name class centre - xjscs.com.au · 1 base 10 and base 2 numbers (p. 5) conversion of base 10...
TRANSCRIPT
Name SOLUTION (TEACHER’S COPY)
Class
Centre
Course Outline Year 7 Term 1
© XJS Coaching School Page 3 2016 version 1
Week 1 Week 2 Week 3 Week 4 Week 5 Week 6 Week 7 Week 8
Sat 30-Jan 6-Feb 13-Feb 20-Feb 27-Feb 5-Mar 12-Mar 19-Mar
Sun 31-Jan 7-Feb 14-Feb 21-Feb 28-Feb 6-Mar 13-Mar 20-Mar
Textbook: Warwick, M., (2013) Understanding Year 8 Maths, Five Senses Education
Week Content Exercises
1 Base 10 and base 2 numbers (p. 5)
Conversion of base 10 numerals to
base 2
Conversion of base 10 to base 2 –
Choose easy and challenging numbers
Conversion of base 2 numerals to base
10
Do 10, 11, 100, 101, 110, 100011 (to
base2)
p. 27 L1, Q. 9
p. 31 L5, Q. 8
Worksheet 1 – Binary
numbers
2 Directed Numbers (p. 17 – 26)
Directed Numbers: Integers,
Graphing an integer line using
number notation, x >, <, and =. Use
of open and closed dots on a number
line.
p. 27 L1, Q. 10
p. 28 L2, Qs. 7, 8, 9
Worksheet 2 - Directed
Numbers 1
3 Continue with Directed Numbers.
Do harder problems: -3 - -2 + -17 – 8
Rules on „+‟, „-„, „x‟ and „ ' of
directed numbers.
Know the words „Ascending‟ and
„Descending‟ order.
p. 29, L3, Q. 9
p. 30, L4, Q. 8
p. 31, L5, Qs. 2, 3, 4,
p. 32, L6, Q. 3
Worksheet 3 – Directed
Numbers 2
Practice Exam 1
4 Geometry (p. 172 – 178)
Types of angles: Acute, Right Angle,
Obtuse, Straight, Reflex, Perigon.
Complementary / Supplementary
Sum of angle of a triangle.
Sum of angle of quadrilateral
Congruent triangles
p. 189 L1, Qs.1 – 4, 6
p. 190 L2, Q. 1
p. 191 L3, Qs. 1, 2
Worksheet 4 –
Geometry 1
5 Continue with Geometry (p. 179 –
180)
Parallel Lines
Corresponding / Alternate angles /
vertically opp. angles
Co-interior angles
p. 191 L3, Q5
p.192 L4, Qs. 3, 4
p. 193, L5, Q.4
Worksheet 5 –
Geometry 2
Course Outline Year 7 Term 1
© XJS Coaching School Page 4 2016 version 1
6 Geometry (p. 184 – 187)
Similar triangles / Congruent
triangles
Congruent triangles have the same
size triangles, identical corresponding
sides and angles
Similar triangles have different size
triangles but of the same shapes, same
corresponding side ratios and same
corresponding angle ratios.
p. 190 L 2, Qs. 1, 2
p. 191 L3, Qs. 1, 2
Work Sheet 6 –
Geometry 3
7 Practice Exam .
8 Exam Do holiday homework
Worksheet 1 – Binary Numbers Year 7 Term 1
© XJS Coaching School Page 5 2016 version 1
Name: SOLUTION____________ Date:_________
1. Complete the following base 10 numbers by converting each individual
digit into powers of 10: The first one has been done for you. The powers
of 10 has been designated with a *.
(a) 123 = (1 × 10*) + (2 × 10*) + (3 × 10*)
= (1 × 102) + (2 × 10
1)) + (3 × 10
0 )
= (1 × 100) + (2 × 10) + (3 × 1)
= 100 + 20 + 3 = 123
(b) 100 = (1 × 10*) + (0 × 10*) + (0 × 10*)
= (1× 102) + (0 × 10
1) + (0 × 10
0)
= (1× 100) + (0 × 10) + (0 × 1)
= 100 + 0 + 0
(c) 216 = (2 × 10*) + (1 × 10*) + (6 × 10*)
= (2 × 102) + (1 × 10
1) + (6 × 10
0)
= (2 × 100) + (1 × 10) + (6 × 1)
= 200 + 10 + 6
(d) 752 = (7 × 102) + (5 × 10
1) + (2 ×10
0)
= 700 + 50 + 2
= 752
(e) 1010 = (1 × 10*) + (0 × 10*) + (1 × 10*) + (0 × 10*)
= (1 × 103) + (0 × 10
2) + (1 × 10
1) + (0 × 10
0)
= (1 × 1000) + (0 × 100) + (1 × 10) + (0 × 1)
= 1000 + 0 + 10 + 0
(f) 1262 = (1 × 103 ) + (2 × 10
2) + (6 × 10
