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Active Maths 2 (Strands 1–5): Ch 18 Solutions
Chapter 18 Exercise 18.1
1
Q. 1. (i) 180° − 37° = 143° (x = 143°)
(ii) 180° − 117° = 63°
(x = 63°)
180° − 90° = 90°
(y = 90°)
(iii) 2x + x + 3 __ 2 x + 45 = 180
4.5x = 135
x = 30°
(iv) 180 − 90 = 2y
90 = 2y
45° = y
66 + (x + y) + 47 = 180
x + y = 67
x = 67 − y
= 67 − 45
x = 22°
Answer: x + y = 67°
2y = 90°
Q. 2. (i) A = 53° Opposite angles
B = 180 − 53
= 127° Interior angles
C = 53° Alternate angles
(ii) A = 180 − 112
= 68° (Interior angles)
C = 180 − 134 = 46° (Straight line)
B = 180 − 46
= 134° (Interior angles)
Answer: A = 68° B = 134° C = 46°
(iii) A = 73° Corresponding angles
B = 41° Opposite angles
C = 180 − 73 − 41
C = 66° Straight line
D = 66° Corresponding angles
E = 180 − 66
E = 114° Straight line
(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)
Q. 3.
(i) A line is a straight line that goes on forever in both directions; it has no endpoints.
A ray is part of a line that has one endpoint; the other end goes on forever.
(ii) Points that lie on the same plane are coplanar.
Points that lie on the same line are collinear.
(iii) A acute angle is one that is less than 90°.
An ordinary angle is one that is less than 180°.
(iv) An axiom is a statement we accept as true even though there is no proof.
A theorem is a statement we accept as true, as there is a proof.
Q. 4. (i) A = 180 − 43 = 137°
B = 180 − 43 − 83 = 54°
(ii) 2A = 180 − 63 = 117
⇒ A = 117 ÷ 2 = 58.5°
C = 180 − 51 − 58.5 = 70.5
B = 180 − 71 − 70.5 = 38.5
2 Active Maths 2 (Strands 1–5): Ch 18 Solutions
(iii) 2C = 180 − 50 = 130 C = 130 ÷ 2 = 65 B = 180 − 46 − 65 B = 69 A = B = 69 A = 69° B = 69° C = 65° (iv) A = 180 − 52 − 81 A = 47° C = 52° (Corresponding) B = 180 − 47 − 52 = 81°
Q. 5. (i) A = 23° (Alternate angles) B = 58° (Opposite angles/rules of
parallelogram) C: 2(C + 23) + 2(58) = 360 2C + 46 = 360 − 116 2C = 244 − 46 2C = 198 C = 99° OR
(C + 23) + 58 = 180 C + 23 = 122 C = 122 − 23 C = 99° (ii) A = 84° (Corresponding) B = A = 84° (Opposite) (C + 14) + B = 180 C + 14 + 84 = 180 C = 180 − 98 C = 82° (iii) A = 31° Alternate
C
yx
31°
31°
= 31° (isosceles
triangle)
by alternate
angles x = 31
Similarly y = c (by isosceles)
⇒ 180 = 31 + x + C + y
180 = 62 + 2C
118 = 2C
59° = C
B = 180 − C − 31
= 180 − 59 − 31
B = 90°
(iv) AC
yxB
102°
A = 180 − 102
A = 78°
by isosceles triangles
C = 180 − 102 = 78°
y = 180 − 2C
= 180 − 2(78)
y = 24
x = 180 − 102 = 78
B = 180 − 78 − 24
B = 78°
Q. 6. (i) By opposite angles A = 49°
(ii) B = 180 − 49 = 131°
(iii) 2C + 131 = 180
2C = 49
C = 24.5°
Q. 7. (ii) ∠3 and ∠7 are corresponding
(iii) ∠8 and ∠7 make up 180° on a straight line
(iv) ∠5 and ∠4 Alternate angles
3Active Maths 2 (Strands 1–5): Ch 18 Solutions
Q. 8. (a) (i) Triangle ABC and triangle FDE are congruent by SSS (all sides on ABC are the same as those on FDE).
(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: TS = QT and PT is common
but PQ ≠ PS and no angles are similar.
(iii) Congruent: BF = FE
(Since CB = BA)
BA = ED |∠DFE| = |∠BFA| (opposite)
|∠FDE| = |∠FAB| (alternating)
|∠FED| = |∠FBA| ⇒ By SAS or ASA (iv) ΔABE is congruent to ΔADE
AE is common to both AD = AB BE = ED (as E is the bisector of BD)
⇒ by SSS [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 ≡ ΔCEDQ. 9.
110°
110°
opposite
opposite
86°
94°
70°
apnml
b
70°70°
110°
180 – 110
= 70180 – 110 = 70 180 – 86
= 94°
180 – 70
= 110°
180 – 70
= 110°
94° 86°
Straight line
86°
Opposite
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.
