CEMKS3 E8 GM1 1 WS ANS - Cambridge...

Cambridge Essentials Mathematics Extension 8 GM1.1 Answers

Original material © Cambridge University Press 2009 1

GM1.1 Answers

Reasons given for answers are examples only. In most cases there are valid alternatives.

1 a x = 45°; alternate angles are equal.

b y = 125°; alternate angles are equal.

c p = 130°; the sum of angles about a point on a straight line is 180°.

q = 130°; p, q are alternate angles ⇒ p = q.

r = 50°; q, r are angles about a point on a straight line ⇒ q + r = 180°.

d j = 112°; corresponding angles are equal.

k = 112°; j, k are alternate angles ⇒ j = k.

l = 112°; k, l are corresponding angles ⇒ k = l.

m = 112°; l, m are alternate angles ⇒ l = m.

n = 68°; j, n are angles about a point on a straight line ⇒ j + n = 180°.

p = 68°; l, p are angles about a point on a straight line ⇒ l + p = 180°.

2 a x = 30°; angle sum of = 180° ⇒ x + 50° + 100° = 180°.

b p = 20°; angle sum of = 180° ⇒ p + 120° + 40° = 180°.

c y = 35°; angle sum of = 180° ⇒ y + 90° + 55° = 180°.

d m = 66°; angle sum of = 180° ⇒ m + 88° + 26° = 180°.

3 a t = 58°; angle sum of = 180° ⇒ t + 32° + 90° = 180°.

b x = 68°; angle sum of = 180° and an isosceles has one pair of equal angles.

4 a q = 60°; angle sum of = 180° and all three angles of an equilateral are equal.

b r = 45°; angle sum of = 180° and an isosceles has one pair of equal angles.



5 a x = 40°; exterior ∠ of = sum of interior opposite angles ⇒ 140° = x + 100°.

y = 40°; angle sum of = 180° ⇒ y + 100° + 40° = 180°.

b m = 55°; sum of angles about a point on a straight line = 180°.

n = 75°; exterior ∠ of = sum of interior opposite angles ⇒ 125° = n + 50°.

c d = 25°; symmetry of an isosceles .

e = 130°; angle sum of = 180° ⇒ e + 25° + 25° = 180°.

f = 50°; exterior ∠ of = sum of interior opposite angles ⇒ f = 25° + 25°.

d v = 59°; angle sum of = 180° ⇒ v + 31° + 90° = 180°.

w = 31°; symmetry of an isosceles .

x = 59°; symmetry of an isosceles .

y = 121°; exterior ∠ of = sum of interior opposite angles ⇒ y = 31° + 90°.

6 a k = 117°; angle sum of quadrilateral = 360° ⇒ k + 95° + 60° + 88° = 360°.

b t = 55°; angle sum of quadrilateral = 360° ⇒ t + 90° + 90° + 125° = 360°.

c r = 135°; sum of angles about a point on a straight line = 180°.

s = 51°; angle sum of quadrilateral = 360° ⇒ s + 135° + 84° + 90° = 360°.

d z = 118°; angle sum of quadrilateral = 360° ⇒ z + 90° + 62° + 90° = 360°.

7 a w = 70°; alternate angles are equal.

x = 65°; alternate angles are equal.

y = 115°; sum of angles about a point on a straight line = 180°.

z = 110°; sum of angles about a point on a straight line = 180°.

b a = 127°; sum of angles about a point on a straight line = 180°.

b = 53°; alternate angles are equal.

c = 53°; b, c are corresponding angles.

d = 127°; c, d are angles about a point on a straight line.

e = 53°; corresponding angles are equal.



c q = 40°; alternate angles are equal and the sum of angles on a straight line is 180°.

r = 140°; vertically opposite angles are equal.

s = 140°; corresponding angles are equal.

t = 40°; angle sum of quadrilateral = 360°.

d a = 44°; vertically opposite angles are equal.

b = 46°; angle sum of = 180° ⇒ a + b = 90°.

c = 136°; a, c are angles about a point on a straight line.

d = 136°; c, d are alternate angles.

e = 46°; b, e are alternate angles.

