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Ecolint Campus des NationsMathématiques
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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 9 · 9.4 Transformations, Congruence and Similarity

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.(3,−4)(3, -4)
2.(−3,4)(-3, 4)
3.(6,2)(6, 2)
4.(0,3)(0, 3)
5.A′(2,2),B′(4,2),C′(2,6)A'(2,2), B'(4,2), C'(2,6)
6.15 cm
7.8 and 10
8.SSS, SAS, ASA, RHS
9.(4,3)(4, 3)
10.64 cm²
Silver
11.10 cm
12.A′(0,0),B′(−2,0),C′(0,2)A'(0,0), B'(-2,0), C'(0,2)
13.A′(3,1),B′(5,1),C′(4,3)A'(3,1), B'(5,1), C'(4,3)
14.A′(2,2),B′(6,2),C′(6,6),D′(2,6)A'(2,2), B'(6,2), C'(6,6), D'(2,6)
15.A′(−1,1),B′(−1,2),C′(−3,1)A'(-1,1), B'(-1,2), C'(-3,1)
16.x=8x = 8
17.729 cm³
18.Both have area 6 — congruent
19.(a) SSS (b) SAS
20.New area 100 (= 25 × 4)
Gold
21.A′(3,3),B′(9,3),C′(6,9)A'(3,3), B'(9,3), C'(6,9)
22.SAS
23.DE = 10, DF = 15
24.50/18=5/3\sqrt{50/18} = 5/3
25.9, 12, 15
26.A′(0,0),B′(0,−4),C′(3,0)A'(0,0), B'(0,-4), C'(3,0)
27.1 : 8
28.(a) 6 × 9 (b) 54 (c) 30
29.A′(2,1),B′(2,4),C′(6,4)A'(2,1), B'(2,4), C'(6,4)
30.Yes; right angle at Q
Platinum
31.40 cm
32.PQR area 72; perim ratio 1:2
33.2\sqrt{2}
34.(−2,−1)(-2, -1)
35.12
36.Similar (AAA); not necessarily congruent
37.150 cm²
38.A′(−1,−1),B′(5,−1),C′(−1,7)A'(-1, -1), B'(5, -1), C'(-1, 7)
39.16 : 49
40.Rotation is an isometry — preserves all distances, hence all areas

Pack B — Answers

Bronze
1.(−2,−5)(-2, -5)
2.(2,5)(2, 5)
3.(3,3)(3, 3)
4.(−4,0)(-4, 0)
5.A′(3,3),B′(6,3),C′(3,9)A'(3,3), B'(6,3), C'(3,9)
6.24 cm
7.24 and 26
8.Same.
9.(−2,5)(-2, 5)
10.81 cm²
Silver
11.9 cm
12.A′(0,0),B′(0,2),C′(2,0)A'(0,0), B'(0,2), C'(2,0)
13.A′(−2,4),B′(0,4),C′(−1,6)A'(-2,4), B'(0,4), C'(-1,6)
14.A′(0.5,0.5),B′(1.5,0.5),C′(1.5,1.5),D′(0.5,1.5)A'(0.5,0.5), B'(1.5,0.5), C'(1.5,1.5), D'(0.5,1.5)
15.A′(−1,−1),B′(−2,−1),C′(−1,−3)A'(-1,-1), B'(-2,-1), C'(-1,-3)
16.x=3x = 3
17.400 cm³
18.Same.
19.Same.
20.324 (= 36 × 9)
Gold
21.A′(2,2),B′(6,2),C′(4,6)A'(2,2), B'(6,2), C'(4,6)
22.Same.
23.DE = 14, DF = 21
24.75/12=5/2\sqrt{75/12} = 5/2
25.10, 24, 26
26.A′(0,0),B′(0,4),C′(−3,0)A'(0,0), B'(0,4), C'(-3,0)
27.33/53\sqrt{3^3/5^3}... see working
28.(a) 10 × 16 (b) 160 (c) 52
29.Same.
30.Yes; right angle at Q
Platinum
31.48 cm
32.PQR 108; ratio 1:3
33.3\sqrt{3}
34.(1,−2)(1, -2)
35.16.8
36.Same.
37.450 cm²
38.A′(0,0),B′(9,0),C′(0,12)A'(0,0), B'(9,0), C'(0,12)
39.25 : 64
40.Same for reflection.

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 6 (b) A′(1,−1),B′(4,−1),C′(1,−5)A'(1,-1), B'(4,-1), C'(1,-5) (c) A′(−1,1),B′(−1,4),C′(−5,1)A'(-1,1), B'(-1,4), C'(-5,1) (d) A′(3,−2),B′(6,−2),C′(3,2)A'(3,-2), B'(6,-2), C'(3,2)
Full working
(a) Base 3, height 4. Area = (1/2)(3)(4)=6(1/2)(3)(4) = 6.

(b) Reflection in xx-axis: (x,y)→(x,−y)(x, y) \to (x, -y).

(c) 90° anti: (x,y)→(−y,x)(x, y) \to (-y, x).

(d) Add (2,−3)(2, -3) to each.
2Problem 2
Answer
(a) 36 cm² (b) 40 cm (c) (2a/3)×(2b/3)(2a/3) \times (2b/3)
Full working
(a) Area sf = (2/3)2=4/9(2/3)^2 = 4/9. 81×4/9=3681 \times 4/9 = 36 cm².

