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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 9 · 9.1 Surds and Pythagoras

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.5 cm
2.525\sqrt{2}
3.252\sqrt{5}
4.Yes — right-angled
5.6
6.Rational: 5, 0.3‾0.\overline{3}, 4=2\sqrt{4} = 2. Irrational: 2,π\sqrt{2}, \pi.
7.≈ 7.1
8.15 cm
9.828\sqrt{2}
10.8 cm
Silver
11.21\sqrt{21}
12.5
13.323\sqrt{2}
14.2132\sqrt{13}
15.5
16.32=42\sqrt{32} = 4\sqrt{2}
17.13
18.24 cm
19.7+437 + 4\sqrt{3}
20.232\sqrt{3}
Gold
21.2132\sqrt{13}
22.61\sqrt{61} ≈ 7.81 cm
23.≈ 16.16
24.Side 525\sqrt{2}; area 50
25.91≈9.54\sqrt{91} \approx 9.54 m
26.4−24 - \sqrt{2}
27.232\sqrt{3}
28.10 cm
29.52+52=50=(52)25^2 + 5^2 = 50 = (5\sqrt{2})^2 ✓
30.434\sqrt{3} m ≈ 6.93
Platinum
31.≈ 21.8°
32.a=c2−b2a = \sqrt{c^2 - b^2}
33.h≈43.82h \approx 43.82, B ≈ 40.99 m
34.144+32=176≈13.27\sqrt{144 + 32} = \sqrt{176} \approx 13.27
35.2−1\sqrt{2} - 1
36.(a) 12 cm (b) 60 cm²
37.253≈43.325\sqrt{3} \approx 43.3 cm²
38.a=6,b=8a = 6, b = 8 (or swap)
39.h=6⋅86+8=4814≈3.43h = \dfrac{6 \cdot 8}{6 + 8} = \dfrac{48}{14} \approx 3.43 m
40.Yes — 8+18=268 + 18 = 26? No: 8+18=26≠328 + 18 = 26 \neq 32. So not right-angled this way. Try: longest side 32\sqrt{32} → squared 32. Other squares 8 + 18 = 26. Not equal. So NOT right-angled.

Pack B — Answers

Bronze
1.10 cm
2.626\sqrt{2}
3.353\sqrt{5}
4.Yes
5.10
6.Same.
7.≈ 5.5
8.17 cm
9.535\sqrt{3}
10.12 cm
Silver
11.34\sqrt{34}
12.525\sqrt{2}
13.737\sqrt{3}
14.555\sqrt{5}
15.10
16.40=210\sqrt{40} = 2\sqrt{10}
17.10
18.30 cm
19.11−6211 - 6\sqrt{2}
20.252\sqrt{5}
Gold
21.555\sqrt{5}
22.116\sqrt{116} ≈ 10.77 cm
23.≈ 18.38
24.Side 727\sqrt{2}; area 98
25.12 m
26.7−337 - 3\sqrt{3}
27.434\sqrt{3}
28.13 cm
29.Same approach.
30.4 m
Platinum
31.≈ 21.8°
32.Same.
33.h=45h = 45, B ≈ 46.90
34.144+50=194≈13.93\sqrt{144 + 50} = \sqrt{194} \approx 13.93
35.5+2\sqrt{5} + 2
36.(a) 24 cm (b) 168 cm²
37.93≈15.599\sqrt{3} \approx 15.59 cm²
38.a=5,b=12a = 5, b = 12
39.Same.
40.Check.

Problem-solving — Worked Solutions

1Problem 1
Answer
(b) ≈ 9.54 m (c) ≈ 2.46 m
Full working
(a) Right-angled triangle with hyp 10, horizontal 3, vertical unknown.

(b) h2=100−9=91⇒h≈9.54h^2 = 100 - 9 = 91 \Rightarrow h \approx 9.54 m.

(c) 12−9.54≈2.4612 - 9.54 \approx 2.46 m.
2Problem 2
Answer
(a) All ✓ (b) ✓ (c) (21, 28, 35) (d) Multiply each by kk
Full working
(a) 9+16=259+16=25, 25+144=16925+144=169, 64+225=28964+225=289. All ✓.

(b) 49+576=625=25249 + 576 = 625 = 25^2 ✓.

(c) (21,28,35)(21, 28, 35). Verify: 441+784=1225=352441 + 784 = 1225 = 35^2 ✓.

(d) (ka)2+(kb)2=k2(a2+b2)=k2c2=(kc)2(ka)^2 + (kb)^2 = k^2(a^2 + b^2) = k^2 c^2 = (kc)^2. ∎
3Problem 3
Answer
(a) 5 m (b) 29≈5.39\sqrt{29} \approx 5.39 m (c) ≈ 21.8°
Full working
(a) Base diagonal =16+9=5= \sqrt{16 + 9} = 5.

(b) Space diagonal =16+9+4=29≈5.39= \sqrt{16 + 9 + 4} = \sqrt{29} \approx 5.39.

(c) Angle: tan⁡θ=2/5\tan \theta = 2/5, so θ=tan⁡−1(0.4)≈21.8°\theta = \tan^{-1}(0.4) \approx 21.8°.
4Problem 4
Answer
(a) 626\sqrt{2} (b) 535\sqrt{3} (c) 10 (d) 424\sqrt{2}
Full working
(a) 72=36×272 = 36 \times 2. 72=62\sqrt{72} = 6\sqrt{2}.

(b) 23+33=532\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}.

(c) 100=10\sqrt{100} = 10.

(d) 82×22=822=42\tfrac{8}{\sqrt{2}} \times \tfrac{\sqrt{2}}{\sqrt{2}} = \tfrac{8\sqrt{2}}{2} = 4\sqrt{2}.
5Problem 5
Answer
(a) a2a\sqrt{2} (b) a3a\sqrt{3} (c) Face 52≈7.075\sqrt{2} \approx 7.07; space 53≈8.665\sqrt{3} \approx 8.66
Full working
(a) Face diagonal: a2+a2=a2\sqrt{a^2 + a^2} = a\sqrt{2}.

(b) Space diagonal: a2+a2+a2=a3\sqrt{a^2 + a^2 + a^2} = a\sqrt{3}.

(c) Face 52≈7.075\sqrt{2} \approx 7.07 cm; space 53≈8.665\sqrt{3} \approx 8.66 cm.
6Problem 6
Answer
(a) 6100≈78.1\sqrt{6100} \approx 78.1 cm (b) ≈ 22.4° (c) Front face: 4500≈67.1\sqrt{4500} \approx 67.1; side face: 2500=50\sqrt{2500} = 50; top: 5200≈72.1\sqrt{5200} \approx 72.1
Full working
(a) Space diagonal =602+402+302=6100≈78.1= \sqrt{60^2 + 40^2 + 30^2} = \sqrt{6100} \approx 78.1 cm.

(b) Base diagonal =602+402=5200≈72.1= \sqrt{60^2 + 40^2} = \sqrt{5200} \approx 72.1. Angle θ=tan⁡−1(30/72.1)≈22.6°\theta = \tan^{-1}(30/72.1) \approx 22.6°.

(c) Front face (60×30): 4500≈67.1\sqrt{4500} \approx 67.1. Side face (40×30): 2500=50\sqrt{2500} = 50. Top face (60×40): 5200≈72.1\sqrt{5200} \approx 72.1.
7Problem 7
Answer
(a) AB = 5, BC = 3, AC = 4 (b) 3-4-5 triple (c) 12
Full working
(a) AB=9+16=5AB = \sqrt{9 + 16} = 5. AC=4AC = 4 (vertical). BC=3BC = 3 (horizontal).

(b) 3, 4, 5 is the classic Pythagorean triple — and our triangle has these exact side lengths (right-angled).

(c) Perim = 12.
8Problem 8
Answer
(a) 100 m (b) ≈ 36.9° (c) 150 m
Full working
(a) 602+802=100\sqrt{60^2 + 80^2} = 100 m.

(b) tan⁡θ=60/80=0.75⇒θ≈36.9°\tan \theta = 60/80 = 0.75 \Rightarrow \theta \approx 36.9°.

(c) Linear sf 1.5 → new diagonal = 100×1.5=150100 \times 1.5 = 150 m.
9Problem 9
Answer
(a) 525\sqrt{2} cm (b) 10 cm (c) 20220\sqrt{2} cm
Full working
(a) s=50=52s = \sqrt{50} = 5\sqrt{2}.

(b) Diagonal = s2=52⋅2=10s \sqrt{2} = 5\sqrt{2} \cdot \sqrt{2} = 10.

(c) P=4s=202≈28.28P = 4s = 20\sqrt{2} \approx 28.28 cm.
10Problem 10
Answer
(a) 13 m (b) ≈ 67.4° (c) 5-12-13
Full working
(a) Hypotenuse =25+144=13= \sqrt{25 + 144} = 13.

(b) tan⁡θ=12/5=2.4⇒θ≈67.4°\tan \theta = 12/5 = 2.4 \Rightarrow \theta \approx 67.4°.

(c) 5-12-13 is a famous Pythagorean triple.
11Problem 11
Answer
(a) 522\tfrac{5\sqrt{2}}{2} (b) 3+12\tfrac{\sqrt{3} + 1}{2} (c) 5−2\sqrt{5} - \sqrt{2}
Full working
(a) Multiply by 2/2\sqrt{2}/\sqrt{2}.

(b) Multiply by (3+1)/(3+1)(\sqrt{3} + 1)/(\sqrt{3} + 1): 3+13−1=3+12\tfrac{\sqrt{3} + 1}{3 - 1} = \tfrac{\sqrt{3} + 1}{2}.

(c) Multiply by (5−2)/(5−2)(\sqrt{5} - \sqrt{2})/(\sqrt{5} - \sqrt{2}): 3(5−2)5−2=5−2\tfrac{3(\sqrt{5} - \sqrt{2})}{5 - 2} = \sqrt{5} - \sqrt{2}.
12Problem 12
Answer
(a) d=3d = 3 (b) Side ≈5.77\approx 5.77 m (c) No — would require side =s3/2≈0.577s= s\sqrt{3}/2 \approx 0.577 s, contradicting s=s = side
Full working
(a) d2=1+4+4=9d^2 = 1 + 4 + 4 = 9, so d=3d = 3.

(b) d=s3=10⇒s=10/3≈5.77d = s\sqrt{3} = 10 \Rightarrow s = 10/\sqrt{3} \approx 5.77.

(c) If d=2sd = 2s: 4s2=3s2⇒s=04s^2 = 3s^2 \Rightarrow s = 0. Only trivial solution — space diagonal can never equal twice the side.