1) + (2 × 10
0)
= (1 × 1000 ) + (2 × 100) + (6 × 10) + (2 × 1)
= 1000 + 200 + 60 + 2
= 1262
(g) 25678 = (2 × 104 ) + (5 × 10
3 ) + (6 × 10
2) + (7 × 10
1) + (8 × 10
0)
= (2 × 10000 ) + (5 × 1000 ) + (6 × 100) + (7 × 10) + (8 × 1)
= 20 000 + 5 000 + 600 + 70 + 8
= 25 678
Key words: Binary Numbers / Base Numbers / Power / Conversion
123
100
216
752
1010
1262
25678
Worksheet 1 – Binary Numbers Year 7 Term 1
© XJS Coaching School Page 6 2016 version 1
2. Complete the following binary numbers by converting each individual digit into
powers of 2, then changing them to base 10 numbers. The first one has been done
for you. Find *
(a) 110two = (1 × 2* ) + (1 × 2*) + (0 × 2*)
= (1 × 22 ) + (1 × 2
1 ) + (0 × 2
0 )
= (1 × 4) + (1× 2) + (0 × 1)
= 4 + 2 + 0
= 6ten
(b) 10two = (1 × 2*) + (0 × 2*)
= (1 × 21) + (0 × 2
0)
= (1 × 2) + (0 × 1)
= 2 + 0
= 2
(c) 11two = (1 × 2*) + (1 × 2*)
== (1 × 21) + (1 × 2
0)
= (1 × 2) + (1 × 1)
= 2 + 1
= 3
(d) 101 two = (1 × 2*) + (0 × 2*) + (1 × 2*)
= = (1 × 22) + (0 × 2
1) + (1 × 2
0)
= (1 × 4) + (0 × 2) + (1 × 1)
= 4 + 0 + 1 = 7
(e) 111 two = (1 × 22) + (1 × 2
1) + (1 × 2
0) (fill the empty space)
= (1 × 22) + (1 × 2
1) + (1 × 2
0)
= (1 × 4) + (1 × 2) + (1 × 1)
= 4 + 2 + 1 = 7
(f) 1101 two = ( × 2*) + ( × 2* ) + ( × 2*) + ( × 2* )
(fill in the empty space and *)
= (1 × 23) + ((1 × 2
2) + (0 × 2
1) + (1 × 2
0)
= (1 × 8) + (1 × 4) + (1 × 2) + (1 × 1)
= 8 + 4 + 0 + 1 = 13
(g) 1001 two = ( × ) + ( × ) + ( × ) + ( × )
(fill in the empty space)
= (1 × 23) + ((0 × 2
2) + (0 × 2
1) + (1 × 2
0)
= (1 × 8) + (0 × 4) + (1 × 2) + (1 × 1)
= 8 + 0 + 0 + 1 = 9
(h) 10101 two = (1 × 24) + (0 × 2
3) + ((1 × 2
2) + (0 × 2
1) + (1 × 2
0)
= (1 × 16) + (0 × 8) + (1 × 4) + (0 × 2) + (1 × 1)
= 16 + 0 + 4 + 0 + 1
= 21
(i) 11111 two = = (1 × 24) + (1 × 2
3) + ((1 × 2
2) + 10 × 2
1) + (1 × 2
0)
= (1 × 16) + (1 × 8) + (1 × 4) + (1 × 2) + (1 × 1)
= 16 + 8 + 4 + 2 + 1
= 31
6
2
3
7
5
13
9
21
31
Worksheet 1 – Binary Numbers Year 7 Term 1
© XJS Coaching School Page 7 2016 version 1
3. Convert the following base 10 numbers to binary numbers
(a) 2
(b) 11
(c) 24
(d) 30
(e) 50
(f) 63
(g) 65
(h) 75
4. Convert the following binary numbers to the base 10 equivalent:
(a) 100two
(b) 1010 two
(c) 1110 two
(d) 1100 two
(e) 10110 two
(f) 11101 two
(g) 101001 two
(h) 111110 two
10TWO
110010TWO
1011TWO
111111TWO
11000TWO
1000001TWO
11110TWO
1001011TWO
4
22
10
29
14
41
12
62
Worksheet 2 Numbers and Fractions Year 7 Term 1
© XJS Coaching School Page 8 2016 version 1
5. Complete the following base-2 table, using the powers of 2 to convert the binary
numbers to base 10 numbers. The first one has been done for you.
Base 2
numbers
28
256
27
128
26
64
25
32
24
16
23
8
22
4
21
2
20
1
Base 10
Numbers
101111 1 0 1 1 1 1
32 8 4 2 1 47
100000 1 0 0 0 0 0
32 32
1000001 1 0 0 0 0 0 1
64 1 65
1000011 1 0 0 0 0 1 1
64 2 1 67
1000111 1 0 0 0 1 1 1
64 4 2 1 71
1001111 1 0 0 1 1 1 1
64 8 4 2 1 79
1011111 1 0 1 1 1 1 1
64 16 8 4 2 1 95
1111111 1 1 1 1 1 1 1
64 32 16 8 4 2 1 127
10101010 1 0 1 0 1 0 1 0
128 32 8 2 170
6. The early Chinese invented an abacus (suan pan) which is capable of doing base 5
calculations. It can calculate faster than a person doing base 10 calculations.
Complete the base 5 calculation below and convert it to base 10 answers. The first
one has been done for you.
