Similarity and scale | Year 11

01 / LEARN

Understand the idea.

Match the figures before calculating

Similar figures have equal corresponding angles and one common ratio between corresponding lengths. Start by matching vertices; the order in $A B C \sim D E F$ gives $A \leftrightarrow D$ , $B \leftrightarrow E$ and $C \leftrightarrow F$ .

Then match the sides joining those vertices. A rotation or reflection can change their appearance without changing the correspondence. Do not choose a ratio because two sides happen to sit in similar positions on the page.

Lengths, areas and scale drawings

Define the direction of a length scale factor: $k = \frac{new length}{original length}$ . Multiply lengths and perimeter by $k$ ; divide by $k$ to work backwards. Convert corresponding measurements to common units before forming a ratio.

Area uses two length dimensions, so its factor is $k^{2}$ . For a plan, scale both dimensions into real lengths before multiplying them, or apply the squared scale factor to the plan area and then convert square units.

\begin{aligned} k & = \frac{new length}{original length} \\ new area & = k^{2} \times original area \end{aligned}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Triangles ABC and DEF are similar in that order. AB=6 cm, BC=9 cm, AC=11 cm and DE=15 cm. Find EF and DF.

Worked solution

1
Match the known side pair.
$A B \leftrightarrow D E, k = \frac{15}{6} = 2.5$
The stated vertex order makes these corresponding sides and defines the factor from ABC to DEF.
2
Scale the side matching BC.
$E F = 9 (2.5) = 22.5 cm$
BC and EF join the second and third corresponding vertices.
3
Scale the side matching AC.
$D F = 11 (2.5) = 27.5 cm$
AC and DF join the first and third corresponding vertices.
4
Check the common ratio.
$\frac{22.5}{9} = \frac{27.5}{11} = 2.5$
Both calculated sides use the same linear multiplier as the known pair.

Answer

EF=22.5 cm; DF=27.5 cm.

EXAMPLE 02

Worked example 2

A plan uses scale 1:400. A rectangular courtyard is 7 cm by 4.5 cm on the plan. Find its real dimensions and area.

Worked solution

1
Apply the length scale in centimetres.
$\begin{aligned} L & = 7 (400) = 2800 cm \\ W & = 4.5 (400) = 1800 cm \end{aligned}$
One centimetre on the plan represents four hundred centimetres in reality.
2
Convert both dimensions to metres.
$L = 28 m, W = 18 m$
There are one hundred centimetres in one metre.
3
Multiply the real dimensions.
$A = 28 (18) = 504 m^{2}$
A rectangle's area is length times width, so both dimensions must be scaled before multiplication.

Answer

28 m by 18 m; area 504 m².

03 / YOUR TURN

Try it yourself.

Check your understanding

Similar figures have length factor 3 from small to large. What is the area factor?

Show answer & reasoning

The area factor is $3^{2} = 9$ . Both length dimensions are tripled, so their product is multiplied by nine.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Similar triangles have corresponding sides 8 cm and 20 cm. A second side of the smaller triangle is 12 cm. Find the matching larger side.
2 marks
A model and real object have scale 1:50. A model length is 14 cm. Find the real length in metres.
2 marks
Two similar shapes have length factor 1.5 from A to B. Shape A has perimeter 32 cm. Find B’s perimeter.
2 marks
A triangle of area $18 {cm}^{2}$ is enlarged with length factor 4. Find the new area.
2 marks
Triangles PQR and XYZ are similar in that order. $P Q = 10$ , $Q R = 14$ , $X Y = 15$ . Find YZ.
2 marks
A student says: “The area doubled, so every side doubled.” Explain and give the actual length factor.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Similar triangles have corresponding sides 8 cm and 20 cm. A second side of the smaller triangle is 12 cm. Find the matching larger side.
2 marks
$k = \frac{20}{8} = 2.5$ .
$12 \times 2.5 = 30 cm$ .
30 cm
A model and real object have scale 1:50. A model length is 14 cm. Find the real length in metres.
2 marks
$14 \times 50 = 700 cm$ .
$700 cm = 7 m$ .
7 m
Two similar shapes have length factor 1.5 from A to B. Shape A has perimeter 32 cm. Find B’s perimeter.
2 marks
Perimeters scale by the length factor.
$32 \times 1.5 = 48 cm$ .
48 cm
A triangle of area $18 {cm}^{2}$ is enlarged with length factor 4. Find the new area.
2 marks
The area factor is $4^{2} = 16$ .
$18 \times 16 = 288 {cm}^{2}$ .
$288 {cm}^{2}$
Triangles PQR and XYZ are similar in that order. $P Q = 10$ , $Q R = 14$ , $X Y = 15$ . Find YZ.
2 marks
PQ matches XY, so $k = \frac{15}{10} = 1.5$ .
QR matches YZ, giving $14 \times 1.5 = 21$ .
21 units
A student says: “The area doubled, so every side doubled.” Explain and give the actual length factor.
3 marks
Area factor $k^{2} = 2$ .
Thus $k = \sqrt{2}$ ; doubling lengths would multiply area by 4.
The length factor is $\sqrt{2}$ , approximately 1.414, not 2.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: practical trigonometry, including similarity, right triangles, triangle area, sine/cosine rules and contextual interpretation.

Official curriculum source ↗

In this lesson

Understand the idea.

Match the figures before calculating

Lengths, areas and scale drawings

See the method in action.

Worked example 1

Match the known side pair.

Scale the side matching BC.

Scale the side matching AC.

Check the common ratio.

Worked example 2

Apply the length scale in centimetres.

Convert both dimensions to metres.

Multiply the real dimensions.

Try it yourself.

Check your understanding

Keep practising