Similar Figures, Scale Factors and Corresponding Measures | Grade 8 Math | LevelKite

In congruence work you used rigid motions: translations, reflections and rotations. They move a figure without changing any length or angle. A dilation is a transformation that can change size. It has a fixed point called the centre and a positive number $k$ called the scale factor. Each point $P$ moves along the ray that starts at the centre and passes through $P$ . It stops at the point $P'$ (read P prime) whose distance from the centre is $k$ times the distance from the centre to $P$ . The centre itself does not move.

If $k > 1$ , the image is an enlargement.
If $0 < k < 1$ , the image is a reduction.
If $k = 1$ , every point stays where it is.

On a coordinate plane, the $x$ -axis runs across and the $y$ -axis runs up. Write the centre as $(a, b)$ and a point of the original figure as $(x, y)$ . The point is $x - a$ units across from the centre and $y - b$ units up from it. A negative value means left or down. A dilation multiplies both of these by $k$ . The point stays on the same ray from the centre and ends up $k$ times as far away, so the image of $(x, y)$ is:

(x,\ y) \longmapsto \big(a + k(x - a),\ b + k(y - b)\big), \qquad k > 0

When the centre is the origin, $a = 0$ and $b = 0$ , so the rule shortens to $(kx, ky)$ . Use that shortcut only when the centre is the origin.

Why does a dilation keep a figure's shape? Take two original points with $x$ -coordinates $s$ and $t$ . The $x$ -coordinates of their images differ by $\big(a + k(t - a)\big) - \big(a + k(s - a)\big) = k(t - s)$ . The $y$ -coordinates behave the same way. So each image side has $k$ times the across-change and $k$ times the up-change of its original side. Because $k > 0$ , the image side points the same way as the original side and is $k$ times as long. The angle between any two sides therefore does not change.

Key concept

Similar figures and the scale factor

Two figures are similar when a sequence of rigid motions and dilations carries one onto the other. A vertex and the vertex it lands on are corresponding vertices. Sides and angles correspond in the same way. For similar figures:

corresponding angles are equal;
corresponding side lengths are proportional: each image length is the same positive number $k$ , the scale factor, times its original length.

For polygons, this also works as a test. If you can pair the vertices in order so that both facts hold, the figures are similar.

k = \frac{\text{image length}}{\text{corresponding original length}}, \quad k > 0

Now identify similar figures from coordinates. In the figure below, triangle $ABC$ has $A(1, 1)$ , $B(4, 1)$ and $C(1, 3)$ . Triangle $DEF$ has $D(5, 3)$ , $E(11, 3)$ and $F(5, 7)$ . Lengths are in grid units. A horizontal side's length is its larger $x$ -coordinate minus its smaller one. For a vertical side, use the $y$ -coordinates in the same way. A horizontal side and a vertical side meet at a right angle. Rely on these coordinate facts, not on how the drawing looks.

Triangle

ABC

has vertices

(1, 1)

,

(4, 1)

and

(1, 3)

. Triangle

DEF

has vertices

(5, 3)

,

(11, 3)

and

(5, 7)

.

Worked example

Are these triangles similar?

Triangle $ABC$ has $A(1, 1)$ , $B(4, 1)$ and $C(1, 3)$ . Triangle $DEF$ has $D(5, 3)$ , $E(11, 3)$ and $F(5, 7)$ . Is triangle $DEF$ similar to triangle $ABC$ ? If it is, give the scale factor from $ABC$ to $DEF$ .

Find the leg lengths. $AB$ and $DE$ are horizontal, so subtract their $x$ -coordinates. $AC$ and $DF$ are vertical, so subtract their $y$ -coordinates. Each triangle has a horizontal leg meeting a vertical leg, so $\angle A$ and $\angle D$ are right angles.
$AB = 4 - 1 = 3,\ AC = 3 - 1 = 2,\ DE = 11 - 5 = 6,\ DF = 7 - 3 = 4$
Match the vertices: right angle to right angle, longer leg to longer leg, shorter leg to shorter leg. This gives $A \leftrightarrow D$ , $B \leftrightarrow E$ and $C \leftrightarrow F$ . Then divide each $DEF$ length by its matching $ABC$ length.
$\frac{DE}{AB} = \frac{6}{3} = 2, \qquad \frac{DF}{AC} = \frac{4}{2} = 2$
Compare the slanted sides $BC$ and $EF$ by their changes. From $B$ to $C$ you go $3$ left and $2$ up. From $E$ to $F$ you go $6$ left and $4$ up. Both changes are doubled, so $EF$ points the same way as $BC$ and is twice as long. All three side ratios equal $2$ , so the sides are proportional.
$\text{across: } \frac{6}{3} = 2, \qquad \text{up: } \frac{4}{2} = 2$
Compare the angles. Each side of $DEF$ points the same way as its matching side of $ABC$ . $DE$ and $AB$ both run right, $DF$ and $AC$ both run up, and $EF$ matches $BC$ (see the previous step). So the angles between matching sides are equal.
$\angle A = \angle D = 90^\circ, \quad \angle B = \angle E, \quad \angle C = \angle F$
Confirm with a sequence of moves. First dilate about the origin with $k = 2$ . The centre is the origin, so $(x, y) \to (2x, 2y)$ . Then translate $3$ units right and $1$ unit up, which is a rigid motion. Each vertex lands exactly on its partner.
$A(1, 1) \to (2, 2) \to (5, 3),\ B(4, 1) \to (8, 2) \to (11, 3),\ C(1, 3) \to (2, 6) \to (5, 7)$

Answer

Yes. A dilation about the origin with scale factor $2$ , followed by a translation $3$ right and $1$ up, carries triangle $ABC$ onto triangle $DEF$ . So the triangles are similar, with scale factor $2$ from $ABC$ to $DEF$ .

