Areas of similar triangles

Two triangles can look like exact scaled copies of each other — same angles, sides in the same proportion — and yet their areas differ far more dramatically than their sides do. Double every side of a triangle and its perimeter doubles, but its area does not merely double: it quadruples. This is the single most exam-tested surprise in the Similarity unit, and it has a clean reason behind it — area is built from a length times a length, so when both of those lengths scale by the same factor $k$ , the area scales by $k \times k = k^2$ . This article builds that idea from the ground up: what the squared-ratio rule says, why it must be true given how area is defined, how it applies to sides, altitudes, medians and any other linear measurement of similar triangles, and how it combines with the Basic Proportionality Theorem when a line parallel to one side splits a triangle into a smaller triangle and a trapezium. Two worked problems — one a direct area-from-side-ratio calculation, one a parallel-line split — cover the two question types you will actually meet in ICSE Class 10 exams.

The core insight

For similar triangles with corresponding sides in ratio $k$ , the ratio of their areas is $k^2$ — areas of similar figures scale with the square of the linear scale factor, never with the factor itself.

The core idea

The statement. If $\triangle ABC \sim \triangle DEF$ , with corresponding sides related by a single scale factor $k = \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD}$ , then the areas obey:

\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = k^2

where: $k$ is the linear scale factor between the two similar triangles (the common ratio of any pair of corresponding sides), and $\text{Area}(\triangle ABC)$ , $\text{Area}(\triangle DEF)$ are the areas of the two triangles in the same square units.

It extends to every linear measurement, not just sides. Because altitudes, medians, angle bisectors and perimeters of similar triangles all scale by the same $k$ as the sides do, the area ratio equals the square of any of these ratios too:

\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{h_1}{h_2}\right)^2 = \left(\frac{m_1}{m_2}\right)^2

where: $h_1, h_2$ are corresponding altitudes and $m_1, m_2$ are corresponding medians of the two triangles.

Why the square appears. Area is a product of two lengths, base and height, halved: $\text{Area} = \tfrac{1}{2}\times\text{base}\times\text{height}$ . In similar triangles both the base and the corresponding height scale by the same factor $k$ , so the product scales by $k \times k = k^2$ . This is the general pattern for any similar plane figures, not triangles alone: lengths scale by $k$ , areas scale by $k^2$ . It is worth internalising this pattern because it recurs across many later topics — for instance, doubling every side of a triangle quadruples its area rather than doubling it, a fact many students get wrong on a first attempt if the squared relationship has not been firmly internalised.

Combining with a parallel line (BPT setting). When a line drawn parallel to one side of a triangle cuts off a smaller triangle at a vertex, that smaller triangle is similar to the original (by AA, since the parallel line creates equal corresponding angles). The squared-ratio rule then compares the small triangle's area either to the whole original triangle, or to the trapezium-shaped remainder — these are two different comparisons, and a question will ask for one specific one.

Figure

Two similar triangles ABC and DEF drawn with DEF scaled up by a factor of two, with side and altitude ratios both equal to two but the shaded area of DEF four times that of ABC

Derivation AThe formula, derived from base times height

Here we derive the areas-of-similar-triangles result directly from the definition of triangle area and the way similarity forces every corresponding length, including altitudes, to scale by the same single factor $k$ , so that the derivation shows exactly why an area ratio must come out as $k^2$ rather than $k$ .

Step 1. Let $\triangle ABC \sim \triangle DEF$ with scale factor $k = \dfrac{BC}{EF}$ . Take $BC$ and $EF$ as the bases of the two triangles, and let $AM$ and $DN$ be the altitudes from $A$ and $D$ onto these bases respectively, so that $\text{Area}(\triangle ABC) = \tfrac{1}{2}\cdot BC \cdot AM$ and $\text{Area}(\triangle DEF) = \tfrac{1}{2}\cdot EF \cdot DN$ .

Step 2. Since $\triangle ABC \sim \triangle DEF$ , the triangles $\triangle ABM$ and $\triangle DEN$ are also similar: angle $B$ equals angle $E$ (given similarity of the parent triangles), and angle $AMB$ equals angle $DNE$ (both are right angles, since $AM$ and $DN$ are altitudes). Two equal angles give AA similarity, so $\dfrac{AM}{DN} = \dfrac{AB}{DE} = k$ .

