Math 8 · 8.G.A.4 · G.SRT.A.2

Similarity transformations

When are two figures the same shape, even if one is bigger, turned or flipped over, and how do you show it?

You need

DilationsMath 8 Rigid motions and congruenceMath 8

Where you are 10 in Similarity

Geometry11

Similarity transformations

What a dilation does to a line

Why this exists

Congruence made “the same shape and the same size” exact: two figures are congruent when slides, turns and flips carry one exactly onto the other. Everyday life is full of a looser relation, the same shape at a different size: a photo and its enlargement, a map and the town it shows, a model car and the car. “The same shape” by eye is not a test, though. Are all rectangles the same shape? In ordinary speech they are, since they are all rectangles. Yet a 1 by 1 square and a 1 by 10 strip are both rectangles, and nobody would call them the same shape.

A dilation changes size and keeps shape, but on its own it cannot turn a figure or flip it over: it sends every side to a parallel side. A copy that is both larger and turned is out of its reach. Putting dilations together with the rigid motions closes that gap. The result is a meaning for “the same shape” that works for every kind of figure, polygons, circles, curves and letters of the alphabet alike. It says what to do to show that two figures are similar, and it gives a way to show that they are not.

This lesson takes two days. Part 1 is the definition and what a similarity transformation keeps and changes, used to decide whether two figures are similar. Part 2 carries out similarity transformations on the coordinate plane: following a sequence of moves, finding one that carries a figure onto another, undoing one, and seeing why the order of the moves matters.

Definition

A similarity transformation is a dilation, a rigid motion, or any sequence of these done one after another. A rigid motion is a move of the plane that keeps every distance (rigid motions defined precisely), for example a translation, a rotation or a reflection; every rigid motion of the plane is one of these three or a sequence of them. Doing moves one after another is composing them.

Two figures F and G are similar when some similarity transformation carries F exactly onto G: every point of F lands on G, and every point of G is landed on. We write

F \sim G

and read it “F is similar to G”. For polygons the vertices are listed in matching order, so \triangle ABC \sim \triangle DEF says that the transformation takes A to D, B to E and C to F.

A similarity transformation multiplies every length by the same positive number k, its scale factor (the Results below say why). A point’s image after one move is written P', after two moves P''.

Figure 1A dilation with center O and scale factor 2, then a reflection in line m. The last triangle is similar to the first.

In Figure 1, \triangle ABC is dilated with center O and scale factor 2 to give \triangle A'B'C', and that triangle is reflected in the line m to give \triangle A''B''C''. So \triangle ABC \sim \triangle A''B''C''. Neither move alone would do it: the dilation gets the size right but leaves the triangle facing the wrong way, and the reflection alone cannot change the size.

Congruent figures are similar too, since a rigid motion is a similarity transformation with no dilation in it. Congruence is the case k = 1.

Intuition

The dilation is the photocopier with a zoom setting: it changes the size of a copy and keeps its shape. Now take the copy and slide it across the desk, turn it, or turn it over. None of that changes its size or its shape. Two figures are similar exactly when one could be such a copy of the other, moved into place.

That picture says what survives. Everything about shape survives: every angle, which lines are parallel, and every ratio of two lengths inside the figure, such as “the long side is twice the short side”. What does not survive is size (lengths and areas), position, and, after a flip, which way round the figure is labeled.

It also gives the two ways to use the definition. To show that two figures are similar, produce the moves. To show that they are not, find one thing about shape that differs, an angle or a ratio of sides. No similarity transformation can change it, so no sequence of moves can work, and there is no need to try them all.

Results

Part 1: what a similarity transformation keeps and changes

Take a similarity transformation whose sequence contains dilations with scale factors k_1, k_2, \dots, k_n, and any number of rigid motions.

Every length is multiplied by one number. For any two points P and Q, with images P' and Q',

P'Q' = k \cdot PQ \qquad\text{where}\qquad k = k_1 k_2 \cdots k_n

With no dilations at all, k = 1 and the transformation is a rigid motion.

Every angle keeps its measure.

Every ratio of two lengths is kept. If PQ = 2 \cdot RS, then P'Q' = 2 \cdot R'S'.

Lines go to lines, and parallel lines go to parallel lines.

Every area is multiplied by k^2, which is the subject of scale factor and area.

Similarity can be undone. If F \sim G with scale factor k, then G \sim F with scale factor \tfrac1k.

Similarity can be chained. If F \sim G with scale factor k and G \sim H with scale factor m, then F \sim H with scale factor km. Every figure is similar to itself, with k = 1.

