Areas of Similar Triangles

The Area-Ratio Theorem

We know similar triangles have proportional sides. How do their areas compare? Not in the same ratio as the sides — areas grow faster.

Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If $\triangle ABC \sim \triangle DEF$, then:

$$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{CA}{FD}\right)^2$$

Key Point: Areas are in the ratio of the square of sides. If sides are in ratio $2:3$, areas are in ratio $4:9$.

[Board Important] Don't confuse: perimeters are in the side ratio (e.g. 2:3), but areas are in the squared ratio (4:9). This is a very common error.

Why the Square?

The idea: area depends on two dimensions (base and height), each of which scales by the side ratio $k$. So area scales by $k \times k = k^2$.

The proof idea

For similar triangles, the corresponding altitudes are also in the ratio $k$. Since $\text{area} = \dfrac{1}{2} \times \text{base} \times \text{height}$, and both base and height scale by $k$: $$\frac{\text{ar}_1}{\text{ar}_2} = \frac{\frac{1}{2} b_1 h_1}{\frac{1}{2} b_2 h_2} = k \times k = k^2.$$

Key Point: Because area uses two scaled lengths, the area ratio is the square of the length ratio.

[Board Important] Corresponding altitudes, medians, and angle bisectors of similar triangles are all in the side ratio $k$ — and the areas in $k^2$.

Equivalent Forms of the Area Ratio

Since all corresponding linear measures of similar triangles are in the ratio $k$, the area ratio equals the square of any of them:

$$\frac{\text{ar}(\triangle 1)}{\text{ar}(\triangle 2)} = \left(\text{sides}\right)^2 = \left(\text{altitudes}\right)^2 = \left(\text{medians}\right)^2 = \left(\text{perimeters}\right)^2$$

Worked outline

If the ratio of perimeters of two similar triangles is $4:5$, the ratio of their areas is $4^2 : 5^2 = 16 : 25$.

Key Point: The area ratio is the square of the ratio of any pair of corresponding linear measures (sides, altitudes, medians, or perimeters).

[Board Important] If a problem gives the ratio of areas and asks for the ratio of sides, take the square root: areas $9:16$ ⇒ sides $3:4$.

Going Both Ways

The theorem works in both directions:

• Sides → areas: square the side ratio.
• Areas → sides: take the square root of the area ratio.

Worked outline

Two similar triangles have areas $50\ \text{cm}^2$ and $98\ \text{cm}^2$. The ratio of areas is $\dfrac{50}{98} = \dfrac{25}{49}$, so the ratio of corresponding sides is $\sqrt{\dfrac{25}{49}} = \dfrac{5}{7}$.

Key Point: Square to go from sides to areas; square-root to go from areas to sides.

[Board Important] Always simplify the area ratio to a perfect-square form (like $\dfrac{25}{49}$) so the square root is clean.

Solved Examples

Example 1: Sides to areas

Two similar triangles have corresponding sides in the ratio $3:4$. Find the ratio of their areas.

Solution:

1. Area ratio $= (\text{side ratio})^2 = 3^2 : 4^2 = 9 : 16$.

Final Answer: $9 : 16$.

Takeaway: Square the side ratio to get the area ratio.

Example 2: Areas to sides

The areas of two similar triangles are in the ratio $16 : 25$. Find the ratio of their corresponding sides.

Solution:

1. Side ratio $= \sqrt{16 : 25} = 4 : 5$.

Final Answer: $4 : 5$.

Takeaway: Take the square root of the area ratio for the side ratio.

Example 3: Find an area

$\triangle ABC \sim \triangle DEF$. ar$(\triangle ABC) = 64\ \text{cm}^2$ and $\dfrac{AB}{DE} = \dfrac{2}{3}$. Find ar$(\triangle DEF)$.