Q. 10.
(i) A rhombus
(all sides equal)
(ii) Three
An equilateral triangle
(iii) Four
A square (all sides equal)
Alternative method:
(i) Rectangle has 2 symmetries
(ii) Triangle has 3 symmetries
(iii) Square has 4 symmetries
4 Active Maths 2 (Strands 1–5): Ch 18 Solutions
Q. 11. (i)
C
(ii)
No centre of symmetry (iii)
C
(iv)
No centre of symmetry (v)
No centre of symmetry
(vi) sC
No axis of symmetry
Q. 12. (i) A = Translation
B = Central symmetry
C = Axial symmetry
(ii) A = Central symmetry
B = Axial symmetry
C = Translation
(iii) A = Axial symmetry
B = Translation
C = Central symmetry
Q. 13. (a) 5y
x
4
3
2
1
0–1–1
–2
–3
–4
1 2 3 4–2
(i) (ii)
(iii)
–3–4
A
(b) (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)
Q. 14. (a) 4
3
2
(i)
(ii)
(iii)
1
0–1–1
–2
–3
–4
1 2 3 4 5–2–3–4–5
B
x
y
(b) (i) (−2,1), (−5,1), (−4,3)
(ii) (−2,−1), (−5,−1), (−4,−3)
(iii) (0,1), (3,1), (2,3)
5Active Maths 2 (Strands 1–5): Ch 18 Solutions
Q. 15. (a)
4
5
3
2
1
0–1–1
–2
–3
–4
–5
1 2 3 4 5–2–3
(i)
(ii)
(iii)
–4–5
C
y
x
(b) (i) (−1,1), (−3,1), (−4,3), (−2,3)
(ii) (−1,3), (1,3), (2,5), (0,5)
(iii) (1,−1), (3,−1), (4,−3), (2,−3)
Q. 16.
A
p
BC
r
–6 –5
(ii)
(iii)
(i)
–4 –3 –2
q
–1 10
0 2 3 4 5 6 7 8 9x
y
–5–6
–4–3–2–1
12345
(b) (i) (4,−1), (6,0), (8,−2)
(ii) (−2,3), (−4,2), (−6,4)
(iii) (−1,−2), (0,−4), (−2,−6)
Q. 17. (i) |QP| = |RS| = 20 (ii) |PS| = |QP| = 16 (iii) |MQ| = |MS| = 15 (iv) |SQ| = |MS| + |MQ| = 30
Q. 18. (i) x + 30 = x + y + 6 30 = y + 6 y = 24° (x + 30) + (x + 2y + 16) = 180 2x + 2y + 46 = 180 2x + 48 + 46 = 180 2x = 180 − 94 2x = 86 x = 43° (ii) x + y + 33 = 5x − 4y − 13 −4x + 5y = −46 simultaneous eq. (5x − 4y − 13) + (2x + 3y + 4)
= 180 7x − y = 189 simultaneous eq.
−4x + 5y = −46 35x − 5y = 945 31x = 899 x = 29° y = 7x − 189 = 203 − 189 y = 14°
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
a + 2b = 13 a − b = −2
−a + b = 2 a − 5 = −2
3b = 15 a = −2 + 5
b = 5 a = 3
a = 3, b = 5
(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°
6 Active Maths 2 (Strands 1–5): Ch 18 Solutions
Exercise 18.2Q. 1. (i) x = 7.5 … theorem 11
(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
⇒ x = 7 cm
(ii) |DE| + |GH| = 40 cm
⇒ (x + x) + (x + x) = 40 cm … theorem 11
⇒ 4x = 40
⇒ x = 10 cm
Q. 4. (i)
34
X U
T
|XU| = √_______
16 − 9 = √__
7
(ii)
68
VY
T
|YV| = √________
64 − 36 = √___
28 (OR 2 √__
7 )
(iii)
912
WZ
T
|TW| = 12 (by similar triangles), so
|ZW| = √_________
144 − 81
= √___
63 (OR 3 √__
7 )
Exercise 18.3
Q. 1. (i) bottom length
_____________ top length
= bottom length
_____________ top length
⇒ x __ 4 = 1 __ 2
⇒ x = 2
(ii) x __ 4 = 1 __ 3
⇒ x = 4 __ 3 = 1 1 __ 3
(iii) x __ 3 = 3 __ 2
⇒ x= 9 __ 2
⇒ x = 4.5
(iv) top length
__________ total length
= top length
__________ total length
⇒ 7 ___ 10 = x ___ 11
⇒ x = 77 ___ 10 = 7.7
Q. 2. (i) bottom length
_____________ total length
= bottom length
_____________ total length
⇒ y __ 4 = 3 __ 5
⇒ y = 12 ___ 5 = 2.4
(ii) y _____ 3.75 = 3 __ 5
⇒ y = 11.25 ______ 5 = 2.25
(iii) y __ 6 = 1 __ 5
⇒ y = 6 __ 5 = 1.2
(iv) y __ 7 = 6 ___ 16
⇒ y = 42 ___ 16
⇒ y = 21 ___ 8 = 2 5 __ 8 = 2.625
7Active Maths 2 (Strands 1–5): Ch 18 Solutions
Q. 3. (i) |PS|
_____ 22.5 = 12 ___ 15
⇒ |PS| = 270 ____ 15 = 18
(ii) |ST| = 22.5 – 18 = 4.5
(iii) |PS| : |ST| = 18 : 4.5 = 4 : 1
Q. 4. (i) |AB| : |BD|
= |AD| – |BD| : |BD|
= 5 – 2 : 2
= 3 : 2
(ii) |CE|
_____ 15 = 2 __ 3
⇒ |CE| = 30 ___ 3 = 10
No: only the ratios are given, not the absolute values
Q. 5. (i) |DE|
______ 23.75 = |CD|
______ 9.5 = 9.5 – 6 _______ 9.5 = 3.5 ___ 9.5
⇒ |DE| = 83.125 _______ 9.5 = 8.75
(ii) |FG|
______ 23.75 = 2 ___ 9.5
⇒ |FG| = 47.5 _____ 9.5 = 5
Q. 6. (i) 5 : 2
(ii) |AC| : |AE| = |AB| : |AD| = 7 : 5
(iii) |EC| : |AC| = |DB| : |AB| = 2 : 7
Q. 7. (i) 3 __ 4
(ii) |AY|
_____ |YC|
= |AX|
_____ |XB|
= 3 __ 4
(iii) |YC|
_____ |AC|
= |XB|
_____ |AB|
= 4 __ 7
Q. 8. |AB| : |BC| = 2.25 : 0.75 = 3 : 1 = |AD| : |AC|
⇒ BD || CE. Answer: Yes
Q. 9. |AY| : |YC| = 3.5 : 2.5 = 7 : 5
|AX| : |BX| = 3 : 2
⇒ |AY| : |YC| ≠ |AX| : |BX|
⇒ XY ||Y |||||||||||||| BC
i.e. not parallel. Answer: No
Q. 10.
Ex
A
D
425250
325
CB
(i) x ____ 250 = 325 ____ 425
425x = 81250
x = 191.176.....