f = 134°; e, f are angles about a point on a straight line.

e l = 115°; corresponding angles are equal

k = 25°; exterior ∠ of = sum of interior opposite angles ⇒ 115° = 90° + k ⇒ k = 25°

m = 65°; l, m are angles about a point on a straight line.

n = 65°; angle sum of = 180° ⇒ n + 25° + 90° = 180°.

f p = 72°; reflection symmetry of a rhombus.

q = 18°; angle sum of = 180° ⇒ q + 90° + 72° = 180°.

r = s = 54°; symmetry of isosceles ; angle sum of = 180° ⇒ r + s + 72° = 180°.

t = 54°; s, t are alternate angles.

u = 36°; angle sum of = 180° ⇒ u + 54° + 90° = 180°.

v = 36°; u, v are vertically opposite angles.

g j = 55°; angle sum of = 180° ⇒ j + 90° + 35° = 180°.

k = 90°; sum of angles about a point on a straight line = 180° ⇒ k + 90° = 180°.

l = 90°; angle sum of quadrilateral = 360° ⇒ l + 90° + 90° + 90° = 360°.

m = 35°; corresponding angles are equal.

n = 90°; sum of angles about a point on a straight line = 180° ⇒ n + 90° = 180°.

p = 125°; exterior ∠ of = sum of interior opposite angles ⇒ p = 90° + 35°.

h d = 35°; alternate angles are equal.

e = 35°; symmetry of an isosceles triangle.

f = 110°; angle sum of = 180° ⇒ f + 35° + 35° = 180°.



8 a Three triangles

b The sum of interior angles is 3 × 180° = 540°.

9 a Four triangles

b The sum of interior angles is 4 × 180° = 720°.

Regular polygon

Number of sides

Sum of interior angles

Size of each interior angle

Size of each exterior

angle

Sum of exterior angles

Equilateral triangle 3 1 × 180°

= 180° 180° ÷ 3

= 60° 180° − 60°

= 120° 120° × 3 = 360°

Square 4 2 × 180° = 360°

360° ÷ 4 = 90°

180° − 90° = 90°

90° × 4 = 360°

Pentagon 5 3 × 180° = 540°

540° ÷ 5 = 108°

180° − 108° = 72°

72° × 5 = 360°

Hexagon 6 4 × 180° = 720°

720° ÷ 6 = 120°

180° − 120° = 60°

60° × 6 = 360°

Heptagon 7 5 × 180° = 900°

900° ÷ 7 = 129°

180° − 129° = 51°

51° × 7 = 360°

Octagon 8 6 × 180° = 1080°

1080° ÷ 8 = 135°

180° − 135° = 45°

45° × 8 = 360°

Nonagon 9 7 × 180° = 1260°

1260° ÷ 9 = 140°

180° − 140° = 40°

40° × 9 = 360°

Decagon 10 8 × 180° = 1440°

1440° ÷ 10= 144°

180° − 144° = 36°

36° × 10 = 360°

10

Dodecagon 12 10 × 180° = 1800°

1800° ÷ 12= 150°

180° − 150° = 30°

30° × 12 = 360°

11 The sum of the interior angles of a (regular) n-sided polygon is (n − 2) × 180°. So each

interior angle is (n − 2) × 180° ÷ n and each exterior angle is 180° – (n − 2) × 180° ÷ n.

The sum of the exterior angles of any (regular) polygon is 360°.

12 a 2° b 180 sides

13 a 25 sides b a = 108°, b = 252° c 4140°



14 a i 144° ii Yes, it will form a decagon. iii 10

b i 120° ii Yes, it will form a hexagon. iii 6

c i 102° ii No, there is no regular polygon that has interior angles of 102°.

d i 90° ii Yes, it will form a square. iii 4



GM1.2 Answers

1 a C

b Rhombuses and squares.

c No, both rhombuses and squares have two pairs of parallel sides.

d No, all shapes with all sides the same length are in A.

e Yes, squares are a type of rectangle.

f No, all rectangles have two pairs of parallel sides.

g No, none of the shapes in B have all sides the same length.

h Parallelogram and rectangle.

