Congruent triangles are perfect copies. But what if you scale a triangle up or down — keeping all angles the same, but changing the size? You get a similar triangle. Similar triangles appear everywhere: in maps, in shadows, in the reason a camera lens can capture a distant mountain on a small sensor.
Key facts about similar triangles:
1. All corresponding angles are equal.
2. All corresponding sides are proportional — they share the same ratio (the scale factor).
3. If you know two triangles have the same angles (AA — Angle-Angle), they are similar.
4. Scale factor $k = \dfrac{\text{side in big triangle}}{\text{corresponding side in small triangle}}$. Find any missing side using $\dfrac{a}{b} = \dfrac{c}{d}$.
Worked Example
Worked Example — Find missing sides using scale factor
Triangle $ABC$ has sides $6$ cm, $8$ cm, $10$ cm. Triangle $DEF$ is similar to $ABC$. Side $DE = 9$ cm (corresponds to $AB = 6$ cm). Find sides $EF$ and $DF$.
Step 1 — Find scale factor:
$$k = \frac{DE}{AB} = \frac{9}{6} = 1.5$$
Step 2 — Multiply each side by $k$:
$$EF = 8 \times 1.5 = 12 \text{ cm}$$
$$DF = 10 \times 1.5 = 15 \text{ cm}$$
Check: Ratios $9:6 = 12:8 = 15:10 = 1.5$ ✓
Warm-Up — Scale factors (ratios given)
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Problem 1
Triangle $PQR$ has sides $4$ cm, $6$ cm, $9$ cm. Triangle $STU$ is similar with scale factor $k = 2$. Find all three sides of $STU$.
Equation: $\text{each side} \times 2 = ?$
multiply → list three sides →
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Problem 2
Two similar triangles: the smaller has sides $5$ cm, $12$ cm, $13$ cm. The larger has its shortest side at $15$ cm. Find the scale factor $k$ and then find the other two sides of the larger triangle.
Equation: $k = \dfrac{15}{5} = ?$
find k → multiply remaining sides →
Core Problems — Set up the proportion, then solve
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Problem 3
Two similar triangles share the same angles. The first has sides $8$ m, $x$, and $20$ m. The second (larger) has corresponding sides $12$ m, $18$ m, and $30$ m. Find $x$ using a proportion.
write proportion → cross-multiply → solve →
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Problem 4
A horse trail map uses scale $1:25\,000$. On the map, the trail from the stable to the watering hole is $6.4$ cm long. How far is this in real life? Give your answer in kilometres.
set up proportion → convert units → answer →
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Problem 5
Mia and a barn are both standing in sunlight. Mia is $1.5$ m tall and casts a shadow $2$ m long. The barn casts a shadow $14$ m long. How tall is the barn? (The sun's rays are parallel, so the triangles are similar.)
draw two triangles (shadow + height) → write proportion → solve →
From here — write the equation yourself, then solve
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Problem 6
In Japan, a traditional shrine gate (torii) casts a shadow $8.4$ m long. At the same time, a $1.2$ m measuring rod casts a shadow $0.8$ m long. How tall is the torii gate?
set up proportion → solve →
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Problem 7
A trail map of a riding route in Bavaria uses scale $1:50\,000$. Two towns appear $4.5$ cm apart on the map.
(a) How far apart are they in real life?
(b) If Mia can ride $15$ km/h on her horse, how many minutes will it take to ride between them?
part (a): scale → real distance · part (b): distance ÷ speed → time in hours → convert to minutes →
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Problem 8 Challenge
Triangle $A$ has sides $3$ cm, $4$ cm, $5$ cm. Triangle $B$ is similar with scale factor $k = 3$.
(a) Find all sides of triangle $B$.
(b) Calculate the area of triangle $A$ and triangle $B$.
(c) What is the ratio of the two areas? What do you notice — does it equal $k$ or $k^2$?
sides of B → area A (½ × base × height) → area B → compare ratios →
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Problem 9 Open
Mia wants to measure the width of a river without crossing it. She marks two points on her side ($A$ and $B$, $10$ m apart), then picks a tree $C$ directly across. From $A$ she walks to point $D$ (not at the river), and adjusts her position until $D$, $A$, and $C$ are all in a straight line. She measures $DA = 4$ m and $DB = 6$ m. Use similar triangles to estimate the river width. Explain your method.
draw diagram → identify similar triangles → write proportion → solve → explain →
Show answers
Problem 1
$4 \times 2 = 8$ cm, $6 \times 2 = 12$ cm, $9 \times 2 = 18$ cm · Sides of $STU$: $8$ cm, $12$ cm, $18$ cm
Problem 2
$k = \frac{15}{5} = 3$ · Other sides: $12 \times 3 = 36$ cm, $13 \times 3 = 39$ cm · Larger triangle: $15$ cm, $36$ cm, $39$ cm
Problem 3
$\frac{x}{18} = \frac{8}{12}$ → $x = \frac{8 \times 18}{12} = 12$ m · (Or: scale factor $= 12/8 = 1.5$, so $x = 18/1.5 = 12$ m) ✓
Problem 4
$6.4 \text{ cm} \times 25\,000 = 160\,000 \text{ cm} = 1\,600 \text{ m} = 1.6 \text{ km}$
Problem 5
$\frac{\text{barn height}}{14} = \frac{1.5}{2}$ → barn height $= \frac{1.5 \times 14}{2} = 10.5$ m
Problem 6
$\frac{h}{8.4} = \frac{1.2}{0.8}$ → $h = \frac{1.2 \times 8.4}{0.8} = 12.6$ m
Problem 7
(a) $4.5 \text{ cm} \times 50\,000 = 225\,000 \text{ cm} = 2.25 \text{ km}$ · (b) $2.25 \div 15 = 0.15 \text{ h} = 9 \text{ minutes}$
Problem 8
(a) Sides of $B$: $9$ cm, $12$ cm, $15$ cm · (b) Area $A = \frac{1}{2}(3)(4) = 6$ cm² · Area $B = \frac{1}{2}(9)(12) = 54$ cm² · (c) Ratio $= 54/6 = 9 = k^2 = 3^2$ ✓ — areas scale by the square of the scale factor!
Problem 9
Triangles $DAC$ and $DBC$ are similar (AA). $\frac{AC}{BC} = \frac{DA}{DB} = \frac{4}{6}$. The river width $= BC - AC$. Alternatively: $\frac{BC}{AB} = \frac{DB}{DA}$? (Method varies — accept any correct proportion setup. River width ≈ $15$ m using $BC = AB \times \frac{DB}{DA} = 10 \times \frac{6}{4} = 15$ m.)
Coming up next → Lesson 16: Why Slope Is Constant
Here's the key connection: every right triangle you draw under a straight line — one corner at the x-axis, one pointing up to the line — is similar to every other such triangle on that same line. Same angles, same ratios. That ratio of rise to run is always identical. That's why slope is constant. Lesson 16 makes this visual and concrete.