x = 191 m
Total distance = 250 + 191 = 441m
(ii) 441 ____ 30 = 14.7
= 14 minutes 42 seconds
Exercise 18.4 Q. 1. (i) x __ 3 = 1 __ 2
⇒ x = 3 __ 2 = 1.5
(ii) x __ 6 = 8 ___ 10
⇒ x = 48 ___ 10 = 4.8
(iii) x ___ 10 = 3 __ 4
⇒ x = 30 ___ 4 = 7.5
(iv) x ___ 10 = 21 ___ 9
⇒ x = 210 ____ 9 = 23 1 __ 3
Q. 2. (i) y __ 4 = 4 __ 5
⇒ y = 16 ___ 5 = 3.2
8 Active Maths 2 (Strands 1–5): Ch 18 Solutions
(ii) y __ 9 = 20 ___ 15
⇒ y = 180 ____ 15 = 12
(iii) y ___ 21 = 12 ___ 9
⇒ y = 252 ____ 9 = 28
(iv) y ___ 42 = 44 ___ 66 = 2 __ 3
⇒ y ___ 42 = 2 __ 3
⇒ y = 84 ___ 3 = 28
Q. 3. (i) x ___ 12 = 4 __ 6
⇒ x = 48 ___ 6 = 8
y __ 3 = 4 __ 2
y __ 3 = 2 ⇒ y = 6
(ii) x __ 8 = 24 ___ 10
⇒ x = 192 ____ 10 = 19.2
16.8 – y
________ 16.8
= 10 ___ 24
⇒ 403.2 – 24y = 168 ⇒ 235.2 = 24y ⇒ y = 9.8
(iii) x ___ 4.5 = x _______ x + 3.5
⇒ 1 ___ 4.5 = 1 _______ x + 3.5
⇒ 4.5 = x + 3.5
⇒ x = 1
y __ 2 = 4.5 ___ 1
⇒ y = 9
(iv) x __ 4 = 7.5 ___ 3
⇒ x = 30 ___ 3 = 10
y ___ 20 = 4 ___ 10
⇒ y = 80 ___ 10 = 8
Q. 4. (i) |∠RST| = |∠PQR| = 90°
|∠SRT| is common
⇒ |∠STR| = |∠RPQ|
∴ ∆RST and ∆PQR are similar
(ii) T
S R
106
12x
P
Q R
Let |PR| = x
x ___ 10
= 12 ___ 6 ⇒ x ___ 10 = 2 ⇒ x = 20
|PR| = 20
(iii) Let |RQ| = y 202 = 122 + y2
400 = 144 + y2
256 = y2
16 = y |RQ| = 16
(iv) Let |RS| = z
z ___ 16
= 10 ___ 20 ⇒ z ___ 16 = 1 __ 2 ⇒ z = 8
|RS| = 8
Q. 5. (i) |EG|
_____ 3 = 1 __ 4
⇒ |EG| = 0.75
(ii) |EF| = 7 __ 8 (4) = 3.5
⇒ |GH|
______ 3.5 = 5 __ 4
⇒ |GH| = 17.5 _____ 4 = 4.375
Q. 6. (i) |BC|2 + 62 = 102
|BC|2 + 36 = 100
|BC|2 = 64
|BC| = 8
9Active Maths 2 (Strands 1–5): Ch 18 Solutions
(ii) |ED|
_____ 6 = 4 ___ 10
⇒ |ED| = 24 ___ 10 = 2.4
(iii) |ED|2 + |DC|2 = |EC|2
(2.4)2 + |DC|2 = (4)2
5.76 + |DC|2 = 16
|DC|2 = 10.24
|DC| = 3.2
(iv) area ΔABC : area ΔEDC
1 __ 2 (8)(6) : 1 __ 2 (3.2)(2.4)
24 : 3.84
6.25 : 1
25 ___ 4 : 1
25 : 4
Q. 7. (i)
22
12
8
D
E A
4B
C A
Is ΔABC similar to ΔADE?
|AC|
_____ |AE|
= 12 ___ 22 = 6 ___ 11 .
|BC|
_____ |DE|
= 4 __ 8 = 1 __ 2
As |AC|
_____ |AE|
≠ |BC|
_____ |DE|
No: ΔABC is not similar to ΔADE
(ii)
12.6 6.4
Q R
P
3 2
P
S T
Is ΔPST similar to ΔPQR?
|PQ|
_____ |PS|
= 12.6 ____ 3
= 4.2
|PR|
_____ |PT|
= 6.4 ___ 2
= 3.2
As |PQ|
_____ |PS|
≠ |PS|
_____ |PT|
i.e. the lengths of matching sides are not in proportion
∴ The triangles are not similar.
No: ΔPST is not similar to ΔPQR
Q. 8. 15
11.25
21
x
y
z
y ___ 21 = 11.25 ______ 15 (×21)
y = 21 × 11.25 ___________ 15
y = 15.75
Q. 9.
1
0.75 m 2.1 mbuildingpole
h
h = height of building
h __ 1 = 2.1 ____ 0.75 ∴ h = 2.8 m
The building is 2.8 m tall.