2 a Square, rhombus

b Square, rectangle

c Square, rectangle, rhombus, parallelogram

d Square, rectangle, rhombus, parallelogram

e Square, rectangle, rhombus, parallelogram

f Trapezium, isosceles trapezium

g Square, rectangle, isosceles trapezium

h Square, kite, arrowhead, rhombus

i Kite, arrowhead, isosceles trapezium

j Rectangle, rhombus

k Square

l Rectangle, rhombus, parallelogram

3 A Square B Rectangle, rhombus C Parallelogram

D Kite, arrowhead, isosceles trapezium E Trapezium



4 a (4, 0) b (−2, 1) c (1, 5) d (2, 5)

e (4, 0), (4, 6), (−2, −4) f (1, −4), (5, 4)

5 (−4, 5), (0, 7) or (0, −3), (4, −1) or (−1, 4), (1, 0)

6 a (2, 0) b (0, −1) c (−0.5, −3.5) d (−0.5, 0.5)

7 a (3, 6) b (3, 2) c (0.5, 1.5)

8 a (9, 4) b (10, −3) c (−1, 0)

9 a PQ is a diameter of the circle.

b

∠PRQ = 90° (Position of R is the pupil’s choice.)

c ∠PRQ is always 90°.

P

Q

R

10 cm



10 a AB is a chord.

b, d

∠ACB = 30° It is acute. (Position of C is the pupil’s choice.) ∠ADB = 150° It is obtuse. (Position of D is the pupil’s choice.)

c ∠ACB is always 30°.

e ∠ADB is always 150°.

f The sum of angles ACB and ADB is 180° for any choice of C and D on opposite arcs.

g The sum of opposite angles is 180°.

Questions 10f and 9 show that this is true when the quadrilateral has a diagonal that is

either 5 cm or 10 cm long. It is reasonable to conjecture that it is always true.

Pupil’s own drawing and measurements of angles to check their conjecture.

A

B

C

D

5 cm



GM1.3 Answers

1 Pupils’ constructions.

2 a, b

c The point of intersection is equidistant (approximately 6.5 units) from A, B and C.

3 From question 2, P lies where the perpendicular bisectors of AB, BC and AC meet. It is 22

53 units east and 22192 units south of A.


5 a, b, c Pupils’ constructions.

Check that the 70° angle is accurately drawn and correctly bisected.

c ∠PRQ is always 90°.

6 a, b

c All three bisectors intersect at a single point.

Y

Z

X

B

C

A



7 a, b

Check that the hedge (dotted line)

bisects QPS into two angles of 50°.

8 a, b Pupils’ constructions of a line bisected and then one of the right angles bisected.


10 a, b

Either of the dashed lines is a

correct answer to part b.

Check that PQ = QR = 5 units and

PQR = 45°.

11 a

b Area = 15 cm2


A

B P

Q

R

R

Q

R

S

P



13 a

b, c, d

c Length = 3.6 cm.

d Length = 3.6 cm.

14 a, b, c, d, e

e Isosceles trapezium

f PS = QR = 5 cm

P QYX

W Z

S R

W

X Y

Z

W

X Y

Z

P

10 cm

6 cm

50°

4 cm

2 cm

4 cm



15

16 a

b The ladder touches the wall at a point 7.75 m above the ground.

It makes an angle of 76° with the ground.

17

8 cm

2 cm



GM2.1 Answers

1 a 16 cm2 b 13.5 cm2 c 20 cm2 d 12.5 cm2

2 a h = 5 cm b h = 10 mm

3 a 12 cm2 b 24 cm2 c x = 8

4 a 6 cm2 b 18 cm2 c 9 cm

5 a 18 cm2 b 72 cm2 c 54 cm2

6 a 32 cm2 b 60 cm2 c 4 cm2 d 20 cm2

7 a 26 cm2 b 14 cm2 c 36 cm2 d 138 cm2

8 a a = 8 b h = 6 c x = 20 d x = 5

9 a 84 cm2 b 76 cm2 c 112 cm2 d 90 cm2

10 a 42 cm2 b 52 cm2 c 250 cm2 d 24 cm2

11 Area = 21 × 14 × 9 − 2

1 × 14 × 4 = 35 cm2

12 Area = 21 × (6 + 11) × 8 + 2

1 × (11 + 8) × 4 = 106 m2

13 a i 38.5 cm2 ii 38.5 cm2

b The answers are the same.

