Lesson 17 ~25 min Unit 3 · Geometry +85 XP

Finding Missing Sides in Similar Figures

When two figures are similar, the scale factor lets us find unknown side lengths. Set up a proportion from the known matching pairs, then solve. This idea powers maps, scale models and even using shadows to measure a tree.

Today's hook: A $2$-metre-tall stick casts a $3$-metre shadow. The flagpole next to it casts a $12$-metre shadow. How tall is the flagpole?

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Think First

warm-up

A toy car model is in the ratio $1:50$ to a real car (the real car is $50$ times bigger). If the toy is $8$ cm long, how long is the real car (in centimetres, then in metres)? Don't worry about being perfect — just write what feels right.

Record your answer in your workbook.

Two Ways to Find an Unknown Side

+5 XP

Once we know two figures are similar, we can find any missing length. There are two common methods:

Method 1 (Scale Factor): Find the scale factor from a known pair of sides, then multiply (or divide) the unknown side by it.
Method 2 (Proportion): Set up an equation: $\dfrac{\text{new}}{\text{old}} = \dfrac{\text{new}}{\text{old}}$ using known pairs, with $x$ in place of the unknown. Cross-multiply to solve.

Both methods give the same answer. The scale-factor method is fastest for simple problems. The proportion method is more flexible for harder problems and matches the way you'll write working in an exam.

$\dfrac{12}{6} = \dfrac{x}{8} \;\Rightarrow\; x = 16$

Match the pairs

Put corresponding sides in the SAME spot of each ratio (top or bottom).

Cross-multiply

$\frac{a}{b} = \frac{c}{d}$ means $ad = bc$.

Check direction

Going BIG → small? Divide by SF. Going small → BIG? Multiply by SF.

What You'll Master

objectives

Know

What a proportion is ($\frac{a}{b} = \frac{c}{d}$)
How to cross-multiply
The "new ÷ old" rule for scale factor

Understand

Why the scale-factor and proportion methods give the same answer
Why both pairs must match in the same direction
How similar triangles power "shadow" problems

Can Do

Find a missing side using the scale-factor method
Set up and solve a proportion
Solve real-world problems with maps, models, shadows

Words You Need

vocabulary

ProportionAn equation of two equal ratios: $\frac{a}{b} = \frac{c}{d}$.

Cross-multiplyFrom $\frac{a}{b} = \frac{c}{d}$, get $a \times d = b \times c$.

Scale factor (SF)Ratio of corresponding sides: $\frac{\text{new}}{\text{old}}$.

Scale (on a map)A ratio like $1:50\,000$ that compares map to real distance.

ModelA small similar copy of a real object.

UnknownThe side we're trying to find — usually called $x$.

Shadow problemTwo objects + sun: similar triangles from same angle of light.

Inverse SF$\frac{1}{\text{SF}}$ — used when going from large to small.

Method 1 · Scale Factor

+5 XP

Three-step routine:

Pick a pair of CORRESPONDING sides where both lengths are known.
Calculate SF $= \dfrac{\text{new}}{\text{old}}$.
Multiply the matching unknown side by SF.

Example: small triangle has sides $4$ and $5$; large has $8$ and $x$. Pick the known pair $4 \to 8$. SF $= \frac{8}{4} = 2$. So $x = 5 \times 2 = 10$.

$x = 5 \times \text{SF} = 5 \times 2 = 10$

Multiply if going up

From small to large: multiply by SF.

Divide if going down

From large to small: divide by SF (or multiply by $\frac{1}{\text{SF}}$).

Same SF everywhere

Once SF is found from one pair, it works for ALL pairs.

Book notes · Scale-factor method

Find SF from a known pair: SF $=$ new ÷ old.
Multiply the corresponding unknown side by SF.
If going BIG → small, divide instead.

Small triangle has sides $3$ and $5$. Similar large triangle has corresponding sides $12$ and $x$. Find $x$.

Method 2 · Proportion + Cross-Multiply

+5 XP

Set up a proportion (two equal ratios) using corresponding sides, then cross-multiply:

$\dfrac{\text{new side}}{\text{old side}} = \dfrac{\text{another new side}}{\text{another old side}}$

Make sure the SAME triangle's sides are always on top — if you flip which side is "new", you'll get the wrong answer.

Example: triangle $ABC \sim$ triangle $DEF$. $AB = 6, BC = 9$. $DE = 10, EF = x$. $\frac{DE}{AB} = \frac{EF}{BC}$, giving $\frac{10}{6} = \frac{x}{9}$. Cross-multiply: $10 \times 9 = 6x$, so $90 = 6x$, $x = 15$.

$\dfrac{10}{6} = \dfrac{x}{9} \;\Rightarrow\; 6x = 90 \;\Rightarrow\; x = 15$

Same shape on top

Always put sides from the same figure on the numerators (and the same on the denominators).

Set $x$ on numerator

It makes the algebra easier — one cross-multiply, one divide.

Always check

Plug your answer back in: does the ratio match?

Book notes · Proportion method

Write $\frac{\text{new}}{\text{old}} = \frac{\text{new}}{\text{old}}$.
Cross-multiply: $ad = bc$.
Solve for $x$ by dividing.

True or false?

In the proportion $\frac{a}{b} = \frac{c}{d}$, the cross-multiplication rule gives $a \times d = b \times c$.

Maps, Models & Shadows

+5 XP

Similarity isn't just for triangles in textbooks. Real applications:

Map scale — "$1:50\,000$" means every $1$ cm on the map is $50\,000$ cm $= 500$ m on the ground.
Models — a $1:24$ scale toy car is $\frac{1}{24}$ the real car's size in every dimension.
Shadow problems — the sun shines at the same angle on both objects, so each object + its shadow form a triangle, and those triangles are similar.

Shadow example: a $2$ m stick casts a $3$ m shadow at the same time a flagpole casts a $12$ m shadow. Both have the SAME sun angle. So the triangles formed are similar: $\frac{\text{flagpole}}{2} = \frac{12}{3}$, giving flagpole $= 2 \times 4 = 8$ m.

$\dfrac{h}{2} = \dfrac{12}{3} \;\Rightarrow\; h = 8$ m

Same sun angle

Both shadows are cast at the SAME moment, so the triangles ARE similar.

Match the units

If the stick is in metres, the shadow must be in metres too.

Maps need conversion

$1:50\,000$: convert your map cm to m or km after multiplying.

Book notes · Real-world similarity

Map scale: real distance $=$ map distance $\times$ scale factor.
Shadow triangles are similar (same sun angle).
Models use a single scale factor in every dimension.

A map has scale $1:1000$. A line on the map is $5$ cm. The real distance is __________ cm.

Watch Me Solve It · 3 examples

Watch Me Solve It · Find $x$ using scale factor

+15 XP per step

PROBLEM

Two similar triangles. The small one has sides $4$ cm and $7$ cm. The corresponding side $4$ cm becomes $12$ cm on the large triangle. Find $x$, the side corresponding to $7$ cm.

1
Find scale factor
SF $= \dfrac{12}{4} = 3$
2
Apply SF to the unknown side
$x = 7 \times 3$
3
Compute
$x = 21$ cm
Check: $\frac{12}{4} = 3 = \frac{21}{7}$ ✓

Answer$x = 21$ cm.

Watch Me Solve It · Proportion method

+15 XP per step

PROBLEM

$\triangle ABC \sim \triangle PQR$. $AB = 8$, $BC = 10$, $PQ = 20$. Find $QR$.

1
Set up proportion
$\dfrac{PQ}{AB} = \dfrac{QR}{BC}$
2
Substitute
$\dfrac{20}{8} = \dfrac{QR}{10}$
3
Cross-multiply & solve
$8 \times QR = 20 \times 10 = 200$, so $QR = 25$
Check SF: $\frac{20}{8} = 2.5$ and $\frac{25}{10} = 2.5$ ✓

Answer$QR = 25$.

Watch Me Solve It · Shadow problem

+15 XP per step

PROBLEM

A $1.5$ m tall student casts a $2$ m shadow. At the same time, a tree's shadow is $14$ m long. How tall is the tree?

1
Recognise similar triangles
Same sun angle ⇒ (student + shadow) $\sim$ (tree + shadow).
2
Set up proportion
$\dfrac{\text{tree height}}{1.5} = \dfrac{14}{2}$
3
Solve
Tree height $= 1.5 \times \dfrac{14}{2} = 1.5 \times 7 = 10.5$ m
The tree is $10.5$ m tall.

AnswerTree height $= 10.5$ m.

Common Pitfalls

heads-up

Flipping which side is "new"

Writing $\frac{\text{old}}{\text{new}}$ on one side and $\frac{\text{new}}{\text{old}}$ on the other gives the WRONG answer (often the reciprocal).

Fix: Keep the SAME figure on top in BOTH ratios.

Multiplying when you should divide

If you're going from BIG to small, you have to divide by SF (or multiply by $\frac{1}{\text{SF}}$).

Fix: Sanity-check — is the unknown side bigger or smaller than the known side? Adjust.

Forgetting units in word problems

Mixing cm and m in a map problem can lead to answers off by a factor of 100.

Fix: Convert everything to the same unit before substituting.

Copy Into Your Books

Method 1

Find SF $= \frac{\text{new}}{\text{old}}$.
Multiply unknown by SF (small → big).
Divide if big → small.

Method 2

$\frac{\text{new}_1}{\text{old}_1} = \frac{\text{new}_2}{\text{old}_2}$
Cross-multiply: $ad = bc$.
Solve for $x$.

Maps

Scale $1:n$: $1$ unit map $=n$ same units real.
Multiply map distance by $n$.
Convert cm $\to$ m by $\div 100$.