10 Active Maths 2 (Strands 1–5): Ch 18 Solutions
Q. 10.
2.510.5 m
1.72
h
h = height of tree in metres.
h ____ 1.72 = 10.5 ____ 2.5 (×1.72)
h = 10.5 × 1.72 _______________ 2.5
h = 7.224 m
The tree is 7.224 m tall
Q. 11.
15 cm
12 cm
Model
R
P
Q
3 m
(300 cm)
Actual
R ′
P′
Q ′
|P’R’|
______ 12 = 300 ____ 15
|P’R’| = 12 × 300 _________ 15
|P’R’| = 240 cm
= 2.4 m
The length of PR on the actual frame is 2.4 m.
Q. 12. (i)
1.2 m
1.45 mgirl
4.2 m
Flagpole
h
(ii) h ___ 5.4 = 1.45 ____ 1.2 (×5.4)
h = 1.45 × 5.4 __________ 1.2
h = 6.525 m
h = 652.5 cm
The flagpole is 653 cm high to the nearest cm.
Q. 13. h
22 m
1.45 m
5 m
h ____ 1.45 = 22 ___ 5
h = 22 × 1.45 _________ 5
h = 6.38 m
The school building is 6.38 m high.
Q. 14.
25 m
12 m
x ___ 25 = 28 ___ 12
∴ x = 28 × 25 ________ 12
x = 58 1 __ 3 m
The river is 58 1 __ 3 m wide
x m
28 m
11Active Maths 2 (Strands 1–5): Ch 18 Solutions
Q. 15. (i) 0.3
2 m1.7 m
1 m
1.7 m 1.7 m60 m
lh
The height of the kite = h + 1.7m
h ___ 0.3 = 60 ___ 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
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)
7 km 2 kmShop
Park
HomeSchool
yx
7 kmy
x
2 km
x
x = distance between shop and park
x __ 2 = 7 __ x
∴ 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, y2 = 72 + ( √
___ 14 )2
y2 = 49 + 14 y2 = 63 y = √
___ 63
y = 7.9372... km × 1000 y = 7937.2... m
The distance between the school and park is 7,937 m to the nearest m
Q. 17. (i)
12 m38 m HouseJames
1.85 m 2.15
Sign
x
12 Active Maths 2 (Strands 1–5): Ch 18 Solutions
(ii) 12 m
2.15 – 1.85 = 0.3 m
38 + 12 = 50 m
x
x ___ 0.3 = 50 ___ 12
x = 50 × 0.3 ________ 12
x = 1.25 m
Height of house = 1.25 m + 1.85 m
= 3.1 m
The house is 3.1 m high.
Q. 18. (i) A, B and C must be collinear also A, E and D must be collinear, [BE] || [CD]
(ii)
x
4857
133 m
A
B
C
D
E
Let x = distance between the two trees.
x
57
A
B E
x + 48
133 m
A
C
D
x ______ x + 48 = 57 ____ 133
133x = 57(x + 48) 133x = 57x + 2736 76x = 2736 x = 36 The distance between the trees is
36 m. (iii)
40
A
B E
36
36 +
9 =
45
y
A
CD
let y be the distance between C and D
∴ y ___ 40 = 45 ___ 36
y = 45 × 40 ________ 36
∴ y = 50 m
(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
|CB| = |FE| = 48 m and
|BE| = |CF| = 57 m
Exercise 18.5Q. 1. (i) h2 = a2 + b2
h = 13, a = 12, b = x
⇒ 132 = 122 + x2
⇒ 169 = 144 + x2
⇒ x2 = 25
⇒ x = 5
(ii) h2 = a2 + b2
h = x, a = 12, b = 16
⇒ x2 = 122 + 162
⇒ x2 = 144 + 256
⇒ x2 = 400
⇒ x = 20
13Active Maths 2 (Strands 1–5): Ch 18 Solutions
(iii) h2 = a2 + b2
h = x, a = √___
11 , b = 5
⇒ x2 = ( √___
11 ) 2 + 52
⇒ x2 = 11 + 25
⇒ x2 = 36
⇒ x = 6
(iv) h2 = a2 + b2
h = 25, a = 24, b = x
⇒ 252 = 242 + x2
⇒ 625 = 576 + x2
⇒ x2 = 49
⇒ x = 7
(v) h2 = a2 + b2
h = x, a = √___
15 , b = 7
⇒ x2 = ( √___
15 ) 2 + 72
⇒ x2 = 15 + 49
⇒ x2 = 64
⇒ x = 8
(vi) x2 = 202 + 162
⇒ x2 = 400 + 256
⇒ x2 = 656
⇒ x = √____
656
(vii) 242 = 102 + x2
⇒ 576 = 100 + x2
⇒ 476 = x2
⇒ x = 2 √____
119
(viii) x2 = 22 + ( 2 √__
3 ) 2
⇒ 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
(ii) x2 = 92 + 122
⇒ x2 = 81 + 144 = 225
⇒ x = 15
172 = 152 + y2
⇒ 289 = 225 + y2
⇒ 64 = y2
⇒ y = 8 (iii) 32 = x2 + 52
⇒ 9 + x2 = 25 ⇒ x2 = 16 ⇒ x = 4 y2 = 42 + ( √
___ 33 ) 2
⇒ y2 = 16 + 33 = 49 ⇒ y = 7 (iv) x2 = 12 + 22
⇒ x2 = 1 + 4 = 5 ⇒ x = √
__ 5
y2 = ( √__
5 ) 2 + 22
y2 = 5 + 4 = 9 ⇒ y = 3
Q. 3. (i) x2 = 102 + 122
x2 = 100 + 144
x2 = 244
x = √____
244
x = 15.620...