Shaded area = area of trapezium − area of unshaded triangle

= 21 × (x + 11) × 7 − 2

1 × x × 7 = 21 × 11 × 7 = 38.5 cm2

So the area does not depend on the value of x (it cancels).

Alternatively, you could consider the areas of the two shaded triangles:

Shaded area = 21 × (base of first triangle) × 7 + 2

1 × (base of second triangle) × 7

= 21 × (base of first triangle + base of second triangle) × 7

= 21 × 11 × 7 = 38.5 cm2

c 132 cm2



14 a 2x b 2x c 4(10 − x) d 40 m2

15 a Measurement depends on size of exercise book: typically between 400 cm2 and 600 cm2.

b Area in mm2 = 100 × area in cm2, from part a.

16 a 1500 mm2 b 250 mm2 c 58 000 mm2 d 3.8 mm2

17 a 250 cm2 b 67 cm2 c 0.37 cm2 d 4.5678 cm2

18 1 m = 100 cm, so 1 m2 = 100 cm × 100 cm = 10 000 cm2.

To convert an area from square metres to square centimetres, multiply the number by 10 000.

For example, 12 m2 = 120 000 cm2.

19 a 15 435 mm2 b 154.35 cm2 c 3% (to the nearest percent)

20 72 mm2 2.61 cm2 582 mm2 68.4 cm2 0.017 m2

21 a 10 000 cm2 b 1 000 000 mm2



GM2.2 Answers

1 a 48 cm3 b 9.5 cm3 c 54 m3 d 125 cm3

2 a 92 cm2 b 40 cm2 c 138 m2 d 150 cm2

3 a x = 4 b h = 8 c w = 0.2 d l = 7

4 a 128 cm2 b 160 cm2 c 1018 cm2 d 294 cm2

5 a 1200 cm3 b 150 cm2

c B: 40 cm2; C: 120 cm2; D: 80 cm2; E: 40 cm2 d 860 cm2

6 a 108 cm3 b 162 cm2

7 a 220 cm3 b 288 cm2

8 a 68 cm3 b 120 cm2

9 a 8 cm3 b 24 cm2

c Volume of B = 64 cm3 = 8 times the volume of A.

d Surface area of B = 96 cm2 = 4 times the surface area of A.

10 a Volume = 4800 cm3; surface area = 3000 cm2

b Volume = 160 cm3; surface area = 194 cm2

c Volume = 576 cm3; surface area = 432 cm2

d Volume = 21 060 cm3; surface area = 6426 cm2

11 a Measurement depends on size of textbook: typically between 200 cm3 and 400 cm3.

b Typically between 200 000 mm3 and 400 000 mm3.

12 a 15 000 mm3 b 2500 mm3 c 580 000 mm3 d 38 mm3

13 a 25 cm3 b 6.7 cm3 c 0.037 cm3 d 0.456 78 cm3

14 a Estimate depends on size of classroom.

b Answer to part a multiplied by 1 000 000.

c Answer to part b multiplied by 1000.

d Answer to part b divided by 8.



GM2.3 Answers

1 1, 4, 6, 7, 8, 10, 11 and 15 are prisms.

2 A 11 B 2, 10 C 3 D 9

E 1 F 4, 6, 7 G 5 H 14

I 8 J 13 K 15 L 12

3 A 11 B 2, 7 C 1, 4, 6 D 3, 5, 12

E 9, 13 F 10 G 8, 15 H 14

4 a i

ii

b i

ii

c i

ii

d i

ii

5 a A, D, E, F, G b A 5, D 6, E 3, F 6, G 3

6 a

b All the nets tessellate.

c Pupils’ drawings showing how one of the nets tessellates.



