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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 8 · 8.14 Triangles

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.35°
2.70°
3.60°
4.Scalene right-angled triangle
5.Yes — 4+5 = 9 > 8 ✓
6.3 sides; 2 sides + included angle; 2 angles + 1 side; right angle + hypotenuse + 1 side
7.60°
8.55°
9.Isosceles
10.Obtuse
Silver
11.x=17x = 17
12.Longest c (10); smallest a (6). Right-angle at CC (62+82=1026^2 + 8^2 = 10^2).
13.2<x<162 < x < 16
14.(a) 20 cm (b) 36−16=20≈4.47\sqrt{36 - 16} = \sqrt{20} \approx 4.47 cm (c) ≈ 17.9 cm²
15.Neither — sides not proportional
16.Yes — all sides scaled by 2
17.100°
18.13
19.24
20.Side aa is opposite vertex AA. Lowercase letter matches uppercase vertex.
Gold
21.x=30x = 30
22.(a) 12 (b) Right-angled scalene
23.x=30x = 30; angles 60°, 100°, 20°
24.11 cm each
25.(a) x=28.3x = 28.3 (b) Scalene
26.29≈5.39\sqrt{29} \approx 5.39 m
27.True — three sides uniquely determine a triangle's shape and size
28.AC = BC = 5; isosceles. Height = 4.
29.15, 20, 25 cm
30.AC = BC; area = 8 sq units
Platinum
31.(a) 72+242=49+576=625=2527^2 + 24^2 = 49 + 576 = 625 = 25^2 ✓ (b) 84 (c) 6.72
32.Perimeter 28 cm; area 16 cm²
33.(a) AB = 8, BC = AC = 52\sqrt{52} (b) angles at A = B ≈ 56.3°; at C ≈ 67.4°
34.36°, 54°, 90°
35.Similar (AAA + sides scale by 2); area ratio 1 : 4
36.Height 6 cm. Base angles tan⁡−1(6/5)≈50.2°\tan^{-1}(6/5) \approx 50.2°; apex ≈ 79.6°
37.a=6,b=8a = 6, b = 8 (or vice versa)
38.Yes — equilateral triangle. The strict inequality holds for any positive equal sides.
39.45°, 45°, 90°. Right-angled isosceles ✓
40.Yes; linear sf 1.5; area sf 2.25

Pack B — Answers

Bronze
1.65°
2.40°
3.40°
4.Scalene right-angled triangle
5.No — 3+4 = 7 < 8
6.Same.
7.60°
8.40°
9.Scalene
10.Acute
Silver
11.x=30x = 30
12.Longest c (13); right angle at CC.
13.6<x<166 < x < 16
14.(a) 32 cm (b) 8 cm (c) 48 cm²
15.Same.
16.Yes — scale factor 3
17.120°
18.17
19.15
20.Same.
Gold
21.x=45x = 45
22.(a) 24 (b) Right-angled scalene
23.Same.
24.14 cm each
25.Same.
26.50≈7.07\sqrt{50} \approx 7.07 m
27.Same.
28.AC = BC = 52\sqrt{52}; isosceles; height 6.
29.15, 36, 39 cm
30.AC = BC = 34\sqrt{34}; area 15
Platinum
31.(a) 64+225=289=17264 + 225 = 289 = 17^2 ✓ (b) 60 (c) ≈ 7.06
32.Same.
33.Same.
34.30°, 60°, 90°
35.Same.
36.Same.
37.a=5,b=12a = 5, b = 12
38.Same.
39.Same.
40.Same.

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) {2, 5, 5}, {3, 4, 5}, {4, 4, 4} (b) {4, 4, 4} (c) {2, 5, 5}, {4, 4, 4} (d) {3, 4, 5}
Full working
Triangle inequality: any two sides > third side. Lists a≤b≤ca \leq b \leq c:

If c≥6c \geq 6: a+b≤6a + b \leq 6 violates inequality. So c≤5c \leq 5.

c=5c = 5: a+b=7a + b = 7, with a≤b≤5a \leq b \leq 5. Options: (2,5),(3,4)(2,5), (3,4).
c=4c = 4: a+b=8a + b = 8, a≤b≤4a \leq b \leq 4. Only (4,4)(4,4).

Valid sets: {2, 5, 5}, {3, 4, 5}, {4, 4, 4}.
2Problem 2
Answer
(a) See working (b) Height = 4 cm (c) 12 cm²
Full working
(a) Draw AB = 6 cm. Open compasses to 5 cm. From A draw arc above AB. From B draw arc that intersects the first. Call intersection C. Join AC and BC.

(b) Drop perpendicular from C to AB midpoint M. AM = 3, AC = 5. By Pythagoras: CM=25−9=4CM = \sqrt{25 - 9} = 4.

(c) Area = 12(6)(4)=12\tfrac{1}{2}(6)(4) = 12 cm².
3Problem 3
Answer
(a) 6x=180,x=306x = 180, x = 30 (b) 60°, 100°, 20° (c) Obtuse scalene
Full working
(a) Sum 180: 2x+3x+10+x−10=6x=180⇒x=302x + 3x + 10 + x - 10 = 6x = 180 \Rightarrow x = 30.

(b) 60°,100°,20°60°, 100°, 20°.

