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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 8 · 8.4 Scale and Similarity

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.800 m
2.50 cm
3.4.5 m
4.21 cm
5.3
6.4 km
7.Yes — equal angles → similar (AAA).
8.3 m by 2 m
9.7 cm
10.Corresponding angles are equal; corresponding sides are in the same ratio.
Silver
11.1200 m by 1800 m
12.10 cm
13.100 cm = 1 m
14.15 cm and 20 cm
15.15 cm by 22.5 cm
16.1 : 500
17.1 cm : 0.5 km
18.2 m
19.w=4.5w = 4.5 cm
20.12.5 m
Gold
21.x=8x = 8 cm
22.(a) 400 cm by 300 cm (4 m by 3 m) (b) 12 m²
23.3 : 5
24.120 cm²
25.(a) 1500 m (b) 1.5 km
26.18 m; similar because both triangles have a vertical line, a horizontal shadow, and the same sun angle.
27.(a) 10 m (b) ≈ 78.5 m²
28.(a) 4 m by 3 m (b) 6 m²
29.64 cm²
30.(a) AB = 6 km, AC = 4.5 km (b) BC = 7.5 km
Platinum
31.(a) 4:14:1 (b) Yes; length ratio 2:12:1, area ratio =4:1= 4:1
32.(a) c=9.4b/ac = 9.4 b/a (b) Area =πr2= \pi r^2, ratio a2:b2a^2:b^2.
33.(a) See working (b) k=3k = 3
34.2≈1.414\sqrt{2} \approx 1.414
35.480 000 m²
36.12
37.0.05 km²
38.(0,0)(0,0), (12,0)(12, 0), (0,8)(0, 8); old area 12, new 48 = 12 × 4 ✓
39.5.625 × 10⁵ m² ≈ 0.5625 km²
40.(a) 216 m³ (b) 216 m²

Pack B — Answers

Bronze
1.900 m
2.40 cm
3.4.8 m
4.15 cm
5.4
6.7 km
7.Yes — AAA.
8.5 m by 3 m
9.8 cm
10.Same.
Silver
11.800 m by 1400 m
12.17.5 cm
13.200 cm = 2 m
14.15 cm and 36 cm
15.20 cm by 30 cm
16.1 : 250 000
17.1 cm : 0.25 km
18.1.8 m
19.w=9.6w = 9.6 cm
20.15 m
Gold
21.x=15x = 15 cm
22.(a) 500 cm by 400 cm (5 m by 4 m) (b) 20 m²
23.4 : 7
24.108 cm²
25.(a) 5000 m (b) 5 km
26.32 m; same reasoning.
27.(a) 4 m (b) ≈ 12.57 m²
28.(a) 5 m by 4 m (b) 10 m²
29.72 cm²
30.Same approach.
Platinum
31.(a) 25:425:4 (b) Yes; length ratio 5:25:2, area ratio 25:425:4
32.See Pack A.
33.See Pack A.
34.3≈1.732\sqrt{3} \approx 1.732
35.80 000 m²
36.36
37.0.0075 km²
38.(0,0)(0,0), (18,0)(18, 0), (0,12)(0, 12); old 12, new 108 = 12 × 9 ✓
39.5 × 10⁵ m² = 0.5 km²
40.(a) 64 m³ (b) 96 m²

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) R1 1200 × 1800 m, R2 400 × 600 m (b) R1 2.16 km², R2 0.24 km² (c) 9 : 1 (d) Linear sf 3 (e) ≈ 2016 — earlier than 2035, so the prediction is plausible.
Full working
(a) R1: 1 cm represents 200 m. 6×200=12006 \times 200 = 1200 m; 9×200=18009 \times 200 = 1800 m. R2: 1 cm represents 100 m. 400400 m × 600600 m.

(b) R1 area =2.16= 2.16 km²; R2 =0.24= 0.24 km².

(c) Ratio 9:19 : 1.

(d) Area sf 9 → linear sf 9=3\sqrt{9} = 3.

(e) Area lost 1960–2010: 1.921.92 km² in 50 years ⇒ 0.0384 km²/yr. Remaining 0.24 km² lasts ≈6.25\approx 6.25 years from 2010 → **2016**. 2035 conservative.
2Problem 2
Answer
(a) 4 m × 3 m (b) 12 m² (c) Real radius 0.2 m → area ≈ 0.126 m² (d) See working
Full working
(a) 8×50=4008 \times 50 = 400 cm = 4 m; 6×50=3006 \times 50 = 300 cm = 3 m.

(b) 4×3=124 \times 3 = 12 m².

(c) Real radius =0.4×50=20= 0.4 \times 50 = 20 cm = 0.2 m. Area =π×0.04≈0.1257= \pi \times 0.04 \approx 0.1257 m².

(d) Linear scale 1 : 50 (cm:cm). Area scale =502=2500= 50^2 = 2500. So plan : real = 1 : 2500.

Verify: plan area of room = 8×6=488 \times 6 = 48 cm² = 0.0048 m². Real 12 m². Ratio =0.0048:12=1:2500= 0.0048 : 12 = 1 : 2500 ✓.
3Problem 3
Answer
(a) Same sun-angle and both vertical → AAA (b) 10.5 m (c) 3.75 m
Full working
(a) The sun is far away → its rays are parallel. Each object is vertical (90° at the ground), and the sun-angle is the same → both triangles share two angles → similar (AAA).