Base 5
Numbers 5
5
3125
5 4
625
5 3
125
5 2
25
5 1
5
5 0
1
Base 10 Numbers
413 4×25=100
1×5=5
3×1=3
108 444 4×25=100
4×5=20
4×1=4
124
1234 1×125=125 2×25=50
3×5=15
4×1=4
194
21301 2×625=1250 1×125=125
3×25=75
0×5=0
1×1=1
1451 30000 3×625=1875
0×125=0 0×25=0
0×5=0
0×1=0
1875
102432 1×3125=3125
0×625=0 2×125=250
4×25=100
3×5=15
2×1=2
3492 210411 2×3125=6250
1×625=625
0×125=0 4×25=100
1×5=5
1×1=1
6981
Worksheet 2 - Directed Numbers 1 Year 7 Term 1
© XJS Coaching School Page 9 2016 version 1
Name: SOLUTION____________ Date:_________
1. Which number from the list is an integer? [Circle your answer(s), if any]
−2, −3.2, 0.5, 2 ½ , −101.5
2. Is it True (T) or False (F) that 3 > −6?
For questions 3 to 5, answer “Yes” (Y) or “No” (N).
3. Can x be equal to - 3 if x < 2?
4. In the following number notation, x ≥ 5, can x be 5 and also 1000?
5. If the number notation is x ≤ - 5, can x be -10?
6. What is the meaning of the following number notation, -6 ≤ x ≤ 7?
7. What is the difference between x = 3, x ≤ 3 and x ≥ 3?
8. Arrange the numbers in descending order.
0, −5, 3, 12, 2, −11, −8
Key words: Integer / Descending / Ascending / Number Line / Directed Numbers / Inequation / Algebraic Equation / Substitution
Y
Y
Y
x is a number greater or equal to -6 but less or equal to 7
x = 3 means x can take one and only one value, that is 3;
x ≤ 3 means x can take any value less than 3 or exactly 3;
x ≥ 3 means x can be 3 or larger.
12, 3, 2, 0, -5, -8, -11
T
Worksheet 2 - Directed Numbers 1 Year 7 Term 1
© XJS Coaching School Page 10 2016 version 1
9. Arrange the following in ascending order
-10, -3, 0, -5, -7, -1, -13
10. List the integers between −6 and 2.
11. Graph each set of integers on a number line.
(a) x > −3
(b) x ≤ 2
(c) −10 ≤ x ≤ −5
(d) -3 ≤ x < 5
12. Describe the integers graphed on each number line in number notation.
(a)
(b)
13. Do the following:
(a) 3 + 2 =
(b) 3 + - 2 =
(c) 3 - + 2 =
(d) - 3 + 2 =
(e) -3 + -2 =
(f) -3 – 2 =
(g) -3 - -2 =
14. Fill in the following rule:
(a) + + =
(b) + - =
(c) - + =
(d) - - =
-13, -10, -7, -5, -3, -1, 0
-5, -4, -3, -2, -1, 0, 1 {Should not include -6 and 2}
x ≤ 2
-4 < x < 1
5 - 5
1
1
- 1
-5
-1
+
-
-
+
Worksheet 2 - Directed Numbers 1 Year 7 Term 1
© XJS Coaching School Page 11 2016 version 1
15. Evaluate:
(a) 3 + 4 + 5 =
(b) 3 + 4 – 5 =
(c) 3 + - 4 + 5 =
(d) -3 + 4 + 5 =
(e) -3 + - 4 + 5 =
(f) -3 + 4 + - 5 =
(g) -3 – 4 – 5 =
(h) 3 – 4 - - 5 =
(i) 3 - -4 - - 5 =
16. Evaluate the following algebraic expression if a = −4, b = −3 and c = 2.
(a) c − a − b
(b) c + a – b
(c) – c – b + a
(d) – c – b – c
17. Evaluate the following.
(a) −2 × – 2 =
(b) -2 × -2 × -2 =
(c) -2 × -2 × -2 × -2 =
18. From the above, the rule for multiplying negative numbers is:-(fill in the space
given):
(a) – × – = positive (enter positive or negative) Multiplying two or even
number of negative numbers will give a positive value.
(b) – × – × – = negative (enter positive or negative) Multiplying three or odd
number of negative numbers will give a negative value.
12
2
4
6
- 4
- 12
4
12
- 2
9
1
- 3
- 1
4
- 8
16
Worksheet 2 - Directed Numbers 1 Year 7 Term 1
© XJS Coaching School Page 12 2016 version 1
19. Evaluate the following:
(a) 3 – 10 × – 2 + 3 =
(b) 2 × – 4 + 6 =
(c) – 5 + - 6 + - 2 × – 4 =
26
- 2
- 3
Worksheet 3 - Directed Numbers 2 Year 7 Term 1
© XJS Coaching School Page 13 2016 version 1
Name: SOLUTION____________ Date:_________
1. Evaluate the following.
(a) (−2)3
× −3 = (b) −4 × 2 × 9 =
2. Evaluate the following algebraic expression if c = −5 and d = −3.
(a) 2 × c2 × d
(b) d 3
(c) 2c × 3d
(d) c2 × d
3. Compute the following;
(a) – 6 ÷ + 2 =
(b) – 6 ÷ - 3 =
(c) – 6 ÷ - 2 ÷ -3 =
4. Fill in the rule for dividing directed numbers:
(a) – ÷ – = positive. Choose either a negative or positive sign.
Division of 2 negative or even number of negative numbers will produce a
positive value.
