Download Chapter 18: Geometry 1

Chapter 18 Exercise 18.1
Q. 1.
(i) 180° − 37° = 143° (x = 143°)
(iv) A = 102° Corresponding
E = 63° Opposite
D = 102° Opposite
B = 180 − E − (180 − A)
= 180 − 63 − (78)
B = 39°
C = 180 − 39
= 141° Straight line
(v) A = 115° (Opposite)
B = 180 − 115 = 65° (Interior)
C = 180 − 75
= 105° (Straight line)
D = 75° (Alternate)
(ii) 180° − 117° = 63°
(x = 63°)
180° − 90° = 90°
(y = 90°)
3
(iii) 2x + x + __ x + 45 = 180
2
4.5x = 135
x = 30°
(iv) 180 − 90 = 2y
90 = 2y
45° = y
66 + (x + y) + 47 = 180
Q. 3.
x + y = 67
x = 67 − y
(i) A line is a straight line that goes on
forever in both directions; it has
no endpoints.
= 67 − 45
x = 22°
A ray is part of a line that has one
endpoint; the other end goes on
forever.
Answer: x + y = 67°
2y = 90°
Q. 2.
(i) A = 53°
(ii) Points that lie on the same plane
are coplanar.
Opposite angles
B = 180 − 53
Points that lie on the same line are
collinear.
= 127° Interior angles
C = 53°
Alternate angles
(iii) A acute angle is one that is less
than 90°.
(ii) A = 180 − 112
= 68° (Interior angles)
An ordinary angle is one that is
less than 180°.
C = 180 − 134 = 46° (Straight line)
B = 180 − 46
(iv) An axiom is a statement we accept
as true even though there is no
proof.
= 134° (Interior angles)
Answer: A = 68° B = 134°
C = 46°
(iii) A = 73°
B = 41°
A theorem is a statement we
accept as true, as there is a proof.
Corresponding angles
Q. 4.
Opposite angles
(i) A = 180 − 43 = 137°
B = 180 − 43 − 83 = 54°
C = 180 − 73 − 41
(ii) 2A = 180 − 63 = 117
C = 66°
Straight line
D = 66°
Corresponding angles
⇒ A = 117 ÷ 2 = 58.5°
E = 180 − 66
C = 180 − 51 − 58.5 = 70.5
E = 114°
B = 180 − 71 − 70.5 = 38.5
Straight line
Active Maths 2 (Strands 1–5): Ch 18 Solutions
1
(iii) 2C = 180 − 50 = 130
Similarly y = c (by isosceles)
C = 130 ÷ 2 = 65
⇒ 180 = 31 + x + C + y
B = 180 − 46 − 65
180 = 62 + 2C
B = 69
118 = 2C
A = B = 69
59° = C
A = 69°
B = 180 − C − 31
B = 69°
= 180 − 59 − 31
C = 65°
B = 90°
(iv) A = 180 − 52 − 81
(iv) C
A
102°
A = 47°
C = 52°
(Corresponding)
y
B = 180 − 47 − 52 = 81°
Q. 5.
x
B
(i) A = 23° (Alternate angles)
B = 58° (Opposite angles/rules of
parallelogram)
C: 2(C + 23) + 2(58) = 360
2C + 46 = 360 − 116
A = 180 − 102
2C = 244 − 46
A = 78°
2C = 198
by isosceles triangles
C = 99°
C = 180 − 102 = 78°
OR
y = 180 − 2C
(C + 23) + 58 = 180
= 180 − 2(78)
C + 23 = 122
(ii) A = 84°
C = 122 − 23
y = 24
C = 99°
x = 180 − 102 = 78
(Corresponding)
B = A = 84°
B = 180 − 78 − 24
(Opposite)
(C + 14) + B = 180
B = 78°
Q. 6.
C + 14 + 84 = 180
(ii) B = 180 − 49 = 131°
C = 180 − 98
(iii) 2C + 131 = 180
C = 82°
2C = 49
(iii) A = 31° Alternate
C = 24.5°
31°
C
(i) By opposite angles A = 49°
Q. 7.
(ii) ∠3 and ∠7 are corresponding
(iii) ∠8 and ∠7 make up 180° on a
straight line
x
31°
(iv) ∠5 and ∠4 Alternate angles
y
= 31° (isosceles
triangle)
by alternate
angles x = 31
2
Active Maths 2 (Strands 1–5): Ch 18 Solutions
Q. 8.
(a)
Q. 10.
(i) Triangle ABC and triangle FDE
are congruent by SSS (all sides
on ABC are the same as those
on FDE).
(i)
(ii) Though it looks like a rhombus
we can’t assume, as they haven’t
given us the information that
sides are equal.
⇒ not congruent:
A rhombus
(all sides equal)
(ii) Three
TS = QT and PT is common
but PQ ≠ PS and no angles are
similar.
(iii) Congruent: BF = FE
(Since CB = BA)
An equilateral triangle
BA = ED
|∠DFE| = |∠BFA|
(opposite)
|∠FDE| = |∠FAB|
(alternating)
|∠FED| = |∠FBA|
⇒ By SAS or ASA
(iii) Four
(iv) ΔABE is congruent to ΔADE
A square (all sides equal)
AE is common to both
AD = AB
BE = ED (as E is the bisector
of BD)
⇒ by SSS
Alternative method:
(i) Rectangle has 2 symmetries
[could also be proved by SAS or
ASA]
(b) Diagram (ii) ΔPQT ≡ ΔSTR
ΔQTR ≡ ΔPTS
Diagram (iv) all triangles are
congruent.
i.e. ΔABE ≡ ΔBEC ≡ ΔBEA ≡ ΔCED
Q. 9.
l
110°
opposite
70°
70° 110°
opposite
110°
m
n
(ii) Triangle has 3 symmetries
p
a
86°
180 – 110 180 – 110 = 70
= 70
Opposite Straight line
94° 86°
70°
86° 94°
b
180 – 86 180 – 70 180 – 70
= 94°
= 110°
= 110°
(iii) Square has 4 symmetries
Lines a and b are parallel
„ Internal angles sum to 180° (70 + 110)
„ Corresponding angles (70° and 110°)
„ Alternate angles (70° and 110°)
Lines l and p also show the same
features.
Active Maths 2 (Strands 1–5): Ch 18 Solutions
3
Q. 11.
Q. 12.
(i)
(i) A = Translation
B = Central symmetry
C = Axial symmetry
C
(ii) A = Central symmetry
B = Axial symmetry
C = Translation
(ii)
(iii) A = Axial symmetry
B = Translation
C = Central symmetry
Q. 13. (a)
y
No centre of symmetry
5
4
(iii)
3
A
2
1
C
–4 –3 –2 –1
(iii)
0 1
–1
2
3
x
4
–2
–3
(iv)
(i)
(b)
(ii)
–4
(i) (−1,−1), (−4,−1),
(−4,−3), (−1,−3)
(ii) (1,−1), (4,−1), (4,−3),
(1,−3)
(iii) (1,1), (4,1), (4,3), (1,3)
No centre of symmetry
Q. 14. (a)
4
(v)
(i)
3
y
(iii)
2
B
1
x
–5
(ii)
0 1
–1
2
3
4
5
–2
–3
No centre of symmetry
–4
(vi)
s
(b)
C
No axis of symmetry
4
–4 –3 –2 –1
Active Maths 2 (Strands 1–5): Ch 18 Solutions
(i) (−2,1), (−5,1), (−4,3)
(ii) (−2,−1), (−5,−1), (−4,−3)
(iii) (0,1), (3,1), (2,3)
Q. 15. (a)
y
5
Q. 17.
(ii)
(ii) |PS| = |QP| = 16
4
(iii) |MQ| = |MS| = 15
3
(i)
2
(iv) |SQ| = |MS| + |MQ| = 30
1
x
–5
–4 –3 –2 –1
0 1
–1
C
–2
2
3
4
5
Q. 18.
(i) x + 30 = x + y + 6
30 = y + 6
(iii)
–3
y = 24°
–4
(x + 30) + (x + 2y + 16) = 180
–5
(b)
(i) |QP| = |RS| = 20
2x + 2y + 46 = 180
2x + 48 + 46 = 180
(i) (−1,1), (−3,1), (−4,3),
(−2,3)
2x = 180 − 94
(ii) (−1,3), (1,3), (2,5), (0,5)
2x = 86
(iii) (1,−1), (3,−1), (4,−3),
(2,−3)
Q. 16.
5 y
x = 43°
(ii) x + y + 33 = 5x − 4y − 13
−4x + 5y = −46 simultaneous eq.