7 Any two correct nets from the following. The side lengths are 1 cm, 2 cm and 4 cm.



GM2.4 Answers

1 a 8 km = 8000 m b 1500 m = 1.5 km c 8.5 m = 850 cm

d 70 cm = 700 mm e 0.65 m = 650 mm f 560 cm = 0.0056 km

g 2 hectares = 20 000 m2 h 5400 cm2 = 0.54 m2 i 875 g = 0.875 kg

j 1.305 kg = 1305 g k 3.5 litres = 3500 cm3 l 950 litres = 0.95 m3

2 Approximate measure Types of measure Item

2 m length door a 1 tonne mass car b 750 ml capacity bottle c 7500 m2 area football pitch d 100 g mass apple e 10 m length tree f 480 mm2 area postage stamp g 200 litres capacity bathtub h 1 kg mass bag of sugar i 50 cm2 area playing card j 15 cm length pencil

3 a i 666 666.7 hours ii 27 777.8 days iii 76.1 years

b 19 or 20 leap years (or 18 leap years if the lifetime spans the turn of a century)

c Depending on the answer given for part b the answer will be

19 leap years; 40 025 520 minutes, 20 leap years; 40 026 960 minutes,

18 leap years; 40 024 080 minutes

The answer is longer than 40 000 000 minutes because there are 366 days in a leap

year so for each leap year you live through your life will be 1440 minutes longer.

4 a 21.6 cm b 3240 cm2 or 0.324 m2

5 a To the nearest 10 km b To the nearest 10 minutes

6 a 12 gallons b 17 gallons c 18 gallons

7 a i 2.2 lb ii 3.3 lb iii 1.1 lb

b i 0.9 kg ii 2.5 kg



c i bananas 73p, mushrooms £1.98, grapes £1.25, apples £1.16, potatoes £1.21

ii bananas 80p, mushrooms £1.95, grapes £1.15, apples £1.13, potatoes £1.33

d Greene’s Grocers

8 The mile is longer by 100 m. (Accept 110 m.)



GM3.1 Answers

1 a There are six distinct possibilities.

The others are shown in part b.

b The minimum perimeter is 12 units.

(1 way)

The maximum perimeter is 18 units.

(2 ways)

2 a There are six distinct possibilities, as shown in part b.


(1 way)

The maximum perimeter is 10 units.

(5 ways)



There are 11 distinct possibilities. The other four are shown in part b. 3 a


(1 way)

The maximum perimeter is 14 units. (3 ways)

4 A, B, D, F, H, J

5 A and F are congruent (SSS). E and G are congruent (SAS).



GM3.2 Answers

1 a b

2 a b

3 a

b



4 a b

5 a Rotation 180° about (0, 2) b Rotation 90° clockwise about (0, 0)

6 a b

7 a

b

This is the difference between the translation X to A and the translation X to B.



8 a b

c d

9 180° rotation about (0, 0)

10 a b



c d

11 180° rotation about (0, 0)

12 a i

ii 180° rotation about (1, −2)

b i

ii 180° rotation about (3, 3)



13 a, b

c AA″ = 2M1M2 d Translation of 2M1M2 to the right

14 a b

15 a b

The order in which the transformations are carried out affects the position of the image in these cases.



16 a

b

17 There are many possible answers for each part; below are some general forms (pupils are expected to give a particular example only).

a Reflection in the line x = 1 − a, followed by translation ⎟⎟⎠

⎞⎜⎜⎝

⎛− 32a

or translation ⎟⎟⎠

⎞⎜⎜⎝

⎛− 32b

followed by reflection in the line x = 1 + b.

b Reflection in the line y = c followed by translation ⎟⎟⎠

⎞⎜⎜⎝

⎛− c233


⎞⎜⎜⎝

⎛− 32

3d

followed by reflection in the line y = d.

c Rotation 90° anticlockwise about (p, q) followed by translation ⎟⎟⎠

⎞⎜⎜⎝

⎛−−+−−42

qpqp

or a translation followed by rotation [there are many possibilities, the simplest being

translation ⎟⎟⎠

⎞⎜⎜⎝

⎛−−

24

followed by rotation 90° anticlockwise about (0, 0)].

d Reflection in the line y = t followed by translation ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−32

3t


⎞⎜⎜⎝

⎛−u3

followed by reflection in the line y = 0.5u + 1.5.