One dilation and one rigid motion are always enough. A similarity transformation with scale factor k, however long its sequence, has the same effect as a single dilation with scale factor k (with any center you like) followed by a single rigid motion.

For polygons: equal angles and proportional sides. Two polygons are similar exactly when their vertices can be matched so that every pair of matching angles is equal and every pair of matching sides is in one ratio. For triangles, \triangle ABC \sim \triangle DEF exactly when \angle A = \angle D, \angle B = \angle E, \angle C = \angle F and

\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA} = k

For polygons with four or more sides, neither half is enough on its own (Examples 2 and 4). For triangles, less is needed: two pairs of equal angles are enough, and so are three proportional pairs of sides.

Part 2: on the coordinate plane

The rules for the moves used in Part 2, each taught in its own lesson (dilations, translations, reflections, including the swap for y = x and rotations):

Move	Rule
Dilation, center the origin, scale factor k	(x, y) \longmapsto (kx,\ ky)
Translation by \langle a, b \rangle	(x, y) \longmapsto (x + a,\ y + b)
Reflection in the x-axis	(x, y) \longmapsto (x,\ -y)
Reflection in the y-axis	(x, y) \longmapsto (-x,\ y)
Reflection in the line y = x	(x, y) \longmapsto (y,\ x)
Rotation 90^\circ counterclockwise about the origin	(x, y) \longmapsto (-y,\ x)
Rotation 180^\circ about the origin	(x, y) \longmapsto (-x,\ -y)
Rotation 90^\circ clockwise about the origin	(x, y) \longmapsto (y,\ -x)

A sequence is applied one move at a time, each move acting on the output of the move before. Feeding one rule into the next gives a single rule for the whole sequence.

The order matters. The same moves in a different order can put the figure somewhere else (Example 8). To undo a sequence, undo its moves in reverse order, the last move first, just as when undoing a composition of rigid motions.

Turning direction. Name a triangle’s vertices so that A \to B \to C runs counterclockwise. Translations, rotations and dilations keep that direction; a reflection reverses it. So when a triangle and its image run in opposite directions, every similarity transformation that carries one onto the other contains a reflection (in fact an odd number of them).

A dilation followed by a translation is a dilation. When k \neq 1, the rule

(x, y) \longmapsto (kx + a,\ ky + b)

is the dilation with scale factor k and center \left(\dfrac{a}{1 - k},\ \dfrac{b}{1 - k}\right).

Why it is true

Lengths, angles and ratios. Follow one segment \overline{PQ} through the sequence. Each rigid motion keeps its length, and each dilation multiplies it by that dilation’s factor. After the whole sequence it has been multiplied by k_1k_2 \cdots k_n, and that product is the same whichever segment you followed. Every rigid motion and every dilation keeps angle measures, and sends lines to lines and parallel lines to parallel lines (what a dilation does to a line), so the whole sequence does too. A ratio survives because both of its lengths are multiplied by the same k:

\frac{P'Q'}{R'S'} = \frac{k \cdot PQ}{k \cdot RS} = \frac{PQ}{RS}

Undoing and chaining. Every move can be undone by a move of the same kind: a translation by the opposite translation, a rotation by turning back, a reflection by the same reflection again, and a dilation with factor k_i by the dilation with the same center and factor \tfrac{1}{k_i}. Undo the last move first, then the one before it, and so on back to the first, as with any composition. The undoing sequence multiplies lengths by \tfrac{1}{k_n} \cdots \tfrac{1}{k_1} = \tfrac1k. To chain, do one sequence and then the other; the scale factors multiply to km.

One dilation and one rigid motion. Let S be a similarity transformation with scale factor k, and let D be the dilation with center O (any point) and scale factor k. Make a new move R: first undo D, then do S. The undoing multiplies every length by \tfrac1k and then S multiplies it by k, so R keeps every distance, and a move of the plane that keeps every distance is a rigid motion by definition (rigid motions defined precisely). Now doing D and then R means doing D, undoing it, and doing S, which has exactly the effect of S.

Polygons. Suppose a similarity transformation carries polygon F onto polygon G. It takes each vertex of F to a vertex of G and each side to a side, which matches them up; by the results above, matching angles are equal and every side is multiplied by k. For the other direction, suppose the vertices can be matched so that the angles are equal and every side of G is k times the matching side of F. Dilate F with scale factor k about any center. The image has the same angles as F, which are the angles of G, and sides k times those of F, which are the sides of G. Polygons with all matching sides and all matching angles equal are congruent, so a rigid motion carries the image onto G. A dilation followed by a rigid motion is a similarity transformation, so F \sim G.