q
(5x − 4y − 13) + (2x + 3y + 4)
= 180
4
3
(ii)
2
p
1
x
A
0
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9
–1
B
–2
(i)
C
–3
r
(b)
(iii)
–5
–6
= 203 − 189
(iii) (−1,−2), (0,−4), (−2,−6)
Q. 19. (i) Equation 1:
4a + b − 2 = 3a − b + 11
(Diagonals)
a + 2b = 13
Equation 2:
2a + b − 1 = a + 2b − 3
(Diagonals)
a − b = −2
b=5
35x − 5y = 945
y = 7x − 189
(ii) (−2,3), (−4,2), (−6,4)
3b = 15
−4x + 5y = −46
x = 29°
(i) (4,−1), (6,0), (8,−2)
−a + b = 2
simultaneous eq.
31x = 899
–4
a + 2b = 13
7x − y = 189
a − b = −2
y = 14°
(ii) (2a + b) + (4a + 20) = 180
6a + b = 160 1
(5a + 70) + (4a + 20) = 180
9a = 90
a = 10°
b = 160 − 6(10)
b = 100°
a − 5 = −2
a = −2 + 5
a=3
a = 3, b = 5
Active Maths 2 (Strands 1–5): Ch 18 Solutions
5
Exercise 18.2
Q. 1.
Exercise 18.3
(i) x = 7.5 … theorem 11
Q. 1.
(ii) x = 3 … theorem 11
Q. 2.
(i) x = 5
y = 6 … theorem 11
(ii) x = 15
y = 20 … theorem 11
(iii) x + 5 = 8 … theorem 11
⇒x=3
y – 3 = 10 … theorem 11
⇒ y = 13
Q. 3.
(i) |AB| + |CD| = 56
⇒ x + x + 3x + 3x = 56
… theorem 11
⇒ 8x = 56
7
x
⇒ ___ = ___
10 11
⇒ x = 7 cm
77
⇒ x = ___ = 7.7
10
(ii) |DE| + |GH| = 40 cm
⇒ (x + x) + (x + x) = 40 cm
… theorem 11
⇒ 4x = 40
Q. 2.
⇒ x = 10 cm
Q. 4.
4
3
U
_______
__
|XU| = √ 16 − 9 = √ 7
(ii) T
6
Y
8
V
________
___
__
|YV| = √ 64 − 36 = √ 28 (OR 2√ 7 )
(iii) T
9
12
Z
W
|TW| = 12 (by similar triangles), so
_________
|ZW| = √ 144 − 81
___
__
= √ 63 (OR 3√ 7 )
6
bottom length bottom length
(i) _____________ = _____________
total length
total length
y 3
⇒ __ = __
4 5
(i) T
X
bottom length bottom length
(i) _____________ = _____________
top length
top length
x
1
⇒ __ = __
4 2
⇒x=2
x
1
(ii) __ = __
4 3
4
1
⇒ x = __ = 1__
3
3
3
x __
__
(iii) =
3 2
9
⇒ x= __
2
⇒ x = 4.5
top length
top length
(iv) __________ = __________
total length total length
Active Maths 2 (Strands 1–5): Ch 18 Solutions
12
⇒ y = ___ = 2.4
5
y
3
(ii) _____ = __
3.75 5
11.25
⇒ y = ______ = 2.25
5
y
1
(iii) __ = __
6 5
6
⇒ y = __ = 1.2
5
y
6
(iv) __ = ___
7 16
42
⇒ y = ___
16
5
21
⇒ y = ___ = 2__ = 2.625
8
8
Q. 3.
|PS| 12
(i) _____ = ___
22.5 15
Q. 9. |AY| : |YC| = 3.5 : 2.5 = 7 : 5
|AX| : |BX| = 3 : 2
270
⇒ |PS| = ____ = 18
15
(ii) |ST| = 22.5 – 18 = 4.5
⇒ |AY| : |YC| ≠ |AX| : |BX|
⇒ XYY || BC
(iii) |PS| : |ST| = 18 : 4.5 = 4 : 1
Q. 4.
(i) |AB| : |BD|
i.e. not parallel.
Q. 10.
A
= |AD| – |BD| : |BD|
250
=5–2:2
=3:2
30
⇒ |CE| = ___ = 10
3
No: only the ratios are given, not
the absolute values
Q. 6.
|DE|
|CD| 9.5 – 6 3.5
(i) ______ = ______ = _______ = ___
23.75
9.5
9.5
9.5
83.125
⇒ |DE| = _______ = 8.75
9.5
|FG|
2
(ii) ______ = ___
23.75 9.5
47.5
⇒ |FG| = _____ = 5
9.5
(i) 5 : 2
(ii) |AC| : |AE| = |AB| : |AD|
=7:5
(iii) |EC| : |AC| = |DB| : |AB|
=2:7
Q. 7.
3
(i) __
4
|AY| |AX| 3
(ii) _____ = _____ = __
|YC| |XB| 4
|YC|
|XB| 4
(iii) _____ = _____ = __
|AC| |AB| 7
Q. 8.
|AB| : |BC| = 2.25 : 0.75 = 3 : 1
= |AD| : |AC|
⇒ BD || CE.
Answer: Yes
425
E
|CE| 2
(ii) _____ = __
3
15
Q. 5.
Answer: No
D
x
325
B
C
325
x
(i) ____ = ____
250 425
425x = 81250
x = 191.176.....
x = 191 m
Total distance = 250 + 191
= 441m
441
(ii) ____ = 14.7
30
= 14 minutes 42 seconds
Exercise 18.4
x
1
(i) __ = __
3 2
3
⇒ x = __ = 1.5
2
8
x
(ii) __ = ___
6 10
48
⇒ x = ___ = 4.8
10
3
x
___
__
(iii)
