Mathematics notes, chapter 5
Congruency, Enlargement and Combined Transformations
Textbook Mathematics Form 5 (DLP), pages 122–171
What this chapter covers
- Differentiate between congruent and non-congruent shapes based on sides and angles (5.1.1)
- Make and verify the conjecture of triangle congruency (SSS, SAS, ASA, AAS) and solve problems involving congruency (5.1.2, 5.1.3)
- Explain similarity of geometric objects and connect it to enlargement, describing enlargement by centre and scale factor (5.2.1, 5.2.2)
- Determine the image and object of an enlargement and use area of image = k² × area of object (5.2.3, 5.2.4, 5.2.5)
- Determine the image and object of a combined transformation and test the commutative law (5.3.1, 5.3.2)
- Describe a combined transformation and solve problems involving it (5.3.3, 5.3.4)
- Explain tessellation and design tessellations using isometric transformations (5.4.1, 5.4.2)
5.1Congruency
Congruent shapes have the same shape and the same size, wherever they are placed and however they are turned. Their corresponding sides are equal and their corresponding angles are equal.
To test two polygons, work out every missing side and angle first, then compare each corresponding pair. If one pair is different, the shapes are not congruent. Congruent triangles are written with the symbol ≅, for example △ABC ≅ △PQR (textbook p. 125-126).
Write the letters in matching order. If △ABC ≅ △PQR, then A matches P, B matches Q and C matches R, so AB = PQ and ∠B = ∠Q.
Conditions for congruent triangles
| Rule | What must be equal | Works alone? |
|---|---|---|
| SSS | All three pairs of sides | Yes |
| SAS | Two pairs of sides and the angle between them | Yes |
| ASA | Two pairs of angles and the side between them | Yes |
| AAS | Two pairs of angles and a side that is not between them | Yes |
| AAA | All three pairs of angles | No: same shape, different size |
| SSA | Two pairs of sides and an angle that is not between them | No |
AAA and SSA only prove congruency when the two triangles are also known to have the same area (p. 128-129). For right-angled triangles, RHS (equal hypotenuse and one other side) is a special case of SSA that does work (p. 130).
Example (based on textbook p. 130): which rule?
Triangles EFG and KLM have EF = KL, ∠FEG = ∠LKM and EG = KM.Two sides and the angle between them are equal, so the rule is SAS.
If instead ∠EGF = ∠KML, ∠EFG = ∠KLM and EF = KL, then two angles and a side that is not between them are equal, so the rule is AAS.
Using "AAA" or "SSA" as a proof. Three equal angles only make triangles similar; a 3-4-5 triangle and a 6-8-10 triangle have the same angles but are not congruent.
Solving problems with congruency
Example (textbook p. 131): show two triangles are congruent
In isosceles △PQR, PR = QR. A and B are the midpoints of PR and QR. Show that △PBR ≅ △QAR.• PR = QR (given)
• ∠PRB = ∠QRA (the same angle R)
• BR = AR (halves of equal sides)
Two sides and the included angle are equal, so △PBR ≅ △QAR by SAS.
Give each fact its reason, then state the rule, then state the conclusion. Once two triangles are congruent you may use any other pair of corresponding sides or angles.
△ABC and △DEF have AB = DE = 7 cm, ∠B = ∠E = 40° and ∠C = ∠F = 65°. Are they congruent? Give the rule.
Show answerHide answer
∠A = 180° − 40° − 65° = 75° and ∠D = 75°. The side AB lies between ∠A and ∠B, so two angles and the side between them are equal: congruent by ASA.
5.2Enlargement
Similar objects
Two objects are similar when all corresponding angles are equal and all ratios of corresponding sides are equal (p. 134).
For triangles, equal angles are enough (the sides then follow). For shapes with more than three sides, check both: two rectangles need equal side ratios as well, because all their angles are 90° anyway (p. 135).
Example: are these rectangles similar?
Rectangle 1 is 6 cm × 4 cm. Rectangle 2 is 15 cm × 10 cm. 15 ÷ 6 = 2.5 and 10 ÷ 4 = 2.5. The ratios are equal, so they are similar.A rectangle 12 cm × 6 cm compared with 6 cm × 4 cm gives 12 ÷ 6 = 2 but 6 ÷ 4 = 1.5. Not similar.
Enlargement, centre and scale factor
An enlargement moves every point of an object away from (or towards) a fixed point, the centre of enlargement, by the same ratio, the scale factor k. The object and its image are similar (p. 138).
Scale factor k = distance of image point from the centre ÷ distance of object point from the centre = length of image side ÷ length of object side.
| Scale factor | Size of image | Position |
|---|---|---|
| k > 1 | Larger | Same side of the centre |
| 0 < k < 1 | Smaller | Same side of the centre |
| k = 1 | Same size | Same place |
| −1 < k < 0 | Smaller | Opposite side, inverted |
| k = −1 | Same size | Opposite side, inverted |
| k < −1 | Larger | Opposite side, inverted |
To find the centre, join each object point to its image point. The lines meet at the centre of enlargement (p. 140). To describe an enlargement, always state both the centre and the scale factor.