(b) – ÷ – ÷ – = negative. Choose either a negative or positive sign.
Division of 3 negative or odd number of negative numbers will produce a
negative value.
5. Evaluate the following:
(a) – 48 ÷ 6 =
(b) – 39 ÷ - 3 =
(c) – 40 ÷ -2 ÷ - 4 =
(d) – 50 ÷ - 2 ÷ -5 ÷ -1 =
Key words: Continue with Directed Numbers - harder problems / Rules on ‘+’, ‘-‘, ‘x’ and ‘ ‚' of directed numbers / Know the words ‘Ascending’ and ‘Descending’ order.
24 -72
- 150 90
- 27 - 75
- 3 - 1
2
- 8
13
- 5
5
Worksheet 3 - Directed Numbers 2 Year 7 Term 1
© XJS Coaching School Page 14 2016 version 1
6. Use BODMAS to evaluate the following.
7.
(a) - 2 × 5 – 5 =
(b) 3 + - 5 × 2 =
(c) 2 × 5 + -2 × 4 =
(d) – 10 ÷ 5 + 5 =
(e) 12 + - 20 ÷ 2 =
(f) 20 × -2 +12 – 40 ÷ 2 =
For the following questions 7 & 8, write the word sentences into number sentences before solving the problem.
8. Jack borrowed $85 from his brother. He repaid $20 but then borrowed another $12.
How much does he now owe his brother?
9. Sharon‟s body temperature was 37°C. After 1 hour it increased by 2 degrees after 2
hours it decreased by 0.5 degrees and then decreased a further 1.5 degrees after 3
hours. What is her final temperature?
10. Arrange in ascending order:
½, 0, - ¼, −5.3, 4.8, −9.2
11. Graph the following sets of directed numbers on a number line
(a) x ≥ − 6.5
(b) −11
4 ≤ x ≤ 2.3
12. Write the same missing number in all the three boxes below.
−
−
- 15 3
- 7 2
2 - 48
$77
He still owes his brother $77 (positive value), though a negative
implies owing, - $77 or $-77 are both unacceptable as it will be a
double negative.
37°C
−9.2, −5.3, -¼, 0, ½,, 4.8,
3
3
3
Worksheet 3 - Directed Numbers 2 Year 7 Term 1
© XJS Coaching School Page 15 2016 version 1
13. Calculate the following.
−2
5 −2
3−1
15
14. Find the answer.
1
2 −1
5− −
1
3
−12
15
19
30
81
9 15
41
19
31 50
81
153
Worksheet 3 - Directed Numbers 2 Year 7 Term 1
© XJS Coaching School Page 16 2016 version 1
1
4
Practice Test 1 Year 7 Term 1
© XJS Coaching School Page 17 2016 version 1
Name: SOLUTION____________ Date:_________
Section A: Multiple Choice Questions 1. What does (7 + 7 3) ( 4)
equal?
A 8
B 4
C 0
D 4
E 7
E
2. - 3 + - 2 × 3 is:
A. - 15
B. 15
C. - 9
D. 9
E. -2
C
3. What does 6 + 12 3 (2 + 12 0)
equal?
A 8
B 4
C 0
D 4
E 8
E
4. Which signs need to be inserted to
make the following equation true?
6 – 6 8 4 = 3
A ,,
B ,,
C ,,
D ,,
E ,,
E
5. What is 0.299 513 7 correct to 3
decimal places?
A 0.300
B 0.295
C 0.3
D 0.299
E 0.30
A
6. Find - 3 - - 2 + 5 + - 10:
a) -6
b) -10
c) -5
d) -14
e) -4
A
7. A number has been rounded to
27.40, which of the following could
the unrounded number have been?
A 27.351 32
B 2.740
C 27.45
D 27.459 89
E 0.2740
A
8. The day temperature started at 13°.
By 12 p.m the temperature had risen
3°. It then increased by another
2°by the end of the day. A cold
change caused a sudden drop in
temperature of 9°. Find the new
temperature.
A 5°
B 18°
C 6°
D 9°
E 8°
D
9. Mary has in her account $40.00. On
Monday she deposited $15 into the
account but withdrew $70.00 from
it on Tuesday. On Wednesday she
deposited $20.00 into the same
account. How much money has she
in her account now?
A $15.00
B $5
C $55
D $50
E $0.00
B
10. Find - 4 × - 3 × – 2 × – 1 + -5 × - 4
A 4
B - 4
C 44
D -44
E 20
C
Practice Test 1 Year 7 Term 1
© XJS Coaching School Page 18 2016 version 1
Section B: Short Answer Questions
1 Remembering to use the correct order of
operations, evaluate these expressions.
(a) 12 4 2 + 3
(b) 6 10 + 12
(c) 12 11
(d) 6 (4 + 2) + 3 2
(a) 12 4 2 + 3 = 12 8 + 3
= 4 + 3 = 7
(b) 6 10 + 12 = 6 +
10 12
= 4 12
= - 8
(c) 12 11 = 132
(d) 6 (4 + 2) + 3 2 = 6
2 + 3 2
= 12
+ 6 = - 6
2 Round the number 469.79518 to:
(a) 4 decimal places
(b) 3 decimal places
(c) 2 decimal places
(d) 1 decimal place
(e) 0 decimal places
(a) 469.7952
(b) 469.795
(c) 469.80
(d) 469.8
(e) 470
3 Use your calculator to evaluate the following
correct to 3 decimal places.