=
10 4
30
⇒ x = ___ = 7.5
4
x
21
(iv) ___ = ___
9
10
210
1
⇒ x = ____ = 23__
3
9
y 4
Q. 2. (i) __ = __
4 5
16
⇒ y = ___ = 3.2
5
Q. 1.
Active Maths 2 (Strands 1–5): Ch 18 Solutions
7
y 20
(ii) __ = ___
9 15
180
⇒ y = ____ = 12
15
y
12
(iii) ___ = ___
9
21
252
⇒ y = ____ = 28
9
y
44 2
(iv) ___ = ___ = __
42 66 3
y
2
⇒ ___ = __
42 3
84
⇒ y = ___ = 28
3
Q. 3.
Q. 4.
|∠SRT| is common
⇒ |∠STR| = |∠RPQ|
∴ ∆RST and ∆PQR are similar
(ii) T
6
x
12
Q
R
Let |PR| = x
x
___
10
12
x
= ___ ⇒ ___ = 2 ⇒ x = 20
6
10
|PR| = 20
(iii) Let |RQ| = y
202 = 122 + y2
400 = 144 + y2
256 = y2
16 = y |RQ| = 16
(iv) Let |RS| = z
⇒ 403.2 – 24y = 168
⇒ 235.2 = 24y
⇒ y = 9.8
z
___
16
10
z
1
= ___ ⇒ ___ = __ ⇒ z = 8
20
16 2
|RS| = 8
Q. 5.
|EG| 1
(i) _____ = __
4
3
⇒ |EG| = 0.75
⇒x=1
y ___
4.5
__
=
2
1
⇒y=9
8
R
P
48
⇒ x = ___ = 8
6
y __
4
__
=
3 2
y
__
=2⇒y=6
3
x
24
(ii) __ = ___
8 10
192
⇒ x = ____ = 19.2
10
16.8 – y ___
10
________
=
24
16.8
7.5
x
(iv) __ = ___
4
3
30
⇒ x = ___ = 10
3
y
4
___
= ___
20 10
80
⇒ y = ___ = 8
10
10
S
x
4
(i) ___ = __
12 6
x
x
(iii) ___ = _______
4.5 x + 3.5
1
1
⇒ ___ = _______
4.5 x + 3.5
⇒ 4.5 = x + 3.5
(i) |∠RST| = |∠PQR| = 90°
7
(ii) |EF| = __(4) = 3.5
8
|GH|
5
⇒ ______ = __
4
3.5
17.5
⇒ |GH| = _____ = 4.375
4
Q. 6.
Active Maths 2 (Strands 1–5): Ch 18 Solutions
(i) |BC|2 + 62 = 102
|BC|2 + 36 = 100
|BC|2 = 64
|BC| = 8
|ED|
4
(ii) _____ = ___
6
10
Is ΔPST similar to ΔPQR?
|PQ|
_____
12.6
= ____
3
|PS|
= 4.2
24
⇒ |ED| = ___ = 2.4
10
(iii) |ED|2 + |DC|2 = |EC|2
|PR|
_____
(2.4)2 + |DC|2 = (4)2
|PT|
5.76 + |DC|2 = 16
= 3.2
|DC|2 = 10.24
|PQ| |PS|
As _____ ≠ _____
|PS|
|PT|
|DC| = 3.2
(iv) area ΔABC : area ΔEDC
1
1
__
(8)(6) : __ (3.2)(2.4)
2
2
24 : 3.84
i.e. the lengths of matching sides
are not in proportion
∴ The triangles are not similar.
No: ΔPST is not similar to ΔPQR
6.25 : 1
25
___
:1
4
25 : 4
Q. 7.
(i)
6.4
= ___
2
Q. 8.
15
D
x
11.25
21
8
z
y
y
___
22
E
11.25
= ______
(×21)
15
21 × 11.25
y = ___________
15
21
A
B
4
C
y = 15.75
A
12
Q. 9.
Is ΔABC similar to ΔADE?
|AC| ___
6
12 ___
_____
=
= .
22 11
|AE|
|BC|
_____
4
__
h
1
__
= =
|DE| 8 2
|AC| |BC|
As _____ ≠ _____
|AE|
|DE|
No: ΔABC is not similar to ΔADE
(ii)
1
0.75 m
pole
2.1 m
building
h = height of building
P
2.1
= ____
∴ h = 2.8 m
1 0.75
The building is 2.8 m tall.
h
__
12.6
Q
6.4
R
P
3
S
2
T
Active Maths 2 (Strands 1–5): Ch 18 Solutions
9
Q. 10.
Q. 12.
(i)
h
h
1.72
2.5
Flagpole
10.5 m
1.45 m
girl
h = height of tree in metres.
10.5
= ____ (×1.72)
2.5
1.72
h = 10.5 × 1.72
_______________
2.5
h
____
4.2 m
1.2 m
1.45
h
(ii) ___ = ____
5.4
1.2
1.45 × 5.4
h = __________
1.2
h = 6.525 m
h = 7.224 m
The tree is 7.224 m tall
Q. 11.
R
h = 652.5 cm
The flagpole is 653 cm high to the
nearest cm.
cm
12
(×5.4)
Model
Q
Q. 13.
h
15 cm
22 m
P
1.45 m
R′
5m
22
= ___
5
1.45
22 × 1.45
h = _________
5
h = 6.38 m
h
____
Actual
Q′
P′
3m
cm)
(300
|P’R’| ____
300
______
=
12
15
12 × 300
|P’R’| = _________
15
|P’R’| = 240 cm
The school building is 6.38 m high.
Q. 14.
= 2.4 m
The length of PR on the actual frame is
2.4 m.
10
Active Maths 2 (Strands 1–5): Ch 18 Solutions
xm
25 m
12 m
28
= ___
12
28 × 25
∴ x = ________
12
1
__
x = 58 m
3
1
The river is 58__ m wide
3
x
___
25
28 m
Q. 15.
(i)
0.3
2m
1.7 m
y
7 km
1m
l
h
1.7 m
x
1.7 m
60 m
The height of the kite
= h + 1.7m
60
h
___
= ___
0.3
1
h = 60 × 0.3.
h = 18 m
∴ Height of kite = 18 + 1.7
= 19.7 m
(ii) l = length of kite string
Using Pythagoras’ Theorem:
l2 = 182 + 602
x
2 km
x = distance between shop and
park
x __
7
__
=x
2
∴ x2 = 14___
x = √ 14
x = 3.7416.. km
∴ Distance between shop and
park is 3,742 m to the nearest m.
(ii) Let y = distance between school
and park
Using Pythagoras’
___ Theorem,
2
2
y = 7 + (√ 14 )2
y2 = 49 + 14
y2 = 63
___
y = √ 63
y = 7.9372... km × 1000
y = 7937.2... m
l2 = 3,924
______
l = √ 3,924
l = 62.641... m
l = 6,264.1... cm
The length of the kite is 6,264 cm
or 62.64 m to the nearest
centimetre.
Q. 16.
(i)
Park
The distance between the school
and park is 7,937 m to the
nearest m
y
x
School
7 km
Shop 2 km Home
Q. 17.
(i)
x
1.85 m
2.15
James
Sign
12 m
38 m
House
Active Maths 2 (Strands 1–5): Ch 18 Solutions
11
57
= ____
133
133x = 57(x + 48)
133x = 57x + 2736
76x = 2736
x = 36
The distance between the trees is
36 m.
x
______
2.15 – 1.85 = 0.3 m
x + 48
12 m
x
(iii)
38 + 12 = 50 m
50
= ___
12
50 × 0.3
x = ________
12
x
___
A
A
36
0.3
B 40
E
C
x = 1.25 m
= 3.1 m
The house is 3.1 m high.
(i) A, B and C must be collinear also
A, E and D must be collinear,
[BE] || [CD]
(ii)
x
|CB| = |FE| = 48 m and
E
57
48
|BE| = |CF| = 57 m
C
Exercise 18.5
133 m
D
Let x = distance between the
two trees.
