GM1.1 Homework 1 Answers - Cambridge Essentialsessentials.cambridge.org/media/CEMKS3_E7_GM1_1_WS_HANS3.pdf · Cambridge Essentials Mathematics Extension 7 GM3.3 Homework 2 Original

Cambridge Essentials Mathematics Extension 7 GM1.1 Homework 1

Original Material © Cambridge University Press 2008 1

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




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




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.




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




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.




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


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


Original material © Cambridge University Press 2008 1


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.


Original material © Cambridge University Press 2008 1


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




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.




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




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




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




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.




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



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'




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



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




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.




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.

× × × ××




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.




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.

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GM1.1 Homework 1 Answers - Cambridge Essentialsessentials.cambridge.org/media/CEMKS3_E7_GM1_1_WS_HANS3.pdf · Cambridge Essentials Mathematics Extension 7 GM3.3 Homework 2 Original