In similar triangles, ratio of areas = square of ratio of sides — prove

Question

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Solution — Step by Step

Let $\triangle ABC \sim \triangle PQR$ .

To prove: $\dfrac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)} = \dfrac{AB^2}{PQ^2} = \dfrac{BC^2}{QR^2} = \dfrac{AC^2}{PR^2}$

Strategy: Express the area of each triangle using base × height, then use the similarity ratio.

Draw $AD \perp BC$ and $PS \perp QR$ (altitudes from A and P to their respective bases).

\text{area}(\triangle ABC) = \frac{1}{2} \times BC \times AD

\text{area}(\triangle PQR) = \frac{1}{2} \times QR \times PS

Therefore:

\frac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)} = \frac{\frac{1}{2} \times BC \times AD}{\frac{1}{2} \times QR \times PS} = \frac{BC \times AD}{QR \times PS}

In $\triangle ABD$ and $\triangle PQS$ :

$\angle ADB = \angle PSQ = 90°$ (both are altitudes)
$\angle ABD = \angle PQS$ (corresponding angles in similar triangles $\triangle ABC \sim \triangle PQR$ )

By AA similarity: $\triangle ABD \sim \triangle PQS$

Therefore: $\dfrac{AD}{PS} = \dfrac{AB}{PQ}$

Since $\triangle ABC \sim \triangle PQR$ : $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ (sides of similar triangles are proportional)

So: $\dfrac{AD}{PS} = \dfrac{BC}{QR}$

From Step 2:

\frac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)} = \frac{BC}{QR} \times \frac{AD}{PS}

Substituting $\dfrac{AD}{PS} = \dfrac{BC}{QR}$ :

\frac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)} = \frac{BC}{QR} \times \frac{BC}{QR} = \frac{BC^2}{QR^2}

Since all corresponding sides of similar triangles are in the same ratio $k = \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR}$ :

\boxed{\frac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)} = \frac{AB^2}{PQ^2} = \frac{BC^2}{QR^2} = \frac{AC^2}{PR^2}}

Hence proved.

Why This Works

Area scales as length squared because area is a two-dimensional quantity. When you scale a figure by factor $k$ (all sides multiplied by $k$ ), each dimension scales by $k$ , so the area scales by $k^2$ .

The proof makes this rigorous: the altitude of a similar triangle scales by the same ratio as the sides (proved via AA similarity on the altitude triangles). Since area = $\frac{1}{2}$ × base × height, and both base and height scale by $k$ , the area scales by $k^2$ .

This result extends to all similar figures: ratio of areas = square of ratio of corresponding lengths.

Alternative Method — Direct Scaling

If $k = \dfrac{AB}{PQ}$ , then all sides of $\triangle ABC$ are $k$ times the corresponding sides of $\triangle PQR$ .

Area of $\triangle ABC = \frac{1}{2} \cdot AB \cdot c\sin A$ (formula using two sides and included angle).

Area of $\triangle PQR = \frac{1}{2} \cdot PQ \cdot c'\sin P$ .

Since $AB = k \cdot PQ$ , $AC = k \cdot PR$ , and $\angle A = \angle P$ (corresponding angles in similar triangles):

\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \frac{\frac{1}{2} \cdot AB \cdot AC \cdot \sin A}{\frac{1}{2} \cdot PQ \cdot PR \cdot \sin P} = \frac{k \cdot PQ \cdot k \cdot PR \cdot \sin P}{PQ \cdot PR \cdot \sin P} = k^2

Common Mistake

Saying the ratio of areas equals the ratio of sides (not the square). This is a very common error. If triangles are similar with ratio $3:5$ , the area ratio is $9:25$ (not $3:5$ ). The word “similar” tells us sides are proportional, but area involves two dimensions — so the ratio squares. Remember: area ∝ length², so area ratio = (length ratio)².

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Scaling

Common Mistake

Try These Next