Mathematics questions, chapter 5

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Congruency, Enlargement and Combined Transformations

40 marks. Use Practice to check each answer as you go, or Exam to answer everything first. Show your working for Section B.

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Question 1

Which statement about two congruent polygons is correct?

Only the corresponding angles are equal.
All corresponding sides and all corresponding angles are equal.
Only the corresponding sides are equal.
The corresponding angles are equal but the sides are in the same ratio.

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Congruent shapes have the same shape and size, so every corresponding side and every corresponding angle is equal. Equal angles with sides in a ratio describe similar shapes.

For the teacher: Remembering, low difficulty, standard 5.1.1

Textbook p. 125

Question 2

Which condition is not enough, by itself, to show that two triangles are congruent?

Three pairs of equal sides (SSS)
Two pairs of equal sides and the angle between them (SAS)
Three pairs of equal angles (AAA)
Two pairs of equal angles and the side between them (ASA)

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AAA only gives triangles of the same shape; they may be different in size. SSS, SAS and ASA each prove congruency on their own.

For the teacher: Understanding, low difficulty, standard 5.1.2

Textbook p. 128

Question 3

In triangles ABC and PQR, AB = PQ = 7 cm, BC = QR = 9 cm and ∠ABC = ∠PQR = 52°. Which rule shows that the triangles are congruent?

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The two pairs of equal sides, AB, PQ and BC, QR, have the equal angle B and Q between them, so the rule is Side-Angle-Side. SSA needs the angle not between the sides.

For the teacher: Applying, low difficulty, standard 5.1.2

Textbook p. 130

Question 4

△ABC ≅ △PQR with AB = PQ. Given ∠B = 50° and ∠C = 70°, what is the size of ∠P?

50°
70°
110°
60°

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∠A = 180° − 50° − 70° = 60°. A corresponds to P, so ∠P = 60°. 50° and 70° are the angles that correspond to Q and R.

For the teacher: Applying, low difficulty, standard 5.1.1

Textbook p. 125

Question 5

Which pair of rectangles is similar?

4 cm × 6 cm and 10 cm × 12 cm
4 cm × 6 cm and 10 cm × 15 cm
4 cm × 6 cm and 8 cm × 14 cm
4 cm × 6 cm and 6 cm × 8 cm

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10 ÷ 4 = 2.5 and 15 ÷ 6 = 2.5, so the sides are in the same ratio. Rectangles always have equal angles, so the ratios must also be equal; the other pairs give 2.5 and 2, 2 and 2.33, 1.5 and 1.33.

For the teacher: Understanding, medium difficulty, standard 5.2.1

Textbook p. 135

Question 6

Diagram 1 shows triangle N, the image of triangle M under an enlargement. What is the centre and scale factor of the enlargement?

Triangle M with vertices (2,3), (3,3), (3,4) and triangle N with vertices (4,5), (7,5), (7,8) on a grid from 0 to 8 — Diagram 1

Centre (2, 3), scale factor 3
Centre (1, 2), scale factor 13
Centre (1, 2), scale factor 3
Centre (0, 0), scale factor 3

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Join each object point to its image point: the lines meet at (1, 2). M has a base of 1 unit and N has a base of 3 units, so k = 3. Using the point (2, 3) as centre is a common slip of picking an object vertex.

For the teacher: Applying, medium difficulty, standard 5.2.2

Textbook p. 140

Question 7

The image of a triangle of area 6 cm² under an enlargement with scale factor −3 has an area of

18 cm²
−54 cm²
−18 cm²
54 cm²

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Area of image = k² × area of object = (−3)² × 6 = 54 cm². An area is never negative; 18 comes from multiplying by 3 instead of 9.

For the teacher: Applying, medium difficulty, standard 5.2.4

Textbook p. 146

Question 8

An enlargement has a scale factor of −12. The image is

smaller than the object and on the opposite side of the centre
smaller than the object and on the same side of the centre
larger than the object and on the opposite side of the centre
larger than the object and on the same side of the centre

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|k| = 1/2 is less than 1, so the image is smaller. The negative sign puts it on the opposite side of the centre, inverted.

For the teacher: Understanding, low difficulty, standard 5.2.2

Textbook p. 139

Question 9

The area of an object is 3 cm² and the area of its image is 75 cm² under an enlargement. Which of these could be the scale factor?

25
−25
−5
15

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k² = 75 ÷ 3 = 25, so k = 5 or −5. 25 is k², not k. The image is larger, so k cannot be 1/5.

For the teacher: Applying, medium difficulty, standard 5.2.4

Textbook p. 146

Question 10

B is a reflection on the x-axis and A is a translation 3−1. What is the image of point P(2, 3) under the combined transformation AB?

(5, 2)
(−1, 4)
(5, 4)
(5, −4)

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AB means B first: (2, 3) → (2, −3). Then A: (2 + 3, −3 − 1) = (5, −4). Doing A first gives (5, 2) and then (5, −2), which is the order BA mistake. (−1, 4) subtracts the vector.

For the teacher: Applying, medium difficulty, standard 5.3.1

Textbook p. 150

Question 11

The image of K under combined transformation VT is (−6, 8). T is an enlargement at the origin with scale factor 2 and V is a reflection on the y-axis. What is K?

(−3, 4)
(3, 4)
(−12, 16)
(6, 8)

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VT: T first, then V. To find K undo V first: (−6, 8) → (6, 8). Then undo T with k = 1/2: (3, 4). (−3, 4) forgets the reflection; (−12, 16) enlarges instead of reducing.

For the teacher: Analysing, high difficulty, standard 5.3.1

Textbook p. 152

Question 12

Which of the following is not an isometric transformation used to design a tessellation?

