Solutions — Full Answer Key

Mathematics

Year 10 · Geometry

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze

1.40 cm²

2.30 cm²

25\pi

cm²

4.5 cm

5.8 cm

8\pi

090°

8.120 cm³

\sin\theta = \dfrac{3}{5}

10.26 cm

Silver

11.32 cm²

12.46 cm²

13.5.00 cm

14.9.19 cm

15.5.60 cm

16.

\theta = 30.0°

17.

140°

18.

160\pi

cm³

19.

\sqrt{13}

20.84 cm²

Gold

21.12.00 cm²

22.East: 43.3 km; North: 25.0 km

23.

67.4°

24.9.04 cm

25.6.24 cm

26.(a)

90°

(b)

AC = \sqrt{244} \approx 15.62

27.23.66 cm

28.13.00 cm

29.7.21 cm

30.

95.7°

Platinum

31.50 km

32.

\approx 52.36

cm²

33.

\approx 19.53

34.

B \approx 53.5°

(or

126.5°

)

35.

108\pi

cm³

36.24 cm² (right triangle)

37.50 km

38.(a) 4 m (b)

\approx 53.1°

39.

\approx 8.85

40.

54\sqrt{3}

cm²

Pack B — Answers

Bronze

1.84 cm²

2.56 cm²

49\pi

cm²

4.13 cm

5.12 cm

12\pi

180°

8.210 cm³

\sin\theta = \dfrac{5}{13}

10.46 cm

Silver

11.42 cm²

12.72 cm²

13.11.49 cm

14.10.00 cm

15.10.00 cm

16.

\theta = 30.0°

17.

180°

18.

200\pi

cm³

19.

\sqrt{41}

20.45 cm²

Gold

21.42.43 cm²

22.East: 28.3 km; North: 28.3 km

23.

53.1°

24.14.62 cm

25.10.45 cm

26.(a)

90°

(b)

AC = \sqrt{164} \approx 12.81

27.30.31 cm

28.13.60 cm

29.13.00 cm

30.

102.6°

Platinum

31.

\sqrt{1300} \approx 36.06

32.

\approx 25.13

cm²

33.

\approx 21.27

34.

B \approx 65.4°

(or

114.6°

)

35.

\dfrac{850\pi}{3}

cm³

36.30 cm² (right triangle)

37.64.03 km

38.(a) 12 m (b)

\approx 67.4°

39.

\approx 7.60

40.

24\sqrt{3}

cm²

Problem-solving — Worked Solutions

1Problem 1

Answer

(a) 5.66 m (b)

\approx 70.5°

Full working

(a) Height =

\sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt{2} \approx 5.66

m.

(b)

\cos\theta = \frac{2}{6} = \frac{1}{3} \Rightarrow \theta = \cos^{-1}(1/3) \approx 70.5°

.

(c) 70.5° is within the [70°, 80°] safe range — **just safe**. If the foot were moved closer to the wall (say 1 m away),

\cos\theta = \frac{1}{6}

, giving

\theta \approx 80.4°

(a bit too steep). The safe range for this ladder corresponds to foot distance

6\cos 80° \leq d \leq 6\cos 70°

, i.e.

1.04 \leq d \leq 2.05

2Problem 2

Answer

(a) See working (b) 10 km (c)

\approx 103°

Full working

(a) At

P

, the walker turns from bearing

050°

140°

. The interior angle at

P

in the triangle

CPQ

180° - (140° - 050°) = 180° - 90° = 90°

.

(b) Right-angled triangle

CPQ

with legs

CP = 6

km and

PQ = 8

km.

CQ = \sqrt{36 + 64} = 10

km.

(c) Bearing of

Q

from

C

: angle clockwise from north. Component east:

6\sin 50° + 8\sin 140° = 4.596 + 5.142 = 9.74

km. Component north:

6\cos 50° + 8\cos 140° = 3.857 - 6.128 = -2.27

km. So

Q

is south-east of

C

. Bearing =

180° - \tan^{-1}(9.74/2.27) = 180° - 76.9° \approx 103°

3Problem 3

Answer

(a)

h \approx 12.59

m (b)

\approx 26.7°

(c)

\approx 21.81

Full working

(a)

\tan 40° = \frac{h}{15}

, so

h = 15 \tan 40° \approx 15 \times 0.839 \approx 12.59

m.

(b) New angle:

\tan\theta = \frac{12.59}{25} = 0.504 \Rightarrow \theta \approx 26.7°

.

(c) Shadow at 30°:

\tan 30° = \frac{12.59}{L} \Rightarrow L = \frac{12.59}{\tan 30°} = \frac{12.59}{0.5774} \approx 21.81

4Problem 4

Answer

(a)

c \approx 11.81

cm (b)

A \approx 40.9°

(c)

\approx 42.50

cm²

Full working

(a)

c^2 = a^2 + b^2 - 2ab\cos C = 64 + 121 - 176\cos 75° \approx 185 - 45.55 \approx 139.45

c \approx \sqrt{139.45} \approx 11.81

cm.

(b) Sine rule:

\frac{\sin A}{a} = \frac{\sin C}{c}

\sin A = \frac{8 \sin 75°}{11.81} \approx \frac{7.73}{11.81} \approx 0.654

A \approx 40.9°

.