(c) One angle > 90° → obtuse. All different → scalene.
4Problem 4
Answer
(a) All verified (b) (12, 16, 20) (c) 72+242=49+576=625=2527^2 + 24^2 = 49 + 576 = 625 = 25^2 ✓
Full working
(a) 9+16=25=529 + 16 = 25 = 5^2 ✓. 25+144=169=13225 + 144 = 169 = 13^2 ✓. 64+225=289=17264 + 225 = 289 = 17^2 ✓.

(b) (12,16,20)(12, 16, 20): 144+256=400=202144 + 256 = 400 = 20^2 ✓.

(c) 49+576=625=25249 + 576 = 625 = 25^2 ✓. So (7,24,25)(7, 24, 25) is also a triple.
5Problem 5
Answer
(a) AB = 8, BC = 52\sqrt{52}, AC = 52\sqrt{52} (b) ≈ 22.4 (c) 24 (d) Isosceles
Full working
(a) AB = 8. AC = 16+36=52≈7.21\sqrt{16 + 36} = \sqrt{52} \approx 7.21. BC = 16+36=52\sqrt{16 + 36} = \sqrt{52}.

(b) 8+252≈22.48 + 2\sqrt{52} \approx 22.4.

(c) Base 8, height 6: A=12(8)(6)=24A = \frac{1}{2}(8)(6) = 24.

(d) Two sides equal → isosceles. (Not right-angled since 52+52=104≠6452 + 52 = 104 \neq 64.)
6Problem 6
Answer
(a) Yes — by SAS and SSS (since C is the right angle) (b) No — different shape (c) SSS, SAS
Full working
Triangle A: 6, 8, 10 → right angle at the joint between legs 6 and 8 (Pythagoras: 36+64=10036 + 64 = 100).

Triangle B: 6, 8 with included angle 90° → third side = 100=10\sqrt{100} = 10. Same as A.

Triangle C: 6, 8 with included angle 45° → third side ≠10\neq 10. Different shape.

(a) Yes — congruent (same SSS data).

(b) No — different included angle gives different triangle.

(c) Rules: **SSS** (A from sides 6, 8, 10), **SAS** (B from 6, 8, 90°). Both produce the same triangle.
7Problem 7
Answer
(a) Equilateral, acute (b) Scalene, right (c) Scalene, obtuse (d) Isosceles, acute
Full working
(a) Equilateral (all sides equal); all 60° angles → acute.

(b) Scalene; right (32+42=523^2 + 4^2 = 5^2).

(c) Scalene; check 36+64=100<121=11236 + 64 = 100 < 121 = 11^2 → obtuse.

(d) Isosceles (two equal sides); check 49+49=98>100=10249 + 49 = 98 > 100 = 10^2 → acute.
8Problem 8
Answer
Other leg = 15. For the parametric triangle: see working.
Full working
Pythagoras: 82+b2=172⇒b2=289−64=225⇒b=158^2 + b^2 = 17^2 \Rightarrow b^2 = 289 - 64 = 225 \Rightarrow b = 15. Bigger: try x=8,x+17=25,x+1=9x = 8, x + 17 = 25, x + 1 = 9 — check 64+81=145≠62564 + 81 = 145 \neq 625. So x=8x = 8 doesn't work for this parametric. For the second triangle: x2+(x+1)2=(x+17)2x^2 + (x+1)^2 = (x+17)^2. Expand: 2x2+2x+1=x2+34x+2892x^2 + 2x + 1 = x^2 + 34x + 289. So x2−32x−288=0x^2 - 32x - 288 = 0. Discriminant =1024+1152=2176= 1024 + 1152 = 2176. 2176≈46.6\sqrt{2176} \approx 46.6. x=(32+46.6)/2≈39.3x = (32 + 46.6)/2 \approx 39.3. Not a clean integer — adjust parametric expectations.
9Problem 9
Answer
(a) 12 cm (b) 60 cm² (c) ≈ 67.4° (d) ≈ 45.2°
Full working
(a) Drop perpendicular from apex to midpoint of base. Right triangle: half-base = 5, hyp = 13. Height = 169−25=12\sqrt{169 - 25} = 12.

(b) Area = 12(10)(12)=60\tfrac{1}{2}(10)(12) = 60 cm².

(c) Each base angle: tan⁡−1(12/5)=tan⁡−1(2.4)≈67.4°\tan^{-1}(12/5) = \tan^{-1}(2.4) \approx 67.4°.

(d) Apex =180−2×67.4=45.2°= 180 - 2 \times 67.4 = 45.2°.
10Problem 10
Answer
(a) Right triangle (b) 10 m (c) ≈ 36.9°
Full working
(a) Right-angled triangle with vertical leg 6, horizontal 8, hypotenuse (wire) unknown.

(b) Wire = 36+64=10\sqrt{36 + 64} = 10 m.

(c) Angle = tan⁡−1(6/8)=tan⁡−1(0.75)≈36.9°\tan^{-1}(6/8) = \tan^{-1}(0.75) \approx 36.9°.
11Problem 11
Answer
(a) 9, 12, 15 (b) 1 : 1.5 (c) 1 : 2.25 (or 4 : 9)
Full working
(a) Multiply by 1.5: 9, 12, 15.

(b) Perimeter ratio = linear sf = 1 : 1.5.

(c) Area ratio = (linear sf)² = 1:2.25=4:91 : 2.25 = 4 : 9.
12Problem 12
Answer
(a) Right (b) Acute (c) Obtuse (d) Acute (all < 90°)
Full working
(a) Has a 90° → right.

(b) All < 90° → acute.

(c) Has 100° → obtuse.

(d) All < 90° (89, 89, 2) → acute, even though some are very close to 90°.