Enlargement
Translation
Reflection
Rotation

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An isometric transformation gives an image that is congruent with the object. Translation, reflection and rotation do, but an enlargement changes the size of the object (unless k = 1 or k = −1).

For the teacher: Remembering, low difficulty, standard 5.4.2

Textbook p. 164

Section B: structured questions

28 marks

Answer all questions. Show your working.

Question 13

State the triangle congruence rule that proves △ABC ≅ △PQR in each case.

[1 mark] (a)AB = PQ, ∠ABC = ∠PQR and BC = QR

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SAS [1]1

Tick each point you got.

[1 mark] (b)∠BAC = ∠QPR, ∠ABC = ∠PQR and BC = QR

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AAS [1]1

The equal side BC is not between the two equal angles A and B.

Tick each point you got.

[1 mark] (c)AB = PQ, BC = QR and AC = PR

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SSS [1]1

Tick each point you got.

For the teacher: Remembering, low difficulty, standard 5.1.2

Textbook p. 128

Question 14

Diagram 2 shows an isosceles triangle ABC with AB = AC. D and E are the midpoints of AB and AC. Show that triangle DBC and triangle ECB are congruent.

Isosceles triangle ABC with D on AB and E on AC at the midpoints; lines BE and CD are drawn — Diagram 2

[4 marks] (a)State the three facts about the triangles that are needed and the rule used.

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DB = EC (halves of equal sides AB and AC) [1]1
∠DBC = ∠ECB (base angles of an isosceles triangle) [1]1
BC is common to both triangles [1]1
SAS, so △DBC ≅ △ECB [1]1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 5.1.3

Textbook p. 131

Question 15

Diagram 3 shows triangle S and its image S′ under an enlargement at centre P(3, 4). S has area 1 unit².

Triangle S with vertices (4,3), (5,3), (5,5) and image S dash with vertices (1,6), (-1,6), (-1,2); centre P at (3,4) — Diagram 3

[3 marks] (a)Describe the enlargement in full.

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Scale factor: S′ base 2 units, S base 1 unit, ratio 2 [1]1
Opposite side of P, so k = −2 [1]1
Enlargement at centre (3, 4), scale factor −2 [1]1

Tick each point you got.

[1 mark] (b)Calculate the area of S′.

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(−2)² × 1 = 4 unit² [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 5.2.2

Textbook p. 140

Question 16

In Diagram 4, triangle ADE is the image of triangle ABC under an enlargement at centre A. AB = 6 cm, BD = 9 cm and the area of the shaded region is 105 cm².

Triangle ABC inside triangle ADE with common vertex A; AB 6 cm and BD 9 cm along the base; the region between the triangles is shaded and marked 105 cm squared — Diagram 4

[1 mark] (a)Calculate the scale factor of the enlargement.

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k = (6 + 9) ÷ 6 = 52 [1]1

Tick each point you got.

[2 marks] (b)Calculate the area of triangle ABC.

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(k² − 1) × area ABC = 105, so (254 − 1) × area ABC = 105 [1]1
Area ABC = 105 × 421 = 20 cm² [1]1

Tick each point you got.

[1 mark] (c)Calculate the area of triangle ADE.

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254 × 20 = 125 cm² [1]1

Check: 125 − 20 = 105.

Tick each point you got.

For the teacher: Applying, high difficulty, standard 5.2.5

Textbook p. 147

Question 17

Diagram 5 shows triangle T on a Cartesian plane. A is a translation −32 and B is a reflection on the line y = x.

Triangle T with vertices (1,1), (4,1) and (4,3) on a grid — Diagram 5

[3 marks] (a)Write the coordinates of the vertices of the image of T under the combined transformation AB.

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B first (swap x and y): (1, 1), (1, 4), (3, 4) [1]1
Then A: add (−3, 2) to each [1]1
(−2, 3), (−2, 6), (0, 6) [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 5.3.1

Textbook p. 153

Question 18

M is a reflection on the y-axis, T is a translation 20. A point has coordinates (3, 1).

[2 marks] (a)Find the image of the point under TM and under MT.

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TM: (−3, 1) then (−1, 1) [1]1
MT: (5, 1) then (−5, 1) [1]1

Tick each point you got.

[1 mark] (b)Does the combined transformation TM satisfy the commutative law? Give a reason.

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No, TM and MT give different images [1]1

Tick each point you got.

For the teacher: Understanding, medium difficulty, standard 5.3.2

Textbook p. 156

Question 19

Diagram 6 shows triangle E and its image E′ under the combined transformation ST. T is a reflection on the line x = 6.

Triangle E with vertices (1,1), (5,1), (5,3) and image E dash with vertices (14,2), (22,2), (14,6); the dashed line x = 6 is shown — Diagram 6

[2 marks] (a)Write the coordinates of the image of E under T.

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Reflection on x = 6: x becomes 12 − x [1]1
(11, 1), (7, 1), (7, 3) [1]1

Tick each point you got.

[2 marks] (b)Describe the transformation S in full.

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Enlargement at the origin [1]1
Scale factor 2, e.g. (7, 1) → (14, 2) [1]1

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 5.3.3

Textbook p. 158

Question 20

Diagram 7 shows a tessellation of equilateral triangles with sides of 2 units. Triangle Q is the image of P under an isometric transformation.

Tessellation of equilateral triangles; triangle P is up-pointing, Q is the down-pointing triangle touching P along edge BC, and R is the up-pointing triangle two units to the right of P — Diagram 7

[2 marks] (a)State two different transformations that map P onto Q.

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Reflection on line BC [1]1
Rotation of 60° clockwise at centre B [1]1

Tick each point you got.

[1 mark] (b)Describe the translation that maps P onto R.

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2 units to the right (parallel to the base) [1]1

Tick each point you got.

For the teacher: Understanding, medium difficulty, standard 5.4.2

Textbook p. 165

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