x = 15.6 to 3 significant figures
(ii)
8
8
16
16
x
x2 = 82 + 162
x2 = 64 + 256
x2 = 320
x = √____
320
x = 17.888...
x = 17.9 to 3 significant figures
14 Active Maths 2 (Strands 1–5): Ch 18 Solutions
(iii) 8
3
11
y x
y2 = 82 + 32
y2 = 64 + 9
y2 = 73
y = √___
73
∴ x2 + ( √___
73 )2 = 112
x2 = 112 − ( √___
73 )2
x2 = 121 − 73
x2 = 48
x = √___
48
x = 6.928...
x = 6.93 to 3 significant figures
(iv)
5
7
3.5 3.5
x
x2 + 3.52 = 52
x2 = 52 − 3.52
x2 = 12.75
x = √______
12.75
x = 3.5707...
x = 3.57 to 3 significant figures
(v)
8
15
9
xy
y2 = 152 + 82
y2 = 289
y = √____
289
y = 17
172 = x2 + 92
289 = x2 + 81
208 = x2
√____
208 = x
∴ x = 14.4 (to three significant figures)
(vi)
1
1a
1st triangle
a2 = 12 + 12
a = √_______
12 + 12
a = √__
2
a
1b
2nd triangle.
b2 = 12 + a2
b2 = 12 + 12 + 12
b = √____________
12 + 12 + 12
b
1c
3rd triangle.
c2 = 12 + b2
c2 = 12 + 12 + 12 + 12
c = √_________________
12 + 12 + 12 + 12
15Active Maths 2 (Strands 1–5): Ch 18 Solutions
Following this pattern on the 7th triangle
x = √___________________________________
12 + 12 + 12 + 12 + 12 + 12 + 12 + 12
x = √__
8
x = 2.8284...
x = 2.83 to 3 significant figures.
Q. 4. (i) Triangle with sides 72, 70 and 21. 722 = 5,184 702 + 212 = 4,900 + 441 = 5,341 No: since 702 + 212 ≠ 722 this is
not a right-angled triangle (ii) Triangle with sides 8.9, 8 and 3.9 8.92 = 79.21 82 + 3.92 = 64 + 15.21 = 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
Longest side squared must equal the sum of the squares of the two shorter sides.
872 = 7,569
602 + 632 = 3,600 + 3,969
= 7,569
Yes: these lengths will form a right-angled triangle since 602 + 632 = 872
(ii) Side lengths 39, 80, 89
892 = 7,921
392 + 802 = 1,521 + 6,400
= 7,921
Yes: these lengths will form a right-angled triangle since
392 + 802 = 892
(iii) Side lengths 55, 130, 148
1482 = 21,904
552 + 1302 = 3,025 + 16,900
= 19,925
No: these lengths will not form a right-angled triangle since 552 + 1302 ≠ 1482
(iv) Side lengths 64, 120, 136
1362 = 18,496
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
B
45
40
75 – 45 = 30
75
40
|AB|2 = 302 + 402
|AB|2 = 2,500
|AB| = √______
2,500
|AB| = 50
The distance AB is 50.
16 Active Maths 2 (Strands 1–5): Ch 18 Solutions
(ii)
112
66
38
74
A
B
|AB|2 = 662 + 1122
= 16,900 |AB| = 130
Q. 7. (i) (2x)2 + (3x)2 = 402
4x2 + 9x2 = 1,600
13x2 = 1,600
x2 = 1,600
______ 13
x = √
______ 1,600 _______
√___
13
x = 11.094...
x = 11.09 to 2 d.p.
(ii) x2 + (2x + 8)2 = 522
x2 + 4x2 + 32x + 64 = 2,704 5x2 + 32x + 64 − 2,704 = 0 5x2 + 32x − 2,640 = 0
a = 5, b = 32, c = −2,640 b2 − 4ac = (32)2 − 4(5)(−2,640)
= 53,824
Using the quadratic formula:
x = −32 ± √
_______ 53,824 _______________ 2 × 5
x = −32 + √
_______ 53,824 _______________ 10
OR
x = −32 − √
_______ 53,824 _______________ 10
x = 20 OR x = −26.4
Since x > 0 reject x = −26.4
∴ x = 20
(iii) (x − 1)2 + (x − 2)2 = x2
x2 − 2x + 1 + x2 − 4x + 4 − x2 = 0
x2 − 6x + 5 = 0
(x − 1)(x − 5) = 0
x − 1 = 0 OR x − 5 = 0
x = 1 OR x = 5
if x = 1 then one of the sides
x − 1 = 1 − 1 = 0
Since we cannot have a side of zero length, x = 1 is rejected.
∴ x = 5
(iv) (x − 1)2 + (x + 1)2 = (x + 5)2
x2 − 2x + 1 + x2 + 2x + 1 = x2 + 10x + 25
2x2 + 2 − x2 − 10x − 25 = 0
x2 − 10x − 23 = 0
a = 1, b = −10, c = −23
b2 − 4ac = (−10)2 − 4(1)(−23)
= 100 + 92 = 192
Using the quadratic formula:
x = −(−10) ± √
____ 192 _______________ 2 × 1
x = 10 + √____
192 __________ 2
OR
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
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.
17Active Maths 2 (Strands 1–5): Ch 18 Solutions
(ii)
225 cm
90 cm
h
h = new height of ladder against the wall
h2 + 1902 = 2252
h2 = 2252 − 1902
h2 = 14,525
h = √_______
14,525
Slip = original height − new height
= 160 cm − √_______
14,525 cm
= 39.480... cm
∴ The ladder slipped 39 cm (to the nearest cm).