Q. 1.
A
(i) h2 = a2 + b2
h = 13, a = 12, b = x
⇒ 132 = 122 + x2
⇒ 169 = 144 + x2
x
⇒ x2 = 25
B
57
E
⇒x=5
(ii) h2 = a2 + b2
A
h = x, a = 12, b = 16
⇒ x2 = 122 + 162
x + 48
⇒ x2 = 144 + 256
C
12
⇒ x2 = 400
133 m
D
(iv) To ensure [BE] was parallel to [CD]
students could have inserted a peg
F into the ground, such that CBEF is
a parallelogram where
A
B
y
let y be the distance between
C and D
y
45
∴ ___ = ___
40 36
45 × 40
y = ________
36
∴ y = 50 m
Height of house = 1.25 m + 1.85 m
Q. 18.
36 + 9 = 45
(ii)
D
Active Maths 2 (Strands 1–5): Ch 18 Solutions
⇒ x = 20
(iii) h2 = a2 + b2
(ii) x2 = 92 + 122
___
h = x, a = √ 11 , b = 5
⇒ x2 = 81 + 144 = 225
⇒ x2 = ( √ 11 ) + 52
⇒ x = 15
⇒ x2 = 11 + 25
172 = 152 + y2
⇒ x2 = 36
⇒ 289 = 225 + y2
⇒x=6
⇒ 64 = y2
___
2
⇒y=8
(iv) h2 = a2 + b2
(iii) 32 = x2 + 52
h = 25, a = 24, b = x
⇒ 9 + x2 = 25
⇒ 252 = 242 + x2
⇒ x2 = 16
⇒ 625 = 576 + x2
⇒x=4
⇒ x2 = 49
⇒x=7
(v)
h2
=
a2
b2
+
⇒
⇒y=7
___
___
= ( √ 15 ) +
2
(iv) x2 = 12 + 22
72
⇒ x2 = 1 + 4 = 5
__
⇒ x2 = 15 + 49
⇒ x = √5
⇒ x2 = 64
y2 = ( √ 5 )2 + 22
⇒x=8
y2 = 5 + 4 = 9
__
⇒y=3
(vi) x2 = 202 + 162
⇒ x2 = 400 + 256
⇒
x2
Q. 3.
(i) x2 = 102 + 122
= 656
x2 = 100 + 144
____
⇒ x = √656
(vii)
242
=
102
+
x2 = 244
____
x2
x = √ 244
⇒ 576 = 100 + x2
x = 15.620...
⇒ 476 = x2
x = 15.6 to 3 significant figures
____
⇒ x = 2 √ 119
__
(viii) x2 = 22 + ( 2√ 3 )
2
(ii)
8
⇒ x2 = 4 + 12
⇒ x2 = 16
⇒x=4
Q. 2.
(i) x2 = 42 + 72
⇒ x2 = 16 + 49 = 65
___
⇒ x = √65
___
( √65 )2 = 82 + y2
⇒ 65 = 64 + y2
⇒ y2= 1
⇒y=1
2
⇒ y2 = 16 + 33 = 49
h = x, a = √15 , b = 7
x2
___
y2 = 42 + ( √33 )
16
16
8
x
x2 = 82 + 162
x2 = 64 + 256
x2 = 320
____
x = √ 320
x = 17.888...
x = 17.9 to 3 significant figures
Active Maths 2 (Strands 1–5): Ch 18 Solutions
13
(iii)
172 = x2 + 92
8
y
3
289 = x2 + 81
x
208 = x2
____
11
√ 208 = x
y2 = 82 + 32
∴ x = 14.4 (to three significant
figures)
y2 = 64 + 9
y2 = 73
(vi)
___
y = √ 73
___
a
∴ x2 + (√ 73 )2 = 112
1
___
x2 = 112 − (√ 73 )2
x2 = 121 − 73
1
x2 = 48
1st triangle
x = √ 48
+ 12
a2 = 12_______
___
a = √ 12 + 12
x = 6.928...
x = 6.93 to 3 significant figures
__
a = √2
(iv)
5
x
3.5
b
3.5
1
7
x2 + 3.52 = 52
a
x2 = 52 − 3.52
2nd triangle.
x2 = 12.75
______
x = √ 12.75
b2 = 12 + a2
x = 3.5707...
b2 = 12____________
+ 12 + 12
x = 3.57 to 3 significant figures
(v)
b = √ 12 + 12 + 12
9
c
15
y
1
x
b
3rd triangle.
c2 = 12 + b2
8
y2 = 152 + 82
y2 = 289
____
y = √ 289
y = 17
14
Active Maths 2 (Strands 1–5): Ch 18 Solutions
c2 = 12_________________
+ 12 + 12 + 12
c = √ 12 + 12 + 12 + 12
x=√
Following this pattern on the 7th
triangle
___________________________________
12
+
12
+
12
__
+
12
+
12
+
12
+
12
+
(ii) Side lengths 39, 80, 89
892 = 7,921
12
392 + 802 = 1,521 + 6,400
x = √8
= 7,921
x = 2.8284...
Yes: these lengths will form a rightangled triangle since
x = 2.83 to 3 significant figures.
Q. 4.
392 + 802 = 892
(i) Triangle with sides 72, 70 and 21.
(iii) Side lengths 55, 130, 148
722 = 5,184
1482 = 21,904
702 + 212 = 4,900 + 441
552 + 1302 = 3,025 + 16,900
= 5,341
= 19,925
No: since 702 + 212 ≠ 722 this is
not a right-angled triangle
No: these lengths will not form a
right-angled triangle since
552 + 1302 ≠ 1482
(ii) Triangle with sides 8.9, 8 and 3.9
8.92 = 79.21
(iv) Side lengths 64, 120, 136
82 + 3.92 = 64 + 15.21
1362 = 18,496
= 79.21
Yes: since 82 + 3.92 = 8.92 this is
a right-angled triangle
(iii) Triangle with sides 162, 134 and
102
1622 = 26,244
1342 + 1022 = 17,956 + 10,404
= 28,360
No: since 1342 + 1022 ≠ 1622
this is not a right-angled triangle
(iv) Triangle with sides 113, 112
and 15.
1132 = 12,769
1122 + 152 = 12,544 + 225
= 12,769
Yes: since 1122 + 152 = 1132 this
is a right-angled triangle
Q. 5.
(i) Side lengths 60, 63, 87
642 + 1202 = 4,096 + 14,400
= 18,496
Yes: these side lengths will form a
right-angled triangle since
642 + 1202 = 1362
Q. 6.
A
75 – 45 = 30
75
40
B
45
40
|AB|2 = 302 + 402
|AB|2 = 2,500
Longest side squared must equal
the sum of the squares of the two
shorter sides.
______
|AB| = √ 2,500
|AB| = 50
The distance AB is 50.
872 = 7,569
602 + 632 = 3,600 + 3,969
= 7,569
Yes: these lengths will form a rightangled triangle since
602 + 632 = 872
Active Maths 2 (Strands 1–5): Ch 18 Solutions
15
(ii)
(iii) (x − 1)2 + (x − 2)2 = x2
B
x2 − 2x + 1 + x2 − 4x + 4 − x2 = 0
38
x2 − 6x + 5 = 0
66
(x − 1)(x − 5) = 0
112
x − 1 = 0 OR
x−5=0
x = 1 OR
x=5
74
if x = 1 then one of the sides
x−1=1−1=0
A
|AB|2
662
Since we cannot have a side of
zero length, x = 1 is rejected.
1122
=
+
= 16,900
|AB| = 130
Q. 7.
∴x=5
(iv) (x − 1)2 + (x + 1)2 = (x + 5)2
x2 − 2x + 1 + x2 + 2x + 1
= x2 + 10x + 25
(i) (2x)2 + (3x)2 = 402
4x2 + 9x2 = 1,600
2x2 + 2 − x2 − 10x − 25 = 0
13x2 = 1,600
x2 =
x2 − 10x − 23 = 0
1,600
______
a = 1, b = −10, c = −23
13
______
x=
b2 − 4ac = (−10)2 − 4(1)(−23)
√1,600
_______
___
= 100 + 92 = 192
√13
x = 11.094...
Using the quadratic formula:
____
x = 11.09 to 2 d.p.
x=
(ii) x2 + (2x + 8)2 = 522
x2 + 4x2 + 32x + 64 = 2,704
5x2 + 32x + 64 − 2,704 = 0
5x2 + 32x − 2,640 = 0
x=
2
x=
____
10 − √ 192
__________
2
x = 11.928... OR x = −1.928...
Since x > 0, x = −1.928... is
rejected
∴ x = 11.93 to 2 d.p.
