Year 9 Mathematics | Victorian Curriculum 2.0
Enlargement and similarity
Topic 13 | Space | Practice

What you will learn

  • describe and perform enlargement transformations using a centre of enlargement and scale factor kkk,
  • identify properties preserved under enlargement (angles equal, sides in proportion),
  • calculate how area scales by k2k^2k2 and volume scales by k3k^3k3,
  • identify similar figures and use proportional reasoning to find unknown side lengths,
  • apply algorithms for geometric constructions involving enlargement.
Why study enlargement and similarity?

Enlargement is the only transformation that changes the size of a figure while keeping its shape. Whenever you zoom in on a photo, scale a recipe, or read a map, you are working with enlargement and similarity. These ideas also underpin trigonometry — two right-angled triangles with the same angles are always similar, which is why the sine, cosine, and tangent ratios work for any size of triangle.

Where you'll see this
  • Architecture: blueprints are similar figures — every length is scaled down by the same factor.
  • Photography: cropping and resizing keeps objects in proportion because the image is enlarged or reduced.
  • Medicine: CT scans produce images at a known scale factor so doctors can measure real tumour sizes from a screen.
  • Engineering: model testing (wind tunnels, bridge models) relies on similarity to predict full-scale behaviour.
Worked example 0 Real-world example: reading a floor plan

A floor plan uses a scale of 1:501:501:50. On the plan, the living room measures 888 cm by 666 cm.

  1. The scale factor from plan to real life is k=50k = 50k=50.
  2. Real length =8×50=400= 8 \times 50 = 400=8×50=400 cm =4= 4=4 m.
  3. Real width =6×50=300= 6 \times 50 = 300=6×50=300 cm =3= 3=3 m.
  4. Real area =4×3=12= 4 \times 3 = 12=4×3=12 m2^22. Alternatively, plan area =48= 48=48 cm2^22 and real area =48×502=48×2500=120 000= 48 \times 50^2 = 48 \times 2500 = 120\,000=48×502=48×2500=120000 cm2=12^2 = 122=12 m2^22.

Key idea: lengths multiply by kkk, but areas multiply by k2k^2k2 — always square the scale factor for area.

1. Enlargement transformation

An enlargement (or dilation) maps every point in a figure to a new position using:

  • a centre of enlargement OOO, and
  • a scale factor kkk (where k>0k > 0k>0).

For each point PPP, its image P′P'P′ lies on the ray OPOPOP such that OP′=k×OPOP' = k \times OPOP′=k×OP.

Image distance
OP′=k×OPOP' = k \times OPOP′=k×OP
  • If k>1k > 1k>1, the image is larger than the original.
  • If 0<k<10 < k < 10<k<1, the image is smaller (a reduction).
  • If k=1k = 1k=1, the image is identical in size (identity).
3 cm4 cm5 cmABCk = 26 cm8 cm10 cmA’B’C’
Two similar triangles: the smaller triangle (left) enlarged by scale factor k = 2 to produce the larger triangle (right). Corresponding sides are labelled.
Worked example 1 Performing an enlargement

Triangle PQRPQRPQR has vertices P(1,1)P(1,1)P(1,1), Q(3,1)Q(3,1)Q(3,1), R(1,4)R(1,4)R(1,4). Enlarge with centre O(0,0)O(0,0)O(0,0) and scale factor k=2k=2k=2.

  1. P′=(2×1,  2×1)=(2,2)P' = (2 \times 1,\; 2 \times 1) = (2, 2)P′=(2×1,2×1)=(2,2).
  2. Q′=(2×3,  2×1)=(6,2)Q' = (2 \times 3,\; 2 \times 1) = (6, 2)Q′=(2×3,2×1)=(6,2).
  3. R′=(2×1,  2×4)=(2,8)R' = (2 \times 1,\; 2 \times 4) = (2, 8)R′=(2×1,2×4)=(2,8).

Each coordinate is multiplied by kkk. The image triangle is the same shape, with every side twice as long.

2. Properties preserved under enlargement

Enlargement preserves:

  • angles — every angle in the image equals the corresponding angle in the original,
  • shape — the figure and its image are similar,
  • parallelism — lines that were parallel remain parallel.

Enlargement does not preserve:

  • side lengths (unless k=1k = 1k=1),
  • area or perimeter (unless k=1k = 1k=1).