18 Pupils’ own words and examples to describe the following transformations.

a A single rotation about the same centre through the sum of the two individual angles of rotation (taking angles of clockwise rotation as negative).

b A single translation that is the vector sum of the two individual translations.

c A translation in a direction perpendicular to the axes of reflection.

d A rotation of 180° about the point of intersection of the two axes of reflection.



19 Number of lines of symmetry 0 1 2

None G D, E, F

Rotation symmetry Order 2 B A, C

20 a i

ii order 4

b i

ii order 3

c i

ii order 6

d i

ii order 2

e i

ii order 1 (no rotation symmetry)

f i

ii order 2

21 a 5 b 8 c infinity

22 a

b



23 a

b

c

d

e

f



24



GM3.3 Answers

1 a enlargement factor 2 b enlargement factor 4 2 a

b

3 a

b

4 a i centre (1, 3) ii enlargement factor 2

b i centre (9, −2) ii enlargement factor 3

5 a A1 (1, 5), A2 (2, 10), A3 (3, 15), A4 (4, 20), A5 (5, 25)

b A6 (6, 30), A10 (10, 50), An (n, 5n)

c B1 (3, 1), B2 (6, 2), B3 (9, 3), B4 (12, 4), B5 (15, 5)

d B6 (18, 6), B10 (30, 10), Bn (3n, n)

e scale factor 15

f Each side in image n is n times the length of the corresponding side in object 1.

g Corresponding angles in object 1 and all the images are the same.



6 a

b

c

d

7 a −2 b −3



GM4.1 Answers

1 a 4.8 m b 2.1 m

2 a 2.5 m b 275 m c 25 m

3 a 24 cm b 13 cm c 72 mm

4 The scale is 1 : 2500000. The distance from Aberdeen to Cambridge is 850 km.

5 a Scale drawing of pool: a 5 cm by 2.5 cm rectangle.

b 12.5 cm2

c i 1250 m2 ii 12500000 cm2

6

7 a, b

R

S

4.8 cm

3.2 cm

0.75 cm



8 a, b

c 12.7 m

9 a 720000 cm2 b 72 m2

10 a 0.04 m2 b 400 cm2

11 4 cm2

12 17.5 km2

13 a i 250 m ii 0.25 km

b i 62500 m2 ii 0.0625 km2

c 0.5 km2

Z

W X

Y

4 cm

4 cm



GM4.2 Answers

1 a i Pupils’ constructions; check that PQ = QR = PR = 6 cm.

ii equilateral triangle

b i Pupils’ constructions; check that XZ = YZ = 6 cm and XY = 4cm.

ii isosceles triangle

c i Pupils’ constructions; check that AB = 10 cm, BC = 12 cm and AC = 5 cm.

ii scalene triangle

2 It is impossible to construct such a triangle because length LM is greater than the sum of

lengths LN and MN.

3 ABC, DEF and MNO can be constructed.

4 a, b Pupils’ constructions c BD = 8.5 cm

5 a, b Pupils’ constructions c XZ = 4.1 cm

6 a, b Pupils’ constructions

7

8 a

b GH = 4.3 cm

c GH = 5.4 m

6.4 cm

5.6 cm

4 cm

4.3 cm

E F

G

H

5 cm

8 cm

2.5 cm



9 a

b XY = 2.4 m

10 Pupils’ nets

11

This is one example of the net.

12 Some shapes consisting of four scalene or isosceles triangles can be made up to form

irregular tetrahedra. To support their reasoning, pupils should construct at least one valid

net for an irregular tetrahedron, and one ‘net’ that cannot be folded to form a solid shape.

For example, this is a valid net for a tetrahedron formed by cutting a corner off a cube.

These are not valid nets for tetrahedra.

In the first, edges A and B would meet, but they are not the same length.

In the second, the lengths of meeting edges match correctly, but the net folds to form a

flat (2-D) shape.