(a) 212
2448
(b) 336
)10(4 23
(c) 23 )30.8(.)2(
(a) 212
2448
= - 3
(b) 2
36
36
10064
336
)10(4 23
= 18
(c) 258)30.8(.)2( 23
= - 33
Could also be - 18 if
you take √(36) = ±6
Practice Test 1 Year 7 Term 1
© XJS Coaching School Page 19 2016 version 1
4 Evaluate the following:
a) 1112930
22
b) - 4 + - 6 - -5
- 5
c) 8 – 14 + 4 – 12
- 14
d). )911(22
22 ÷ 2 = 11
e) 270480
20 + 35 = 55
f) 9915
- 3
g) )53(717
17 – 7 × - 2 = 17 + 14 = 31
5 Insert signs to make the following equation
true.
2 3 6 5 = –11
2 – 3 × 6 + 5
6 Evaluate.
(a) 10
9
4
3
5
4
(b) 6
5
10
9
(c) -21
20
7
10
(a) 20
17
20
181516
(b) 4
3
(c) -2
3
20
21
7
10
7 If a = -5, b = -3 and c = 3, evaluate the
following:
(a) 3a (3 × b)
(b) a × b × c
(c) a × 3b + 5c
(d) ab + bc – a
(a) 3 × - 5 ÷ (3 × - 3) = - 15 ÷ -
9 = 3
21
3
5
(b) - 5 × - 3 × 3 = 45
(c) - 5 × 3 × - 3 + 5 × 3 = -
15 × -3 + 15
= 45 + 15 = 60
(d) - 5 × – 3 + - 3 × 3 - - 5
= 15 – 9 + 5 = 11
- + ×
Practice Test 1 Year 7 Term 1
© XJS Coaching School Page 20 2016 version 1
8 Graph each of the integers on a number line
(a) x < 5
(b) x > - 4
(c) 24 x
Worksheet 4 Geometry 1 Year 7 Term 1
© XJS Coaching School Page 21 2016 version 1
Name: SOLUTION____________ Date:_________
1. Find the complementary angle of the following angles:
(a) 20°
(b) 55°
(c) 63°
(d) 89°
2. Find the supplementary angles of the following angles:
(a) 125°
(b) 95°
(c) 150°
(d) 145°
3. Solve the following:
(a) x + 10° = 90°
(b) x – 25° = 90°
(c) x + 60° = 180°
(d) 2y + 45° = 90°
(e) 3a – 30° = 90°
(f) 5h + 100° = 180°
4. Find the pronumeral of the following equations.
(a) x + x + 10 = 40
(b) x + 1 + x + 2 = 63
(c) x + 2x + 60 = 120
Key words: Pronumeral / Exterior angle / Types of angles: Acute,
Right Angle, Obtuse, Straight, Reflex, Perigon /
Complementary / Supplementary / Sum of angle of a triangle /
Sum of angle of quadrilateral
70°
35°
27°
1°
55°
85°
30°
35°
80°
115°
22.5°
40°
120° 16°
15
30
20
Worksheet 4 Geometry 1 Year 7 Term 1
© XJS Coaching School Page 22 2016 version 1
5. Find the value of the pronumeral in each triangle.
(a)
(b)
(c)
6. Find the pronumeral of the following triangles
(a)
(b)
(c)
7. Find the value of the interior angle x in the triangle below:-
8. Find the value of the exterior angles of the triangles below.
(a
)
(b)
x = 136°
y = 160°
x = 85°
x = 40° x = 50° x = 30°
48° 51° 10°
Worksheet 4 Geometry 1 Year 7 Term 1
© XJS Coaching School Page 23 2016 version 1
9. The value of the exterior angle of a triangle = sum of 2 opposite __interior__
angles
10. Find the value of the pronumeral:
(a)
(b)
(c)
(d)
11. The sum angle of a quadrilateral is ______________
12. Find the sum angle of a heptagon.(Use the rule a = 180°× (n – 2), where a is the
angle and n=no. of sides)
Sum angle of heptagon = 180(7 -2) = 180 × 5 = 900°
x = 118° x = 30°
x = 92°
360°
900°
57° 112°
x° 73°
4x °
x °
2x °
5x °
x = 138°
Worksheet 4 Geometry 1 Year 7 Term 1
© XJS Coaching School Page 24 2016 version 1
Worksheet 5 Geometry 2 Year 7 Term 1
© XJS Coaching School Page 25 2016 version 1
Name: SOLUTION____________ Date:_________
1. Find the value of the pronumerals.
2. Use the following rules(corresponding angles, alternate angles, vertically opposite
angles, straight angles, co- interior angles) to describe the following equations
(a) y = z Rule:
(b) z = 69° Rule:
(c) x + 69° = 180° Rule:
(d) y = 69°
Rule:
(e) x + y = 180°
Rule:
3. Are the following statements true (T) or false (F)?
(a) Angles a and d are alternate.
(b) Angles c and f are corresponding.