Q. 8.
(i) x = distance of foot of ladder from
the wall
∴ x2 + 1.62 = 2.252
OR
_______
√
−32 − 53,824
_______________
10
x = 20 OR x = −26.4
Since x > 0 reject x = −26.4
∴ x = 20
16
10 + √ 192
__________
OR
= 53,824
x=
2×1
____
a = 5, b = 32, c = −2,640
b2 − 4ac = (32)2 − 4(5)(−2,640)
Using the quadratic
formula:
_______
−32 ± √ 53,824
x = _______________
2 ×_______
5
−32 + √ 53,824
x = _______________
10
−(−10) ± √ 192
_______________
Active Maths 2 (Strands 1–5): Ch 18 Solutions
x2 = 2.252 − 1.62
x2 = 2.5025
_______
x = √ 2.5025
x = 1.5819... m
x = 158.19... cm
The foot of the ladder is 158 cm
(to the nearest cm) from the wall.
There are no whole number roots to
this equation (p has a decimal value
on using the quadratic formula)
(ii)
22
5c
m
h
As p is not a whole number
{p + 1, p + 6, p − 3} cannot be
positive integers and ∴ do not form
a Pythagorean triple.
90 cm
h = new height of ladder against the
wall
h2 + 1902 = 2252
h2 = 2252 − 1902
h2 = 14,525
Q. 10.
80 cm
_______
l
h = √ 14,525
Slip = original height − new height
_______
= 160 cm − √ 14,525 cm
= 39.480... cm
39 cm
∴ The ladder slipped 39 cm (to the
nearest cm).
Q. 9.
Diameter of drum = 2 × 39
= 78 cm
l = length of wire
∴ l2 = 782 + 802
(i) (p − 2)2 + (p + 1)2 = (p + 4)2
p2 − 4p + 4 + p2 + 2p + 1
= p2 + 8p + 16
l2 = 12,484
2p2 − 2p + 5 − p2 − 8p − 16 = 0
p2
_______
l = √ 12,484
− 10p − 11 = 0
l = 111.7|3...
(p + 1)(p − 11) = 0
∴p+1=0
The longest wire is 111.7 cm to 1 d.p.
OR p − 11 = 0
p = −1 OR
p = 11
Q. 11.
x
40 cm
if p = −1 then p + 1 = −1 + 1
=0
1200 cm
Cannot have side of zero length
x2 = 1,2002 + 402
∴ p = −1 is rejected
x2 = 1,441,600
__________
∴ p = 11
x = √ 1,441,600
(ii) Show that
x = 1,200.66...
{p + 1, p + 6, p − 3} does not
form a Pythagorean triple.
(p + 1)2 + (p − 3)2 = (p + 6)2
p2 + 2p + 1 + p2 − 6p + 9
= p2 + 12p + 36
The insects are 1,201 cm apart to the
nearest cm.
Q. 12.
y
2p2 − 4p + 10 − p2 − 12p − 36
=0
D
o
iag
n
al
3
x
p2 − 16p − 26 = 0.
3
3
Active Maths 2 (Strands 1–5): Ch 18 Solutions
17
1st find the length of the diagonal x of
the base of the cube
Q. 14. Cube of side lengths 12 cm.
= 120 mm
x2 = 32 + 32
x2 = 9 ___
+9
x = √ 18
2nd use Pythagoras’ Theorem again (on
the blue triangle).
___
∴ y2 = 32 + (√ 18 )2
y2 = 9 ___
+ 18
y = √ 27
y = 5.196... cm
The diagonal is 5.2 cm to 1 d.p.
Q. 13.
Radius of cone = 120 ÷ 2
= 60 mm
Height of cone = 120 mm
let l = slant height of cone
l
h
(i)
6m
r
w
r = 60 mm
h = 120 mm
4.8 m
w = width of garden
w2 + 4.82 = 62
w2 = 62 − 4.82
w2 = 12.96
______
w = √ 12.96
w = 3.6 m
Width of garden is 3.6 m
(ii) Perimeter = 2l + 2w
= 2 × 4.8 + 2 × 3.6
∴ l2 = h2 + r2
l2 = 1202 + 602
l2 = 18,000
_______
l = √ 18,000
l = 134.16... mm
The slant height to the nearest mm is
134 mm
Q. 15.
6 cm
= 16.8 m.
x
6 cm
Perimeter of garden is 16.8 m
6 cm 3
(iii)
l
18 cm
(1 – 2√5) m
Let x = height of truncated cone
(3 + √5) m
l = length of__ drainage pipe__
l2 = (3 + √ 5__)2 + (1 − 2√ 5 )2__
l2 = 9 + 6√ 5 + 5 + 1 − 4√ 5
+ 4__ × 5
2
l = 2_________
√ 5__+ 35
l = √ 2√ 5 + 35
l = 6.282... m
l = 628.2... cm
The pipe is 628 cm to the nearest
cm.
18
Active Maths 2 (Strands 1–5): Ch 18 Solutions
∴ x2 + 32 = 62
x2 = 62 − 32
x2 = 27
___
x = √ 27
6
27
h
3
9
Two similar triangles in the cone where
h = height of original cone.
9
h
___
∴ ____
= __
3
√ 27
___
9 × √ 27
________
h=
3
h = 15.588... cm
h = 155.88... mm
Height of original cone is 156 mm to
nearest mm.
Q. 16.
G
5
P
2x2 = 400
x2 = 200
____
x = √ 200
(i)
|AE|2
l
E
B
D
82
20
A
8m
10 m
x
30 cm
(Pyramid height)
3m
C
H
A
x
∴ x2 + x2 = 202
F
5
8
B
Q. 17. Note diagonals of a
square bisect each other
at right angles.
l = length of ____
wire
2
= 30 + (√ 200 )2
l2 = 900 + 200
l2 = 1,100
______
l = √ 1,100
l = 33.166... cm
∴ Length of wire to connect point A to
B is 33 cm to nearest cm
l2
102
= +
= 164
____
|AE| = √ 164
|AE| = 12.806... m
|AE| = 1,280.6 cm
|AE| = 12.81 m (to nearest cm)
(ii) |AF|2 = |AF|2 + 32
|AF|2 = 164 + 9
|AF|2 = 173
____
|AF| = √ 173
|AF| = 13.152... m
|AF| = 1,315.2... cm
∴ |AF| = 13.15 m (to nearest cm)
(iii) |BP|2 = 52 + 82
|BP|2 = 89___
|BP| = √ 89
|BP| = 9.433... m
|BP| = 943.3... cm
∴ |BP| = 9.43
m (to nearest cm)
___
2
2
(iv) |AP| = (√ 89 ) + 32
|AP|2 = 89___
+9
|AP| = √ 98
|AP| = 9.899...m
|AP| = 989.9... cm
∴ |AP| = 9.90 m (to nearest cm)
√200
|AE|2
Q. 18.
A
B
18 km
12 km
9 km
5 km
H 4 km
8 km
C
(i) |AH|2 = 182 + 92
|AH|2 = 405
____
|AH| = √ 405
AH = 20.12... km
Ship A is 20 km from the harbour to
2 significant figures.
Active Maths 2 (Strands 1–5): Ch 18 Solutions
19
(ii) |BH|2 = 52 + 122
|BH|2 = 169
____
|BH| = √ 169
|BH| = 13 km
∴ Ship B is 13 km from the
harbour.
A
(v)
18 – 12 = 6
(iii)
|BC|2 = 401
____
|BC| = √ 401
|BC| = 20.02... km
∴ Ships B and C are
20 km apart to
2 significant figures. C
A
|AB|2 = 62 + 142
|AB|2 = 232
____
|AB| = √232
|AB| = 15.23... km
∴ Ships A and B are 15 km apart
to 2 significant figures.