Turning direction. A translation or a rotation moves a figure without lifting it off the page, and a dilation with a positive factor slides each point straight out from or in toward its center, so none of them can make the vertices run the other way round. A reflection turns the figure over, and a path that runs counterclockwise is seen in a mirror running clockwise. Each reflection reverses the direction once, so an even number of them leaves it as it was.

Dilation then translation. A dilation has exactly one point that stays put, its center, so look for one. The point (x, y) stays put when kx + a = x and ky + b = y, that is when x = \dfrac{a}{1 - k} and y = \dfrac{b}{1 - k}, which needs k \neq 1. Call that point (c, d). The dilation with center (c, d) and factor k sends x to

c + k(x - c) = kx + (1 - k)c = kx + a

and in the same way sends y to ky + b, which is the given rule.

Worked examples

Examples 1 to 5 go with Part 1, deciding about similarity from what a similarity transformation keeps. Examples 6 to 10 go with Part 2, the coordinate plane.

Example 1

What a long sequence does to a triangle

A triangle has sides of 3 cm, 4 cm and 5 cm, with a right angle between the two shorter sides. It is reflected in a line, then dilated with scale factor 2, then rotated 90^\circ about some point, then dilated with scale factor \tfrac13 about a different center. Describe the image.

Solution

The reflection and the rotation are rigid motions, so they change no lengths. Only the two dilations change size, and the scale factor of the whole sequence is their product:

k = 2 \cdot \tfrac13 = \tfrac23

Every side is multiplied by \tfrac23:

\tfrac23 \cdot 3 = 2 \qquad \tfrac23 \cdot 4 = \tfrac83 \qquad \tfrac23 \cdot 5 = \tfrac{10}{3}

Every angle keeps its measure, so the image is a right triangle with sides 2 cm, \tfrac83 cm and \tfrac{10}{3} cm, and its right angle is between the two shorter sides.

Check. The Pythagorean theorem must hold for the image: 2^2 + \left(\tfrac83\right)^2 = 4 + \tfrac{64}{9} = \tfrac{100}{9} = \left(\tfrac{10}{3}\right)^2.

The mirror line, the centers and the point of rotation were never given, and the answer did not need them. They decide where the image sits, not its size or its shape.

Example 2

Which rectangles are similar

Three rectangles measure 6 by 9, 10 by 15 and 9 by 12. Which pairs of them are similar?

Solution

All the angles of a rectangle are right angles, and a similarity transformation carries them to right angles, so the angles never stand in the way. The question is the sides.

A similarity transformation multiplies every length by the same positive k, so it keeps the order of sizes: the shorter sides of one rectangle must go to the shorter sides of the other. Since every ratio inside a figure is kept, compare short side to long side in each rectangle:

\frac{6}{9} = \frac23 \qquad \frac{10}{15} = \frac23 \qquad \frac{9}{12} = \frac34

The 6 by 9 and the 10 by 15 rectangles have the same ratio, and they are similar with scale factor \tfrac{10}{6} = \tfrac{15}{9} = \tfrac53: all angles equal and all sides in one ratio, which is the result for polygons. The 9 by 12 rectangle has a different ratio, so it is similar to neither. No sequence of moves can change a ratio inside a figure, so there is nothing more to try.

Directly, for the 6 by 9 and the 9 by 12: the short sides would need k = \tfrac96 = \tfrac32 and the long sides k = \tfrac{12}{9} = \tfrac43, and one transformation has only one scale factor.

Example 3

Any two circles are similar

Circle C_1 has center (1, 2) and radius 3. Circle C_2 has center (-4, 0) and radius 7.5. Find a similarity transformation that carries C_1 onto C_2, and explain why any two circles are similar.

Solution

First move the center into place. The translation by \langle -5, -2 \rangle takes (1, 2) to (-4, 0). A translation keeps every distance, so it carries C_1 to the circle with center (-4, 0) and radius 3.

Now change the size. The dilation with center (-4, 0) and scale factor

k = \frac{7.5}{3} = 2.5

moves each point at distance 3 from the center along its own ray to distance 7.5, which is a point of C_2. Every point of C_2 is reached, because a point of C_2 is the image of the point on the same ray at distance 3. So the image is all of C_2, and C_1 \sim C_2.

The same two steps work for any two circles: translate one center onto the other, then dilate about that center by the ratio of the radii. A circle has no sides and no corners, so “matching angles and proportional sides” has nothing to say about it, while the definition by moves handles it at once.

