Areas of similar triangles

If two triangles are similar in the ratio $k$ , then every length scales by $k$ , sides, medians, altitudes, perimeters. But what about area?

Statement

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

In symbols, if $\triangle ABC \sim \triangle DEF$ , then $\frac{[\triangle ABC]}{[\triangle DEF]} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{CA}{FD}\right)^2.$

The same square-of-ratio rule holds for corresponding medians, altitudes, angle bisectors, and perimeters when squared.

Proof

Drop the altitude from $A$ to $BC$ in $\triangle ABC$ ; call its length $h_A$ . Drop the altitude from $D$ to $EF$ in $\triangle DEF$ ; call it $h_D$ .

Area: $[\triangle ABC] = \frac{1}{2} \cdot BC \cdot h_A$ and $[\triangle DEF] = \frac{1}{2} \cdot EF \cdot h_D$ .

Ratio: $\frac{[\triangle ABC]}{[\triangle DEF]} = \frac{BC \cdot h_A}{EF \cdot h_D}.$

Now we need to show $h_A / h_D = BC/EF$ (equivalently, the ratio of altitudes equals the ratio of corresponding sides).

Consider the right triangles formed by the altitudes and corresponding sides. They are similar by AA (they share an angle of the original triangles, and both have a $90^\circ$ ). Hence $h_A / h_D = AB/DE$ . And from the similarity, $AB/DE = BC/EF$ . So $h_A/h_D = BC/EF$ .

Substitute: $\dfrac{[\triangle ABC]}{[\triangle DEF]} = \dfrac{BC}{EF} \cdot \dfrac{BC}{EF} = \left(\dfrac{BC}{EF}\right)^2$ . $\blacksquare$

Why squared?

Area has units of length-squared. Scaling lengths by $k$ scales areas by $k^2$ . This is a universal feature of geometry: $n$ -dimensional measure scales by the $n$ -th power of the scale factor. (Volumes scale by $k^3$ , which is exactly the surface-area-and-volume chapter ahead.)

Worked examples

Example 1. Two similar triangles have areas $36$ cm² and $100$ cm². Find the ratio of their corresponding sides.

Ratio of areas $= 36/100 = 9/25$ . Ratio of sides $= \sqrt{9/25} = 3/5$ .

Example 2. Two similar triangles have corresponding sides $4$ cm and $9$ cm. The area of the smaller is $32$ cm². Find the area of the larger.

Ratio of sides $= 4/9$ , so ratio of areas $= 16/81$ .

Area of larger $= 32 \cdot 81/16 = 2 \cdot 81 = 162$ cm².

Example 3. $\triangle ABC \sim \triangle DEF$ with the ratio of their perimeters $3 : 4$ . If $[\triangle DEF] = 80$ cm², find $[\triangle ABC]$ .

Ratio of perimeters $=$ ratio of sides $= 3/4$ . So ratio of areas $= 9/16$ .

$[\triangle ABC] = 80 \cdot 9/16 = 45$ cm².

Example 4. In $\triangle ABC$ , $D$ and $E$ are points on $AB$ and $AC$ with $DE \parallel BC$ , $AD : DB = 1 : 2$ . Find the ratio of areas of $\triangle ADE$ and trapezium $DECB$ .

Since $DE \parallel BC$ , $\triangle ADE \sim \triangle ABC$ (AA). Side ratio $= AD/AB = 1/3$ . Area ratio $= 1/9$ .

So $[\triangle ADE] / [\triangle ABC] = 1/9$ , hence $[\triangle ADE] : [DECB] = 1 : 8$ .

Example 5. In $\triangle ABC$ , an isoceles triangle is constructed on $BC$ such that the new triangle is similar to $\triangle ABC$ . If $[\triangle ABC] = 64$ cm² and one side of $\triangle ABC$ has length $8$ cm while the corresponding side of the new triangle is $10$ cm, find the area of the new triangle.

Ratio of corresponding sides $= 8/10 = 4/5$ . Area ratio $= 16/25$ .

$[\triangle ABC] / [\triangle \text{new}] = 16/25$ , so $[\triangle \text{new}] = 64 \cdot 25/16 = 100$ cm².

Try it yourself

Ratio of sides of similar triangles is $3 : 5$ . Ratio of areas?
Areas of two similar triangles are $25$ and $144$ cm². Ratio of corresponding altitudes?
$\triangle ABC \sim \triangle PQR$ with $BC = 6, QR = 10$ , $[\triangle ABC] = 18$ cm². Find $[\triangle PQR]$ .
In $\triangle ABC$ , $DE \parallel BC$ with $AD = 6, DB = 4$ . Find $[\triangle ADE] : [\triangle ABC]$ .
Two similar triangles with perimeters $48$ cm and $60$ cm; area of smaller is $32$ cm². Area of larger?
The areas of two similar triangles are $81$ and $49$ . Ratio of corresponding medians?
The ratio of areas is $9 : 16$ . If one side is $12$ cm, find the corresponding side of the other.
Two equilateral triangles have side ratio $2 : 3$ . Find the ratio of their (a) perimeters, (b) areas.
Show that if two triangles are similar and an altitude of one is $h_1$ and of the other $h_2$ , then $h_1^2 / h_2^2$ equals the ratio of their areas.
The areas of two similar triangles are in ratio $4 : 9$ . If the perimeter of the larger is $36$ cm, find the perimeter of the smaller.

Pitfalls / Insight

Square (or square-root) the ratio when moving between sides and areas , never forget.
Altitudes, medians, angle bisectors, perimeters all scale linearly. Only area (and any "second-order" quantity) is square.
Identify the smaller and larger carefully when reading the problem.

Insight. Area scales like the square of length, a fact you will see again with volume (cube of length) in chapter 12. This is the geometry of dimensions in miniature.