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Lesson 11 ~35 min Unit 3 · Geometry +85 XP

Similarity and Scale Factors

Same shape, different size. Learn how scale factors connect lengths, areas, and volumes -- and why a scale factor of 2 makes the area four times bigger.

Today's hook: A model car is built to 1:24 scale. The real car is 4.8 metres long. How long is the model? If the real car's bonnet has area 2.4 m², what is the model bonnet's area? Why is it not simply 2.4 / 24?
0/5QUESTS
Think First
warm-up

A photocopier enlarges a photo so every dimension doubles. The original photo is 10 cm by 15 cm. What are the new dimensions? What happens to the area? Does the area also double? Explain your reasoning.

Record your answer in your workbook.
1
The Big Idea
+5 XP to read

Similar figures have the same shape but not necessarily the same size. Every length is multiplied by the same number -- the scale factor. But areas scale by the square of the scale factor, and volumes scale by the cube. This is one of the most powerful ideas in geometry.

If the scale factor is $k$, then lengths become $k$ times longer, areas become $k^2$ times larger, and volumes become $k^3$ times larger. A scale factor greater than 1 is an enlargement. Less than 1 is a reduction.

A 2 3 4A 4 6 k = 2 lengths ×2, area ×4
length $\times k$  |  area $\times k^2$  |  volume $\times k^3$
Scale factor = ratio
Always divide new by original: $k = \frac{\text{new}}{\text{original}}$.
Area scales by k²
Doubling lengths quadruples area. Halving lengths quarters area.
Volume scales by k³
Doubling lengths makes volume 8 times larger.
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What You'll Master
objectives

Know

  • Similar figures have the same shape but not necessarily the same size
  • The three similarity tests: AAA, SSS ratio, SAS ratio + angle
  • How scale factor relates corresponding lengths

Understand

  • Why area scales by $k^2$ and volume by $k^3$
  • The difference between enlargement and reduction
  • Why congruent is a special case of similar ($k = 1$)

Can Do

  • Calculate scale factors from corresponding lengths
  • Use scale factors to find unknown lengths
  • Calculate area and volume of similar figures using $k^2$ and $k^3$
3
Words You Need
vocabulary
SimilarSame shape, proportional sides, equal corresponding angles. Symbol: $\sim$ or $\|$.
Scale factorThe ratio of corresponding lengths in similar figures. $k = \frac{\text{new}}{\text{original}}$.
EnlargementA transformation with scale factor $k > 1$. The image is larger than the original.
ReductionA transformation with scale factor $0 < k < 1$. The image is smaller than the original.
CorrespondingMatching parts in the same relative position in similar figures.
ProportionalIn the same ratio. Corresponding sides of similar figures are proportional.
4
Spot the Trap
heads-up

Wrong: Similar triangles must be the same size.

Right: Similar triangles have the same shape (equal angles) but can be different sizes. The ratio of corresponding sides is the scale factor.

Wrong: The scale factor is the difference between corresponding sides.

Right: The scale factor is the ratio of corresponding sides, not the difference. Scale factor = new length / original length.

5
The Three Similarity Tests
+5 XP

Unlike congruence which has four tests, similarity has three. The key difference: similarity only requires the same shape, not the same size. So sides can be in proportion rather than equal.

AAA (Angle-Angle-Angle): all corresponding angles equal. Since angles fix the shape, the sides must be in proportion. SSS ratio: all three pairs of sides in the same ratio. SAS ratio + angle: two sides in ratio with the included angle equal.

AAA all angles equal shape is fixed SSS sides in ratio same proportion SAS two sides in ratio + included angle equal ratio and angle fix the shape
AAA  |  SSS ratio  |  SAS ratio + angle
AAA needs only two angles
The third angle is fixed by 180°. So two equal angles is enough.
SSS ratio = proportion
Check: $\frac{a}{d} = \frac{b}{e} = \frac{c}{f}$.
Congruent is similar
When $k = 1$, similar becomes congruent.
6
Scale Factor
+5 XP

The scale factor $k$ is the multiplier that takes every length in the original figure to the corresponding length in the similar figure. It is always a ratio, never a difference.

If $k > 1$, the image is an enlargement -- bigger than the original. If $0 < k < 1$, the image is a reduction -- smaller. If $k = 1$, the figures are congruent. The scale factor formula is always $k = \frac{\text{image length}}{\text{original length}}$.

k = 0.5 reduction k = 1 congruent k = 2 enlargement k = image / original
$k = \frac{\text{image}}{\text{original}}$
Always divide new by old
$k = \frac{\text{new}}{\text{original}}$ never the other way around.
Check with a second pair
Verify $k$ is the same for all corresponding sides.
k = 1 means congruent
Congruent triangles are a special case of similar.
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Area and Volume Scale Factors
+5 XP

When lengths scale by $k$, areas scale by $k^2$ and volumes scale by $k^3$. This is because area depends on two dimensions (length and width) and volume depends on three (length, width, and height).