(b) tree height1.5=71\frac{\text{tree height}}{1.5} = \frac{7}{1} ⇒ tree = 10.5 m.

(c) Stick: 1.5/2=0.751.5 / 2 = 0.75 (vertical/shadow). Flag: 0.75×5=3.750.75 \times 5 = 3.75 m.
4Problem 4
Answer
(a) 2 (b) 20 cm × 30 cm (c) 16 chf
Full working
(a) Small area =150= 150 cm². Area sf =600/150=4= 600/150 = 4. Linear sf =2= 2.

(b) Dimensions × 2: 20 cm × 30 cm.

(c) Cost ∝ area, so cost sf = 4. Larger cost = 4×4=164 \times 4 = 16 chf.
5Problem 5
Answer
(a) 20 cm × 7.5 cm × 6.25 cm (b) Area ratio 1:5761 : 576; model SA =625= 625 cm² (c) Volume ratio 1:138241 : 13824; model volume ≈868\approx 868 cm³
Full working
(a) Divide by 24: 4.8/24=0.24.8/24 = 0.2 m = 20 cm. 1.8/24=0.0751.8/24 = 0.075 m = 7.5 cm. 1.5/24=0.06251.5/24 = 0.0625 m = 6.25 cm.

(b) Area sf = 1:242=1:5761 : 24^2 = 1 : 576. Model SA = 36 m2/576=0.062536 \text{ m}^2 / 576 = 0.0625 m² = 625 cm².

(c) Volume sf =1:243=1:13824= 1 : 24^3 = 1 : 13824. Model volume = 12/13824≈8.68×10−412 / 13824 \approx 8.68 \times 10^{-4} m³ ≈868\approx 868 cm³.
6Problem 6
Answer
(a) (2, 2), (10, 2), (2, 8) (m) (b) 24 m² (c) 0.25 cm
Full working
(a) Multiply each coordinate by 2: (2,2),(10,2),(2,8)(2, 2), (10, 2), (2, 8) in metres.

(b) Real base =10−2=8= 10 - 2 = 8 m. Real height =8−2=6= 8 - 2 = 6 m. Area =12×8×6=24= \tfrac{1}{2} \times 8 \times 6 = 24 m².

(c) Drawing scale: 1 cm = 2 m, so 0.5 m = 0.25 cm wide.
7Problem 7
Answer
(a) 2 (b) 60 cm (c) 6 cm
Full working
(a) Area sf = 4 ⇒ linear sf = 2.

(b) Perimeter is linear → multiply by 2: 30×2=6030 \times 2 = 60 cm.

(c) Trim length is linear → multiply by 2: 3×2=63 \times 2 = 6 cm.
8Problem 8
Answer
(a) 6 000 m = 6 km (b) 24 cm (c) 64 cm²
Full working
(a) 12×50000=60000012 \times 50000 = 600000 cm = 6000 m = 6 km.

(b) On a 1 : 25 000 map (twice as detailed), the same real distance: 600000/25000=24600000 / 25000 = 24 cm.

(c) Area sf =500002=2.5×109= 50000^2 = 2.5 \times 10^9. Real area =16= 16 km² =1.6×1011= 1.6 \times 10^{11} cm². Map area =1.6×1011/2.5×109=64= 1.6 \times 10^{11} / 2.5 \times 10^9 = 64 cm².
9Problem 9
Answer
(a) 7.5 cm (b) 20 cm × 15 cm (c) ≈ 4.6 g
Full working
(a) 90/12=7.590 / 12 = 7.5 cm.

(b) 240/12=20240 / 12 = 20 cm; 180/12=15180 / 12 = 15 cm.

(c) Volume sf =1/123=1/1728= 1/12^3 = 1/1728. Mass =8000= 8000 g /1728≈4.63/ 1728 \approx 4.63 g.
10Problem 10
Answer
(a) k3Vk^3 V (b) k=2k = 2 (c) 800 cm²
Full working
(a) Volume scales as the cube of the linear scale factor: Vlarger=k3VV_{\text{larger}} = k^3 V.

(b) k3=2000/250=8k^3 = 2000/250 = 8, so k=2k = 2.

(c) Surface area sf =k2=4= k^2 = 4. 200×4=800200 \times 4 = 800 cm².
11Problem 11
Answer
(a) ≈ 54.9° (b) ≈ 1.19 m (c) See working
Full working
(a) tan⁡(elevation)=541/380≈1.4237\tan(\text{elevation}) = 541/380 \approx 1.4237. Angle ≈54.9°\approx 54.9°.

(b) Pedestrian shadow = 1.7×380/541≈1.191.7 \times 380/541 \approx 1.19 m. (Same ratio.)

(c) Both triangles share the sun-angle, both have a vertical (90°) side and a horizontal shadow. AAA → similar. Corresponding sides have the same ratio (height : shadow ratio).
12Problem 12
Answer
(a) 1.5 (b) 7.5 cm (c) ≈ 5.06 L (= 1.5 × 3.375)
Full working
(a) Linear sf =30/20=1.5= 30/20 = 1.5.

(b) Depth =5×1.5=7.5= 5 \times 1.5 = 7.5 cm.

(c) Volume sf =1.53=3.375= 1.5^3 = 3.375. New volume =1.5×3.375=5.0625= 1.5 \times 3.375 = 5.0625 L. (Note: this assumes the larger tin is geometrically similar — the rule k³ for volume).