What Is the Area of Similar Triangles Theorem?
The area of similar triangles theorem states that the ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of corresponding sides. For similar triangles $\triangle ABC \sim \triangle DEF$,
$$\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{AC}{DF}\right)^2$$
Similar triangles have equal corresponding angles and corresponding sides in the same proportion. That common proportion is the scale factor $k$. The theorem says: whatever $k$ is for the sides, the area ratio is $k^2$. Because the three side ratios are all equal to $k$, you may use any corresponding pair.
Double the Sides, And the Area Does Not Double
Double every side of a triangular sail and the cloth it needs does not double, it quadruples. That surprise is the whole theorem in one sentence: when two triangles have the same shape but different sizes, their areas do not grow in step with their sides - they grow with the square of the side ratio. A shape twice as wide holds four times the space, and a triangle is the cleanest place to see why.
Why Is the Area Ratio the Square of the Side Ratio?
The square is not an arbitrary rule; it follows directly from how area is built. Area of a triangle is $\tfrac{1}{2} \times \text{base} \times \text{height}$, and in similar triangles both the base and the corresponding height are in the same ratio $k$.
Let $\triangle ABC \sim \triangle DEF$ with scale factor $\dfrac{AB}{DE} = k$. Drop the altitude to the corresponding base in each triangle; call them $h_1$ and $h_2$. Because the triangles are similar, the altitudes are in the same ratio as the sides, so $\dfrac{h_1}{h_2} = k$ as well.
Now take the ratio of areas:
$$\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{\tfrac{1}{2} \times AB \times h_1}{\tfrac{1}{2} \times DE \times h_2} = \frac{AB}{DE} \times \frac{h_1}{h_2} = k \times k = k^2$$
Two factors are each in ratio $k$ - the base and the height - so their product is in ratio $k^2$. That is the whole reason area scales with the square: area is a product of two lengths, and both lengths scale together. The same logic extends to any similar figures, which is why two similar squares or circles also have areas in the ratio $k^2$.
Does the Same Square Rule Apply to Perimeters, Altitudes, and Medians?
Here is a distinction worth getting right. In similar triangles, the linear measures all share the single ratio $k$ - corresponding sides, perimeters, altitudes, medians, and angle bisectors are each in ratio $k$. Only the area, a two-dimensional measure, jumps to $k^2$.
Quantity | Ratio in similar triangles |
|---|---|
Corresponding sides | $k$ |
Perimeters | $k$ |
Corresponding altitudes | $k$ |
Corresponding medians | $k$ |
Areas | $k^2$ |
So if the perimeters of two similar triangles are in ratio $3 : 2$, their sides are in ratio $3 : 2$ and their areas in ratio $9 : 4$. Keep one dimension versus two dimensions straight and the theorem never confuses you.
Examples of Area of Similar Triangles
Six examples, running the theorem forward (sides to area) and backward (area to sides).
Example 1
Two similar triangles have corresponding sides in the ratio $3 : 5$. Find the ratio of their areas.
The area ratio is the square of the side ratio.
$\left(\frac{3}{5}\right)^2 = \frac{9}{25}$
Final answer: the areas are in ratio $9 : 25$.
Example 2
Two similar triangles have sides in ratio $2 : 3$. A student says the areas are in ratio $2 : 3$ as well. Where does this go wrong?
The tempting move assumes area scales the same way as the sides, so the area ratio is read straight off as $2 : 3$.
That treats area like a length. Area is a product of two lengths, and in similar triangles both the base and the height scale by the side ratio, so the area scales by the ratio squared.
The correct method squares the side ratio:
$\left(\frac{2}{3}\right)^2 = \frac{4}{9}$
Final answer: the areas are in ratio $4 : 9$, not $2 : 3$. The side ratio never equals the area ratio unless the triangles are congruent.
Example 3
$\triangle ABC \sim \triangle DEF$ with $AB = 4$ cm and $DE = 6$ cm. If the area of $\triangle ABC$ is $32$ cm², find the area of $\triangle DEF$.
Side ratio $\dfrac{AB}{DE} = \dfrac{4}{6} = \dfrac{2}{3}$, so the area ratio is $\left(\tfrac{2}{3}\right)^2 = \tfrac{4}{9}$.
$$\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{4}{9} \quad\Rightarrow\quad \frac{32}{\text{Area of } \triangle DEF} = \frac{4}{9}$$
$\text{Area of } \triangle DEF = \dfrac{32 \times 9}{4} = 72$
Final answer: $72$ cm². The larger triangle has the larger area, as expected.
Example 4
The areas of two similar triangles are in the ratio $196 : 625$. Find the ratio of their corresponding sides.
Sides are the square root of the area ratio.
$\frac{\text{side}_1}{\text{side}_2} = \sqrt{\frac{196}{625}} = \frac{14}{25}$
Final answer: the corresponding sides are in ratio $14 : 25$. Going from area to sides means taking a square root, not dividing.