Think of a square with side $s$. Its area is $s^2$. If the side becomes $ks$, the area becomes $(ks)^2 = k^2 s^2$ -- that is $k^2$ times the original area. For a cube, volume goes from $s^3$ to $(ks)^3 = k^3 s^3$ -- $k^3$ times the original.

LENGTH × k AREA × k² VOLUME × k³ each dimension multiplies by k
$A_{\text{new}} = k^2 \cdot A_{\text{old}}$
Square for area
If $k = 3$, area is $3^2 = 9$ times larger.
Cube for volume
If $k = 3$, volume is $3^3 = 27$ times larger.
Work backwards too
From area ratio, $k = \sqrt{\text{area ratio}}$.
Watch Me Solve It · Find missing sides
+15 XP per step
Q1
PROBLEM
$\triangle ABC \sim \triangle DEF$. $AB = 6$ cm, $BC = 8$ cm, $AC = 10$ cm, and $DE = 9$ cm. Find $EF$ and $DF$.
  1. 1
    Find the scale factor
    $k = \frac{DE}{AB} = \frac{9}{6} = 1.5$
    Scale factor = image length / original length. $DE$ corresponds to $AB$.
  2. 2
    Apply the scale factor to the other sides
    $EF = BC \times k = 8 \times 1.5 = 12$ cm
    $DF = AC \times k = 10 \times 1.5 = 15$ cm
    Every side in the image is 1.5 times the corresponding original side.
  3. 3
    Verify by checking ratios
    $\frac{DE}{AB} = \frac{9}{6} = 1.5$, $\frac{EF}{BC} = \frac{12}{8} = 1.5$, $\frac{DF}{AC} = \frac{15}{10} = 1.5$
    All three ratios equal 1.5. The sides are proportional -- confirming similarity.
Answer$EF = 12$ cm, $DF = 15$ cm
Watch Me Solve It · Photo enlargement
+15 XP per step
Q2
PROBLEM
A photograph is enlarged so that its dimensions are tripled. By what factor does the area increase? Explain your reasoning.
  1. 1
    Identify the linear scale factor
    $k = 3$ (dimensions are tripled)
    "Tripled" means every length becomes 3 times longer.
  2. 2
    Apply the area scale factor rule
    Area scale factor = $k^2 = 3^2 = 9$
    Area depends on two dimensions. Both length and width are multiplied by 3, so area is multiplied by $3 \times 3 = 9$.
  3. 3
    State the conclusion
    The area increases by a factor of 9
    A concrete example: original 10 cm by 15 cm has area 150 cm². Tripled: 30 cm by 45 cm has area 1350 cm². $1350 / 150 = 9$.
AnswerArea increases by a factor of 9
Watch Me Solve It · Working backwards from area
+15 XP per step
Q3
PROBLEM
Two similar triangles have areas 36 cm² and 144 cm². The shortest side of the smaller triangle is 4 cm. Find the shortest side of the larger triangle.
  1. 1
    Find the area scale factor
    $\frac{\text{area}_{\text{large}}}{\text{area}_{\text{small}}} = \frac{144}{36} = 4$
    Divide the larger area by the smaller area to get the area ratio.
  2. 2
    Find the linear scale factor
    $k^2 = 4$ → $k = \sqrt{4} = 2$
    Since area scales by $k^2$, the linear scale factor is the square root of the area ratio.
  3. 3
    Find the unknown side
    Shortest side of larger triangle = $4 \times 2 = 8$ cm
    Multiply the original side by the linear scale factor $k = 2$.
Answer$8$ cm
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Common Pitfalls
heads-up
Using difference instead of ratio
Calculating scale factor as the difference between corresponding sides (e.g., $9 - 6 = 3$) instead of the ratio ($9 / 6 = 1.5$). Scale factor is always a multiplicative relationship, never additive.
Fix: always divide new by original. $k = \frac{\text{new}}{\text{original}}$.
Forgetting that area scales by $k^2$
When dimensions double, claiming the area also doubles. Area depends on two dimensions, so it quadruples. This error is extremely common in exam questions.
Fix: write "length $\times k$, area $\times k^2$, volume $\times k^3$" on every similar figure problem.
Mixing up corresponding sides
Pairing the shortest side of one triangle with the longest side of the other. This gives an inconsistent scale factor and incorrect answers for the remaining sides.
Fix: always match shortest to shortest, middle to middle, longest to longest. Verify $k$ is the same for all pairs.
Copy Into Your Books

Similarity Tests

  • AAA -- all angles equal
  • SSS ratio -- sides proportional
  • SAS -- two sides in ratio + included angle

Scale Factor

  • $k = \frac{\text{new}}{\text{original}}$
  • $k > 1$ = enlargement
  • $k < 1$ = reduction
  • $k = 1$ = congruent

Area and Volume

  • Length: $\times k$
  • Area: $\times k^2$
  • Volume: $\times k^3$

Working Backwards

  • From area ratio: $k = \sqrt{\text{ratio}}$
  • From volume ratio: $k = \sqrt[3]{\text{ratio}}$

How are you completing this lesson?