5–4=1
18 + 8 = 26
C
9 + 4 = 13
∴ |AC|2 = 262 + 132
|AC|2 = 845
____
|AC| = √ 845
|AC| = 29.06... km
∴ Ships A and C are 29 km apart
to 2 significant figures.
Exercise 18.6
(i) 2|∠A| = 80° … angle at the center of a circle
⇒ |∠A| = 40°
(ii) |∠A| = 2(50°) = 100° … angle at the centre of a circle
(iii) |∠A| = 90° … angle in a semi circle
1
(iv) |∠A| = __(110°) = 55° … angle
2
at the centre of a circle
1
(v) |∠A| = __(66°) = 33° … angle
2
at the centre of a circle
1
(vi) |∠A| = __(80°) = 40° … angle at
2
the centre of a circle
(vii) Smaller angle at 0 = 360° − 260° = 100°
1
∴ |∠A| = __(100°)
2
= 50°
(viii) Larger angle at 0 = 360° − 80° = 280°
1
∴ |∠A| = __(280°)
2
= 140°
20
12 + 8 = 20
B
9 + 5 = 14
Q. 1.
B
(iv) |BC|2 = 12 + 202
Active Maths 2 (Strands 1–5): Ch 18 Solutions
Q. 2.
(i) |∠A| = 34° … angles standing on the same arc.
|∠B| = 56° … angles standing on the same arc.
(ii) |∠B| = 2(110°) = 220° … angle at the centre of a circle
|∠A| = 360° – 220° = 140° … angles at a point
(iii) |∠A| = 180° – 36° = 144° … opposite angle in cyclic quadrilateral
|∠B| = 180° – 107° = 73°
(iv) |∠A| = 180° – 51° = 129° … opposite angle in cyclic quadrilateral
|∠B| = 2(129°) = 258° … angle at the centre of a circle
(v) |∠A| = 180° – 104° = 76° … straight angle
|∠B| = 180° – 76° = 104° … opposite angle is a cyclic quadrilateral
(vi) |∠A| = 48° … angles standing on the same arc
|∠B| = 180° – 48° = 132° … opposite angle is a cyclic quadrilateral
Q. 3.
(i) |∠A| = 2(44°) = 88° … angle at the centre of a circle
|∠B| + |∠B| + 88° = 180° … angle is a Δ, isosceles Δ
⇒ 2|∠B| = 92°
⇒ |∠B| = 46°
(ii)
A
angle at a point
360° – B
O
30° B 30°
|∠B| = 2|∠A| … angle at the centre of a circle
|∠A| + 30° + (360° – |∠B|) + 20° = 360° … quadrilateral
⇒ |∠A| + 410° – 2|∠A| = 360°
⇒ |∠A| = 50°
∴ |∠B| = 100°
A
36° B
(iii) 2|∠A| = 140°
⇒ |∠A| = 70°
Adding in the radius makes life easier (mark in the
equal radius and you’ll see the isosceles triangles)
3 6°
O
140°
B
⇒ |∠B| + 36° = 70°
⇒ |∠B| = 34°
(iv) |∠B| = 90° – 32° = 58° … angle is a semi circle, isosceles Δ
|∠A| = 180° – 2(58°) … angles is a Δ
⇒ |∠A| = 180° – 116°
⇒ |∠A| = 64°
Active Maths 2 (Strands 1–5): Ch 18 Solutions
21
(v)
B
O
44°
A
same arc
angle in a semi circle
|∠A| = 180° – (90° + 44°) = 46° … angles in a ∆
|∠B| = 46° … angles standing on the same arc.
(vi)
60°
standing on the
same arc
B
60°
A
O
30 °
|∠A| = 90° – 60° = 30° … angle is a semicircle and angles standing on the same arc.
|∠B| = 30° … standing on the same arc
Q. 4.
(i) |∠A| = 41° as both are angles at the circle being subtended by the same arc.
|∠B| = 2(41°) = 82°
(ii) |∠A| = 2(25°) = 50°
|∠B| = |∠A| = 25° as both are angles at the circumference subtended by the same
arc.
(iii) Reproduce diagram & label points P, Q, R, S, T & U as shown.
|QO| = |RO| as both radii
U
∴ |∠OQR| = |∠ORQ|
as base angles of a Δ.
90°
as US is a tangent
O
B
A
|∠ORU| = |∠ORS| = 90°
90° R
P
|∠OQT| = |∠OQS| as TS is a tangent.
90°
∴ |∠OQS| = |∠ORS|
⇒ |∠C| = |∠QRS|
∴ ΔQRS is isosceles.
180 − 61° 119
So |∠C| = __________ = ____ = 59.5°
2
2
|∠OQR| = 90° − 59.5° = 30.5°
∴ |∠B| = 180° − (2)(30.5°)
= 180° − 61°
= 119°
1
∴ |∠A| = __ (119°) = 59.5°
2
22
Active Maths 2 (Strands 1–5): Ch 18 Solutions
C
90°
T
Q
61°
S
(iv) |∠B| = 2|∠A|
|∠B| + 2|∠C| = 180°
A
⇒ 2|∠A| + 2|∠C| = 180°
O
B
⇒ |∠A| + |∠C| = 90°
34°
C
|∠C| = 90° − |∠A|.
24°
C
|∠A| + 34° + |∠C| + 24° + |∠C| = 180°
|∠A| + 58° + 2|∠C| = 180°
|∠A| + 58° + 2(90° − |∠A|) = 180°
|∠A| + 58° + 180° − 2|∠A| = 180°.
238° − |∠A| = 180°
∴ |∠A| = 58°
⇒ |∠B| = 2(58°) = 116°
⇒ |∠C| = 90° − 58° = 32°
(v) As shown in (iv) |∠C| = 90° − |∠A|
A
∴ |∠A| + 13° + 38° + 2(90° − |∠A|) = 180°
|∠A| + 13° + 38° + 180° − 2|∠A| = 180°
13° + 38° − |∠A| = 0
O
B
38°
13°
13° + 38° = |∠A|
C
C
51° = |∠A|
∴ |∠B| = 2(51°) = 102°
Q. 5.
180° − 38° 142°
(iii) |∠ADB| = __________ = _____
2
2
= 71°
(i) Yes. As opposite angles add to
180°
i.e. 50° + 130° = 180°
(iv) |∠DAC| = 180° − 90° − 38°
= 52°
and 85° + 95° = 180°
(ii) No. As opposite angles do not
add to 180°
Q. 7.
ΔAOE ≡ ΔBOE by RHS.
i.e. 80° + 90° = 170°
∴ |∠AOD| = |∠BOD|
and 115° + 75° = 190°
(iii) No. As opposite angles don’t add
to 180°
i.e. 120° + 50° = 170°
|∠AOD| = 180° − 90° − 31°
= 59°
⇒ |∠AOB| = 2(59°) = 118°
(ii) ΔOCA is isosceles as
|OC| = |OA|
(iv) Yes. As opposite angles add to
180°
|∠COA| = 180° − 59° = 121°
i.e. 108° + 72° = 180°
Q. 6.
(i) |∠AOB|;
(i) |∠ACD| = 38° (both angles at
circle subtended by same arc)
180° − 121°
∴ |∠ACO| = ___________
2
= 29.5°
(ii) |∠CDA| = 90° (angle subtended
by diameter)
Active Maths 2 (Strands 1–5): Ch 18 Solutions
23
(iii) As in (ii) ΔOCB is isosceles
|∠COB| = 121°
180° − 121°
|∠OBC| = ___________
2
= 29.5°
Q. 10. Join the centre of the circle to each
point of the star. This gives 5 equal
angles which are 72° each.
