GM1.1 Homework Answers - Cambridge Essentialsessentials.cambridge.org/media/CEMKS3_E8_GM1_WS_HANS4.pdf · g = 66° (an isosceles triangle has one pair of equal angles) h = 48° (angles

Cambridge Essentials Mathematics Extension 8 GM1.1 Homework Answers

Original material © Cambridge University Press 2009 1

GM1.1 Homework Answers

1 a a = 58° (corresponding angles are equal)

b = 58° (alternate angles are equal or vertically opposite angles are equal)

c = 122° (angles on a straight line add up to 180°)

b d = 57° (angles in a triangle add up to 180°)

e = 123° (angles on a straight line add up to 180° or

an exterior angle of a triangle equals the sum of the interior opposite angles)

c f = 66° (angles on a straight line add up to 180°)

g = 66° (an isosceles triangle has one pair of equal angles)

h = 48° (angles in a triangle add up to 180°)

2 m = 36° (alternate angles are equal)

n = 96° (angles on a straight line add up to 180°)

p = 48° (angles in a triangle add up to 180°)

3 a = 55° (angles in a triangle add up to 180°)

b = 125° (angles on a straight line add up to 180°)

c = 30° (angles in a triangle add up to 180°)

4 x = 60° (the angles of equilateral triangle ABC are equal and add up to 180°)

y = 25° (the angles of triangle ACD add up to 180°; ∠CAD = ∠CDA)

z = 35° (triangle ABC is equilateral, so ∠BAC = 60° = z + y)

5 p = 133° (angles in a quadrilateral add up to 360°; angles on a straight line add up to 180°)

6 a = 60° (the angles of an equilateral triangle are equal and add up to 180°)

b = 10° (each angle in a square is 90°, so b + 60° + 20° = 90°)

c = 70° (the angles of a triangle add up to 180°, so c + 20° + 90° = 180°)

d = 70° (vertically opposite angles are equal)



7 p = 55° (p and 35° together make the right angle)

q = 102° (angles round a point add up to 360°)

r = 151° (angles on a straight line add up to 180°)

s = 52° (angles of in quadrilateral add up to 360°)

8 Sum of the interior angles of a pentagon = (5 – 2) × 180° = 540°

So a = 540 ÷ 5 = 108°.

Sum of angles on a straight line = 180° = a + b

So b = 180° – 108° = 72°.

(Or b is an exterior angle or a regular pentagon, so b = 360° ÷ 5 = 72°.)

Sum of angles in a triangle = 180° = b + b + c (isosceles triangles have two equal angles).

So c = 180° – (2 × 72°) = 36°.




1 a Name of

quadrilateral Do the diagonals always intersect at right angles?

Number of axes of symmetry

Order of rotational symmetry

rectangle no 2 2

rhombus yes 2 2

parallelogram no 0 2

kite yes 1 1

square yes 4 4

trapezium no 0 1

b Name of quadrilateral symmetry rotational oforder

symmetryofaxesofnumber

rectangle 1

rhombus 1

parallelogram 0

kite 1

square 1

trapezium 0

The ratio is always 0 or 1. The number of axes of symmetry is either zero or it is equal

to the order of rotational symmetry.

c i

Isosceles trapezium

ii One axis of symmetry

iii Order 1



2 a E (3, 3) b F (1, 0) c G (1, 5) or (5, –9) or (–3, 1)

3 a (5, 7) b (–1, –3)

4 (3, 4)

5 (9, –10)

6 Pupils’ drawings




1 a–d A

B C

X

e AX = 3.2 cm

2 M

NL

3 a–c X

ZY35°

P

Q

d YQ = 4.3 cm



4

F H

G

5 Construction of circumcircle of triangle of pupil’s choice.

6

4 cm

5 cm

12 cm



GM2.1 Homework 1 Answers

1 a 12 cm2 b 25 cm2 c 13.5 cm2 d 27.5 cm2

2 18 cm

3 a 6 cm2 b 30 cm2 c 12 cm

d i True. Area of R = Area of Q = half the area of the 5 cm by 12 cm rectangle

ii False. Area of S = (4 × 12) cm2 − area of P = (48 − 6) cm2 = 42 cm2

iii True. P of AreaS of Area =

2

2

cm6

cm 42 = 7

iv True. Q and P of Area

rectangle of Area = 2

2

cm36

cm 108 = 3

4 a i 35 cm2 ii 36 cm2

b i 70 cm2 ii 30 cm2

5 a x = 7; the marked length is 7 cm.

b x = 6; the marked length is 6 cm.

c x = 2; the marked lengths are 2 cm and 10 cm.

