Measurement – AP Book 8, Part 1: Unit 6
1.
2.
There are more circles
because the ratio 94 : 93
indicates that for every 94
circles, there are 93
triangles and 94 > 93.
a)
1; 2 : 1
b)
5; 2 : 5
c)
3; 2 : 3
d)
5; 4 : 5
e)
4; 3 : 4
AP Book ME8-2
page 165
f)
4; 2 : 4
1.
g)
3; 1 : 3
h)
2; 3 : 2
a)
ii)
1:3
iii)
b)
c)
3.
6.
a)
a)
3:2 = 6:4
b)
1:4 = 2:8
c)
2 : 5 = 4 : 10
a)
4:2
3:1
b)
6:3
iv)
2:1
c)
8:4
v)
2:3
d)
10 : 5
vi)
3:5
e)
20 : 10
ii)
4:8
a)
iii)
3:8
Answers will vary –
teacher to check.
iv)
1:3
v)
1:4
vi)
1:5
2.
3.
b)
c)
Circle:
“squares to polygons”
and “triangles to
polygons”
ii)
iii)
3
8 ; shapes; light
iv)
1
3 ; circles; dark
v)
1
4 ; squares; light
vi)
1
5 ; dark shapes;
4.
5.
a)
5.
a)
part-to-whole ratio
b)
part-to-part ratio
c)
part-to-part ratio
d)
part-to-whole ratio
e)
part-to-part ratio
f)
part-to-whole ratio
If it was considered as a
fraction, the denominator
wouldn’t be the “whole” –
it would be a second
“part”, like the numerator.
That’s like saying this pie
3
3
is 1 instead of 4 :
4
blue : all = 4 : 7 = 7
3
red : all = 3 : 7 = 7
d)
7
blue : all = 7 : 9 = 9
2
red : all = 2 : 9 = 9
8.
5
They won 7 of the games.
9.
4
11 are rock CDs
6.
7.
5:7
b)
7:2
c)
5:3
d)
2:3
e)
28 : 9
f)
6:5
b)
2 : 3; 9
c)
3 : 5; 27
a)
4 : 5; 24
b)
7 : 10; 21
c)
2 : 3; 16
d)
1 : 4; 4
e)
1 : 2; 10
f)
14; 2 : 1
AP Book ME8-3
page 167
g)
10 : 3; 15
h)
10; 2 : 5
1.
i)
4 : 1; 28
2.
3.
a)
16
8.
3:2
b)
5
c)
8
d)
30
e)
12
INVESTIGATION
f)
8
A.
a)
100
ii)
6, 4
b)
25
iii)
9, 6
c)
25
iv)
12, 8
d)
9
v)
15, 10
a)
2
i)
2
b)
7
ii)
4
c)
30
iii)
6
d)
15
iv)
8
a)
i)
25
v)
10
The ratio has simply been
reversed.
B.
i)
3, 2
a)
9 : 12 = 12 : 16
b)
6 : 14 = 9 : 21
= 12 : 28
ii)
223
C.
Yes
iii)
235
D.
10 : 16 = 15 : 24
= 20 : 32
iv) 3  5  5
girls : boys = 1 : 2, so
girls : students = 1 : 3
1
 3 are girls.
E.
Ms. X – 2 : 3
d)
7.
c)
a)
6:9
c)
6.
5.
b)
circles
4.
No – to have a 1 : 1
ratio, there must be
an equal number
of vowels and
consonants.
5
blue : all = 5 : 8 = 8
3
red : all = 3 : 8 = 8
Answers will vary –
teacher to check.
This is impossible
with a 9-letter word
since the letter count
needs to be even.
4
8 ; shapes;
squares
b)
4.
b)
i)
2
6 : 20 = 9 : 30
= 12 : 40
ii)
2  5 = 10
iii)
5
a)
12 girls
iv) 2  3 = 6
b)
35 are blue
v)
c)
9 L are needed
a)
3
blue : all = 3 : 5 = 5
2
red : all = 2 : 5 = 5
3
vi) 3  5 = 15
c)
i)
5:6
ii)
1:3
iii)
2 : 15
Mr. Y – 5 : 3
F.
Mr. Y’s class
You can tell because:
* fraction – the numerator
> half the denominator;
* ratio – the first part (girls)
> the second part (boys)
iv) 2 : 5
v)
4 : 25
vi) 5 : 2
K‐30 AnswerKeysforAPBook8.1
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AP Book ME8-1
page 163
Measurement – AP Book 8, Part 1: Unit 6
AP Book ME8-4
page 169
r cm
7.
2r cm
Teacher to check drawing.
1.
27
2.
12
3.
15 minutes
4.
24
5.
120
6.
490
7.
80
8.
6.25 cups
INVESTIGATION
9.
27
A.
10.
5
11.
a)
18
b)
50
12.
8.
9.
B.
a)
2.5 cm
b)
3.5 cm
(continued)
b)
KM, LN
c)
OA = 34 which is
less than 6; inside
d)
OB =
41, outside;
OC =
32, inside;
OD =
34, inside;
OE =
40, outside
Teacher to check drawing.
a)
1.1 cm
b)
1.4 cm
Teacher to check.
Teacher to check drawing.
e)
Teacher to check.
f)
Teacher to check.
radius = 2 cm
AP Book ME8-6
page 174
OA = 1.4 cm
1.
a), b)
OB = 0.9 cm
4 cm
8 cm
12 cm
4:1
8:2
12 : 3
6 cm
12 cm
18 cm
6:2
12 : 4
18 : 6
24 cm
12 cm
36 cm
24 : 8
12 : 4
36 : 12
OC = 1.5 cm
4:5
OD = 1.7 cm
These lengths are all
shorter than the radius.
