Chapter 18 Exercise 18shevlinbiology.webs.com/ch 18 solutions.pdf · Active Maths 2 (Strands...

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


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)


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°


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


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


(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


(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.


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


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


(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


(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...



(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...


(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...


(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


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.


(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.


(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


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


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


(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°


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°


(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°


(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°


(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


(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°


(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


(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


(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


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


|∠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


(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


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


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°


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

Chapter 18 Exercise 18shevlinbiology.webs.com/ch 18 solutions.pdf · Active Maths 2 (Strands...

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Transcript of Chapter 18 Exercise 18shevlinbiology.webs.com/ch 18 solutions.pdf · Active Maths 2 (Strands...