 2.4 + 10.4
4 2. Calculate (6 − 0.24) ÷ 3 Correct to two decimal places
3 5.76 × 4
K1
1.44 × 3
= 4.32
N1
SQUARES, SQUARE ROOTS • CUBES AND CUBES ROOTS
1. a. Find the value
4 5 9 49 9
7 3
or
1 2 3
P1
1. b. Calculate the value 3
0.064 - (-1)2
0.4 − 1 − 0.6
K1
N1
2. a Find the value of
0.36
= 0.6 b. Calculate
P1
(5 + − 27 ) 3
[(5 + (−3)]
4
2
2
K1
N1
3a. Find the value of
0.3
3
= 0.3 × 0.3 × 0.3
= 0.027
P1
3.a. find the value of
(−2) × 64 3
− 8×8 − 64
b. Calculate
P1
7 4− 1 9
4 4− 3 8 3
K1
N1
INDICES • MARKING SCHEME
1. a. Simplify
−5 −3
(q )
÷ (q )
4 2
15−8
q 7 =q b.
9×3
3 ×3 2
x−2
x −2
P1
= 27
=3
3
K1
2+ x−2 =3
x=3
N1
2. a. Simplify
(m ) ÷ m 2 4
5
8−5
m3 m b.
P1
16 × 2 = 2 4 m 2 2 ×2 = 2 4+m = 2 m = −2 m
2
K1
N1
3. a .simplify
3 3
4 4
16( x ) ( y ) 10 4 xy 9 16 16 x y = 10 4 xy 8
4x y
6
P1
b.
5
5
x−2
x −2
x −2
125 = x 5 3
5 = x 5
3− x
5 =5 x − 2 = 3− x
K1
2x = 5 x=5
2
N1
ALGEBRAIC EXPRESSIONS • EXPANSION • FACTORIZATION
8. Simplify (a) 2(n + 5) − 3 = 2n + 10 − 3 N1 = 2n + 7 (b) 3(4m – 3k) – (5k – m) K1 12m – 9k – 5k + m N1 13m – 14k
1. Factorise completely a.
− 5 y + 25 y 2
− 5 y ( y − 5)or 5 y (− y + 5) b.
P1
− 3qst + 8qr − 3 pst + 8 pr
− 3qst − 3 pst + 8qr + 8 pr − 3st (q + p ) + 8r (q + p ) K1
(−3st + 8r )(q + p )
N1
2, Factorise completely a.
4 pq + 6qr
2q (2 p + 3r ) b.
4 + x( x + 5)
4 + x + 5x 2 x + 5x + 4 2
( x + 1)( x + 4)
K1
N1
P1
− 5k − 4(k − 2) − 5k − 4k + 8
3. Simplify to its simplest form a.
8 − 9k
b.
P1
( y − 2) + 4( y − 1) 2
y − 4y + 4 + 4y − 4 2
y
2
N1
K1
7 Factorise completely (a) 2y + 6 (b) 12 – 3x2
S T E P S T O B E S H O W N
7 (a) 2y + 6 = 2(y + 3) P1 7 (b) 12 – 3x2 =3(4 – x2) K1 =3(2 + x)(2 – x) N1
8
Expand (a) q(2 + p) (b) (3m – n)2
S T E P S T O B E S H O W N
8 (a) q(2 + p) = 2q + pq P1 8 (b) (3m – n)2 = (3m)2 – 2(3m)(n)+(n)2 K1 = 9m2 – 6mn + n2 N1
Factorize completely 5h(k – 3) – 2(3 – k)
S T E P S
5h(k – 3) – 2(3 – k)
T O
5h(k – 3) + 2(k – 3)
B E S H O W N
5h(k – 3) – 6 + 2k 5h(k – 3) + 2k – 6 (5h + 2)(k – 3)
N1
K1
1.
4 1 − 3q 4q 16 3 − 12q 12q
13 12q
K1
N1
2.
1 4 + 2 p + q 3(2 p + q )
3+ 4 3(2 p + q )
K1
7 3(2 p + q )
N2
Express
5 2 − 3w + 3n 6n as a single fraction in its simplest form.
S T E P S T O
5 2 − 3w + 3n 6n 2( 5) 2 − 3w = + 2( 3n ) 6n
B E
10 2 − 3w = + 6n 6n
S H O W N
10 + 2 − 3w = 6n
12 − 3w = 6n
K1
K1
1
3( 4 − w ) = 2 6n 4−w = 2n
N1
17. Express as a single fraction in its simplest form 2 b −2 − 3b 6b
4 b− 2 = − 2 x3b 6b 4 - b+ 2 = 6b 6-b = 6b
K1 K1 N1
17.