Cambridge Essentials Mathematics Extension 8 GM2.1 Homework Answers 2



1 a 57 cm2 b 222 cm2 c 87 cm2 d 41 cm2

2 a 900 mm2 b 459 mm2 c 47 000 mm2 d 2.9 mm2

3 a 160 cm2 b 49 cm2 c 0.55 cm2 d 3.8217 cm2

4 Jodie is not correct. 1 m = 100 cm, but 1 m2 = 1002 cm2 = 10 000 cm2

5 a i 39 600 mm2 ii 396 cm2

b i 88 250 mm2 ii 882.5 cm2

c Sarah is not correct. letter large a of area maximumletter normal a of area maximum = 396

882.5 ≈ 0.42 < 0.5




1 a 105 cm3 b 64 cm3 c 6 cm3

2 a x = 5; marked side length is 5 cm.

b x = 5; marked side lengths are both 5 cm.

c x = 5; marked side length is 5 cm.

3 a 158 cm2 b 90 cm2 c 150 cm2

4 a Volume = 120 cm3

b Area of A = 30 cm2

c Area of B = 40 cm2

Area of C = 8 cm2

Area of D = 20 cm2

d Total surface area = 196 cm2

5 a 729 cm3 b 486 cm2 c PofvolumeQ of volume = 27 d

P of area surfaceQ of area surface = 9

6 a Cross-sectional area = 21 × 10 cm × 12 cm = 60 cm2

Volume = 60 cm2 × 40 cm = 2400 cm3

Surface area = 2 × 60 cm2 + 40 cm × (13 cm + 13 cm + 10 cm) = 1560 cm2

b Cross-sectional area = 36 cm × 5 cm + 6 cm × 5 cm = 210 cm2

Volume = 210 cm2 × 20 cm = 4200 cm3

Surface area = 2 × (210 cm2 + 20 cm × 36 cm + 10 cm × 36 cm) = 2260 cm2

7 a 9000 mm3 b 1800 mm3 c 340 000 mm3 d 42 mm3

8 a 36 cm3 b 8.3 cm3 c 0.055 cm3 d 0.82561 cm3

9 a Pupils’ own estimates.

b Answer to part a × 1000000

c Answer to part b × 1000




1 A 3 B 2 C 3 D 5 E 2 F 6

G 3 H 4 I 4 J 7 K 1

2 a F, K b C c A d H

3 a i

ii

iii

or

b i

ii

iii

4 There are many possible nets; two examples are shown below.

2 cm

3 cm

5 cm

2 cm

3 cm

5 cm



5 Yes, he can draw the following net on to the 10 cm by 10 cm card.

3 cm2 cm

9 cm

10 cm

5 cm

6 a, b

Base

Back

Top

Front

Side Side

X

X

X




1 a 3 km = 3000 m b 1800 m = 1.8 km c 4 m = 400 cm

d 320 cm = 3.2 m e 30 cm = 300 mm f 0.26 m = 260 mm

g 780 cm = 0.0078 km h 38 mm = 3.8 cm i 625 mm = 0.625 m

j 1.6 km = 160000 cm k 2.1 m = 2100 mm l 0.481 km = 481 m

2 a 4 kg = 4000 g b 2 tonnes = 2000 kg c 275 g = 0.275 kg

d 5.81 kg = 5810 g e 3.5 tonnes = 3500 g f 1384 g = 1.384 kg

3 a 3 m2 = 30 000 cm2 b 4 cm2 = 400 mm2 c 1.5 hectares = 15 000 m2

d 175 mm2 = 1.75 cm2 e 4.2 litres = 4200 ml f 36 000 m2 = 3.6 hectares

g 2 cm3 = 2000 mm3 h 5.26 litres = 5260 cm3 i 5 m3 = 5000 litres

4 a metres b tonnes c hectares d cubic centimetres e millilitres

f kilograms g litres h centimetres i square millimetres

5 a 63 years 7 months 2 days

b i 763.1 months ii 23211 days

6 a 43.75 m2

b 656.25 m3

c 656 250 litres

7 120 km/h ≈ 75m.ph, so Fabrice is breaking the speed limit.