X

Y

8 cm

6.9 cm

A

B



13 a a = 7.5 cm, b = 4 cm, c = 8.5 cm

b Pupils’ own choices of triangle. c right-angles triangles

14 a i

ii

iii

2.8 cm

10.5 cm

9.1 cm

7.5 cm

2 cm

6.5 cm

6.5 cm

7 cm

7.5 cm

a

b c



b ABC could be either of the triangles drawn in parts i or ii. In order for a triangle to be

completely described by the lengths of two sides and the size of one (acute) angle, the

angle must be enclosed by the two given sides (SAS).

c DEF could be either of the triangles drawn in parts ii or iii. The sizes of all three

angles alone do not describe a triangle fully: the lengths of the sides can all be scaled

by the same factor to give another triangle with the same three angles.



GM4.3 Answers

1

2

3

4

A 5 cm 2 cm

3 cm

O

X

5 cm

A

B C

D

3 cm

3 cm



5 a, b

6 a, b

7

8

9

c The point of intersection is also equidistant from B and C.

5 cm

5.2 cm

ba

A

C

B

M

L

P

Q

P

Q R

SX

2 cm

1 cm

Y

X Z

3 cm



10 a–d

e equilateral triangle

11

12 Pupil’s construction of the angle bisected.

O

A B

4 cm



13 a, b, d, e

c XD = 3 cm

d YD = 1.1 cm

14

15

C

DA

B

X

Y

45° 22.5°

22.5°

2 cm

4 cm 2 cm 2 cm

5 cm

7 cm

4 cm



16 a

b

5 cm

Y

1 cm

7 cm

3 cm

5 cm

7 cm

X

7 cm

3 cm

5 cm

5 cm7 cm



17 a, b

18 a, b

c BC = 6.6 m

19 a

6 cm

2 cm

1.5 cm

1.5 cm

A B

D C



b i PT must be at most 6 km so it is within the transmitter’s range.

ii 4.8 km

iii

c i 10 km ii 16 km

iii



GM4.4 Answers

1 a 135° b 075° c 212° d 214°

2 a 315° b 255° c 032° d 034°

3 a Pupils’ diagrams. b 117° c 297° d 180°

4 a Pupils’ diagrams. b 060° c 240° d 180°

5 a Pupils’ diagrams. b 200°

6 a N

J

K

3 cm

322°

b 142°

7 a

N

L

K

6 cm

M

45°

22°

b i KM = 10.9 cm, LM = 5.8 cm

ii KM = 13.6 km, LM = 7.2 km

c ∠MLK = 180° − 45° = 135°

(angles about a point on a straight line)

∠LMK = 180° − ∠MLK − ∠LKM

(angle sum of triangle)

∠LMK = 180° − 135° − 22° = 23°

d 225°



8 a

X Y

Z

8 cm 155°

225°

NN

b i XZ = 6.0 cm, YZ = 7.7 cm ii XZ = 6.0 km, YZ = 7.7 km

c ∠YXZ = 65°

d ∠XZY = 70°

9 a, b, c

8 cm

V

W

U

U

N

N

N

120°

d 252°, 348°

e 168°, 072°

10 a

b 047°

c 2.9 cm

d 3.8 km

BA

C

N

5 cm

3.5 cm

125°



11 a

7 cm

4.5 cm

Ahmed

Brian

Carlos

300°

b 040°

c 184 m

12 a, b

6.5 cm L M

B

010°

315°

c LB = 5.6 cm

MB = 7.8 cm

d 28 km from L

39 km from M

13 a

6 cm

P

Q

X

Y

3.2 cm

5.5 cm

b XY = 1.7 cm

c 2.6 km

d 010°

e 020°, 7.4 km

f 108°, 8.4 km



14

A

B

trawler

27° 60°

60°

100°

27°

5 km

15

S 200°225°

225°

14 km

8 km

315°

N

N

N

N

A

B

T

The two possible positions of the group are labelled A and B in the diagram. From measurements of the diagram, AT = 6.8 km and BT = 18.0 km. If the group were to walk at a steady 4 km/h, the least time they would take to reach T is 1.7 hours = 1 hour 42 min and the most time they would take to reach T is 4.5 hours = 4 hours 30 min.

Date post:	03-Jan-2020
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