Scaling rules

Lengths

Image length=k×original length\text{Image length} = k \times \text{original length}Image length=k×original length

Perimeter

Image perimeter=k×original perimeter\text{Image perimeter} = k \times \text{original perimeter}Image perimeter=k×original perimeter

Area

Image area=k2×original area\text{Image area} = k^2 \times \text{original area}Image area=k2×original area

Volume

Image volume=k3×original volume\text{Image volume} = k^3 \times \text{original volume}Image volume=k3×original volume

Worked example 2 Area and volume scaling

A model car is built at scale 1:181:181:18. The real car’s windscreen has area 1.21.21.2 m2^22 and the engine bay has volume 0.150.150.15 m3^33.

  1. Scale factor from real to model: k=118k = \dfrac{1}{18}k=181​.
  2. Model windscreen area =1.2×(118)2=1.2×1324≈0.0037= 1.2 \times \left(\dfrac{1}{18}\right)^2 = 1.2 \times \dfrac{1}{324} \approx 0.0037=1.2×(181​)2=1.2×3241​≈0.0037 m2≈37^2 \approx 372≈37 cm2^22.
  3. Model engine volume =0.15×(118)3=0.15×15832≈0.0000257= 0.15 \times \left(\dfrac{1}{18}\right)^3 = 0.15 \times \dfrac{1}{5832} \approx 0.0000257=0.15×(181​)3=0.15×58321​≈0.0000257 m3≈25.7^3 \approx 25.73≈25.7 cm3^33.

3. Similar figures

Two figures are similar if one can be mapped to the other by a combination of enlargement (and possibly reflection, rotation, or translation). In practice, this means:

  • all corresponding angles are equal, and
  • all corresponding sides are in the same ratio.
Worked example 3 Finding unknown lengths using similarity

Triangles ABCABCABC and DEFDEFDEF are similar with ∠A=∠D\angle A = \angle D∠A=∠D, ∠B=∠E\angle B = \angle E∠B=∠E, ∠C=∠F\angle C = \angle F∠C=∠F. Given AB=6AB = 6AB=6 cm, BC=9BC = 9BC=9 cm, AC=12AC = 12AC=12 cm, and DE=4DE = 4DE=4 cm. Find EFEFEF and DFDFDF.

  1. Scale factor k=DEAB=46=23k = \dfrac{DE}{AB} = \dfrac{4}{6} = \dfrac{2}{3}k=ABDE​=64​=32​.
  2. EF=k×BC=23×9=6EF = k \times BC = \dfrac{2}{3} \times 9 = 6EF=k×BC=32​×9=6 cm.
  3. DF=k×AC=23×12=8DF = k \times AC = \dfrac{2}{3} \times 12 = 8DF=k×AC=32​×12=8 cm.
Match the right sides

When using similarity, always match corresponding vertices. If △ABC∼△DEF\triangle ABC \sim \triangle DEF△ABC∼△DEF, then ABABAB corresponds to DEDEDE, BCBCBC to EFEFEF, and ACACAC to DFDFDF. Writing vertices in the wrong order leads to wrong answers.

Worked example 4 Shadow problem

A 1.61.61.6 m tall student casts a 2.42.42.4 m shadow at the same time that a flagpole casts a 999 m shadow. Find the height of the flagpole.

  1. The sun’s rays create similar triangles (same angles).
  2. height of poleheight of student=pole shadowstudent shadow\dfrac{\text{height of pole}}{\text{height of student}} = \dfrac{\text{pole shadow}}{\text{student shadow}}height of studentheight of pole​=student shadowpole shadow​.
  3. h1.6=92.4\dfrac{h}{1.6} = \dfrac{9}{2.4}1.6h​=2.49​, so h=1.6×92.4=1.6×3.75=6h = 1.6 \times \dfrac{9}{2.4} = 1.6 \times 3.75 = 6h=1.6×2.49​=1.6×3.75=6 m.

The flagpole is 666 m tall.

4. Constructing enlarged figures

To construct an enlargement by hand:

  1. Mark the centre of enlargement OOO.
  2. Draw rays from OOO through each vertex of the original figure.
  3. Measure the distance from OOO to each vertex and multiply by kkk.
  4. Mark the image vertex on the ray at the new distance.
  5. Join the image vertices.

This algorithm can also be performed on the Cartesian plane by multiplying each coordinate relative to the centre.


