Same shape, different size — and the ratios tell the whole story.
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 $\dfrac{a}{b} = \dfrac{c}{d}$.
Same three angles → same shape. Every side of the big triangle is the small one × the scale factor (here k = 1.5).
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 $DF$.
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 →
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
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 →
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 →
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.)
Height + shadow make a right triangle; the sun's parallel rays make Mia's triangle similar to the barn's.
From here — write the equation yourself, then solve
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 →
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 →
Problem 8 Challenge
Triangle $A$ is a right-angled triangle with sides $3$ cm, $4$ cm, $5$ cm — the right angle sits between the $3$ cm and $4$ cm sides, so use those as its base and height. Triangle $B$ is similar to $A$ with scale factor $k = 3$. (a) Find all three sides of triangle $B$. (b) Find the area of each triangle (area of a right triangle $= \tfrac{1}{2} \times \text{base} \times \text{height}$). (c) Divide the big area by the small area. Does that ratio equal $k$, or $k^2$?
Sides scale by 3, but area scales by 3² = 9. Area always grows by the square of the scale factor.
sides of B → area A (½ × base × height) → area B → compare ratios →
Problem 9 Open
Mia measures a river's width without crossing it, using similar triangles. A tree $T$ stands on the far bank. She marks point $A$ on her own bank directly opposite the tree, so the straight line from $A$ across to $T$ is the width $W$ she wants. Then she paces a straight line along her bank: from $A$ to a post $B$ is $10$ m, and from $B$ to $C$ is a further $4$ m. From $C$ she turns and walks straight inland, away from the river, stopping the instant the post $B$ lines up exactly with the tree $T$ — she has walked $6$ m to point $D$. Now the two right-angled triangles $\triangle TAB$ and $\triangle DCB$ are similar (equal angles at $B$; right angles at the bank). Write the proportion and find the width $W = AT$.
Two similar right triangles: △TAB reaches across the river, △DCB sits on land. Measure the small one → the proportion gives W.
write $\frac{AT}{DC} = \frac{AB}{CB}$ → solve for AT →
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) ✓
(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
$\triangle TAB \sim \triangle DCB$ (AA — equal angles at $B$, right angles at $A$ and $C$). Corresponding sides give $\frac{AT}{DC} = \frac{AB}{CB}$ → $\frac{AT}{6} = \frac{10}{4}$ → $AT = \frac{10 \times 6}{4} = 15$. The river is about 15 m wide.
Coming up next → Lesson 17: 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 17 makes this visual and concrete.