A
(iv) ACBD is a cyclic quadrilateral
∴ |∠ADB| + |∠ACB| = 180°
|∠ACB| = 2(29.5°) = 59°
A
∴ |∠ADB| = 180° − 59° = 121°
Q. 8.
(i) |∠QOS| = 2(58°) = 116°.
Angle A is the angle at the circle being
subtended by the same arc as the 72°
(ii) QPSR is a cyclic quadrilateral
∴ |∠QRS| = 180° − 58°
= 122°
angle at the centre and is therefore
1
__
(72°) = 36°.
2
(iii) ΔOSQ is isosceles
180° − 116°
∴ |∠OSQ| = ___________
2
= 32°
Q. 11.
ΔRSQ is isosceles
180° − 122°
∴ |∠RSQ| = ___________ = 29°
2
⇒ |∠RSO| = 32 + 29 = 61°
(iv) P, O, R are collinear
⇒ PR is a diameter as O is centre
5
6
1
2
3
4
5
6
7
Daniel could sit in Row 2, Seat 1 or
Row 2, Seat 7 and have the same
viewing angle.
Revision Exercises
24
4
⇒ |∠PQO| = 90° − |∠RQO|
A parallelogram inscribed in a circle
must have opposite angles which
add to 180°. Therefore every angle
must be 90°, as opposite angles in a
parallelogram are equal. Therefore,
only a rectangle or a square can be
inscribed in a circle.
(a)
3
Seats
= 29°
Q. 1.
2
⇒ |∠PQR| = 90°
= 90° − 61°
Q. 9.
S T A G E
Row
1
(i) 3x = 150° … vertically opposite
⇒ x = 50°
y = 180° – 150° = 30° … straight angle
(ii) x = 180° – 130° = 50° … straight angle
y = 180° – 110° = 70° … straight angle
Active Maths 2 (Strands 1–5): Ch 18 Solutions
Reasoning: two angles subtended by
the same arc (stage), touching the
circumference, are equal in measure.
Another possible answer: Row 5, seat 5.
(iii) 2x = 180° – 100° = 80° … straight angle
⇒ x = 40°
4y = 100° … corresponding angle
⇒ y = 25°
(iv) 4x + x + 90° = 180° … straight angle
⇒ 5x = 90°
⇒ x = 18°
y = 180° – 95° = 85° … straight angle
(v) 2x + 115° = 180° … angle in a Δ, isosceles Δ
⇒ 2x = 65°
⇒ x = 32.5°
isosceles triangles
115º
x y
x
y
30º
(32.5° + y) + (32.5° + y) + 30° = 180° … angles in a Δ
⇒ 2y + 95° = 180°
⇒ 2y = 85°
⇒ y = 42.5°
(b) |AC|:|YC|
⇒ |AY| + |YC| : |YC|
⇒ |AX| + |XB| : |XB|
⇒3+5:5
⇒8:5
Q. 2.
(a)
(i) 7x + 2x + 3x = 180° … angles in a Δ
y = 2x + 3x … exterior angle
⇒ 12x = 180°
⇒ x = 15°
⇒ y = 5x = 75°
(ii) x + 2x + y = 180° … angles in a ∆
⇒ 3x + y = 180°
2y + 20° = 2x + y … exterior angle
⇒ 2x – y = 20°
3x + y = 180°
5x = 200°
⇒ x = 40°
3(40°) + y = 180°
⇒ y = 60°
Active Maths 2 (Strands 1–5): Ch 18 Solutions
25
(iii) (x + y) + (x + 22°) + x = 180° … angles in a Δ
⇒ 3x + y = 158°
x + 10y = x + y + x + 22° … exterior angle
⇒ –x + 9y = 22°
3x +
(× 3)
y = 158°
⇒ –3x + 27y = 66°
3x +
y = 158°
28y = 224°
3x + 8° = 158°
⇒ 3x = 150°
⇒ y = 8°
⇒ x = 50°
(iv) 4y + 4° + x + y + 6y – 4° = 180° … angles in a Δ
⇒ x + 11y = 180°
2x + 6y – 4° = 180° … straight angle
⇒ 2x + 6y = 184°
⇒ x + 3y = 92°
x + 11y = 180°
x + 3(11°) = 92°
8y = 88°
(2nd – 1st)
⇒ x + 33° = 92°
⇒ y = 11°
(b)
⇒ x = 59°
(i) b = 8, a = 6
(ii) a = 7.5
b=5
Q. 3. (a)
(i) x + 65° = 2x + 30° … alternate
⇒ x = 35°
x + 65° + 2y = 180° … straight angle
⇒ 35° + 65° + 2y = 180°
⇒ 2y = 80°
⇒ y = 40°
(ii) x + 2y = 9x – 2y … opposite angles
⇒ 8x – 4y = 0
x + 2y + 7x + 3y = 180° … parallelogram
⇒ 8x + 5y = 180°
–8x + 4y = 0°
8x + 5(20°) = 180°
9y = 180°
⇒ 8x = 80°
⇒ y = 20°
⇒ x = 10°
(iii) 7x + 4y + 8x + 4y = 180 … cyclic quadrilateral
⇒ 15x + 8y = 180
1
2x +8y +8x + 4y = 180 … cyclic quadrilateral
10x + 12y = 180
2
30x + 16y = 360
1 ×2
10x + 12(9) = 180
30x + 36y = 540
2 ×2
10x + 108 = 180
–20y = –180
y=9
26
Active Maths 2 (Strands 1–5): Ch 18 Solutions
10x = 72
x = 7.2
(b)
(i) (2y + 2)2 + (4y)2 = (5y)2 … right-angled Δ
⇒ 4y2 + 8y + 4 + 16y2 = 25y2
⇒ 5y2 – 8y – 4 = 0
(5y + 2)(y – 2) = 0
⇒ 5y + 2 = 0
2
⇒ y ≠ –__
5
⇒ Height of flagpole = 4y = 8
or
y–2=0
y=2
(ii) (2x + 1)2 = (2x)2 + (2x – 1)2
⇒ 4x2 + 4x + 1 = 4x2 + 4x2 – 4x + 1
⇒ 4x2 – 8x = 0
⇒x≠0
x=2
∴ Height = 2x – 1 = 3
1
(iii) No, as 3 ≠ __(8)
2
Q. 4. (a)
(i) |∠B| = 50° … isosceles Δ
|∠A| = 180° – (50° + 50°) = 80° … angles in a Δ
|∠C| = 50° + 80° = 130° … exterior angle
(ii) |∠A| = 42° … corresponding
|∠B| = 180° – 56° = 124° … corresponding angle and straight angle
|∠C| = 124° … alternate
(iii) |∠B| = 180° – 62° = 118° … straight angle
1
Angles in isosceles Δ = __(180° – 30°) = 75°
2
⇒ |∠A| = 180° – 75° = 105°
|∠C| = 360° – (105° + 105° + 62°) … quadrilateral + vertically opposite
angle
= 88°
(iv) |∠B| + 75° = 180° … parallelogram
⇒ |∠B| = 105°
|∠C| + 65° = 105° … opposite
⇒ |∠C| = 40°
|∠A| = |∠C| … alternate
⇒ |∠A| = 40°
(v) |∠A| = 180° – 114° = 66° … straight line
|∠B| = 180° – 2(66°) = 48° … angles in a Δ + vertically opposite
|∠C| = 90° … straight angle and square
Active Maths 2 (Strands 1–5): Ch 18 Solutions
27
(b)
(i)
A
A
3
2
B
C
D
3
4
E
(ii) |∠ABC| = |∠ADE| … corresponding
|∠ACB| = |∠AEC| … corresponding
|∠BAC| is common
⇒ ΔABC and ΔADE are equiangular.
|EC| 1
(iii) _____ = __
3
2
3
⇒ |EC| = __ = 1.5
2
|BC| 3
(iv) _____ = __
4
2
12
⇒ |BC| = ___ = 6
2
Q. 5. (a)
9
x
(i) __ = __
7 6
63
⇒ x = ___ = 10.5
6
8
x
(ii) ____ = ___
6.25 10
50
⇒ x = ___ = 5
10
x
14
(iii) ___ = ___
8
12
168
⇒ x = ____ = 21
8
10
x
(iv) __ = ___
8
6
(b)
h
___
15
y
___
12
72
⇒ y = ___ = 8
9
y
8
______
= ___
13.75 10
110
⇒ y = ____ = 11
10
y ___
14
__
=
3
6
42
⇒ y = ___ = 7
6
y + 6 ___
10
_____
=
⇒
6
10
80
⇒ x = ___
6
2
⇒ y + 6 = 16__
3
1
x = 13__
3
2
y = 10__
3
5
= ___
25
75
⇒ h =___ = 3 m
25
28
6
= __
9
Active Maths 2 (Strands 1–5): Ch 18 Solutions
100
y + 6 = ____
6
Q. 6.
(a)
5
8
(i) Is __ = ___?