Express as a single fraction in its simplest form.
2 2 −b − 3a 9ab
2 × 3b - ( 2 - b ) = 9ab
6b - 2 + b = 9ab
7b -2 = 9ab
K1
K1
N1
LINEAR EQUATION • SOLVE THE VALUE
1. a.
4( p − 3) − 6 = p 4 p − 12 − 6 = p
b.
3 p = 18 p=6 3 − 7k 2k = 5
10k = 3 − 7 k
K1
N1
K1
17 k = 3
3 k= 17
N1
2. a
2n = 3n − 4 − n = −4
n=4
2.b.
P1
3n − 2 = n−3 4
3n − 2 = 4n − 12
K1
3n − 4n = −12 + 2
n = 10
N1
3 a.
3.b.
5x − 3 = 4 x + 6 5x − 4 x = 6 + 3 x=9
P1
3 f + (6 − 4 f ) = −31 2
2 f + 18 − 12 f = −62
K1
− 10 f = −62 − 18
10 f = 80
f =8
N1
4. a.
b.
k = −14 − k
2k = −14 k = −7
P1
p − 3( p + 1) = 9 p − 3p − 3 = 9
− 2 p = 12 p = −6
K1
N1
TRANSFORMATIONS • • • •
TRANSLATION ROTATION REFLECTION ENLARGEMENT
TRANSFORMATIONS Describe the transformations Writing technique Spelling
12 10 P’
8 6 4
P
2 0
2
4
6
8
10
12
P’ is the image of P under transformation M. Describe in full transformation M.
S T E P S T O B E S H O W N
12 10 P’
8 6 4
P
P
2 0
2
4
6
8
10
12
Translation
− 4 5
5. Draw and label of 90o anticlock wise rotation from origin B’ A B C’ 5 4 3
A’
C
2
D’
N1
D
1 -5
-4
-3
-2
-1
1 -1 -2 -3 -4
2
K1
3
4
5
Diagram 2 in the answer space shows triangle PQR drawn on a grid of equal squares. Draw the image of triangle PQR under a clockwise rotation of 90o with T as the centre of rotation. [2 marks]
R
Q
P
T 5 2 DIAGRAM 2
R
Q
Q’ P’
P T
R’
In Diagram 2, the kites ABCD in the answer space was drawn on a grid of equal squares. On the diagram, draw and label A’B’C’D’ the image of ABCD under a reflection at line MN
Diagram 4 in the answer space shows triangle PQR drawn on a grid of equal squares.
5. Draw and label the image A
B
M C D
D’
K1
A’ N B’
C’
N1
6. Draw and label the image under reflection on the line MN M P
K1 P’
N1 N
(iii)
Reflection
Transformations
Eg : Diagram 3 in te answer space shows two quadrilaterals, ABCD and A’B’C’D’, drawn on a grid of equal squares. A’B’C’D’ is the image of a reflection. On the diagram in the answer space, draw the axis of reflection. Answer :
[2 marks]
A B C
D D’ A’ C’
B’
DIAGRAM 3
(iii)
Reflection
Eg : Diagram 3 in te answer space shows two quadrilaterals, ABCD and A’B’C’D’, drawn on a grid of equal squares. A’B’C’D’ is the image of a reflection. On the diagram in the answer space, draw the axis of reflection. Answer :
[2 marks]
A B C
D D’ A’ C’
B’
DIAGRAM 3
Diagram 2 in the answer space shows two quadrilaterals. JKLM and J’K’L’M’, drawn on the grid of equal squares. J’K’L’M’is the image of JKLM under an enlargement. Mark the P as the center of enlargement
Question 6 K
K’ K’
J
J’ J’
M L’ M’ M’
L’
L
P2
Question 6 P
K’ J’
P2
M’
L’
y 6 5 4
B’ B
3 2 A’ 1 –1 0
C’
D A
C •K’ •K
E 1 2 3 4
5 6
x
On the grid, draw the image of triangle PQR Under an enlargement with scale factor 2 at centre M
7. Draw an enlargement image P’
Q’
K1 P Q
M R
R’
N1
21. Diagram 2, in the answer space shows a triangle L drawn on a grid of equal squares. On the diagram in the answer space, draw the image of triangle L under an enlargement with centre Q and scale factor 3.
