Areas of Similar Triangles

When two triangles are similar, their corresponding sides are proportional and their corresponding angles are equal. But similarity does more than just link side lengths — it creates a precise, predictable relationship between the areas of the two triangles. Understanding this relationship is one of the most powerful tools in Euclidean geometry, appearing in everything from architectural scaling to map-making and computer graphics. The key theorem states: if two triangles are similar with a ratio of similarity (scale factor) of k, then the ratio of their areas is k². In other words, area scales as the square of the linear scale factor. If one triangle has sides twice as long as another similar triangle, its area is four times as large — not twice. This quadratic relationship surprises many students at first, but it follows directly from how area depends on two linear dimensions simultaneously. This concept connects deeply to the broader principle that when you scale any two-dimensional figure by a factor of k, all lengths scale by k while all areas scale by k². This is a fundamental result of Euclidean geometry with applications across mathematics, science, and engineering. Mastering the area ratio theorem for similar triangles builds the foundation for understanding similar figures in general, including circles, polygons, and composite shapes.

Key takeaways

Set Up Two Similar Triangles
Write the Area Formula for Each Triangle
Show the Heights Also Scale by k
Compute the Ratio of Areas
State the General Theorem

Set Up Two Similar Triangles

Let triangle ABC be similar to triangle DEF, written △ABC ~ △DEF. This means their corresponding angles are equal: ∠A = ∠D, ∠B = ∠E, ∠C = ∠F. Their corresponding sides are proportional with scale factor k: DE/AB = EF/BC = FD/CA = k.

Write the Area Formula for Each Triangle

The area of any triangle is (1/2) × base × height. For △ABC, choose base BC with corresponding height h₁ from vertex A. So Area(ABC) = (1/2) · BC · h₁. For △DEF, choose base EF with corresponding height h₂ from vertex D. So Area(DEF) = (1/2) · EF · h₂.

Show the Heights Also Scale by k

Since △ABC ~ △DEF with scale factor k, we know EF = k · BC. Now consider the two right triangles formed by dropping the altitudes. The altitude triangles share the same angle at the apex (∠A = ∠D) and have a right angle. By AA similarity, these altitude triangles are also similar with the same ratio k. Therefore h₂ = k · h₁. Both the base and the height scale by the same factor k.

Compute the Ratio of Areas

Now divide the two areas: Area(DEF) / Area(ABC) = [(1/2) · EF · h₂] / [(1/2) · BC · h₁] = (EF/BC) · (h₂/h₁) = k · k = k². Therefore the ratio of the areas equals the square of the scale factor.

State the General Theorem

If △ABC ~ △DEF with similarity ratio k = DE/AB, then Area(DEF) / Area(ABC) = k². Equivalently, if the sides are in ratio m : n, then the areas are in ratio m² : n². This holds regardless of which pair of similar triangles you consider — the quadratic scaling of area under similarity is universal for all two-dimensional shapes, not just triangles.

The Key Equation

\frac{\text{Area}(\triangle DEF)}{\text{Area}(\triangle ABC)} = \left(\frac{DE}{AB}\right)^2 = k^2

\text{Area}(\triangle DEF) — Area of triangle DEF (square units): The area of the larger (or smaller) of the two similar triangles.
\text{Area}(\triangle ABC) — Area of triangle ABC (square units): The area of the reference triangle to which the other is being compared.
DE / AB — Ratio of corresponding sides (dimensionless): The scale factor formed by any pair of corresponding side lengths from the two similar triangles.
k — Scale factor (similarity ratio) (dimensionless): The common ratio of all pairs of corresponding sides. Areas scale as k² because area involves two linear dimensions.

Worked Example

Problem. Triangle PQR is similar to triangle XYZ. The sides of △PQR are 6 cm, 8 cm, and 10 cm. The longest side of △XYZ is 25 cm. Find (a) the scale factor k from △PQR to △XYZ, and (b) the area of △XYZ if the area of △PQR is 24 cm².

k = \frac{5}{2}, \quad \text{Area}(\triangle XYZ) = 24 \times \left(\frac{5}{2}\right)^2 = 24 \times \frac{25}{4} = 150 \text{ cm}^2

Answer. The scale factor is k = 5/2 and the area of triangle XYZ is 150 cm².

Figure

Diagram

Origins · 1 minEuclid of Alexandria and how this was discovered

Euclid of Alexandria

-325–-265 · Greek (Hellenistic)

Euclid systematized geometry in his 13-volume work Elements, written around 300 BCE. Book VI of the Elements contains propositions on similar figures, including Proposition 19, which states that similar triangles are to one another in the duplicate ratio (i.e., the squared ratio) of their corresponding sides. This is precisely the theorem about areas of similar triangles studied today. Euclid's deductive method — building theorems from axioms and previously proved results — remains the model for mathematical proof. When King Ptolemy I asked Euclid if there was a shorter path to learning geometry than studying the Elements, Euclid reportedly replied: 'There is no royal road to geometry.' This saying has echoed through classrooms for over two thousand years.

Similar Triangles — The area ratio theorem applies specifically to similar triangles, so understanding the definition and conditions for similarity (AA, SAS, SSS) is an essential prerequisite.
Scale Factor and Ratio of Similarity — The scale factor k is the core quantity in the area ratio formula k². Understanding how to identify and compute k from corresponding sides is fundamental.
Area of a Triangle — The derivation of the area ratio theorem relies on the formula Area = (1/2) × base × height applied to both triangles.
Perimeters of Similar Figures — Perimeters of similar figures scale by k (not k²), which makes a useful contrast: linear measures scale linearly while areas scale quadratically.
Areas of Similar Polygons — The k² area ratio extends from triangles to all similar polygons, since any polygon can be divided into triangles. This is the natural generalization of the concept.