Step 3. This shows the altitudes scale by exactly the same factor $k$ as the sides do. Now form the ratio of the areas:

\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \frac{\tfrac{1}{2}\cdot BC \cdot AM}{\tfrac{1}{2}\cdot EF \cdot DN} = \frac{BC}{EF}\cdot\frac{AM}{DN} = k \cdot k = k^2

Step 4. So the area ratio is the product of two separate factors of $k$ — one from the base, one from the altitude — confirming $\dfrac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = k^2 = \left(\dfrac{AB}{DE}\right)^2 = \left(\dfrac{BC}{EF}\right)^2 = \left(\dfrac{CA}{FD}\right)^2$ , matching the general statement.

Worked example

$\triangle ABC \sim \triangle PQR$ . The area of $\triangle ABC$ is $50\ \text{cm}^2$ and the area of $\triangle PQR$ is $18\ \text{cm}^2$ . If $BC = 5\ \text{cm}$ , find $QR$ .

Step 1. Set up the area ratio in terms of the corresponding sides $BC$ and $QR$ :

\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{BC}{QR}\right)^2

Step 2. Substitute the known areas and side:

\frac{50}{18} = \left(\frac{5}{QR}\right)^2 \implies \frac{25}{9} = \left(\frac{5}{QR}\right)^2

Step 3. Take the square root of both sides, keeping only the positive root since lengths are positive:

\frac{5}{3} = \frac{5}{QR} \implies \mathbf{QR = 3\ cm}

Competitive stretch

Now extend the same triangle with a BPT twist. In $\triangle ABC$ , $D$ lies on $AB$ and $E$ lies on $AC$ with $DE \parallel BC$ and $AD : DB = 2 : 3$ . Since $DE \parallel BC$ , $\triangle ADE \sim \triangle ABC$ by AA. The ratio $AD:AB$ works out to $2:5$ (because $AD + DB = 2+3 = 5$ parts in total), so $\dfrac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \left(\dfrac{2}{5}\right)^2 = \dfrac{4}{25}$ . Taking Area $(\triangle ABC)$ as $25$ convenient units gives Area $(\triangle ADE) = 4$ units, so the trapezium $DBCE$ has area $25 - 4 = 21$ units, and $\dfrac{\text{Area}(\triangle ADE)}{\text{Area}(DBCE)} = \dfrac{4}{21}$ — a different ratio from $4:25$ , so always check exactly which comparison a question wants.

Common traps

Using the side ratio directly as the area ratio without squaring it — this is the single most common error in this topic.
Forgetting to take the square root when going the other way, that is, an area ratio is given and a side ratio is wanted.
In parallel-line problems, comparing the small triangle's area to the whole triangle when the trapezium remainder was actually asked for, or vice versa.
Confusing a part-to-part ratio such as $AD:DB$ with a part-to-whole ratio such as $AD:AB$ — convert carefully before squaring, since the two give very different area ratios.

Ask the tutor: want to work through a trickier area-ratio problem that mixes BPT with the squared-ratio rule, step by step, with the AI tutor?

Why does area scale as the square of the side ratio and not just the same ratio as the sides

Area is not a single length, it is a product of two lengths (base and height) halved. In similar triangles both the base and the corresponding height scale by the same factor $k$ , so the area, being a product of the two, scales by $k$ from the base and another $k$ from the height, giving $k \times k = k^2$ overall. A single length like a side or perimeter only picks up one factor of $k$ , which is why sides scale by $k$ but area scales by $k^2$ .

Does the squared-ratio rule work for any two triangles that share an angle

No. The two triangles must first be established as genuinely similar, using AA, SSS, or SAS similarity criteria, before the squared-ratio rule can be applied. Sharing one angle, or even one equal side, is not enough on its own — you need matching angles all around (or proportional sides all around) confirmed by one of the standard similarity tests before comparing areas this way.

In a BPT question, how do I know whether to compare the small triangle to the whole triangle or to the trapezium

Read the question wording closely — it will name the two regions being compared explicitly, such as area of triangle ADE to area of trapezium DBCE, or area of triangle ADE to area of triangle ABC. These give different numeric answers (for example 4:21 versus 4:25 in a 2:3 split), so compute the small-triangle-to-whole-triangle ratio first using $k^2$ , and only then subtract to get the trapezium's share if that is what was asked.

Areas of similar triangles

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads