Area of similar triangles

What is the Area of Similar Triangles Theorem? When two triangles are similar, their sides are scaled by a fixed factor. But area is two-dimensional, so it does not scale by the same factor. The theorem states: the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Think of painting a square wall. If you double both the length and the width, you do not need twice as much paint — you need four times as much, because area depends on two lengths multiplied together. The same squaring effect governs similar triangles.

Statement. If $\triangle ABC\sim\triangle PQR$, then

$$\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)}=\left(\dfrac{AB}{PQ}\right)^2=\left(\dfrac{BC}{QR}\right)^2=\left(\dfrac{CA}{RP}\right)^2.$$

Proof. Draw altitudes $AM\perp BC$ and $PN\perp QR$. Then $\text{ar}(\triangle ABC)=\tfrac12\,BC\cdot AM$ and $\text{ar}(\triangle PQR)=\tfrac12\,QR\cdot PN$, so $\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)}=\dfrac{BC\cdot AM}{QR\cdot PN}$. Now in $\triangle ABM$ and $\triangle PQN$, $\angle B=\angle Q$ (from the similarity) and $\angle M=\angle N=90^\circ$, so $\triangle ABM\sim\triangle PQN$ (AA). Hence $\dfrac{AM}{PN}=\dfrac{AB}{PQ}=\dfrac{BC}{QR}$. Substituting, $\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)}=\dfrac{BC}{QR}\cdot\dfrac{AM}{PN}=\left(\dfrac{BC}{QR}\right)^2$.

Why corresponding altitudes share the side ratio. The step $\triangle ABM\sim\triangle PQN$ in the proof above is worth dwelling on. It shows that the corresponding altitudes of two similar triangles are in exactly the same ratio as their corresponding sides. The same reasoning, applied to the triangles formed by a median or an angle bisector, shows those segments too are in the side ratio. So every linear measurement of similar triangles scales by the side ratio $k$, while every area measurement scales by $k^2$. This single idea — linear quantities scale by $k$, areas by $k^2$ — is the heart of the whole topic.

A second way to see the squaring. Take any similar triangles with side ratio $k=\dfrac{AB}{PQ}$. Their areas can be written as $\tfrac12\times\text{base}\times\text{height}$. The base scales by $k$ and the height scales by $k$ too, so the product — the area — scales by $k\times k=k^2$. This is exactly why doubling the sides multiplies the area by four and tripling the sides multiplies the area by nine.

Extension to other line segments. Because corresponding altitudes, medians and angle-bisectors of similar triangles are in the same ratio as the corresponding sides, the area ratio also equals the square of the ratio of their corresponding:

• altitudes (heights),
• medians,
• angle bisectors,
• and even their perimeters — though note the perimeter ratio itself equals the simple side ratio, and the area ratio equals the square of the perimeter ratio.

Useful corollaries. (i) If two similar triangles have equal areas, the side ratio is 1, so the triangles are congruent. (ii) To go from an area ratio back to a side, altitude or median ratio, always take the square root. (iii) The areas of two similar triangles formed by a line parallel to one side (as in BPT figures) compare as the squares of the parts of the divided side.

Triangles between the same parallels. A closely related fact used throughout this chapter: two triangles on the same base and between the same pair of parallel lines have equal areas, because they share the base and have the same height. This is the engine behind the proof of the Basic Proportionality Theorem and behind many trapezium problems. Combined with the area theorem, it lets us compare a small similar triangle with the trapezium left behind when a line is drawn parallel to one side.

Common mistakes. (i) Writing the area ratio as the plain side ratio instead of its square. (ii) Forgetting to take the square root when the question gives areas and asks for a side, altitude or median. (iii) Mixing up which triangle is numerator and which is denominator — keep the same triangle on top throughout. (iv) Applying the theorem to triangles that are not actually similar. (v) Confusing the trapezium area with the whole triangle area when a line is drawn parallel to one side — subtract the small triangle from the whole, do not square the trapezium directly.

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