Q. 9. (i) (p − 2)2 + (p + 1)2 = (p + 4)2
p2 − 4p + 4 + p2 + 2p + 1 = p2 + 8p + 16
2p2 − 2p + 5 − p2 − 8p − 16 = 0
p2 − 10p − 11 = 0
(p + 1)(p − 11) = 0
∴ p + 1 = 0 OR p − 11 = 0
p = −1 OR p = 11
if p = −1 then p + 1 = −1 + 1
= 0
Cannot have side of zero length
∴ p = −1 is rejected
∴ p = 11
(ii) Show that
{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
2p2 − 4p + 10 − p2 − 12p − 36 = 0
p2 − 16p − 26 = 0.
There are no whole number roots to this equation (p has a decimal value on using the quadratic formula)
As p is not a whole number {p + 1, p + 6, p − 3} cannot be positive integers and ∴ do not form a Pythagorean triple.
Q. 10.
80 cm
39 cm
l
Diameter of drum = 2 × 39 = 78 cm l = length of wire ∴ l2 = 782 + 802
l2 = 12,484
l = √_______
12,484
l = 111.7|3...
The longest wire is 111.7 cm to 1 d.p.
Q. 11. 40 cm
1200 cm
x
x2 = 1,2002 + 402
x2 = 1,441,600
x = √__________
1,441,600
x = 1,200.66...
The insects are 1,201 cm apart to the nearest cm.
Q. 12.
3
33
y
xDiagonal
18 Active Maths 2 (Strands 1–5): Ch 18 Solutions
1st find the length of the diagonal x of the base of the cube
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. (i)
4.8 m
6 m w
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 = 16.8 m. Perimeter of garden is 16.8 m
(iii)
l
(3 + √5) m
(1 – 2√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
l2 = 2 √__
5 + 35 l = √
_________ 2 √
__ 5 + 35
l = 6.282... m l = 628.2... cm The pipe is 628 cm to the nearest
cm.
Q. 14. Cube of side lengths 12 cm.
= 120 mm
Radius of cone = 120 ÷ 2
= 60 mm
Height of cone = 120 mm
let l = slant height of cone
lh
r
r = 60 mmh = 120 mm
∴ 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
6 cm 3
6 cmx
18 cm
Let x = height of truncated cone
∴ x2 + 32 = 62
x2 = 62 − 32
x2 = 27
x = √___
27
27
3
6
9
h
19Active Maths 2 (Strands 1–5): Ch 18 Solutions
Two similar triangles in the cone where h = height of original cone.
∴ h ____ √
___ 27 = 9 __ 3
h = 9 × √___
27 ________ 3
h = 15.588... cm h = 155.88... mm Height of original cone is 156 mm to
nearest mm.
Q. 16. 58
5
3 m
8 m10 m
G P F
E
DA
B CH
(i) |AE|2 = 82 + 102
|AE|2 = 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) (iv) |AP|2 = ( √
___ 89 )2 + 32
|AP|2 = 89 + 9 |AP| = √
___ 98
|AP| = 9.899...m |AP| = 989.9... cm ∴ |AP| = 9.90 m (to nearest cm)
Q. 17. Note diagonals of a square bisect each other at right angles.
∴ x2 + x2 = 202
2x2 = 400 x2 = 200 x = √
____ 200
B
A
l
√200
30 cm(Pyramid height)
l = length of wire l2 = 302 + ( √
____ 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
Q. 18. A
B
C
H
12 km
18 km
8 km
5 km4 km9 km
(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.
x x
20
20 Active Maths 2 (Strands 1–5): Ch 18 Solutions
(ii) |BH|2 = 52 + 122
|BH|2 = 169 |BH| = √
____ 169
|BH| = 13 km ∴ Ship B is 13 km from the
harbour.
(iii) 18
– 1
2 =
6
9 + 5 = 14
A
B
|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.
(iv) |BC|2 = 12 + 202
|BC|2 = 401 |BC| = √
____ 401
|BC| = 20.02... km ∴ Ships B and C are
20 km apart to 2 significant figures.
(v)
18 + 8 = 26
9 + 4 = 13C
A
∴ |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.
5 – 4 = 1
12 + 8 = 20
C
B
Exercise 18.6Q. 1. (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
(iv) |∠A| = 1 __ 2 (110°) = 55° … angle
at the centre of a circle
(v) |∠A| = 1 __ 2 (66°) = 33° … angle
at the centre of a circle
(vi) |∠A| = 1 __ 2 (80°) = 40° … angle at
the centre of a circle
(vii) Smaller angle at 0 = 360° − 260° = 100°
∴ |∠A| = 1 __ 2 (100°)
= 50°
(viii) Larger angle at 0 = 360° − 80° = 280°
∴ |∠A| = 1 __ 2 (280°)
= 140°
21Active 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)
B
A
30° 30°
angle at a point
360° – BO
|∠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°
(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)
⇒ |∠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°
BO
BA
140°36°
36°
22 Active Maths 2 (Strands 1–5): Ch 18 Solutions
(v)
O
A
B
44°
same arc
angle in a semi circle
|∠A| = 180° – (90° + 44°) = 46° … angles in a ∆
|∠B| = 46° … angles standing on the same arc. (vi)
O
B A
30°
60°
60°standing on the
same arc
|∠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
∴ |∠OQR| = |∠ORQ|
as base angles of a Δ.
|∠ORU| = |∠ORS| = 90°
as US is a tangent
|∠OQT| = |∠OQS| as TS is a tangent.
∴ |∠OQS| = |∠ORS|
⇒ |∠C| = |∠QRS|
∴ ΔQRS is isosceles.
So |∠C| = 180 − 61° __________ 2 = 119 ____ 2 = 59.5°
|∠OQR| = 90° − 59.5° = 30.5°
∴ |∠B| = 180° − (2)(30.5°)
= 180° − 61°
= 119°
∴ |∠A| = 1 __ 2 (119°) = 59.5°
61°
A
B
C
R
U
QT
S
P
O
90°90°
90°90°
23Active Maths 2 (Strands 1–5): Ch 18 Solutions
(iv) |∠B| = 2|∠A|
|∠B| + 2|∠C| = 180°
⇒ 2|∠A| + 2|∠C| = 180°
⇒ |∠A| + |∠C| = 90°
|∠C| = 90° − |∠A|.
|∠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| + 13° + 38° + 2(90° − |∠A|) = 180°
|∠A| + 13° + 38° + 180° − 2|∠A| = 180°
13° + 38° − |∠A| = 0
13° + 38° = |∠A|
51° = |∠A|
∴ |∠B| = 2(51°) = 102°
34°
24°B
A
CC
O
13°38°B
CC
A
O
Q. 5. (i) Yes. As opposite angles add to 180°
i.e. 50° + 130° = 180°
and 85° + 95° = 180°
(ii) No. As opposite angles do not add to 180°
i.e. 80° + 90° = 170°
and 115° + 75° = 190°
(iii) No. As opposite angles don’t add to 180°
i.e. 120° + 50° = 170°
(iv) Yes. As opposite angles add to 180°
i.e. 108° + 72° = 180°
Q. 6. (i) | ∠ACD| = 38° (both angles at circle subtended by same arc)
(ii) |∠CDA| = 90° (angle subtended by diameter)
(iii) |∠ADB| = 180° − 38° __________ 2 = 142° _____ 2
= 71°
(iv) |∠DAC| = 180° − 90° − 38° = 52°
Q. 7. (i) |∠AOB|;
ΔAOE ≡ ΔBOE by RHS.
∴ |∠AOD| = |∠BOD|
|∠AOD| = 180° − 90° − 31° = 59°
⇒ |∠AOB| = 2(59°) = 118°
(ii) ΔOCA is isosceles as |OC| = |OA|
|∠COA| = 180° − 59° = 121°
∴ |∠ACO| = 180° − 121° ___________ 2
= 29.5°
24 Active Maths 2 (Strands 1–5): Ch 18 Solutions
(iii) As in (ii) ΔOCB is isosceles
|∠COB| = 121°
|∠OBC| = 180° − 121° ___________ 2 = 29.5°
(iv) ACBD is a cyclic quadrilateral
∴ |∠ADB| + |∠ACB| = 180°
|∠ACB| = 2(29.5°) = 59°
∴ |∠ADB| = 180° − 59° = 121°
Q. 8. (i) |∠QOS| = 2(58°) = 116°.
(ii) QPSR is a cyclic quadrilateral
∴ |∠QRS| = 180° − 58° = 122°
(iii) ΔOSQ is isosceles
∴ |∠OSQ| = 180° − 116° ___________ 2 = 32°
ΔRSQ is isosceles
∴ |∠RSQ| = 180° − 122° ___________ 2 = 29°
⇒ |∠RSO| = 32 + 29 = 61°
(iv) P, O, R are collinear
⇒ PR is a diameter as O is centre
⇒ |∠PQR| = 90°
⇒ |∠PQO| = 90° − |∠RQO|
= 90° − 61°
= 29°
Q. 9. 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.
Q. 10. Join the centre of the circle to each point of the star. This gives 5 equal angles which are 72° each.
A
A
Angle A is the angle at the circle being subtended by the same arc as the 72°
angle at the centre and is therefore
1 __ 2 (72°) = 36°.
Q. 11. Row
1
2
3
4
5
6
S T A G E
1 2 3 4 5 6 7Seats
Daniel could sit in Row 2, Seat 1 or Row 2, Seat 7 and have the same viewing angle.
Reasoning: two angles subtended by the same arc (stage), touching the circumference, are equal in measure.
Another possible answer: Row 5, seat 5.
Revision ExercisesQ. 1. (a) (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
25Active Maths 2 (Strands 1–5): Ch 18 Solutions
(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°
x
x
y
y
30º
115º
isosceles triangles
(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° 3(40°) + y = 180° ⇒ x = 40° ⇒ y = 60°
26 Active Maths 2 (Strands 1–5): Ch 18 Solutions
(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° (× 3) 3x + y = 158°⇒ –3x + 27y = 66° 3x + y = 158° 3x + 8° = 158° 28y = 224° ⇒ 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° ⇒ x = 59°
(b) (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 = 0x + 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 12x +8y +8x + 4y = 180 … cyclic quadrilateral
10x + 12y = 180 230x + 16y = 360 1 × 2 10x + 12(9) = 180
30x + 36y = 540 2 × 2 10x + 108 = 180
–20y = –180 10x = 72
y = 9 x = 7.2
27Active Maths 2 (Strands 1–5): Ch 18 Solutions
(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 or y – 2 = 0
⇒ y ≠ – 2 __ 5 y = 2
⇒ Height of flagpole = 4y = 8
(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
(iii) No, as 3 ≠ 1 __ 2 (8)
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
Angles in isosceles Δ = 1 __ 2 (180° – 30°) = 75°
⇒ |∠A| = 180° – 75° = 105°
|∠C| = 360° – (105° + 105° + 62°) … quadrilateral + vertically opposite = 88° angle
(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
28 Active Maths 2 (Strands 1–5): Ch 18 Solutions
(b) (i)
3
A
B C
2 3
4
A
D E
(ii) |∠ABC| = |∠ADE| … corresponding
|∠ACB| = |∠AEC| … corresponding
|∠BAC| is common
⇒ ΔABC and ΔADE are equiangular.
(iii) |EC|
_____ 3 = 1 __ 2
⇒ |EC| = 3 __ 2 = 1.5
(iv) |BC|
_____ 4 = 3 __ 2
⇒ |BC| = 12 ___ 2 = 6
Q. 5. (a) (i) x __ 7 = 9 __ 6 y ___ 12 = 6 __ 9
⇒ x = 63 ___ 6 = 10.5 ⇒ y = 72 ___ 9 = 8
(ii) x ____ 6.25 = 8 ___ 10 y ______ 13.75 = 8 ___ 10
⇒ x = 50 ___ 10 = 5 ⇒ y = 110 ____ 10 = 11
(iii) x ___ 12 = 14 ___ 8 y __ 3 = 14 ___ 6
⇒ x = 168 ____ 8 = 21 ⇒ y = 42 ___ 6 = 7
(iv) x __ 8 = 10 ___ 6 y + 6
_____ 10 = 10 ___ 6 ⇒ y + 6 = 100 ____ 6
⇒ x = 80 ___ 6 ⇒ y + 6 = 16 2 __ 3
x = 13 1 __ 3
y = 10 2 __ 3
(b) h ___ 15 = 5 ___ 25
⇒ h = 75 ___ 25 = 3 m
29Active Maths 2 (Strands 1–5): Ch 18 Solutions
Q. 6. (a) (i) Is 5 __ 2 = 8 ___ 2.4 ?
2.5 = 3.3? No
∴ BC || DE
(ii) Is 3 __ 5 = 8 _____ 13 1 __ 3
?
Is 0.6 = 0.6? Yes
∴ RQ || TS
(b) (i) Is 7.5 ____ 31.5 = 12 ____ 50.4 ?
5 ___ 21 = 5 ___ 21 ? Yes
∴ Triangles are similar.
(ii) Is 2 ___ 11 = 10 ___ 15 ?
2 ___ 11 = 2 __ 3 ? No
∴ 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) (i) |∠B| = 1 __ 2 (180° – 42°) … radii + angles in a Δ
= 1 __ 2 (138°) = 69°
|∠A| = 90° – 69° = 21° … angles in a semicircle and isosceles Δ
(ii) |∠A| = 180° – 2(41°) = 98° … angles in a Δ and isosceles Δ.
Top angle = 1 __ 2 (98°) = 49° … angle at the centre
AB
B
41º41º
20º
49º
|∠B| + 20° = 49°
|∠B| = 29°
(iii) 180° – 70° = 110°
|∠A| + 20° = 1 __ 2 (110°) … isosceles Δ
30 Active Maths 2 (Strands 1–5): Ch 18 Solutions
|∠A| + 20° = 55°
|∠A| = 35°
|∠B| + 35° + 70° = 180°
|∠B| + 105° = 180°
|∠B| = 75°
(iv)
OA
B51º
1
15º
|∠1| = 180° – 51° = 129° … straight angle
|∠A| = 1 __ 2 (180° – 129°) = 25.5° … isosceles Δ
Similarly |∠B| = 2(15°) = 30°
(b) x ___ 1.6 = 10 ___ 4
x = 1.6 ( 10 ___ 4 ) = 4 m
Q. 8. (a) (i) x2 +72 = 252 72 + (x + 5)2 = y2
⇒ x2 + 49 = 625 ⇒ 72 + (29)2 = y2
⇒ x2 = 576 ⇒ 49 + 841 = y2
⇒ x = 24 ⇒ y2 = 890 ⇒ y = √
____ 890
(ii) x2 + 32 = 72 y2 + 4.52 = ( √___
40 ) 2
⇒ x2 + 9 = 49 ⇒ y2 + 20.25 = 40
⇒ x2 = 40 ⇒ y = √______
19.75
⇒ x = √___
40
(iii) x2 + 52 = ( √___
41 ) 2 (2x)2 + 52 = y2
⇒ x2 + 25 = 41 ⇒ 82 + 52 = y2
⇒ x2 = 16 ⇒ 64 + 25 = y2
⇒ x = 4 ⇒ 89 = y2
⇒ y = √___
89
31Active Maths 2 (Strands 1–5): Ch 18 Solutions
(iv) x2 = 22 + 22 Label hypotenuse in middle ⇒ x2 = 4 + 4 = 8 triangle as z:
⇒ x = √__
8 z2 = 22 + x2
⇒ z2 = 4 + 8
⇒ z = √___
12
⇒ 22 + z2 = y2
⇒ 4 + 12 = y2
⇒ y2 = 16
⇒ y = 4
(b)
h650 m
1
Slope = rise ___ run = 1 __ 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)
24
725
In a right-angled triangle, Pythagoras’ Theorem holds.
Is 252 = 242 + 72?
625 = 576 + 49?
625 = 625 ?
True
(iv)
24
20
h
32 Active Maths 2 (Strands 1–5): Ch 18 Solutions
Missing hypotenuse
h2 = 242 + 202
h2 = 576 + 400
h2 = 976
h = √____
976
h = 31.24 cm
(b)
120 m
x
200 m
80 m
A B C
D
E
(i) ΔABE is similar to ΔACD
∴ we could use the similar triangles theorem to calculate ED as
120 _______ 120 + x = 80 ____ 200 .
(ii) 80(120 + x) = 200(120)
9600 + 80x = 24,000
80x = 14,400
x = 180 m
Q. 10. (a)
28 28x
r = 35
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
33Active 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°
34 Active Maths 2 (Strands 1–5): Ch 18 Solutions
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
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