Example 4

Proportional sides are not enough

A rhombus has sides of 4 and one angle of 50^\circ. A square has sides of 6. Every side of the square is \tfrac32 times every side of the rhombus. Are they similar?

Solution

No. A rhombus is a parallelogram, so its opposite angles are equal and its neighboring angles add to 180^\circ. Its angles are 50^\circ, 130^\circ, 50^\circ and 130^\circ. Every angle of the square is 90^\circ.

Any similarity transformation keeps angle measures, so it would have to carry the 50^\circ angle of the rhombus to a 50^\circ angle of the square, and the square has none. No sequence of moves can carry the rhombus onto the square.

Example 2 showed the other half of the story: rectangles of different ratios have all their angles equal and are still not similar. For polygons with four or more sides, similarity needs equal angles and proportional sides.

Example 5

Families that are always similar

Decide whether each is always true, and say why. (a) Any two squares are similar. (b) Any two isosceles triangles are similar. (c) Any two regular hexagons are similar.

Solution

(a) True. Take squares with sides a and b. Every angle of each is 90^\circ, and every side of the second is \tfrac{b}{a} times every side of the first. Equal angles and proportional sides make two polygons similar.

(b) False, and one pair is enough to show it. An isosceles triangle with a 40^\circ angle at its apex has base angles \tfrac{180^\circ - 40^\circ}{2} = 70^\circ, so its angles are 40^\circ, 70^\circ and 70^\circ. One with a 100^\circ apex has base angles \tfrac{180^\circ - 100^\circ}{2} = 40^\circ, so its angles are 100^\circ, 40^\circ and 40^\circ. The first has no 100^\circ angle, and a similarity transformation keeps every angle, so these two are not similar.

(c) True. The angles of a hexagon add to (6 - 2) \cdot 180^\circ = 720^\circ, as the angle sum of a polygon gives, so each angle of a regular hexagon is \tfrac{720^\circ}{6} = 120^\circ. With sides a and b, every side of the second is \tfrac{b}{a} times every side of the first, and every angle is 120^\circ in both. The same argument shows that any two regular polygons with the same number of sides are similar.

Example 6

Following a sequence on a grid

Triangle PQR has P(1, 1), Q(3, 1) and R(1, 2). Dilate it with center the origin and scale factor 2, then reflect the result in the y-axis, then translate that by \langle 1, -3 \rangle. Find the final image \triangle P'Q'R' and a single rule for the whole sequence.

Solution

Apply the moves one at a time, each to the output of the one before.

The dilation doubles both coordinates: (1, 1) \longmapsto (2, 2), (3, 1) \longmapsto (6, 2) and (1, 2) \longmapsto (2, 4).

The reflection in the y-axis changes the sign of the first coordinate: (-2, 2), (-6, 2) and (-2, 4).

The translation adds 1 across and -3 up:

P' = (-1, -1) \qquad Q' = (-5, -1) \qquad R' = (-1, 1)

Figure 2Example 6. The dashed triangles are the two stages in between.

For a single rule, feed the general point (x, y) through the same three moves:

\begin{aligned} (x, y) &\longmapsto (2x,\ 2y) \\ &\longmapsto (-2x,\ 2y) \\ &\longmapsto (-2x + 1,\ 2y - 3) \end{aligned}

Check. The rule gives P' = (-2 + 1,\ 2 - 3) = (-1, -1), as before. \overline{PQ} has length 2 and \overline{P'Q'} has length 4, the scale factor 2. The right angle at P is still a right angle at P', since \overline{P'Q'} is horizontal and \overline{P'R'} is vertical. And P \to Q \to R runs counterclockwise (right, then up and back) while P' \to Q' \to R' runs clockwise, as one reflection should make it.

Example 7

Finding a sequence

In Figure 3, find a similarity transformation that carries \triangle ABC, with A(2, 2), B(6, 2) and C(2, 8), onto \triangle DEF, with D(5, -1), E(7, -1) and F(5, -4), taking A to D, B to E and C to F.

Figure 3Example 7. Which moves carry \triangle ABC onto \triangle DEF?

Solution

The scale factor. AB = 4 and DE = 2; AC = 6 and DF = 3. Both ratios are \tfrac12, so k = \tfrac12. The third sides agree, by the distance formula: \displaystyle BC = \sqrt{4^2 + 6^2} = \sqrt{52} = 2\sqrt{13} and \displaystyle EF = \sqrt{2^2 + 3^2} = \sqrt{13}, half of it.