The rectangle from

(1, 1)

to

(5, 3)

is

5 - 1 = 4

by

3 - 1 = 2

. The rectangle from

(6, 1)

to

(12, 5)

is

12 - 6 = 6

by

5 - 1 = 4

.

To find the scale factor of a dilation, use the key idea: divide an image length by the length of its corresponding original side. Measure both lengths in the same unit, so the units cancel and $k$ is a plain number. In the figure below, a dilation with centre $(2, 2)$ maps triangle $PQR$ onto triangle $P'Q'R'$ . The vertices are $P(10, 2)$ , $Q(10, 10)$ and $R(6, 10)$ . Their images are $P'(6, 2)$ , $Q'(6, 6)$ and $R'(4, 6)$ . Each labelled point in the figure shows its own coordinates.

The centre is

(2, 2)

.

P

,

Q

and

R

are at

(10, 2)

,

(10, 10)

and

(6, 10)

.

P'

and

Q'

are at

(6, 2)

and

(6, 6)

. The unlabelled vertex

R'

is at

(4, 6)

.

Worked example

Scale factor of a dilation not centred at the origin

A dilation with centre $(2, 2)$ maps $P(10, 2)$ , $Q(10, 10)$ and $R(6, 10)$ to $P'(6, 2)$ , $Q'(6, 6)$ and $R'(4, 6)$ . Find the scale factor $k$ .

$PQ$ and $P'Q'$ are vertical, because the two endpoints of each side share an $x$ -coordinate. So each length is the larger $y$ -coordinate minus the smaller one.
$PQ = 10 - 2 = 8, \qquad P'Q' = 6 - 2 = 4$
Divide the image length by the corresponding original length.
$k = \frac{P'Q'}{PQ} = \frac{4}{8} = \frac{1}{2}$
Check another pair of sides. $QR$ and $Q'R'$ are horizontal, so subtract their $x$ -coordinates. This ratio matches the first one.
$\frac{Q'R'}{QR} = \frac{6 - 4}{10 - 6} = \frac{2}{4} = \frac{1}{2}$
Confirm the centre with the rule, using $a = 2$ , $b = 2$ and $k = \tfrac{1}{2}$ . $R$ lands exactly on $R'$ . Likewise, $P \to (2 + 4,\ 2 + 0) = (6, 2)$ and $Q \to (2 + 4,\ 2 + 4) = (6, 6)$ . Halving the coordinates of $R$ would give $(3, 5)$ instead, not $R'$ . The $(kx, ky)$ shortcut needs the centre at the origin.
$R(6, 10) \to \left(2 + \tfrac{1}{2}(6 - 2),\ 2 + \tfrac{1}{2}(10 - 2)\right) = (2 + 2,\ 2 + 4) = (4, 6)$

Answer

The scale factor is $k = \tfrac{1}{2}$ . Because $0 < k < 1$ , the dilation is a reduction: each side of triangle $P'Q'R'$ is half as long as its corresponding side in triangle $PQR$ .

To find a missing measure in similar figures, you need the scale factor $k$ and the vertex correspondence. A side joins two vertices, so its corresponding side joins their partners. Then:

image length $= k \times$ corresponding original length;
original length $=$ corresponding image length $\div k$ . You may divide by $k$ because $k > 0$ , so $k \neq 0$ ;
an angle has the same measure as its corresponding angle. Angles are not multiplied by $k$ , because rigid motions and dilations do not change angle measures.

Worked example

Missing side and angle from a scale factor

Triangle $JKL$ is similar to triangle $MNP$ , with $J \leftrightarrow N$ , $K \leftrightarrow P$ and $L \leftrightarrow M$ . In triangle $JKL$ , $JK = 6$ cm and $\angle K = 50^\circ$ . In triangle $MNP$ , $NP = 9$ cm and $PM = 12$ cm. Find the scale factor from $JKL$ to $MNP$ , the length $KL$ , and $\angle P$ .

Pair the parts using the correspondence. $JK$ joins $J$ and $K$ , so it matches the side joining $N$ and $P$ . In the same way, $KL$ matches $PM$ , and $\angle K$ matches $\angle P$ .
$JK \leftrightarrow NP, \quad KL \leftrightarrow PM, \quad \angle K \leftrightarrow \angle P$
Find $k$ from the pair where both lengths are known: image length over original length. The centimetres cancel.
$k = \frac{NP}{JK} = \frac{9 \text{ cm}}{6 \text{ cm}} = \frac{3}{2} = 1.5$
Since $PM = k \times KL$ , divide both sides by $k$ . This is allowed because $k \neq 0$ . Check: $1.5 \times 8 = 12$ , which matches $PM$ .
$KL = \frac{PM}{k} = \frac{12 \text{ cm}}{1.5} = 12 \times \frac{2}{3} \text{ cm} = 8 \text{ cm}$
$\angle P$ corresponds to $\angle K$ . Corresponding angles of similar figures are equal, so the measure stays the same.
$\angle P = \angle K = 50^\circ$

Answer

The scale factor from triangle $JKL$ to triangle $MNP$ is $1.5$ , side $KL$ is $8$ cm long, and $\angle P$ measures $50^\circ$ .

Takeaway: Similar figures are linked by a sequence of rigid motions and dilations. Their corresponding angles are equal, and every corresponding length is multiplied by the same scale factor $k > 0$ . You find $k$ by dividing an image length by its corresponding original length. With $k$ and the vertex correspondence, you multiply or divide lengths by $k$ and keep angle measures the same. Next, you will study dilations on the coordinate plane and write the rule for a dilation.