Shadows

Same sun ⇒ similar triangles.
$\frac{\text{tall thing}}{\text{short thing}} = \frac{\text{long shadow}}{\text{short shadow}}$

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Missing Sides

4 problems

Four quick drills on finding unknowns in similar figures.

1 Small triangle sides $5, 6$. Large triangle sides $20, x$. Find $x$.

SF $= 20/5 = 4$. $x = 6 \times 4$.$x = 24$
2 $\frac{x}{6} = \frac{15}{9}$. Find $x$.

Cross: $9x = 90$.$x = 10$
3 A $1$ m post casts a $1.6$ m shadow. A nearby tree's shadow is $8$ m. Tree height?

$\frac{h}{1} = \frac{8}{1.6}$.$h = 5$ m
4 Map scale $1:200$. A house is $3$ cm wide on the map. Real width?

$3 \times 200 = 600$ cm.$6$ m

Complete in your workbook.

Quick Check · 5 questions

Small triangle sides $5$ and $8$. Similar large triangle has corresponding sides $15$ and $x$. Find $x$.

+10 XP

Solve $\dfrac{x}{7} = \dfrac{3}{4}$.

+10 XP

A $1.5$ m student casts a $2$ m shadow. At the same time a tree casts a $12$ m shadow. The tree height is:

+10 XP

A map has scale $1:25\,000$. Two towns are $4$ cm apart on the map. The real distance is:

+10 XP

A large triangle has sides $24$ and $18$. A similar small triangle has corresponding side $8$ and $x$. Find $x$.

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

Apply Easy 3 MARKS

Q6. Two similar quadrilaterals. Small one: sides $3, 5, 4, 6$. Large one: $9, ?, 12, ?$ corresponding.
(a) Find the scale factor.
(b) Find the two missing sides.
(c) Confirm by checking with the third pair.

Answer in your workbook.

Apply Medium 3 MARKS

Q7. $\triangle ABC \sim \triangle DEF$. $AB = 6$, $BC = 8$, $AC = 10$. $DE = 9$.
(a) Find the scale factor from $ABC$ to $DEF$.
(b) Find $EF$.
(c) Find $DF$.

Answer in your workbook.

Reason Hard 3 MARKS

Q8. A map has scale $1:5000$.
(a) A road is $7$ cm on the map. Find the real length in metres.
(b) A school oval is $200$ m long in real life. How long is it on the map (in cm)?
(c) Explain why a scale of $1:1$ would be useless for a city map.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — SF $= 3$, $x = 24$.

2. B — $4x = 21$, $x = 5.25$.

3. A — $h = 1.5 \times 6 = 9$ m.

4. D — $4 \times 25\,000 = 100\,000$ cm $= 1$ km.

5. B — SF $= \frac{1}{3}$, $x = 6$.

Show Your Working Model Answers

Q6 (3 marks): (a) SF $= \frac{9}{3} = 3$ [1]. (b) Missing sides: $5 \times 3 = 15$ and $6 \times 3 = 18$ [1]. (c) Third pair $\frac{12}{4} = 3$ ✓ [1].

Q7 (3 marks): (a) SF $= \frac{9}{6} = 1.5$ [1]. (b) $EF = 8 \times 1.5 = 12$ [1]. (c) $DF = 10 \times 1.5 = 15$ [1].

Q8 (3 marks): (a) $7 \times 5000 = 35\,000$ cm $= 350$ m [1]. (b) $200$ m $= 20\,000$ cm; map $= 20\,000 / 5000 = 4$ cm [1]. (c) A $1:1$ map is the same size as the city — useless because you can't fit a real-size city on a piece of paper [1].

Stretch Challenge · +25 XP, +10 coins

The Mirror Trick

To measure a tall tree, you put a flat mirror on the ground $20$ m from the tree's base. You stand $2$ m on the other side of the mirror and adjust until you see the top of the tree in the mirror. Your eye is $1.6$ m above the ground. (Light bounces off a mirror at the same angle it hits, so the two right-triangles formed are similar.) (a) Draw a diagram showing the two similar triangles. (b) Set up a proportion. (c) How tall is the tree?

Reveal solution

(a) Two right-triangles share the angle at the mirror (mirror reflection). Triangle 1: eye-height $1.6$, base $2$. Triangle 2: tree-height $h$, base $20$. (b) $\frac{h}{20} = \frac{1.6}{2}$. (c) $h = 20 \times \frac{1.6}{2} = 20 \times 0.8 = 16$ m. The tree is $16$ m tall.

Quick Review

Scale factor method

SF $=$ new $\div$ old. Multiply (or divide) the unknown side.

Proportion method

$\frac{\text{new}_1}{\text{old}_1} = \frac{\text{new}_2}{\text{old}_2}$. Cross-multiply.

Maps

$1:n$ — real $=$ map $\times n$ (same units).

Models

A single SF applies to every length.

Shadows

Same sun angle gives similar triangles.

Big to small

Divide by SF (or multiply by $\frac{1}{\text{SF}}$).

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