AP Book ME8-5
page 170
1.
Teacher to check.
2.
a)
Teacher to check.
b)
c)
C.
OF = 3.3 cm
OG = 3.4 cm
c)
They are equivalent.
Teacher to check.
OH = 2.5 cm
d)
Hexagon: 3, 3;
OA = OB
These lengths are all
longer than the radius.
They are both radii.
3.
4.
5.
6.
a)
Teacher to check.
b)
Teacher to check.
c)
Teacher to check.
d)
The triangle is
equilateral: BA = BC
= AC = radius of the
congruent circles.
a)
10 cm
b)
2m
c)
13 mm
d)
1.5 cm
a)
108 mm
b)
94 cm
c)
42 m
d)
2.8 cm
Radius
4 cm
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
OE = 2.5 cm
12 cm
D.
10.
11.
12.
Yes, should get the same
result – teacher to check.
a)
inside
b)
outside
c)
on
b)
AB = 5
c)
AB = 29 which is
between 5 and 6.
AB = 45 which is
between 6 and 7.
a)
Teacher to check
centre.
radius = 5 squares
c)
2
=
16 + 9
8 cm
=
25
=5
5 mm
10 mm
8m
16 m
2.7 cm
5.4 cm
0.5 m
1m
0.9 cm
1.8 m
AnswerKeysforAPBook8.1
d)
outside; longer than;
OR =
13.
B.
b)
2
5 +4
Teacher to check.
B.
Teacher to check.
C.
Teacher to check.
D.
radius
E.
1
fraction = 2
F.
base ≈ C ÷ 2
G.
radius  circumference ÷ 2
H.
r  C ÷ 2;
≈ r  2 r ÷ 2
=r2r÷2
=rr2÷2
2
=r
1.
27 mm;
10 mm;
80 mm;
≈ 80 mm
32 mm;
12 mm;
96 mm;
≈ 96 mm
64 : 22 = 2.91 : 1
80 : 27 = 2.96 : 1
96 : 32 = 3 : 1
ratio ≈ 3
C.
3.14 : 1
All are equal.
2
Diameter
24 cm
A.
3.14 : 1
4 +3
A.
INVESTIGATION
on; equal to;
OQ =
INVESTIGATION
Octagon: 3, 3
c)
d)
AP Book ME8-7
page 176
2.
a)
21.98 m
b)
37.68 m
c)
31.4 m
d)
28.26 cm
314 cm
b)
78.5 km
c)
153.86 m
d)
≈ 34.19 mm
25 + 16
=
41 > 5
e)
Teacher to check.
a)
OK, OL, OM, ON
2
2
2
2
e)
113.04 m
f)
907.46 cm
2
2
g)
≈ 0.79 m
h)
≈ 660.19 mm
2
AP Book ME8-8
page 177
1.
2.
2
b)
25.12 m
c)
226.08 cm
a)
62.8 cm
b)
1
2 ; 31.4 cm
c)
diameter; 2r
d)
31.4 + 20 = 51.4 cm
e)
Taking half the
circumference only
gives the length of
the curved side of
the semi-circle.
2
To find the complete
distance around the
semi-circle, you must
add the length of the
circle’s diameter,
i.e. the straight edge.
2
=
2
a)
f)
b) 20.56 m
c) 61.68 cm
K‐31
Measurement – AP Book 8, Part 1: Unit 6
3.
a)
A = 113.04 km
2
iii)
>1m
D = 37.68 km
b)
A ≈ 31.79 cm
2
A = 37.68 m
2
D = 26.84 m
d)
NOTE: The first step
is to make the units
the same – we have
used metres.
A ≈ 2.08 m
2
7.
≈ 706.5 m
8.
2
2
≈ 14.13 cm ;
≈ 6 + 14.13 = 20.13 cm
9.
2
a)
175.84 cm
b)
In 2 seconds, Karen
will go ≈ 3 × 175.84
≈ 527.52 cm.
 in 10 seconds,
she’ll go ≈ 2 637.6 cm
or 26.376 m.
 in 1 minute, she’ll
go ≈ 15 825.6 cm or
158.256 m.
In 1 hour, she’ll go
≈ 949 536 cm or
9 495.36 m.
c)
2
2
x +x =6
b)
From solving a),
we find that:
18 cm
1
A=2 b × h
x=
1
= 2 ( 18 × 18)
D ≈ 0.231 3 m
5.
2
a)
2
By looking at 3 b), you
can see that you must
divide by 10 000 in order
to convert a measurement
2
2
in cm to m .
2
6 cm ;
Using metres, the radius
is 0.045 m.
 A ≈ 0.003 179 m
2
3 cm;
D ≈ 6.96 m
4.
2
2
RECALL: 1 m
2
= 10 000 cm
D = 23.13 cm
c)
≈ 10 318.47 cm
(continued)
= 9 cm
10.
11.
2
c)
r
A= 2
a)
(13 + 12) cm
b)
(8 + ) m
c)
  + 4 cm2
2 
a)
i)
10.26 cm
2
ii)
15.25 cm
2
b)
2
9
2
= 2 cm
2
2
Teacher to check
drawings but, in both
i) and ii), the shaded
area is less than the
area of the triangle.
Yes, she will keep
up: Karen can only
go ≈ 9.5 km/hour on
her bike.
6.
a)
b)
i)
≈ 1.72 m
ii)
≈ 0.32 m
iii)
≈ 114.65 cm
i)
≈ 2.32 m
>1m
ii)
K‐32 2
2
≈ 0.08 m
<1m
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BONUS
His wheels should rotate
≈ 4 581.8 times in 1 hour,
or ≈ 1.27 times per second.
2
2
AnswerKeysforAPBook8.1