Finding missing lengths by similarity | Year 9

01 / LEARN

The idea, step by step

Find the multiplier once

When similarity is given, choose a pair of known corresponding lengths and divide image by original. Multiply every original length by the same factor to obtain its image. If a missing length belongs to the original figure, divide instead.

Write the paired side names beside your ratio. This prevents using a hypotenuse with a shorter side merely because their printed numbers happen to be close.

image length = k \times original length

Indirect measurements need geometric reasons

For a vertical object on level ground and a second vertical object measured at the same time, parallel sunlight produces equal angles. Their right triangles are similar, so height/shadow length is constant.

State these assumptions. If one shadow is on a steep slope or the observations are at different times, the matching-angle argument may fail. A correct proportion is only as useful as the model that justifies it.

A common trap

If the larger triangle is 1.5 times as large, subtract 1.5 to get a smaller length.

Instead: A scale factor is multiplicative. Divide an image length by 1.5 to reverse the enlargement.

02 / WORKED EXAMPLES

See the thinking

Example 1

Find an image length

In the supplied similar triangles, find $x$ .

AB corresponds to DE, so $k = 12 / 8 = 1.5$ .
AC corresponds to DF, so $x = 6 \times 1.5 = 9$ .

9 cm

Example 2

Use shadows

At the same time on level ground, a 1.5 m pole casts a 2 m shadow. A tree casts a 12 m shadow. Find the tree height.

The vertical pole/tree and parallel sunlight make similar right triangles.
Shadow scale factor is $12 / 2 = 6$ .
Tree height $= 1.5 \times 6 = 9$ m.

9 m

Check your understanding

An original side is 7 cm. Its corresponding image side is 21 cm. Another original side is 5 cm. Find its image.

Show answer

15 cm

$k = 21 / 7 = 3$ , so $5 \times 3 = 15$ cm.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Two similar triangles have corresponding sides 5 cm and 15 cm. Find the scale factor from the first to the second.
2 marks
With scale factor 3, an original side is 8 cm. Find its image length.
2 marks
A 28 cm image side was made using scale factor 4. Find its original length.
2 marks
Similar rectangles measure 8 cm by 5 cm and 20 cm by $w$ cm. Find $w$ .
2 marks
A 1.8 m person casts a 1.2 m shadow. A flagpole casts a 6 m shadow at the same time on level ground. Find its height.
3 marks
Similar triangles have perimeters 24 cm and 60 cm. A side of the larger is 22.5 cm. Find the corresponding smaller side.
3 marks

Read the worked solutions

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

Two similar triangles have corresponding sides 5 cm and 15 cm. Find the scale factor from the first to the second.
2 marks
$k = 15 / 5 = 3$ .
3
With scale factor 3, an original side is 8 cm. Find its image length.
2 marks
$8 \times 3 = 24$ cm.
24 cm
A 28 cm image side was made using scale factor 4. Find its original length.
2 marks
$28 \div 4 = 7$ cm.
7 cm
Similar rectangles measure 8 cm by 5 cm and 20 cm by $w$ cm. Find $w$ .
2 marks
$k = 20 / 8 = 2.5$ .
$w = 5 \times 2.5 = 12.5$ cm.
12.5 cm
A 1.8 m person casts a 1.2 m shadow. A flagpole casts a 6 m shadow at the same time on level ground. Find its height.
3 marks
Shadow scale factor $= 6 / 1.2 = 5$ .
Height $= 1.8 \times 5 = 9$ m.
9 m
Similar triangles have perimeters 24 cm and 60 cm. A side of the larger is 22.5 cm. Find the corresponding smaller side.
3 marks
Length factor is $60 / 24 = 2.5$ .
Smaller side $= 22.5 / 2.5 = 9$ cm.
9 cm

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Year 9 proportional modelling, rates, scale, similarity and enlargement. Assumptions, interpretation and model limitations are taught alongside calculation.

Official curriculum source ↗

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