Solution:

1. $\dfrac{\text{ar}(ABC)}{\text{ar}(DEF)} = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}$.
2. $\dfrac{64}{\text{ar}(DEF)} = \dfrac{4}{9} \Rightarrow \text{ar}(DEF) = \dfrac{64 \times 9}{4} = 144\ \text{cm}^2$.

Final Answer: $144\ \text{cm}^2$.

Takeaway: Use the squared side ratio to relate the two areas.

Example 4: Perimeters and areas

The perimeters of two similar triangles are in the ratio $4:5$. Find the ratio of their areas.

Solution:

1. Perimeter ratio = side ratio = $4:5$.
2. Area ratio $= 4^2 : 5^2 = 16 : 25$.

Final Answer: $16 : 25$.

Takeaway: Perimeter ratio equals side ratio; area ratio is its square.

Example 5: Altitudes

Two similar triangles have corresponding altitudes in the ratio $5:7$. Find the ratio of their areas.

Solution:

1. Altitudes are in the side ratio, $5:7$.
2. Area ratio $= 5^2 : 7^2 = 25 : 49$.

Final Answer: $25 : 49$.

Takeaway: Altitudes follow the side ratio; areas the square.

Example 6: Areas to a length

$\triangle ABC \sim \triangle PQR$ with ar$(\triangle ABC) = 25\ \text{cm}^2$, ar$(\triangle PQR) = 49\ \text{cm}^2$. If $BC = 5$ cm, find $QR$.

Solution:

1. Side ratio $= \sqrt{\dfrac{25}{49}} = \dfrac{5}{7}$.
2. $\dfrac{BC}{QR} = \dfrac{5}{7} \Rightarrow \dfrac{5}{QR} = \dfrac{5}{7} \Rightarrow QR = 7$ cm.

Final Answer: $QR = 7$ cm.

Takeaway: Get the side ratio from area ratio, then find the length.

Example 7: Medians given

Two similar triangles have corresponding medians 6 cm and 9 cm. Find the ratio of their areas.

Solution:

1. Median ratio $= \dfrac{6}{9} = \dfrac{2}{3}$ = side ratio.
2. Area ratio $= \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}$.

Final Answer: $4 : 9$.

Takeaway: Medians scale like sides; areas like the square.

Example 8: Equal areas ⇒ congruent

If two similar triangles have equal areas, show they are congruent.

Solution:

1. Equal areas ⇒ area ratio $= 1$.
2. Side ratio $= \sqrt{1} = 1$, so corresponding sides are equal.
3. Equal corresponding sides ⇒ the triangles are congruent.

Final Answer: They are congruent.

Takeaway: Similar + equal areas ⇒ congruent (scale factor 1).

Example 9: Ratio from a shared figure

In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$, $E$ on $AC$, and $\dfrac{AD}{AB} = \dfrac{1}{3}$. Find $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle ABC)}$.

Solution:

1. $DE \parallel BC \Rightarrow \triangle ADE \sim \triangle ABC$ (AA), with side ratio $\dfrac{AD}{AB} = \dfrac{1}{3}$.
2. Area ratio $= \left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}$.

Final Answer: $\dfrac{1}{9}$.

Takeaway: A parallel line creates similar triangles; areas in the squared ratio.

Example 10: Find the larger area

$\triangle ABC \sim \triangle DEF$ with $AB = 6$ cm, $DE = 9$ cm. If ar$(\triangle ABC) = 48\ \text{cm}^2$, find ar$(\triangle DEF)$.

Solution:

1. $\dfrac{\text{ar}(ABC)}{\text{ar}(DEF)} = \left(\dfrac{6}{9}\right)^2 = \dfrac{4}{9}$.
2. $\dfrac{48}{\text{ar}(DEF)} = \dfrac{4}{9} \Rightarrow \text{ar}(DEF) = \dfrac{48 \times 9}{4} = 108\ \text{cm}^2$.

Final Answer: $108\ \text{cm}^2$.

Takeaway: Larger triangle ⇒ larger area, found via the squared ratio.