(c) Area

= \frac{1}{2}ab \sin C = \frac{1}{2}(8)(11)\sin 75° = 44 \times 0.966 \approx 42.50

cm².

5Problem 5

Answer

52 cm²

Full working

Drop perpendiculars from the ends of the shorter parallel side to the longer. The horizontal projection on each side:

\frac{16 - 10}{2} = 3

cm. Height:

\sqrt{5^2 - 3^2} = \sqrt{16} = 4

cm.

Area

= \frac{1}{2}(10 + 16)(4) = \frac{1}{2}(26)(4) = \mathbf{52}

cm².

6Problem 6

Answer

(a) 14 cm (b)

\approx 59.0°

Full working

(a) Space diagonal

d = \sqrt{4^2 + 6^2 + 12^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14

cm.

(b) Base diagonal:

\sqrt{16 + 36} = \sqrt{52} \approx 7.21

cm. Angle:

\tan\theta = \frac{12}{\sqrt{52}}

\theta = \tan^{-1}(\frac{12}{\sqrt{52}}) \approx \tan^{-1}(1.664) \approx 59.0°

7Problem 7

Answer

(a)

\sin = 3/5, \cos = 4/5, \tan = 3/4

(b) ✓ (c) ✓

Full working

(a) Opposite to

\theta

is 3; adjacent is 4; hypotenuse is 5.
-

\sin\theta = 3/5

\cos\theta = 4/5

\tan\theta = 3/4

(b)

\sin^2\theta + \cos^2\theta = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1

✓.

(c)

\frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} = \frac{3}{4} = \tan\theta

✓.

8Problem 8

Answer

(a) 120° (b)

\approx 60.83

km (c)

\approx 185°

Full working

(a) Angle between bearings

150° - 030° = 120°

.

(b) After 2 hr:

|OA| = 40

km,

|OB| = 30

km, angle

AOB = 120°

. Cosine rule:

|AB|^2 = 40^2 + 30^2 - 2(40)(30)\cos 120° = 1600 + 900 - 2400(-0.5) = 2500 + 1200 = 3700

|AB| = \sqrt{3700} \approx 60.83

km.

(c) Position of

A

(40\sin 30°, 40\cos 30°) = (20, 34.64)

. Position of

B

(30\sin 150°, 30\cos 150°) = (15, -25.98)

. Vector from

A

B

(-5, -60.62)

— south and slightly west of

A

. Bearing (clockwise from north): pointing south is

180°

; the vector deviates slightly west of south by

\tan^{-1}(\frac{5}{60.62}) \approx 4.7°

. So bearing

= 180° + 4.7° \approx 185°

(since vector points west-of-south).

9Problem 9

Answer

(a) All check out (b) Scaling preserves the relation (c)

(15, 20, 25)

Full working

(a) Checks:
-

3^2 + 4^2 = 9 + 16 = 25 = 5^2

✓
-

5^2 + 12^2 = 25 + 144 = 169 = 13^2

✓
-

8^2 + 15^2 = 64 + 225 = 289 = 17^2

✓

(b) If

a^2 + b^2 = c^2

, then

(ka)^2 + (kb)^2 = k^2 a^2 + k^2 b^2 = k^2(a^2 + b^2) = k^2 c^2 = (kc)^2

✓.

(c) Scale up

(3, 4, 5)

k = 5

(15, 20, 25)

. Check:

225 + 400 = 625 = 25^2

✓.

10Problem 10

Answer

(a) See working (b)

150\sqrt{3}

cm² ≈ 259.81 (c)

100\pi \approx 314.16

— hexagon ≈ 82.7% of circle

Full working

(a) Connect the centre to each vertex. This divides the hexagon into 6 congruent isoceles triangles, each with two radii of length

r

and central angle

60°

(i.e.

360°/6

). With two equal sides and a

60°

included angle, the third side (a side of the hexagon) must also equal

r

(forming equilateral triangles).

(b) Six equilateral triangles of side 10. Each area:

\frac{\sqrt{3}}{4}(10)^2 = 25\sqrt{3}

. Total:

6 \times 25\sqrt{3} = 150\sqrt{3} \approx 259.81

cm².

(c) Circle area:

\pi r^2 = 100\pi \approx 314.16

cm². Ratio:

\frac{150\sqrt{3}}{100\pi} \approx \frac{259.81}{314.16} \approx 82.7\%

. The hexagon fills about 83% of the circumscribing circle.

11Problem 11

Answer

(a) 12 cm (b)

100\pi

cm³ (c)

90\pi

cm²

Full working

(a) Height

h = \sqrt{13^2 - 5^2} = \sqrt{144} = 12

cm.

(b) Volume =

\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(25)(12) = 100\pi

cm³.

(c) Total surface area = base + lateral =

\pi r^2 + \pi r \ell = 25\pi + \pi(5)(13) = 25\pi + 65\pi = 90\pi

cm².

12Problem 12

Answer

(a) 5 m each (b)

\approx 36.9°

Full working

(a) Each slope is the hypotenuse of a right triangle with horizontal

4

m and vertical

3

m. Slope length =

\sqrt{16 + 9} = 5

m.

(b) Angle:

\tan\theta = \frac{3}{4} \Rightarrow \theta = \tan^{-1}(0.75) \approx 36.9°

.

(c) Each rectangular slope =

5 \times 12 = 60

m². Two slopes total =

\mathbf{120}

m².