(b) Perim sf =2/3= 2/3. 60×2/3=4060 \times 2/3 = 40 cm.

(c) Multiply both by 2/32/3.
3Problem 3
Answer
(a) 8 (b) 200 cm³ (c) 120 cm²
Full working
(a) Linear sf = 2. Volume sf = 23=82^3 = 8.

(b) 25×8=20025 \times 8 = 200 cm³.

(c) Area sf = 22=42^2 = 4. 30×4=12030 \times 4 = 120.
4Problem 4
Answer
(a) SAS (b) AC=PRAC = PR (equal) (c) Yes — congruent ⇒ similar (sf 1)
Full working
(a) SAS (two sides and the included angle).

(b) AC must equal PR (since the triangles are congruent).

(c) Congruent triangles are similar with sf = 1.
5Problem 5
Answer
Step 1: (−2,3)(-2, 3). Step 2: (−3,−2)(-3, -2). Step 3: (−2,−4)(-2, -4).
Full working
Step 1: Reflect (2,3)(2, 3) in yy-axis: (−2,3)(-2, 3).

Step 2: Rotate 90° anti about origin: (−y,x)=(−3,−2)(-y, x) = (-3, -2).

Step 3: Translate by (1,−2)(1, -2): (−3+1,−2−2)=(−2,−4)(-3 + 1, -2 - 2) = (-2, -4).
6Problem 6
Answer
(a) Both have a vertical line, horizontal shadow, same sun angle (AAA) (b) 21 m (c) ≈ 71.6°
Full working
(a) Sun-rays are parallel. Both triangles have a vertical, a horizontal, and the same sun-angle.

(b) flag7=124=3\tfrac{\text{flag}}{7} = \tfrac{12}{4} = 3. Flag = 21 m.

(c) tan⁡=12/4=3⇒θ≈71.6°\tan = 12/4 = 3 \Rightarrow \theta \approx 71.6°.
7Problem 7
Answer
(a) 64:34364:343 (b) 343 cm³ (c) 80 cm²
Full working
(a) Volume sf = 43:73=64:3434^3 : 7^3 = 64 : 343.

(b) Larger vol =64×343/64=343= 64 \times 343/64 = 343.

(c) Area sf = 16:4916:49. Smaller SA =245×16/49=80= 245 \times 16/49 = 80.
8Problem 8
Answer
(a) A′(2,−1),B′(5,−1),C′(2,−4)A'(2,-1), B'(5,-1), C'(2,-4) (b) A′′(−2,1),B′′(−5,1),C′′(−2,4)A''(-2,1), B''(-5,1), C''(-2,4) (c) Reflection in the yy-axis
Full working
(a) (x,y)→(x,−y)(x, y) \to (x, -y).

(b) (x,y)→(−x,−y)(x, y) \to (-x, -y). After step (a): apply to A′(2,−1)A'(2,-1) → (−2,1)(-2, 1), etc.

(c) Combined: (x,y)→(−x,y)(x, y) \to (-x, y) — that's a reflection in the yy-axis.
9Problem 9
Answer
(a) 2 × 10⁶ m² (b) 2 km²
Full working
Area sf =500002=2.5×109= 50000^2 = 2.5 \times 10^9. Real area =8×2.5×109= 8 \times 2.5 \times 10^9 cm² =2×1010= 2 \times 10^{10} cm² =2×106= 2 \times 10^6 m² =2= 2 km².
10Problem 10
Answer
(a) Both fully determined (b) Only if B's third side = 7 (c) Use cosine rule to find B's third side: c2=25+36−60cos⁡40°≈61−45.96≈15.04c^2 = 25 + 36 - 60\cos 40° \approx 61 - 45.96 \approx 15.04, c≈3.88c \approx 3.88. Not 7.
Full working
(a) A is SSS (uniquely defined). B is SAS (uniquely defined).

(b) They are congruent only if both produce the same triangle, which requires B's third side = 7.

(c) Cosine rule: c2=52+62−2(5)(6)cos⁡40°≈25+36−45.96≈15.04c^2 = 5^2 + 6^2 - 2(5)(6)\cos 40° \approx 25 + 36 - 45.96 \approx 15.04. c≈3.88c \approx 3.88. NOT 7 → A and B are different triangles.
11Problem 11
Answer
(a) A′(4,1),B′(13,1),C′(7,10)A'(4, 1), B'(13, 1), C'(7, 10) (b) 40.5
Full working
(a) Formula: A′=P+k(A−P)=(1,1)+3((2,1)−(1,1))=(1,1)+(3,0)=(4,1)A' = P + k(A - P) = (1, 1) + 3((2, 1) - (1, 1)) = (1, 1) + (3, 0) = (4, 1). Similarly for B, C.

(b) Area sf = 32=93^2 = 9. 4.5×9=40.54.5 \times 9 = 40.5.
12Problem 12
Answer
(a) 6 (b) 6 (c) Triangle: order 3, 3 lines
Full working
(a) A regular hexagon maps onto itself when rotated by 360°/6=60°360°/6 = 60°. Order of rotational symmetry = 6.

(b) 6 lines of reflective symmetry: 3 through opposite vertices, 3 through opposite sides.

(c) Equilateral triangle: order of rotational symmetry 3; 3 lines of reflective symmetry (each through a vertex and the midpoint of the opposite side).