[2 marks]
L Q
• DIAGRAM 2
•
L’ • L Q
• DIAGRAM 2
•
STATISTICS • • • •
CONSTRUCT PIE CHART CONSTRUCT BAR CHART CONSTRUCT LINE GRAPH CONSTRUCT PICTOGRAM
STATISTICS The following data shows the number of stamps each student have. Abu 30
Bakar Chua David Ellan Fadil 50
20
40
60
40
Construct a pie chart for the above data.
First, find the value of represents the number according to the table. Abu Bakar Chua David Ellan Fadil
30 ⇒ × 3600 240 50 ⇒ × 3600 240 20 ⇒ × 3600 240 40 ⇒ × 3600 240 60 ⇒ × 3600 240 40 ⇒ × 3600 240
angles that of stamps = 45o = 75o = 30o = 60o = 90o = 60o
Using a protractor, draw the angles and label the sectors on the pie chart. Title : Number of stamps each student have.
Abu
Fadil 60o 90o Ellan
60
45o
Bakar 75o
o
David
30o Chua
STATISTICS 1.
Line Graph i)
Label both axes uniformly and correctly.
ii)
Mark all points correctly.
iii)
Join all points with straight line.
iv)
Use a ruler to join all the points.
v)
Do not start from 0.
2.
Rainfall(mm)
300 x
270 240 210
x
180
x x
150 120
x
90 60 30 M
A
M
J
J
Month
GRAPH OF FUNCTION • DRAW THE AXIS • MARK THE LOCATIONS • JOIN THE POINTS
GRAPH OF FUNCTIONS. 1. Use the real graph paper not a square grid paper. 2. Always remember that the big square of the graph paper measures 2 cm. 3. Identify the lowest and the highest value of the x-axis and y-axis. 4. Label the x-axis and the y-axis correctly and uniformly. 5. Relate the number on the value table with coordinates. [eg. ( - 3,15), (-2,5) ‌.] 6. Then mark all the points correctly using small cross or dots. 7. Draw smooth curve that passes through all the
Use graph paper provided to answer this question. Table 2 shows the values of two variables x and y, of a function.
x
-3
-1
0
1
2
y
-32 -11 -2
1
4
13 34
-2
3
Draw the graph of the function using a scale of 2 cm to 1 unit at the x-axis and 2 cm to 10 units at the y –axis.
y
40
1.
⊗
30 20 ⊗
10 ⊗ ⊗
-3
-2 ⊗
-1⊗
0 -10 -20
⊗
-30 -40
1
2
3
X
y
S T E P S
40
20
T O B E S H O W N
×
30
10 -3
-2
×
×
-1
× 0
× ×
-10 -20
×
-30 -40
1
2
3
x
2.
x y
-3 -2 -1 15 5 -1
0 1 2.5 4 5 -3 -1 9.5 29 47
50
×
y
45 40 35
×
30 25 20
×
15
×
10
× -3
-2
5 -1×
0×
1×
-5
-10
2
3
4
5
x
X Y
-4
-2.5
-1
0
1
2
3
-27
-7.5
3
5
3
-3
-13
4
5
-27
-45
y
× -5
-3
-4
-2
×
-1
10
×5
0
× 1
2
×
3
4
5
-5 -10
×
-15 -20
×
-25
×
-30 -35 -40 -45 -50
×
x
y
12 ×
×
× 8 ×
× 4 × -3
-2
-1
0
×
×
1
2
3
4
SOLID GEOMETRY Net and layout Surface area Volume of solid geometry and combinations Basic characteristics
6 . Diagram 3 shows a prism with a rectangle base.
3 units 8 units 4 units
DIAGRAM 3
Draw a full scale net of the prism on the grid in the answer space. The grid has equal squares with sides of 1 unit. [3 marks]
6 . Diagram 3 shows a prism with a rectangle base.
3 units 8 units 4 units
DIAGRAM 3
Draw a full scale net of the prism on the grid in the answer space. The grid has equal squares with sides of 1 unit. [3 marks]
7. Draw the net of the prism 3 unit
K2
8 unit
3 unit
3 unit
4 unit
5 unit
N1
7.
Solid Geometry
Eg : Diagram 4 shows a cuboid. 3 units 1 unit 2 units DIAGRAM 4
Draw a full scale the net of the pyramid on the grid in the answer space. The grid has equal squares with sides of 1 unit.
WHAT YOU THINK IS WHAT YOU ARE Notes: All the guides given is not a guarantee that you will achieve an excellent result in mathematics. What is important is that you have the determination to succeed, prepared to work hard and consistently practise past year’s questions and forever asking when confused. This is the real key to success. You determine your own success.
Practice makes perfect Be the Best and Beat the Rest