(c) Angles b and f are co-interior
Key words: Complementary / Supplementary / Pronumeral /
Exterior angle Parallel Lines / Corresponding / Alternate angles /
vertically opp. Angles / Co-interior angles
x =
y =
z =
111°
69°
69°
Vertically opposite
Corresponding angles
Straight angles (or Supplementary)
Alternate angles
Co-interior angles
True
False
False
Worksheet 5 Geometry 2 Year 7 Term 1
© XJS Coaching School Page 26 2016 version 1
4. Find the value of the pronumerals given.
5. Find the value of the pronumerals
6. Find the value of the pronumeral:
7. Find the value of the pronumerals given:
8. Find the pronumeral given:
a 60°
b c
d e
f g
115°
x°
x°
y°
120°
30°
35°
h°
a =
b =
c =
d =
e =
f =
g =
x =
y =
z =
x =
y =
x =
h =
120°
120°
60°
120°
120°
60°
60°
33°
57°
57°
65°
120°
60°
65°
Worksheet 5 Geometry 2 Year 7 Term 1
© XJS Coaching School Page 27 2016 version 1
9. Find the exterior angle of an equilateral triangle.
Exterior angle = 360 ÷ 3 = 120°
10. Find the exterior angle of an isosceles triangle in line with the base angle which
equals 60°
11. In a triangle, the first angle is designated x, the second angle is twice that of the first
and the third angle is three times that of the second angle. What angle is x?
x + 2x + 3(2x) = 180, 9x = 180, x = 180/9 = 20
12. Find out the sum of all three exterior angles of a triangle
13. How many times is the 3 exterior angles of a triangle bigger than its three interior
angles?
120°
120°
If the base is 60°, it is possible an equilateral
20°
360°
The sum of all exterior angles of any polygon is 360°
2 times
Any polygon has 360° as the sum of exterior angles including triangle. The sum of
interior angles is 180°
165°
Worksheet 5 Geometry 2 Year 7 Term 1
© XJS Coaching School Page 28 2016 version 1
Challenge Questions
1. A car travelled 281 kilometres in 4 hours 41 minutes. What was the average speed
of the car in kilometres per hour?
281 ÷ (4 × 60 + 41) × 60
2. The length of a rectangle is four times its width. If the area is 100 m2 what is the
length of the rectangle?
4w × w = 100 w2 = 25 w = 5
Length = 4 w = 20 m
3. The length of a rectangle is increased to 2 times its original size and its width is
increased to 3 times its original size. If the area of the new rectangle is equal to
1800 square meters, what is the area of the original rectangle?
Length 2 times, width 3 times, hence Area will be 6 times.
1800 ÷ 6 = 300 m2.
4. Each dimension of a cube has been increased to twice its original size. If the new
cube has a volume of 64,000 cubic centimetres, what is the area of one face of the
original cube?
Volume would be increased by 23 = 8 times, hence original volume = 64000 ÷ 8 =
8000 cm3. Each side =√ 000
20 Each face = 20
2 = 400 cm
2.
5. Pump A can fill a tank of water in 5 hours. Pump B can fill the same tank in 8
hours. How long does it take the two pumps working together to fill the tank?
(Round your answer to the nearest minute).
Assume capacity = x Rate of Pump A = x/5, Rate of Pump B = x/8
Combined rate = x/5 + x/8 = 13x/40. Time = x ÷ 13x/40
40/13 = 3.07692 hours = 3 hours 5 minutes.
60 km/hour
20 m
300 m2
400 cm2
3 hours 5 minutes
Worksheet 6 Geometry 3 Year 7 Term 1
© XJS Coaching School Page 29 2016 version 1
Name: SOLUTION____________ Date:_________
1 Find the value of the pronumeral a in this pair
of similar triangles.
cm7
210
1814
8
14
8
18
a
a
a
2 Find the value of the pronumeral b in this pair
of similar triangles.
cm11
921
811
30
11
30
8
b
b
b
3 Find the value of the pronumeral c in the figure
below.
3
211
1059
2401359
1615159
9
16
15
15
c
c
c
c
c
4 Find the value of the pronumeral d in the figure
below.
7
19
1614
8
14
8
16
d
d
d
5 Which of the following triangles are congruent.
Give a reason for your answer.
PNG HDJ (SAS)
Key words: Similar Triangles / Congruent Triangles / Pronumeral
Worksheet 6 Geometry 3 Year 7 Term 1
© XJS Coaching School Page 30 2016 version 1
6 A flagpole casts a shadow 7 m long. A metre
ruler at the same time casts a shadow 65 cm
long. How high is the pole?
m11
m13
1010
10065
7
65
7
100
x
x
x
x
The height of the flagpole is
about 11 m.
7 Prove that the diagonal of a rectangle divides
the figure into two congruent triangles.
In ABD and CDB
AB = CD (opposite sides of
rectangle)
AD = CB (opposite sides of
rectangle)
DB is common
(SSS)
ABD CDB
So the diagonal divides the
rectangle into two congruent
triangles.
8 Given that PQR PSR, find the values of
the pronumerals shown.
PQR PSR
So QR = SR that is,
a = 2 cm
Also PQR = PSR that is,
b = 55
In PSR
c + 90 + 55 = 180
c + 145 = 180
c + 145 145 = 180 45
c = 35
9 Find the value of the pronumerals in the
following pair of congruent triangles.
