Enlargement and similarity - Year 9 Mathematics

What you will learn

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

Worked example 0 Real-world example: reading a floor plan

A floor plan uses a scale of $1:50$ . On the plan, the living room measures $8$ cm by $6$ cm.

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

Key idea: lengths multiply by $k$ , but areas multiply by $k^2$ — 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 $O$ , and
a scale factor $k$ (where $k > 0$ ).

For each point $P$ , its image $P'$ lies on the ray $OP$ such that $OP' = k \times OP$ .

Image distance

OP' = k \times OP

If $k > 1$ , the image is larger than the original.
If $0 < k < 1$ , the image is smaller (a reduction).
If $k = 1$ , the image is identical in size (identity).

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

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

Each coordinate is multiplied by $k$ . 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 $k = 1$ ),
area or perimeter (unless $k = 1$ ).

Scaling rules

Lengths

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

Perimeter

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

Area

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

Volume

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

Worked example 2 Area and volume scaling

A model car is built at scale $1:18$ . The real car’s windscreen has area $1.2$ m $^2$ and the engine bay has volume $0.15$ m $^3$ .

Scale factor from real to model: $k = \dfrac{1}{18}$ .
Model windscreen area $= 1.2 \times \left(\dfrac{1}{18}\right)^2 = 1.2 \times \dfrac{1}{324} \approx 0.0037$ m $^2 \approx 37$ cm $^2$ .
Model engine volume $= 0.15 \times \left(\dfrac{1}{18}\right)^3 = 0.15 \times \dfrac{1}{5832} \approx 0.0000257$ m $^3 \approx 25.7$ cm $^3$ .

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 $ABC$ and $DEF$ are similar with $\angle A = \angle D$ , $\angle B = \angle E$ , $\angle C = \angle F$ . Given $AB = 6$ cm, $BC = 9$ cm, $AC = 12$ cm, and $DE = 4$ cm. Find $EF$ and $DF$ .

Scale factor $k = \dfrac{DE}{AB} = \dfrac{4}{6} = \dfrac{2}{3}$ .
$EF = k \times BC = \dfrac{2}{3} \times 9 = 6$ cm.
$DF = k \times AC = \dfrac{2}{3} \times 12 = 8$ cm.

Worked example 4 Shadow problem

A $1.6$ m tall student casts a $2.4$ m shadow at the same time that a flagpole casts a $9$ m shadow. Find the height of the flagpole.

The sun’s rays create similar triangles (same angles).
$\dfrac{\text{height of pole}}{\text{height of student}} = \dfrac{\text{pole shadow}}{\text{student shadow}}$ .
$\dfrac{h}{1.6} = \dfrac{9}{2.4}$ , so $h = 1.6 \times \dfrac{9}{2.4} = 1.6 \times 3.75 = 6$ m.

The flagpole is $6$ m tall.

4. Constructing enlarged figures

To construct an enlargement by hand:

Mark the centre of enlargement $O$ .
Draw rays from $O$ through each vertex of the original figure.
Measure the distance from $O$ to each vertex and multiply by $k$ .
Mark the image vertex on the ray at the new distance.
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

A triangle has vertices $A(2,1)$ , $B(4,1)$ , $C(2,3)$ . Enlarge with centre $O(0,0)$ and $k=3$ . State the image vertices.
A rectangle is enlarged by scale factor $k = 2$ . If the original has length $5$ cm and width $3$ cm, find the image length and width.
The original perimeter of a square is $20$ cm. What is the perimeter after enlargement with $k = 4$ ?
A shape has area $18$ cm $^2$ . After enlargement with $k = 3$ , what is the new area?
A cube has volume $27$ cm $^3$ . It is enlarged by factor $k = 2$ . What is the new volume?
Two similar triangles have corresponding sides $5$ cm and $15$ cm. State the scale factor.
Triangle $PQR \sim$ triangle $XYZ$ . If $PQ = 8$ , $QR = 12$ , and $XY = 6$ , find $YZ$ .
A map has scale $1:25\,000$ . Two towns are $8$ cm apart on the map. Find the real distance in km.
State whether each property is preserved under enlargement: (a) angle size, (b) side length, (c) area, (d) shape.
A photo measuring $10$ cm $\times$ $15$ cm is enlarged by factor $k = 1.5$ . Find the new dimensions.

Reasoning

Tier 2: mixed practice

Triangle $ABC$ has $AB = 10$ cm, $BC = 8$ cm, $CA = 6$ cm. Triangle $DEF$ is similar with $DE = 15$ cm. Find $EF$ and $FD$ .
A model bridge is built at scale $1:200$ . The real bridge is $340$ m long. How long is the model?
Two similar rectangles have areas $50$ cm $^2$ and $200$ cm $^2$ . Find the scale factor of their sides.
A $2$ m post casts a $3$ m shadow. At the same time, a tree casts a $12$ m shadow. How tall is the tree?
Enlarge the point $(5, -2)$ with centre $(1, 0)$ and scale factor $k = 3$ . Find the image coordinates.
A solid sphere has radius $4$ cm. A similar sphere has radius $12$ cm. How many times greater is the volume of the larger sphere?
Explain why two circles are always similar to each other.
A rectangular garden is $12$ m by $8$ m. A scale drawing uses $k = \dfrac{1}{100}$ . Find the area of the garden on the drawing in cm $^2$ .