D
Brain Trainer · Scale factor drill
4 problems

Four quick problems on scale factors. Work each one, then reveal the answer.

  1. 1 $\triangle ABC \sim \triangle DEF$. $AB = 4$ cm, $DE = 12$ cm. Find the scale factor.

    $k = \frac{DE}{AB} = \frac{12}{4} = 3$$k = 3$
  2. 2 $\triangle ABC \sim \triangle DEF$. $BC = 6$ cm, scale factor $k = 3$. Find $EF$.

    $EF = BC \times k = 6 \times 3 = 18$ cm$EF = 18$ cm
  3. 3 $\triangle ABC \sim \triangle DEF$. $AC = 9$ cm, $DF = 3$ cm. Find the scale factor and state whether it is an enlargement or reduction.

    $k = \frac{DF}{AC} = \frac{3}{9} = \frac{1}{3}$. Since $k < 1$, this is a reduction.$k = \frac{1}{3}$, reduction
  4. 4 Two similar cubes have side lengths 2 cm and 6 cm. By what factor does the volume increase?

    $k = \frac{6}{2} = 3$. Volume scale factor = $k^3 = 3^3 = 27$. Small volume = $8$ cm³, large volume = $216$ cm³. $216 / 8 = 27$.Volume increases by factor of 27
Complete in your workbook.
MC1
Definition of similar
+10 XP

Two figures are similar if:

MC2
Calculating scale factor
+10 XP

The scale factor from a 3 cm side to a 9 cm corresponding side is:

MC3
Area scale factor
+10 XP

If the scale factor is 2, the area scale factor is:

MC4
Reduction or enlargement
+10 XP

A scale factor less than 1 produces:

MC5
Corresponding angles
+10 XP

Corresponding angles in similar triangles are:

Q6
Similar triangle calculations
+15 XP
Q6
SHORT ANSWER
$\triangle ABC \sim \triangle DEF$. $AB = 5$ cm, $BC = 7$ cm, $AC = 9$ cm, and $DE = 15$ cm. (a) Find the scale factor. (b) Find the lengths of $EF$ and $DF$. (c) If the area of $\triangle ABC$ is 14 cm², what is the area of $\triangle DEF$?
Write your working in your book.
Q7
Photo enlargement
+15 XP
Q7
SHORT ANSWER
A photograph is enlarged so that its dimensions are tripled. By what factor does the area increase? Explain your reasoning.
Write your working in your book.
Q8
Congruent vs similar
+15 XP
Q8
SHORT ANSWER
Explain the difference between congruent and similar triangles. Use diagrams to support your answer.
Write your working in your book.
S
Stretch Challenge · Model village
+20 XP
S
STRETCH
A model village is built to a scale of 1:50. (a) A real house is 12 m tall. How tall is the model in centimetres? (b) The model house has a roof area of 80 cm². What is the real roof area in square metres? (c) The real house requires 150 litres of paint for its exterior. How much paint would the model require? Explain why this might not be practical.
Record in your book -- full marks require clear working.
Reveal solution

(a) $k = \frac{1}{50}$. Model height = $12 \times \frac{1}{50} = 0.24$ m = 24 cm.

(b) Area scale factor = $k^2 = \frac{1}{2500}$. Real area = $80 \times 2500 = 200{,}000$ cm² = 20 m².

(c) Volume (paint) scale factor = $k^3 = \frac{1}{125{,}000}$. Model paint = $150 \times \frac{1}{125{,}000} = 0.0012$ litres = 1.2 mL. This is not practical because paint does not scale -- a minimum thickness is needed for coverage, and surface texture effects do not shrink proportionally.

R
Quick Review

Similar

Same shape, proportional sides

Scale factor

$k = \text{new} / \text{original}$

Enlargement

$k > 1$

Reduction

$0 < k < 1$

Area

Scales by $k^2$

Volume

Scales by $k^3$

Interactive: Circle Explorer

Explore circle geometry and properties in the interactive below. While this lesson focuses on similarity, understanding circles deepens your geometric reasoning.

Consolidation Game -- Doodle Jump Quiz
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Jump your way to the top by answering questions on similarity, scale factors, and area/volume scaling. The higher you climb, the harder the questions.

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