Practice

Fluency

Tier 1: basic skills

    1. A triangle has vertices A(2,1)A(2,1)A(2,1), B(4,1)B(4,1)B(4,1), C(2,3)C(2,3)C(2,3). Enlarge with centre O(0,0)O(0,0)O(0,0) and k=3k=3k=3. State the image vertices.
    2. A rectangle is enlarged by scale factor k=2k = 2k=2. If the original has length 555 cm and width 333 cm, find the image length and width.
    3. The original perimeter of a square is 202020 cm. What is the perimeter after enlargement with k=4k = 4k=4?
    4. A shape has area 181818 cm2^22. After enlargement with k=3k = 3k=3, what is the new area?
    5. A cube has volume 272727 cm3^33. It is enlarged by factor k=2k = 2k=2. What is the new volume?
    6. Two similar triangles have corresponding sides 555 cm and 151515 cm. State the scale factor.
    7. Triangle PQR∼PQR \simPQR∼ triangle XYZXYZXYZ. If PQ=8PQ = 8PQ=8, QR=12QR = 12QR=12, and XY=6XY = 6XY=6, find YZYZYZ.
    8. A map has scale 1:25 0001:25\,0001:25000. Two towns are 888 cm apart on the map. Find the real distance in km.
    9. State whether each property is preserved under enlargement: (a) angle size, (b) side length, (c) area, (d) shape.
    10. A photo measuring 101010 cm ×\times× 151515 cm is enlarged by factor k=1.5k = 1.5k=1.5. Find the new dimensions.
Reasoning

Tier 2: mixed practice

    1. Triangle ABCABCABC has AB=10AB = 10AB=10 cm, BC=8BC = 8BC=8 cm, CA=6CA = 6CA=6 cm. Triangle DEFDEFDEF is similar with DE=15DE = 15DE=15 cm. Find EFEFEF and FDFDFD.
    2. A model bridge is built at scale 1:2001:2001:200. The real bridge is 340340340 m long. How long is the model?
    3. Two similar rectangles have areas 505050 cm2^22 and 200200200 cm2^22. Find the scale factor of their sides.
    4. A 222 m post casts a 333 m shadow. At the same time, a tree casts a 121212 m shadow. How tall is the tree?
    5. Enlarge the point (5,−2)(5, -2)(5,−2) with centre (1,0)(1, 0)(1,0) and scale factor k=3k = 3k=3. Find the image coordinates.
    6. A solid sphere has radius 444 cm. A similar sphere has radius 121212 cm. How many times greater is the volume of the larger sphere?
    7. Explain why two circles are always similar to each other.
    8. A rectangular garden is 121212 m by 888 m. A scale drawing uses k=1100k = \dfrac{1}{100}k=1001​. Find the area of the garden on the drawing in cm2^22.
Reasoning

Tier 3: explain and apply

    1. A triangle is enlarged by factor k=−1k = -1k=−1. Describe the transformation. How does this differ from a rotation of 180°180°180° about the centre?
    2. Prove that the ratio of the areas of two similar figures equals the square of the ratio of corresponding sides.
    3. Two similar cylinders have heights 101010 cm and 252525 cm. The smaller has volume 200200200 cm3^33. Find the volume of the larger cylinder.
    4. An architect’s model uses a scale of 1:501:501:50. A room in the model has floor area 484848 cm2^22. Find the real floor area in m2^22.
    5. On a coordinate plane, △ABC\triangle ABC△ABC has vertices A(0,0)A(0,0)A(0,0), B(6,0)B(6,0)B(6,0), C(3,4)C(3,4)C(3,4). The triangle is enlarged with centre AAA and scale factor 12\dfrac{1}{2}21​. Find the image vertices and verify that all sides are halved.

Challenge

Reasoning

Harder reasoning

    1. Two similar cones have surface areas in the ratio 4:254:254:25. If the smaller cone has volume 808080 cm3^33, find the volume of the larger cone.
    2. A photograph is enlarged by factor kkk so that its area triples. Find the exact value of kkk.
    3. Point P(8,6)P(8, 6)P(8,6) is enlarged with centre C(2,3)C(2, 3)C(2,3) and scale factor kkk. The image is P′(17,10.5)P'(17, 10.5)P′(17,10.5). Find kkk.
    4. Two similar solids have masses m1m_1m1​ and m2m_2m2​, made of the same material. Show that m1m2=(l1l2)3\dfrac{m_1}{m_2} = \left(\dfrac{l_1}{l_2}\right)^3m2​m1​​=(l2​l1​​)3 where l1l_1l1​ and l2l_2l2​ are corresponding lengths.
Year 9 Mathematics study companion | Practice