Congruency, Enlargement and Combined Transformations · Form 5
Congruency, Enlargement and Combined Transformations: Worked Examples (KBAT)
Real-situation, multi-step problems where the transformation is hidden inside the story, ink cost from an enlargement, a game token under a combined transformation, and describing two reflections as one single equivalent transformation.
Worked example 1
A designer's logo is a triangle with area 6 cm². For a banner she enlarges the logo so that every side is 5 times as long, keeping the same shape.
The logo is filled with ink that costs RM0.80 per cm². Find how much more the ink for the enlarged logo costs than the ink for the original.
- Every side is 5 times as long, so the scale factor k = 5.
- Area of enlarged logo = k² × object area = 5² × 6 = 25 × 6 = 150 cm².
- Extra area to be inked = 150 − 6 = 144 cm².
- Extra ink cost = 144 × RM0.80 = RM115.20.
Worked example 2
In a puzzle game, a token starts on a grid at A(2, 3). Clearing Level 1 slides the token by the vector (3, −1) (3 units right, 1 unit down).
Clearing Level 2 then reflects the token in the x-axis. The prize sits at (5, −2).
Determine, with working, whether the token lands exactly on the prize after both levels.
- Level 1 translation: add (3, −1) to A(2, 3) → (2 + 3, 3 − 1) = (5, 2).
- Level 2 reflection in the x-axis: (x, y) → (x, −y), so (5, 2) → (5, −2).
- Compare the final position (5, −2) with the prize (5, −2): they are the same point.
Worked example 3
On a coordinate plan of a garden, a decorative shape is reflected in the vertical line x = 1, and its image is then reflected in the vertical line x = 4. To give the builder one clear instruction, describe fully the single transformation that maps the original shape onto the final image.
- Two reflections in parallel lines give a translation, and neither reflection changes size, so the shapes stay congruent.
- The translation is horizontal (perpendicular to the two vertical mirror lines).
- Its distance is twice the gap between the lines: 2 × (4 − 1) = 2 × 3 = 6 units.
- The direction is from the first mirror (x = 1) toward the second (x = 4), i.e. the positive x-direction. Check: a point at x maps to 2(1) − x = 2 − x, then to 2(4) − (2 − x) = 6 + x, confirming x → x + 6.
Worked example 4
In a design, a motif is first reflected in the x-axis and then reflected in the line y = x. By tracking a general point (x, y), show that these two reflections together are equivalent to a single transformation, and describe it fully.
- Reflection in the x-axis: (x, y) → (x, −y)
- Reflection in y = x swaps the coordinates: (x, −y) → (−y, x)
- Net effect: (x, y) → (−y, x)
- This is exactly a 90° anticlockwise rotation about O, which sends (x, y) → (−y, x)
- Check (2, 0): x-axis → (2, 0); y = x → (0, 2); and the rotation also gives (0, 2) ✓
Worked example 5
A photo measuring 15 cm by 10 cm is enlarged so that its area becomes 600 cm². Find the scale factor of the enlargement, the new dimensions, and the percentage increase in the perimeter.
- Original area = 15 × 10 = 150 cm²
- Area scale factor = 600 ÷ 150 = 4, so k² = 4 and k = 2
- New dimensions = 15 × 2 by 10 × 2 = 30 cm by 20 cm
- Original perimeter = 2(15 + 10) = 50 cm; new perimeter = 2(30 + 20) = 100 cm
- Percentage increase = (100 − 50) ÷ 50 × 100% = 100%
Worked example 6
A floor is tiled with congruent triangular tiles, each of area 50 cm². A decorative version of the tile is made with every side 1.5 times as long.
Find how many times bigger the decorative tile's area is, its actual area, and the total area covered by 200 such decorative tiles, in m².
- Area scale factor = (1.5)² = 2.25, so the decorative tile is 2.25 times the area
- Decorative tile area = 50 × 2.25 = 112.5 cm²
- 200 tiles = 200 × 112.5 = 22500 cm²
- Convert to m²: 22500 ÷ 10000 = 2.25 m²
Worked example 7
A mural artist paints a triangular design of area 4 m² on a small wall. For a bigger wall, she enlarges the design, keeping the same shape, so that each side becomes 2.5 times as long.
Special mural paint costs RM45 per tin, and one tin covers exactly 4.5 m². (a) Find the area of the enlarged mural.
(b) Find the number of tins of paint she must buy, given that a partly-used tin still counts as one full tin, and the total cost of the paint.
- Area scale factor = k² = 2.5² = 6.25
- Area of enlarged mural = 4 × 6.25 = 25 m²
- Number of tins needed = 25 ÷ 4.5 = 5.56 (to 2 d.p.)
- Since a partly-used tin still counts as a full tin, round up: 6 tins are needed
- Total cost = 6 × RM45 = RM270
Worked example 8
A ceiling fan design has one blade rotated 90° clockwise about the centre of the fan to create a second blade, and that second blade is then rotated a further 90° clockwise about the same centre to create a third blade. By tracking how a general point P(x, y) on the first blade moves under each step, show that the third blade's position can be obtained from the first blade by a single transformation, and describe that transformation fully.
- First 90° clockwise rotation about the centre: (x, y) → (y, −x)
- Apply the same rule again to (y, −x): (y, −x) → (−x, −y)
- So overall, (x, y) → (−x, −y) after the two rotations
- (x, y) → (−x, −y) is the rule for a single rotation of 180° about the centre
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
Why do KBAT questions on transformations feel so different from normal exercises?
KBAT items here rarely say "find the image" directly. Instead they embed transformations in a scenario, a map, a design, or a sequence of moves, and expect you to identify which transformation(s) occurred, or reason about the effect on area and shape.
You need to interpret the situation before applying any rule.
What do examiners reward in a combined-transformation answer?
Full marks need every transformation stated with its type, centre or line, and scale factor or angle, applied in the given order, with the intermediate image shown clearly before the next transformation acts on it. A correct final diagram without labelled working earns little; the reasoning is what's assessed.
What mistakes cost the most marks in this topic?
Two are common: applying transformations in the wrong order (combined transformations are not commutative, so reversing the sequence changes the image), and mixing up length scale factor k with area scale factor k². Losing track of which shape is the object versus the image after each step also causes errors.