Congruent triangles means all
corresponding sides and all
corresponding angles equal.
Therefore:
a = 4
b = 3
10 Find the value of the pronumerals in the
following pair of congruent triangles.
Congruent triangles means all
corresponding sides and all
corresponding angles equal.
Therefore both triangles are
equilateral triangles. Therefore
a = 60°
b = c = d = e = 4:
Pactice Test 2 Year 7 Term 1
© XJS Coaching School Page 31 2016 version 1
Name: SOLUTION____________ Date:_________
Multiple Choice Questions:
1 The value of the angle which is
complementary to 27° is
A 72°
B 63°
C 153°
D 163°
E None of the above
B
2 The value of the angle, a, in the
figure below is:
A 111°
B 69°
C 79°
D 89°
E None of the above
B
3 The value of the angle, a, in the
figure below is:
A 28°
B 62°
C 79°
D 69°
E 59°
E
4 The value of the angle which is
supplementary to 34° is:
A 43°
B 56°
C 146°
D 156°
E None of the above
C
5 The values of the angles a and
b, in the figure below are:
A a = 108° , b = 68°
B a = 72°, b = 68°
C a = 72° , b = 22°
D a = 72° , b = 40°
E a = 108° , b = 4°
D
6 An obtuse-angled triangle has:
A 1 angle equal to 90°
B 1 angle less than 90°
C 1 angle more than 90°
D 2 equal angles
E no equal angles
C
7 The co-interior angles of 2
parallel lines adds up to:
A. 60°
B. 90°
C 180°
D 120°
E 30°
C
8 The value of the angle, a, in the
figure below is:
A 18°
B 26°
C 62°
D 118°
E 128°
D
Pactice Test 2 Year 7 Term 1
© XJS Coaching School Page 32 2016 version 1
9 An acute-angled triangle has:
A 1 angle less than 90°
B 2 angles less than 90°
C 3 angles less than 90°
D 1 angle more than 90°
E None of the above
C
10 The 2 triangles in the figure
below are congruent. Therefore
we would write:
A ABC = PQR
B ABC PQR
C ABC = RQP
D ABC RPQ
E All the above are correct
D
11 In the figure below the 2
triangles are similar.
Which of the following
statements is correct?
A DE = 15
B EF = 15
C EF = 35
D DE = 35
E None of the above
B
12 Which of the following
statements is true?
A All similar triangles are
congruent.
B Some similar triangles
are congruent.
C Some congruent triangles
are not similar.
D All right–angled triangles
are similar.
E All equilateral triangles
are congruent.
B
13 The value of the pronumeral b
in the figure below is:
A 9
4
B 4
111
C 5
4
D 2
12
E 4
12
E
15 The value of the pronumerals c
and d, respectively, in the pair
of congruent triangles in the
figure below is:
A 11 and 50
o
B 5.5 and 25o
C 5.5 and 80o
D 5.5 and 40o
E 11 and 40o
A
Practice Test 2 Year 7 Term 1
© XJS Coaching School Page 33 2016 version 1
Section B: Short Answer Questions:
1. Determine the supplementary angles to:
a) 34 b) 124 c) 91 d) 179
Ans: 180 – 34 = 146 56 89 1
2. Find the value of the angle, a, in the figure below.
a = 90 – 28 = 62
3. Find the values of the angles a, b and c as shown in the figure below
a = 180 – 115 = 65
c = 38
b = 180 – 65 – 38 = 77
d = d = 77
4. Find the angles of the pronumerals listed.
a + a + 20 = 180
2a = 180 – 20 = 160
a = 160/2 a= 80
b = 80 + 20 b= 100
c = 180 – 100 c= 80
5. Find the value of „a‟ and thus the 3 angles of the triangle.
a + 2a + 3a = 180
6a = 180
a = 180/6 = 30°
2a = 60°
3a = 90°
a
3a
2a
Practice Test 2 Year 7 Term 1
© XJS Coaching School Page 34 2016 version 1
6. Find the sum of interior angle of an octagon.
180 × (8 – 2) = 180 × 6
= 1080
7. If 3 angles of a quadrilateral equal 290°, what will the magnitude of the fourth
angle?
360 – 290 = 70
8. Find the following pronumerals.
a) x + 20 = 180° b) x +2 +x +3 = 45°
x = 180 – 20 x = 160° 2x + 5 = 45
2x = 45 – 5 = 40
x = 40/2 x = 20°
c) x – 10 +2x + 20 = 130° d) 2x + x + 5 + 2x + 10 = 115°
3x + 10 = 130 5x + 15 = 115
3x = 130 – 10 = 120 5x = 115 – 15 = 100
x = 120/3 x = 40° x = 100/5 x = 20°
9. Determine the value of the angle a in the figure
below.
90 + a + a = 180
90 + 2a = 180
2a = 90
a = 45°
10. Find the values of the angles a and b in the
figure below, using the exterior angle theorem.
b + 70 = 125
b = 55°
a + b + 70 = 180 (angle sum of a triangle)
a + 55 + 70 = 180
a + 125 = 180
a = 55°
11. Find the height, h, of the building in the figure
below.
The small triangle is similar to the large
triangle, therefore their sides are in equal ratio.