Example 5
The perimeters of two similar triangles are $36$ cm and $24$ cm. Find the ratio of their areas.
Perimeters share the side ratio, so $\dfrac{36}{24} = \dfrac{3}{2}$ is the side ratio too.
Square it for the area ratio:
$\left(\frac{3}{2}\right)^2 = \frac{9}{4}$
Final answer: the areas are in ratio $9 : 4$. Perimeter is linear, so it behaves like the sides.
Example 6
Two similar right triangles have hypotenuses $10$ cm and $14$ cm. The smaller triangle has area $24$ cm². Find the area of the larger.
Corresponding sides (here the hypotenuses) give the ratio $\dfrac{10}{14} = \dfrac{5}{7}$.
Area ratio is $\left(\tfrac{5}{7}\right)^2 = \tfrac{25}{49}$.
$\frac{24}{\text{Area}{\text{large}}} = \frac{25}{49} \quad\Rightarrow\quad \text{Area}{\text{large}} = \frac{24 \times 49}{25} = 47.04$
Final answer: $47.04$ cm². The reasoning step is recognising the hypotenuses as corresponding sides.
Where Does the Area of Similar Triangles Show Up?
This theorem is the quiet engine behind every scale drawing, map, and model - anywhere a shape is enlarged or shrunk while keeping its proportions.
Maps and scale drawings. A region drawn at half scale occupies a quarter of the paper area, because area follows $k^2$. A map's area scale is the square of its length scale.
Scale models. A scale model bridge built at $\tfrac{1}{50}$ the length has $\tfrac{1}{2500}$ the surface, which is why paint and material estimates cannot use the length scale directly.
Similar figures in coordinate geometry. Once triangles are placed on axes, the area of a triangle in coordinate geometry confirms the same square relationship numerically.
What Are the Most Common Mistakes With Area of Similar Triangles?
Three errors account for most wrong answers, and all come from mishandling the square.
Mistake 1: Using the side ratio as the area ratio
Where it slips in: The moment a student reads a side ratio and reports it as the area ratio unchanged.
Don't do this: Saying sides $2 : 3$ means areas $2 : 3$.
The correct way: Square the side ratio for the area ratio: $2 : 3$ sides give $4 : 9$ areas. Students first meeting the theorem almost always forget the square, because "ratio" feels like it should stay the same. The fix is to remember area is two-dimensional, so its ratio is the side ratio squared.
Mistake 2: Forgetting to take the square root going from area to sides
Where it slips in: Working backward from a known area ratio to find the side ratio.
Don't do this: Reporting the side ratio as $196 : 625$ when that is the area ratio.
The correct way: Sides are the square root of the area ratio: $\sqrt{196 : 625} = 14 : 25$. The rusher who divides area figures directly lands here. Forward means square; backward means square-root.
Mistake 3: Pairing non-corresponding sides
Where it slips in: Matching a side of one triangle to the wrong side of the other.
Don't do this: Dividing $AB$ by $EF$ instead of by its true corresponding side $DE$.
The correct way: Use the similarity statement $\triangle ABC \sim \triangle DEF$ to pair sides in order — $AB$ with $DE$, $BC$ with $EF$, $AC$ with $DF$. The second-guesser who matches by size instead of by correspondence gets a wrong ratio. Read the vertex order.
Conclusion
For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides, $k^2$.
It follows from area being $\tfrac{1}{2} \times \text{base} \times \text{height}$, with both base and height scaling by $k$.
Going forward (sides to area) means squaring; going backward (area to sides) means taking a square root.
Sides, perimeters, altitudes, and medians stay in ratio $k$; only area uses $k^2$.
To take similar triangles further with a teacher, explore Bhanzu's geometry tutor, a high school math tutor, or math classes online.
Practice These to Solidify Your Understanding
Work through these, then check your answers:
Two similar triangles have sides in ratio $4 : 7$. Find the ratio of their areas. (Answer to Question 1: $16 : 49$.)
The areas of two similar triangles are $81$ cm² and $144$ cm². Find the ratio of their corresponding sides. (Answer to Question 2: $9 : 12 = 3 : 4$.)
$\triangle ABC \sim \triangle PQR$ with $AB = 5$ cm, $PQ = 10$ cm. If area of $\triangle ABC$ is $20$ cm², find area of $\triangle PQR$. (Answer to Question 3: $80$ cm².)
If Question 2 tripped you, revisit Example 4 and the square-root step. Want a trainer to walk similarity proofs through with your child? Book a free demo class.
Read More
Triangles — the foundation the similarity results build on.
Congruence in triangles — the special case where the scale factor is exactly 1.
Pythagoras theorem — the right-triangle relation used in Example 6.
Properties of a triangle — the base facts behind altitudes and medians.
Midpoint theorem — a classic source of similar triangles inside one figure.
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