Cambridge Essentials Mathematics Extension 7 GM1.1 Homework 1

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GM1.1 Homework 1 Answers

1 a 6.3 cm b 2.2 cm c 6.7 cm

2 a Accurate drawings

b i 5.7 cm ii 5.7 cm iii 5.8 cm iv 5.8 cm

c If drawn accurately: i Triangle CDE is an isosceles triangle as ED = EC.

ii Triangle JKL is an equilateral triangle as JK = JL = KL.

3 a 14 cm b 9 mm c 4 m d 21 m e 143 cm

4 Yes. Draw a line 14.8 cm long. The 8.7 cm and 9.4 cm sides will meet above the line.

5 a 15.6 cm b 102 mm c 4.8 cm d 5

Cambridge Essentials Mathematics Extension 7 GM1.1 Homework 2

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GM1.1 Homework 2 Answers

1 a 18 cm b 42 cm

2 48 cm

3 a 76 cm b 13.4 m

4 18 cm

5 If the side lengths are a, a and b, then a + a + b = 2a + b = 18 cm, so a < 9 cm.

The sum of any two sides must be greater than the third side, so 2a > b.

b = 18 cm − 2a, so 2a > (18 cm − 2a) and 4a > 18.

Therefore a > 4.5 cm. There are four possibilities:

(a, a, b) = (5 cm, 5 cm, 8 cm); (6 cm, 6 cm, 6 cm); (7 cm, 7 cm, 4 cm); (8 cm, 8 cm, 2cm)

6 50 cm

Cambridge Essentials Mathematics Extension 7 GM1.2 Homework 1

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GM1.2 Homework 1 Answers

1 a 13.5 cm2 b 3.6 cm2 or 360 mm2

c 2.16 m2 or 21 600 cm2

2 a 144 cm2 b 48 cm

3 a 72 mm2 b 0.72 cm2

4 100 cm2

5 25 cm2

6 a 33 cm2 b 77 m2

c 224 mm2 d 129 cm2

7 a D

b The diagonal of a square is longer than the side of a square. The shape with the greatest perimeter is made by drawing as many diagonals of squares as possible. Here is one example.

Many other shapes will have the same perimeter.

Cambridge Essentials Mathematics Extension 7 GM1.2 Homework 2

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GM1.2 Homework 2 Answers

1 a 99 m2 b 139 m2 c 196 m2

2 a 12 cm2 b 84 mm2 c 71.5 m2

3 a 26 cm2 b 121.5 cm2

4 a 40 cm2 b 93 cm2

5 Approximately 75 km2

6 31.5 cm2

Cambridge Essentials Mathematics Extension 7 GM2.1 Homework 1

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GM2.1 Homework 1 Answers

1 a i 120° ii 70° iii 90° iv 310° v 38° vi 154°

b i obtuse ii acute iii right angle iv reflex v acute vi obtuse

2 a 127°

More pupils measured the angle as 127° than any other size and two pupils measured it only 1° above or below 127°. It is difficult to measure accurately to the exact number of degrees using a protractor.

b Toby must have used the wrong scale on the protractor, because 180° – 52° gives 128°, which is very nearly the correct size of 127°.

3 a 7 b 150°

4 a i 132° ii 48° iii 73°

iv 59° v 59°

b 180°

c 180°. They add up to 180° or 2 right angles.

d They are the same.

Cambridge Essentials Mathematics Extension 7 GM2.1 Homework 2

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GM2.1 Homework 2 Answers

1 a =133° b = 144° c = 58° d = 122°

e = 122° f = 41° g = 139° h = 62°

i = 62° j = 56° k = 304° l = 65°

m = 115° n = 30°

2 a = 34° b = 112° c = 56° d = 56°

e = 27° f = 27° g = 29° h = 61°

i = 29° j = 61° k = 54° m = 56°

n = 34° p = 85°

3 a = 15°

4 x = 12° y = 72° z = 48°

Cambridge Essentials Mathematics Extension 7 GM2.2 Homework

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GM2.2 Homework Answers

1 a CF, DE, AH

b DC, DE, AB, AH, EF, CF, BG, HG

2 a Equilateral b 3 c Pupil’s construction

d A rhombus

3 a It is a trapezium.

b It is a kite.

4

5 a Pattern with exactly one line of symmetry.

b Pattern with exactly four lines of symmetry.

6 a False b True c False d True e False

7 a

b (2, –2)

c (–4, 4)

0

Cambridge Essentials Mathematics Extension 7 GM3.1 Homework 1

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GM3.1 Homework 1 Answers

1 a 500 litres b 5 ml c 8 litres d 250 ml

2 a 145 g b 40 g c 77 kg d 1.5 kg e 4 kg

3 Pupils’ own lists of items.

4 Pupils’ own lists of items.

Cambridge Essentials Mathematics Extension 7 GM3.1 Homework 2

Original material © Cambridge University Press 2008 1

GM3.1 Homework 2 Answers

1 a 5 b 20 c 35 d 34 e 38 f 1.636

g 2.6 h 3.1 i 3.9 j 1.624 k 1.632 l 1.636

2 44 mph

3 a 14 cm = 140 mm b 35 mm = 3.5 cm c 165 cm = 1.65 m

d 5.7 m = 570 cm e 3.28 m = 328 cm f 0.6 m = 60 cm

4 a 350 ml = 0.35 litres b 58 cl = 580 ml c 593 ml = 59.3 cl

d 5.7 litres = 5700 ml e 0.8 litres = 80 cl f 285 cl = 2.85 litres

5 a 685 g = 0.685 kg b 5.4 kg = 5400 g c 70 g = 0.07 kg

6 Yes, the bus can go under the bridge. 12 feet 10 in = 3.95 m to 3 s.f.

7 £10 806

Cambridge Essentials Mathematics Extension 7 GM3.2 Homework 1

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GM3.2 Homework 1 Answers

1 a A, D

b B

c B, C

d D

e A

2 AC = 4.5 cm, angle BAC = 86°

3 a i 7.5 cm

ii 3.1 cm

b Area = 5 × 3.1 = 16 cm2 to the nearest cm2.

Cambridge Essentials Mathematics Extension 7 GM3.2 Homework 2

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GM3.2 Homework 2 Answers

1 b i AB = 11.7 cm

ii AC = 17.2 cm

iii XY = 10.0 cm

2 ML = 4.9 cm, MN = 3.3 cm

3 Angle PQR = 77°

4

The two circles don’t intersect. There is no point that is 4.8 cm from U and 5.4 cm from V. This means that the triangle can’t be drawn.

U V

Cambridge Essentials Mathematics Extension 7 GM3.3 Homework 1

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GM3.3 Homework 1 Answers

1 a

b D

c F

d 4 cm × 3 cm × 2 cm

e 52 cm2

2 a Triangular prism

b i 6.5 cm ii 6 cm iii 6 cm

c 165 cm2

d i 10 cm × 6 cm × 2.5 cm ii 200 cm2

Cambridge Essentials Mathematics Extension 7 GM3.3 Homework 2

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GM3.3 Homework 2 Answers

1 a Trapezium

b V = 8, F = 6, E = 12

V + F – E = 8 + 6 – 12

= 14 – 12

= 2

2 a 72 cm

b 180 cm2

3 a There are only two possible nets for a tetrahedron.

b If this were a net for a square-based pyramid, the edges labelled p and q in the

diagram below would meet.

However they are different lengths, so this is not a net for a square-based pyramid.

p

q

Cambridge Essentials Mathematics Extension 7 GM3.4 Homework

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GM3.4 Homework Answers

1 a Triangular prism

b Cube

c Cuboid

d Tetrahedron

e Square-based pyramid

2

3

4 No. There are three places where two faces meet, so there are six covered faces.

The surface area is given by 4 × 6 cm2 − 3 × 2 × 1 cm2 = 24 cm2 – 6 cm2 = 18 cm2.

Cambridge Essentials Mathematics Extension 7 GM4.1 Homework

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GM4.1 Homework Answers

1 a

b

c

d

2 b, c

c y = x

d C(2, 6), C′(6, 2). The coordinates use the same numbers but they are reversed (x→ y and y → x).

x

8

7

6

5

4

3

2

1

y

1 2 3 4 5 6 7 8

B

C

D

0

B′

A

Cambridge Essentials Mathematics Extension 7 GM4.1 Homework

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3

4 a (5, –4)

b (–8, 5)

c (–2, –6)

5 a x = 2 b y = –1 c (–5, –3)

6 Pupil’s own answer.

–8 –7 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 8

8

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

x

y

y = –x

0

A B

CD

A'B'

D'C'

Cambridge Essentials Mathematics Extension 7 GM4.2 Homework

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GM4.2 Homework Answers

1 a

b

c

2

rotation of 90° clockwise rotation of 180° rotation of 90° anticlockwise

3

5

4

3

2

1

–1

–2

–3

–4

–5

A

y

–5 –4 –3 –2 –1 1 2 3 4 50

B

C

x

B A ×

A

B×

A

B

×

×P

×P

× P

Cambridge Essentials Mathematics Extension 7 GM4.2 Homework

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4 a

parallelogram

b

parallelogram

c

rectangle

5 a

b P′(–2, 3)

c Rotation through 90° clockwise with centre (–1, 1)

Q

P R

Cambridge Essentials Mathematics Extension 7 GM4.3 Homework

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GM4.3 Homework Answers

1 a, b

c A′(1, –3), B′(3, –1), C′(5, –3), D′(3, –5) d 6 squares to the left and 7 squares up.

2 a 3 squares to the right and 5 squares up.

b B′(2, 6), C′(5, 7)

3 a 5 squares to the left and 7 squares down.

b (–4, –9)

c (8, 6)

4 a, b, c c ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

65

5 Pupil’s own pattern.

Cambridge Essentials Mathematics Extension 7 GM5.1 Homework

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GM5.1 Homework Answers

1 a 3 b 4 c 1 d 6

2 a

b There are many possibilities; this is an example.

3 There are several possibilities; these are some examples.

a i

ii

b i

ii

4 Number of lines of symmetry

0 1 2 3 4 1 C

2 F 3 D A

Order of rotational symmetry

4 B E 5 a 6 right

b Rotation of 180° about any of the points marked with a cross in this diagram.

6 a i Rotation of 90° clockwise about (0, 0) ii Translation 6 squares right.

b There are many possible answers including:

i rotation of 180° about (0, 0) followed by translation 6 squares right.

ii rotation of 90° clockwise about (–2, –2) followed by translation 4 squares up.

iii rotation of 90° clockwise about (2, –2) followed by translation 4 squares left.

× × × ××

Cambridge Essentials Mathematics Extension 7 GM5.2 Homework 1

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GM5.2 Homework 1 Answers

1

Size of large cube

No. of small cubes with no yellow faces

No. of small cubes with 1 yellow face

No. of small cubes with 2 yellow faces

No. of small cubes with 3 yellow faces

Total no. of small cubes

3 × 3 × 3 1 6 12 8 27

4 × 4 × 4 8 24 24 8 64

5 × 5 × 5 27 54 36 8 125

6 × 6 × 6 64 96 48 8 216

7 × 7 × 7 125 150 60 8 343

⋮ ⋮ ⋮ ⋮ ⋮ ⋮

2

n × n × n (n – 2)3 6(n – 2)2 12(n – 2) 8 n3 3

Any kite with diagonals a and b can be divided along the line of symmetry into two equal triangles. Dividing along AC, the area of each triangle is

21 × base × height

= 21 × b ×

2a = 1

4ab

Since the kite is made of two such triangles the area of the kite is

21 ab or half the product

of the diagonals.

Cambridge Essentials Mathematics Extension 7 GM5.2 Homework 2

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GM5.2 Homework 2 Answers

1 w = 33° w + 27° = 360° ÷ 6 because the shape has rotational symmetry of order 6.

so w + 27° = 60°

2 No, Jason cannot draw the triangle. When the 12 cm side is drawn at 30° to the base line,

the other end is more than 5 cm above the base line.

If you reflect the triangle in its base line it forms

a triangle in which angles x and y must be equal.

But the third angle in the triangle is 60°,

so x = y = 60°. This means that the triangle is an

equilateral triangle and the vertical side is

therefore 12 cm. Two lines of 5 cm each are

together less than 12 cm, so the triangle is

impossible to draw.

3 A square be divided into 4, 6, 7, 8, 9 … n squares.

Once you have any number of squares, say n, you can make n + 3 squares by dividing one

of the squares into 4.

Since you can make 6, 7, or 8 squares, you can therefore make any numbers of squares

above that.

You cannot make 2 or 3 squares, because one small square would have to have at least 2

of its corners at the corners of the big square. Then it would be the same size as the big

square.

For 5 squares you would have to have a different square in each corner of the big square,

but this cannot leave one single space for 1 more square. So you cannot make 5 squares.