9 a £5.04

b 49p

c 42p




1 There are four other ways.

2 a

b

c One possibility is

d One possibility is

3 A and J

F and H

I and K

L and O

Cambridge Essentials Mathematics Extension 8 GM3.2 Homework 1 Answers



1

2



3 a N b K c M d L

4

A′(2, 1), B′(1, 6), C′(7, 4)

A′′(–2, 3), B′′(–1, 8), C′′(–7, 6)

A′′′(3, –6), B′′′(8, –7), C′′′(6, –1)




1 a, b

c Reflection in the line y = x

2 a, b c Reflection in the line x = 4

ab



3

c (–6, –1)

4 Number of lines of symmetry

0 1 2 3 4 6 None D A Order 2 J G, I Order 3 E B Order 4 F

Rot

atio

n Sy

mm

etry

Order 6 C H

5 Examples of correct answers are given below.

a

b

c

6




1 Scale factor 3

2

3 a, b c A′ (–2, –2), B′ (10, –2),

C′ (2, 10), D′ (–2, 6)

4 a Scale factor 3 b (2, 10)

5

P




1 a i 4.5 m ii 3.15 m

b The dishwasher will not fit.

The dishwasher width is 60 cm; actual length of gap = 1.8 × 30 = 54 cm.

2 a 1.2 m b 2.1 m c 10 200 m

3 a 2 cm b 20 cm c 1.6 cm

4 a 600 000 cm2 b 60 m2

5 Please check that lengths on pupils’ diagrams match the given dimensions.

8.4 cm

7.8 cm

2.4 cm 2.4 cm

3.3 cm

2.1 cm

2.1 cm




Check measurements on pupils’ diagrams are as given throughout. 1 a i

3 cm 4 cm

5 cmA B

C

ii Right-angled triangle

b i

X

Y

Z

7 cm 4 cm

10 cm

ii Scalene triangle

iii Obtuse angle

2

A B

C must lie on thearc of points 6 cmfrom point A.

C must lie on thearc of points 5 cmfrom point B.

It is impossible to construct triangle ABC, since the two arcs in the diagram do not

intersect: there is no point that is both 6 cm from A and 5 cm from B.

This is because AB > AC + BC; that is, 12 cm > 6 cm + 5 cm.



3 a, b

8 cm

5 cm5 cm

5 cm 5 cm

P

Q

R

S

c Rhombus

d QS = 6 cm

4

A




1

7 cm

3 cm

5 cm

P Q

2

X2 cm

4 cm



3 a

P

X

YZ

b XP = 4.6 cm

4 Check pupil’s diagrams. The line OZ

bisects the angle XOY.

The diagram shown is an illustrative

example and OX and OY are not the

correct length.

X

YO

Z

5

10 cm

8 cm

6 cm 4.6 cm

Goat Atetheredhere

Goat Btetheredhere

a

b

b

a The area shaded in grey can be reached by both goats.

b Areas shaded with checks can be reached by neither goat. They lie outside both arcs.



6 NewcastleUpon Tyne

Sheffield

4 cm3 cm

5 cm4 cm

7.6 cm

7 a, b

A

O B

X

Peter’sroute

Archie’sroute

7 cm

4 cm

c On the scale diagram the distance Peter had walked when he meets Archie is 7.8 cm.

The actual distance is 10.9 km.



8

sea

P

Q

R

location of boat




1 a 128° b 247° c 232° d 324°

2 a 308° b 067° c 052° d 144°

3 a P = 106°

b Bearing of Y from X = 254°

c Bearing of X from Y = 180° – 106° = 074°

d 180°

4 a Check pupils’ diagrams. Measurements should be as indicated. This diagram is not

actual size.

b AC = 64 mm = 6.4 cm

BC = 103 mm = 10.3 cm

c ∠CAB = 180° – 80° = 100°

d Bearing of A from C = 180° + 80° = 260°



5 a, b

5 cm

5 cm

3 cm

8 cm

165°

30°

N

M

E

B

The area coveredby Manchester Airport radar

is the interior of this circle.

The area coveredby Exeter Airportradar is the interiorof this circle.

b ii Yes. B lies within the circle of radius 5 cm, centre M.

c i 5.6 cm

ii 224 km



6 a

X

Y

L P70°

N

300°

N

15°

3 cm

6 cm

4 cm

b LP = 3.4 cm

c 510 m

d 271°

e L is 705 m from Y on a bearing of 329°.

f P is 945 m from X on a bearing of 082°.

Documents

GM1.1 Homework Answers - Cambridge Essentialsessentials.cambridge.org/media/CEMKS3_E8_GM1_WS_HANS4.pdf · g = 66° (an isosceles triangle has one pair of equal angles) h = 48° (angles