Reasoning

Tier 3: explain and apply

A triangle is enlarged by factor $k = -1$ . Describe the transformation. How does this differ from a rotation of $180°$ about the centre?
Prove that the ratio of the areas of two similar figures equals the square of the ratio of corresponding sides.
Two similar cylinders have heights $10$ cm and $25$ cm. The smaller has volume $200$ cm $^3$ . Find the volume of the larger cylinder.
An architect’s model uses a scale of $1:50$ . A room in the model has floor area $48$ cm $^2$ . Find the real floor area in m $^2$ .
On a coordinate plane, $\triangle ABC$ has vertices $A(0,0)$ , $B(6,0)$ , $C(3,4)$ . The triangle is enlarged with centre $A$ and scale factor $\dfrac{1}{2}$ . Find the image vertices and verify that all sides are halved.

Challenge

Reasoning

Harder reasoning

Two similar cones have surface areas in the ratio $4:25$ . If the smaller cone has volume $80$ cm $^3$ , find the volume of the larger cone.
A photograph is enlarged by factor $k$ so that its area triples. Find the exact value of $k$ .
Point $P(8, 6)$ is enlarged with centre $C(2, 3)$ and scale factor $k$ . The image is $P'(17, 10.5)$ . Find $k$ .
Two similar solids have masses $m_1$ and $m_2$ , made of the same material. Show that $\dfrac{m_1}{m_2} = \left(\dfrac{l_1}{l_2}\right)^3$ where $l_1$ and $l_2$ are corresponding lengths.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

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Tier 1

$A'(6,3)$ , $B'(12,3)$ , $C'(6,9)$ .
Image length $= 10$ cm, image width $= 6$ cm.
$20 \times 4 = 80$ cm.
$18 \times 3^2 = 18 \times 9 = 162$ cm $^2$ .
$27 \times 2^3 = 27 \times 8 = 216$ cm $^3$ .
$k = \dfrac{15}{5} = 3$ .
$k = \dfrac{6}{8} = \dfrac{3}{4}$ . So $YZ = \dfrac{3}{4} \times 12 = 9$ .
$8 \times 25\,000 = 200\,000$ cm $= 2$ km.
(a) preserved, (b) not preserved, (c) not preserved, (d) preserved.
$15$ cm $\times$ $22.5$ cm.

Tier 2

$k = \dfrac{15}{10} = 1.5$ . $EF = 1.5 \times 8 = 12$ cm. $FD = 1.5 \times 6 = 9$ cm.
$340 \div 200 = 1.7$ m.
Area ratio $= \dfrac{200}{50} = 4$ . Side ratio $= \sqrt{4} = 2$ .
$\dfrac{h}{2} = \dfrac{12}{3}$ , so $h = 8$ m.
Vector from centre to point: $(5-1, -2-0) = (4, -2)$ . Multiply by $3$ : $(12, -6)$ . Image $= (1+12, 0-6) = (13, -6)$ .
$k = \dfrac{12}{4} = 3$ . Volume ratio $= 3^3 = 27$ . The larger sphere’s volume is $27$ times greater.
Any circle can be mapped to any other by a single enlargement (centred at any point) with scale factor equal to the ratio of the radii. Since all angles in a circle are determined by its curvature and all circles have the same shape, they are always similar.
Drawing dimensions: $12 \times \dfrac{1}{100} = 0.12$ m $= 12$ cm, and $8 \times \dfrac{1}{100} = 0.08$ m $= 8$ cm. Area $= 12 \times 8 = 96$ cm $^2$ .

Tier 3

A scale factor of $k = -1$ produces an image that is the same size but on the opposite side of the centre, with each point reflected through the centre. The result is identical to a $180°$ rotation about the centre — a negative scale factor combines enlargement with a half-turn.
Let two similar figures have corresponding sides in ratio $k:1$ . Divide each figure into the same small unit squares (or use the same triangulation). Each unit in the larger figure has sides $k$ times as long, so its area is $k^2$ times as large. Summing over all units, total area scales by $k^2$ .
Height ratio $= \dfrac{25}{10} = 2.5$ . Volume ratio $= 2.5^3 = 15.625$ . Larger volume $= 200 \times 15.625 = 3125$ cm $^3$ .
Scale factor from model to real: $k = 50$ . Real area $= 48 \times 50^2 = 48 \times 2500 = 120\,000$ cm $^2 = 12$ m $^2$ .
Centre is $A(0,0)$ , so multiply coordinates by $\dfrac{1}{2}$ . $A' = (0,0)$ , $B' = (3,0)$ , $C' = (1.5, 2)$ . $AB = 6 \to A'B' = 3$ . $AC = \sqrt{9+16} = 5 \to A'C' = \sqrt{2.25+4} = 2.5$ . $BC = \sqrt{9+16} = 5 \to B'C' = \sqrt{2.25+4} = 2.5$ . All sides halved.

Challenge

Surface area ratio $= 4:25$ , so side ratio $= \sqrt{4}:\sqrt{25} = 2:5$ . Volume ratio $= 2^3:5^3 = 8:125$ . Larger volume $= 80 \times \dfrac{125}{8} = 1250$ cm $^3$ .
New area $= k^2 \times$ original area $= 3 \times$ original area, so $k^2 = 3$ and $k = \sqrt{3}$ .
$P'_x - C_x = k(P_x - C_x)$ : $17 - 2 = k(8 - 2)$ , so $15 = 6k$ , giving $k = 2.5$ . Check: $P'_y - C_y = 10.5 - 3 = 7.5 = 2.5 \times (6 - 3) = 2.5 \times 3 = 7.5$ . Confirmed $k = 2.5$ .
Since the solids are similar with corresponding length ratio $\dfrac{l_1}{l_2}$ , volumes scale as $\left(\dfrac{l_1}{l_2}\right)^3$ . With the same material (same density $\rho$ ), mass $= \rho \times V$ , so $\dfrac{m_1}{m_2} = \dfrac{\rho V_1}{\rho V_2} = \dfrac{V_1}{V_2} = \left(\dfrac{l_1}{l_2}\right)^3$ .

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