2 2.4
2.5 = 3.3?
No
∴ BC || DE
3
8
(ii) Is __ = _____?
5
1
13__
3
Is 0.6 = 0.6? Yes
∴ RQ || TS
(b)
7.5
12
(i) Is ____ = ____?
31.5 50.4
5
5
___
= ___? Yes
21 21
∴ Triangles are similar.
10
2
(ii) Is ___ = ___?
11 15
2
2
___
= __? No
11 3
∴ Δs are not similar.
(c)
(i) Question 1 is incorrect as Δs are congruent and therefore angles marked 50°
and 55° should be equal.
(ii) Question 2:
A = 60 − 26 = 34°.
Q. 7. (a)
1
(i) |∠B| = __(180° – 42°) … radii + angles in a Δ
2
1
= __(138°) = 69°
2
|∠A| = 90° – 69° = 21° … angles in a semicircle and isosceles Δ
(ii) |∠A| = 180° – 2(41°) = 98° … angles in a Δ and isosceles Δ.
1
Top angle = __(98°) = 49° … angle at the centre
2
49º
B
A
B
41º
20º
41º
|∠B| + 20° = 49°
|∠B| = 29°
(iii) 180° – 70° = 110°
1
|∠A| + 20° = __(110°) … isosceles Δ
2
Active Maths 2 (Strands 1–5): Ch 18 Solutions
29
|∠A| + 20° = 55°
|∠A| = 35°
|∠B| + 35° + 70° = 180°
|∠B| + 105° = 180°
|∠B| = 75°
(iv)
A
1
51º O
B
15º
|∠1| = 180° – 51° = 129° … straight angle
1
|∠A| = __(180° – 129°) = 25.5° … isosceles Δ
2
Similarly |∠B| = 2(15°) = 30°
10
x
(b) ___ = ___
4
1.6
( )
10
x = 1.6 ___
4
=4m
Q. 8.
(a)
(i) x2 +72 = 252
⇒ x2 + 49 = 625
⇒ x2 = 576
72 + (x + 5)2 = y2
⇒ 72 + (29)2 = y2
⇒ 49 + 841 = y2
⇒ x = 24
⇒ y2 = 890
____
⇒ y = √890
___
y2 + 4.52 = ( √ 40 )
(ii) x2 + 32 = 72
⇒ x2 + 9 = 49
⇒ y2 + 20.25 = 40
⇒ x2 = 40
⇒ y = √ 19.75
______
___
⇒ x = √ 40
___
(iii) x2 + 52 = ( √41 )
(2x)2 + 52 = y2
⇒ x2 + 25 = 41
⇒ 82 + 52 = y2
⇒ x2 = 16
⇒ 64 + 25 = y2
⇒x= 4
⇒ 89 = y2
2
___
⇒ y = √ 89
30
2
Active Maths 2 (Strands 1–5): Ch 18 Solutions
(iv) x2 = 22 + 22
⇒ x2 = 4 + 4 = 8
Label hypotenuse in middle
triangle as z:
__
⇒ x = √8
z2 = 22 + x2
⇒ z2 = 4 + 8
___
⇒ z = √ 12
⇒ 22 + z2 = y2
⇒ 4 + 12 = y2
⇒ y2 = 16
⇒y= 4
(b)
650
h
m
1
rise
1
Slope = ___ = __
run
5
∴ Ratio = h : 5h
Using Pythagoras’ Theorem:
(h)2 + (5h)2 = (650)2
h2 + 25h2 = 422,500
26h2 = 422,500
h2 = 16,250
h = 127.475
⇒ h ≈ 127 m
Q. 9.
(a)
(i) No, as no two strips are equal
(ii) No, as opposite sides of a parallelogram are equal and no two strips are equal.
(iii)
25
7
24
In a right-angled triangle, Pythagoras’ Theorem holds.
Is 252 = 242 + 72?
625 = 576 + 49?
625 = 625 ?
True
(iv)
24
h
20
Active Maths 2 (Strands 1–5): Ch 18 Solutions
31
Missing hypotenuse
h2 = 242 + 202
h2 = 576 + 400
h2 = 976
____
h = √ 976
h = 31.24 cm
(b)
D
x
200 m
E
12
0
m
80 m
A
B
C
(i) ΔABE is similar to ΔACD
∴ we could use the similar triangles theorem to calculate ED as
120
_______
80
= ____.
120 + x 200
(ii) 80(120 + x) = 200(120)
9600 + 80x = 24,000
80x = 14,400
x = 180 m
Q. 10. (a)
r = 35
x
28
28
x2 = 352 − 282
x2 = 441
x = 21
∴ Depth of oil = 35 − 21 = 14 cm
(b)
(i) 40 × 2.54 = 101.6 cm
(ii) (9x)2 + (16x)2 = 101.62
81x2 + 256x2 = 10,322.56
337x2 = 10,322.56
32
Active Maths 2 (Strands 1–5): Ch 18 Solutions
x2 = 30.63
x = 5.534
⇒ Length = 16 x 5.534 = 88.55 = 89 to nearest cm
⇒ Height = 9 x 5.534 = 49.81 = 50 to nearest cm
(iii) (4x)2 + (3x)2 = 101.62
16x2 + 9x2 = 10,322.56
25x2 = 10,322.56
x2 = 412.9024
x = 20.32
⇒ Length = 4 x 20.32 = 81.28 = 81 to nearest cm
⇒ Height = 3 x 20.32 = 60.96 = 61 to nearest cm
(81 x 61) – (89 x 50) = 491cm2
The area of the second TV screen (units aspect ratio 4 : 3) is 491 cm2 larger.
Q. 11. (a)
(i)
OR
any (rectangle)
(ii)
(any equilateral triangle)
(iii)
(any square)
(b) A - central symmetry
B - axial symmetry
C - translation
D - axial symmetry
(c) 2a + 73° + 60° = 180°
2a + 133° = 180°
2a = 47°
a = 23.5°
Active Maths 2 (Strands 1–5): Ch 18 Solutions
33
a + b + 73° = 180°
23.5° + b + 73° = 180°
b + 96.5° = 180°
b = 83.5°
l1 || l2 ⇒ a and g are alternate
∴g=a
g = 23.5°
(d)
(i) |AB|2 = 112 + 22
|AB|2 = 121 + 4
|AB|2 = 125
____
__
|AB| = √ 125 = 5√ 5
(ii) |AD| = 2|BD|
__
∴ x2 + (2x)2 = (5√ 5 )2
x2 + 4x2 = 125
5x2 = 125
x2 = 25
x=5
34
Active Maths 2 (Strands 1–5): Ch 18 Solutions