4
7 =
h
277
7h = 4(7 + 27)
7h = 136
h = 19.43 m
Practice Test 2 Year 7 Term 1
© XJS Coaching School Page 35 2016 version 1
12. Find the value of the pronumeral a in the pair
of similar triangles.
11
113
811
18
11
18
8
a
a
a
13. The width of a river can be determined using
similar triangles. How wide is the river?
m31
m4
131
5040
25
40
25
50
w
w
w
w
The width of river is about 31 m.
14. Find the value of the pronumerals d and e in the
following pair of congruent triangles.
Congruent triangles means all corresponding
sides and all corresponding angles are equal.
Therefore:
d = 7
e = 9
Practice Test 2 Year 7 Term 1
© XJS Coaching School Page 36 2016 version 1
Holiday Homework (Last Year) Year 7 Term 1
© XJS Coaching School Page 37 2016 version 1
1. Sands of time.
The problem is to find the most efficient way of using 2 sand glasses to time the
cooking of a 15 minute hard-boiled egg. You only have an 11 minute timer and a 7
minute timer to judge cooking time. How would you use them?
Ans: Start the 11 min and the 7 minute sand glasses at the same time. Start
boiling the egg when the 7 minute sand glass has just finished flowing.
The 11 minute sand glass now has 4 minutes of time left. When the sand
stop flowing in the 11 minute sand glass, turn it around and start timing
the 11 minute sand glass again. You will have another 11 minutes. Total
time now will be 4 + 11 = 15 minutes
2 Lily Pad.
A frog sits in the middle of a lily pond which has a diameter of 30 metres. It can
hop 8 m after which it is so tired it can only hop half that length. Its next hop is half
the length he hopped before and so on. It goes on hopping half its last hop. How
many hops does it take for the frog to get to the edge of the pond?
Ans: Since it sits in the of the lily pond, it will only need to jump 15 m to reach the
edge.
Its first jump is 8m, giving it another 15 – 8 = 7 m to jump.
2nd
jump is 4m, giving it 3 more m to jump,
3rd
jump is 2m, giving it another 1m to jump
4th
jump is 1m which will help it to reach the edge of the lily pad
Its needs 4 jumps to reach the edge of the lily pad.
3. Dividing the clock.
Draw a straight line across the centre of the clock so that the sum of the numbers on
each half of the clock is the same.
Ans:
12 and 1 (being next to each other) will be on the same side – to balance up) so
directly opposite of 6 and 7(total also 13) will be on the other side. Likewise, 11
and 2 balances up 8 and 5 (total 13) and 10 and 3 as well as 9 and 4.
Draw a line between 9 and 10 passing through a point between 3 and 4.
Therfore 10 + 11 + 12 + 1 + 2 + 3 = 4 + 5 + 6 + 7 + 8 =+9 = 39
Holiday Homework (Last Year) Year 7 Term 1
© XJS Coaching School Page 38 2016 version 1
4. Capacity of Football Stadium
In a football grand match, blocks of seats are red, yellow, blue or green. Each block
is labelled in rows A to Z, then rows AA, BB, to ZZ. (Note: There‟s no row marked
AB, AC….). Each row has 100 seats in it.
a) How many people can be seated at the match?
b) In an attempt to determine how many people attended a particular match, the
manager counted the empty seats. He noticed that none of the rows VV to ZZ, were
occupied in the yellow zone, and all rows XX to ZZ were vacant in the green zone.
Red zone was full except for 10 seats and blue except for 15 seats. How many
people attended the match?
Ans:
a) A to Z means 26 × 4 (colors) = 104 rows of seats.
AA to ZZ means another 26 × 4 = 104 rows of seats. Total number of rows = 2 ×
104 = 208
Each row is capable of seating 100 seats.
Capacity = 100 × 208 = 20800 seats, i.e., it can accommodate 20800 people.
b) VV to ZZ = 5 rows empty i.e., 500 empty seats
XX to ZZ means 3 rows empty or 300 empty seats.
There are a further 10 seats empty in the red zone and 15 seats in the blue zone.
Number of people who attended the match = 20800 – 500 – 300 – 10 – 15 = 19975
5. Diet
The number of Kj (kilojoules) provided by each type of food is given inside the
brackets below:
1 boiled egg (335), 1 sliced bread (335),
100ml milk (538), 1 lamp chop (1460),
100g potato (370) 100g carrots (108),
100g cheese (1833), 100g fish (400),
100g peas (368)
Grandma can only have 4500Kj. Her normal diet is an egg, a slice of bread, 100g of
fish and 50ml. of milk.
a) How many Kj is this?
Ans: Number of calories = 335 + 335 + 400 + 538/2 = 1339Kj
b) How many Kj is left from her daily allowance?
Ans: Number of calories left = 4500 – 1339 = 3161Kj
c) If she decides to forgo her normal diet and instead have 2 lamp chops for dinner,
how much cheese is she allowed making it to 4500Kj?
Ans: 2 lamp chops = 2 × 1460 = 2920kj.
She will then have 4500 – 2920 = 1580 kj left over
Since 100g of cheese = 1833kj, 1580 kj = 1580 / 1833 × 100 g of cheese = 86.20 g
So, grandma is only allowed roughly 86 g of cheese