Example (based on textbook p. 140): describe the enlargement
Object side AB = 4 units, image side A′B′ = 6 units, image on the same side of the centre. k = 6 ÷ 4 = 3/2.If a side of 3 units becomes 6 units but the image is on the opposite side, k = −(6 ÷ 3) = −2.
Forgetting the sign. A scale factor of 2 and a scale factor of −2 give images of the same size, but the second is on the other side of the centre. Also, k is measured from the centre, not from one object point to another.
Area of the image
Area of image = k² × area of object. The area changes by the square of the scale factor (p. 146).
Example (based on textbook p. 146)
(a) Object 5 cm², image 45 cm²: k² = 45 ÷ 5 = 9, so k = ±3.(b) Object 100 m², k = −4: image = (−4)² × 100 = 1 600 m². The negative sign disappears when squaring.
(c) k = 1/2, image 7 cm²: object = 7 ÷ (1/4) = 28 cm².
Lengths multiply by k, areas multiply by k². Doubling the sides makes the area 4 times as big, not 2 times.
Example (textbook p. 147 idea): area between object and image
Triangle ADE is the image of triangle ABC at centre A with AB = 6 cm and AD = 15 cm. So k = 15 ÷ 6 = 5/2 and k² = 25/4. The shaded region is the image minus the object: (25/4 − 1) × area ABC = 21/4 × area ABC. If the region is 105 cm², area ABC = 105 × 4/21 = 20 cm² and area ADE = 125 cm².The image of a 12 cm² triangle under an enlargement has area 27 cm². Find the scale factor if the image is on the same side as the object.
Show answerHide answer
k² = 27 ÷ 12 = 9/4, so k = 3/2 (positive because the image is on the same side). The negative value −3/2 is rejected.
5.3Combined transformation
A combined transformation AB means transformation B first, then transformation A (p. 150). AA is written A².
| Transformation | What to give | Property of the image |
|---|---|---|
| Translation | Vector | Congruent, same orientation |
| Reflection | Line (mirror line) | Congruent, inverted |
| Rotation | Centre, angle, direction | Congruent, turned |
| Enlargement | Centre, scale factor | Similar, different size |
Example: image of a point under AB
P(2, 3). B = reflection on the x-axis, A = translation (3, −1).B first: P(2, 3) → P′(2, −3).
Then A: P′ → (2 + 3, −3 − 1) = (5, −4).
Finding the object
To find the object when the image is given, reverse the order and undo each step: translation by the opposite vector, rotation in the opposite direction, enlargement with the reciprocal scale factor. A reflection in the same line undoes itself (p. 150-153).
Example: object of a point
The image of K under VT is (−6, 8), where T = enlargement at the origin, k = 2, and V = reflection on the y-axis. T is done first, then V, so undo V first: (6, 8). Then undo T with k = 1/2: (3, 4). Check: (3, 4) → (6, 8) → (−6, 8).Doing the steps in the wrong order. For a combined transformation AB, B happens first. When finding the object of AB, undo A first and then B.
Commutative law
Combined transformation AB satisfies the commutative law if AB and BA give the same image. This is true only for some pairs (p. 155-156).
Example: does it commute?
Point (3, 1). M = reflection on the y-axis, T = translation (2, 0).TM: M gives (−3, 1), then T gives (−1, 1).
MT: T gives (5, 1), then M gives (−5, 1).
The images are different, so TM does not satisfy the commutative law.
In contrast, a rotation of 180° at the origin and M give (3, −1) in either order, so they commute.
Describing a combined transformation
- Congruent, same orientation, every point moved the same distance: translation.
- Congruent, inverted: reflection. Congruent, turned: rotation.
- Similar but a different size: enlargement.
- For an image under ST, describe T (the first step) before S.
Not every combination equals a single transformation. When it does, check the result by following one vertex through the steps (p. 157-158).
Triangle G has area 4 unit². Under the combined transformation PQ, where Q is a reflection and P is an enlargement with k = −3, find the area of the image.
Show answerHide answer
A reflection does not change the area. The enlargement gives (−3)² × 4 = 36 unit².
5.4Tessellation
A tessellation is a pattern of repeated shapes that fills a plane with no gaps and no overlaps (p. 161).
- A tessellation may use one shape or a combination, for example squares with regular octagons.
- Extra check (not in the textbook): shapes meeting at a point must have angles adding up to 360°. Regular triangles (60°), squares (90°) and regular hexagons (120°) work alone. A regular pentagon (108°) does not, because 360 ÷ 108 is not a whole number.
- A tessellation can be built from one shape using the isometric transformations: translation, reflection and rotation. These give congruent images, so every shape in the pattern is congruent to the others (p. 164).
Example (based on textbook p. 164): transformations in a triangle tessellation
In a tessellation of equilateral triangles, let P be one triangle and Q the triangle that shares an edge with it. Q is the image of P under a reflection on the shared edge, or under a rotation of 60° at the end of that edge, in the direction that turns P onto Q.The next triangle with the same orientation in the same row is the image of P under a translation of one side length along the row.
Saying a pattern is a tessellation when one shape is left out or when shapes overlap. Check that every piece repeats and that no gaps remain.
Notes follow Mathematics Form 5 (DLP), Chapter 5, p. 124-171. Figures are redrawn with new numbers.