Math Core

A photo and its enlargement, a map and the land it shows, a scale model and the real car: each pair has the same shape but a different size. Geometry calls this relationship similarity, and it lets you find lengths you can't measure directly by comparing them with lengths you can.

Same shape, different size

In the transformations unit you saw that a dilation stretches or shrinks a figure by a scale factor $k$ . A dilation keeps every angle the same, and it multiplies every length by $k$ . Rigid motions (translations, reflections, rotations) move a figure without changing it at all. Put those together and you get the definition of similar figures.

Definition

Similar figures

Two figures are similar if one can be mapped onto the other by a dilation followed by a sequence of rigid motions. The symbol for "is similar to" is $\sim$ .

For polygons, that transformation definition boils down to two checks you can do with numbers.

Similar polygons

Two polygons are similar exactly when, for some way of matching their vertices,

every pair of corresponding angles is congruent, and
every pair of corresponding sides is proportional: all the ratios of matching sides are equal.

That common ratio is the scale factor.

Both conditions matter. A $2 \times 2$ square and a $2 \times 5$ rectangle have all their angles equal ( $90^\circ$ each), but they aren't similar, because $\dfrac{2}{2} \ne \dfrac{2}{5}$ . A square and a rhombus with side $2$ have all their sides in ratio $1$ , but a rhombus with $60^\circ$ angles isn't the same shape as a square. You need both matching angles and proportional sides.

Similarity statements

Just like a congruence statement, a similarity statement lists the vertices in matching order. The statement $\triangle ABC \sim \triangle DEF$ tells you:

$\angle A \cong \angle D$ , $\angle B \cong \angle E$ , $\angle C \cong \angle F$ ;
$\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$ .

To find the side that matches $\overline{BC}$ , take the second and third letters of the other name: $\overline{EF}$ . You never have to guess from the picture.

Triangle ABC is similar to triangle DEF, with A, B, C matching D, E, F.

Scale factor

The scale factor from one figure to another is the ratio of a length in the new figure to the matching length in the original. In the figure above, $\overline{DE}$ matches $\overline{AB}$ , so the scale factor from $\triangle ABC$ to $\triangle DEF$ is

$\frac{DE}{AB} = \frac{12}{8} = \frac{3}{2}.$

Every side of $\triangle DEF$ is $\tfrac32$ times as long as the matching side of $\triangle ABC$ . Going the other way, from $\triangle DEF$ to $\triangle ABC$ , the scale factor is $\tfrac23$ . A scale factor greater than $1$ means an enlargement; less than $1$ means a reduction.

Worked example: Finding missing sides

In the figure, $\triangle ABC \sim \triangle DEF$ with $AB = 8$ , $BC = 7$ , $AC = 5$ and $DE = 12$ . Find $x = EF$ and $y = DF$ .

The scale factor from $\triangle ABC$ to $\triangle DEF$ is $\dfrac{12}{8} = \dfrac32$ . Side $\overline{EF}$ matches $\overline{BC}$ , and $\overline{DF}$ matches $\overline{AC}$ :

$x = \frac32 \cdot 7 = 10.5, \qquad y = \frac32 \cdot 5 = 7.5.$

You can also set up a proportion, such as $\dfrac{EF}{BC} = \dfrac{DE}{AB}$ , which gives $\dfrac{x}{7} = \dfrac{12}{8}$ and $8x = 84$ , so $x = 10.5$ .

Worked example: Are the rectangles similar?

Is a $4 \times 6$ rectangle similar to a $6 \times 9$ rectangle? What about a $4 \times 6$ rectangle and a $6 \times 8$ rectangle?

All angles of a rectangle are $90^\circ$ , so only the sides need checking. Match short side with short side and long side with long side.

$\dfrac{4}{6} = \dfrac23$ and $\dfrac{6}{9} = \dfrac23$ . The ratios are equal, so the rectangles are similar, with scale factor $\tfrac32$ from small to large.
$\dfrac{4}{6} = \dfrac23$ but $\dfrac{6}{8} = \dfrac34$ . The ratios differ, so these rectangles are not similar. The second one is "squarer."

Worked example: Using angles from a similarity statement

$\triangle PQR \sim \triangle XYZ$ , $m\angle P = 40^\circ$ and $m\angle Y = 75^\circ$ . Find $m\angle R$ .

From the statement, $\angle Q$ matches $\angle Y$ , so $m\angle Q = 75^\circ$ . The angles of $\triangle PQR$ add to $180^\circ$ :

$m\angle R = 180^\circ - 40^\circ - 75^\circ = 65^\circ.$

Common mistake

Match sides using the similarity statement, not the way the figures sit on the page. One figure may be rotated or flipped. Also keep your ratios consistent: if you write $\dfrac{\text{new}}{\text{original}}$ on one side of a proportion, write $\dfrac{\text{new}}{\text{original}}$ on the other side too.

Perimeters of similar polygons

If every side is multiplied by $k$ , the sum of the sides is multiplied by $k$ too. So the ratio of the perimeters of two similar polygons equals the scale factor.

Worked example: Perimeter and algebra

Quadrilateral $ABCD \sim$ quadrilateral $JKLM$ . $AB = 6$ , $JK = 10$ , $BC = x + 2$ and $KL = 25$ .

a. Find $x$ . Side $\overline{KL}$ matches $\overline{BC}$ , so

$\frac{x + 2}{25} = \frac{6}{10} \implies 10(x + 2) = 150 \implies x + 2 = 15 \implies x = 13.$

b. If $ABCD$ has perimeter $30$ , what is the perimeter of $JKLM$ ? The scale factor from $ABCD$ to $JKLM$ is $\dfrac{10}{6} = \dfrac53$ , so the perimeter is $\dfrac53 \cdot 30 = 50$ .

Tip

To check a proportion quickly, cross-multiply. $\dfrac{a}{b} = \dfrac{c}{d}$ exactly when $ad = bc$ . For example, $\dfrac{4}{6} = \dfrac{6}{9}$ because $4 \cdot 9 = 36 = 6 \cdot 6$ .

Practice

Practice 1

Which statement is true for every pair of similar polygons?

Practice 2

Quadrilateral $ABCD \sim$ quadrilateral $EFGH$ with $AB = 12$ and $EF = 9$ . What is the scale factor from $ABCD$ to $EFGH$ ?