Whether a flip is needed. From A, the side to B runs right and the side to C runs up, so A \to B \to C runs counterclockwise. From D, the side to E runs right and the side to F runs down, so D \to E \to F runs clockwise. The direction has reversed, so the sequence needs a reflection. The reflection in the x-axis turns “up” into “down” and leaves “right” alone, which is exactly the change needed.

The moves. Dilate with center the origin and factor \tfrac12, then reflect in the x-axis:

\begin{aligned} A(2, 2) &\longmapsto (1, 1) \longmapsto (1, -1) \\ B(6, 2) &\longmapsto (3, 1) \longmapsto (3, -1) \\ C(2, 8) &\longmapsto (1, 4) \longmapsto (1, -4) \end{aligned}

The triangle now has the right size and faces the right way, and (1, -1) must still move to D(5, -1): a translation by \langle 4, 0 \rangle. It sends (3, -1) to (7, -1) = E and (1, -4) to (5, -4) = F.

So the dilation by \tfrac12, the reflection in the x-axis and the translation by \langle 4, 0 \rangle carry \triangle ABC onto \triangle DEF, and \triangle ABC \sim \triangle DEF. Other sequences work as well, and any one that works is an answer.

Example 8

Order matters, and a hidden dilation

(a) Dilate with center the origin and scale factor 2, then translate by \langle 3, 0 \rangle. (b) Translate by \langle 3, 0 \rangle first, then dilate with center the origin and scale factor 2. Find the rule for each and compare them. (c) Show that the sequence in (a) is a single dilation.

Solution

(a) (x, y) \longmapsto (2x,\ 2y) \longmapsto (2x + 3,\ 2y).

(b) (x, y) \longmapsto (x + 3,\ y) \longmapsto (2x + 6,\ 2y).

The same two moves in the other order land every point 3 further to the right. Translating first means the slide is doubled by the dilation that follows. To match (a) with the translation first, the slide must be half as long: 2\left(x + \tfrac32\right) = 2x + 3, so translate by \left\langle \tfrac32, 0 \right\rangle.

(c) A dilation leaves exactly one point where it is, so look for one. 2x + 3 = x gives x = -3, and 2y = y gives y = 0. The dilation with center (-3, 0) and factor 2 sends x to -3 + 2(x + 3) = 2x + 3 and y to 0 + 2(y - 0) = 2y, which is rule (a).

So a dilation followed by a translation was a dilation all along, about a different center. The lesson on dilations found the reverse surprise: two dilations whose factors multiply to 1 can make a translation.

Example 9

Undoing a similarity transformation

A similarity transformation S dilates with center the origin and scale factor 3, then rotates 90^\circ counterclockwise about the origin, then translates by \langle 2, -1 \rangle. Find a single rule for S and the image of (1, 2). Then describe a sequence that undoes S, and give its rule.

Solution

Feed (x, y) through the three moves:

\begin{aligned} (x, y) &\longmapsto (3x,\ 3y) \\ &\longmapsto (-3y,\ 3x) \\ &\longmapsto (-3y + 2,\ 3x - 1) \end{aligned}

so (1, 2) \longmapsto (-6 + 2,\ 3 - 1) = (-4, 2).

To undo S, undo its moves in reverse order: translate by \langle -2, 1 \rangle, then rotate 90^\circ clockwise, (x, y) \longmapsto (y, -x), then dilate with factor \tfrac13. Following (-4, 2) back:

(-4, 2) \longmapsto (-6, 3) \longmapsto (3, 6) \longmapsto (1, 2)

which is where it started. For the rule, start from a point (X, Y):

\begin{aligned} (X, Y) &\longmapsto (X - 2,\ Y + 1) \\ &\longmapsto (Y + 1,\ 2 - X) \\ &\longmapsto \left(\frac{Y + 1}{3},\ \frac{2 - X}{3}\right) \end{aligned}

The undoing has scale factor \tfrac13.

Check. Put X = -3y + 2 and Y = 3x - 1 into the undoing rule: \dfrac{3x - 1 + 1}{3} = x and \dfrac{2 - (-3y + 2)}{3} = y. Every point comes back.

The order is not optional. Undoing in the same order as S (dilate by \tfrac13, then rotate clockwise, then translate by \langle -2, 1 \rangle) sends (-4, 2) to \left(-\tfrac43, \tfrac23\right), then \left(\tfrac23, \tfrac43\right), then \left(-\tfrac43, \tfrac73\right), not (1, 2).

Example 10

Narrow and wide parabolas are similar

Show that the dilation with center the origin and scale factor \tfrac13 carries the parabola y = x^2 onto the parabola y = 3x^2. Then show that y = x^2 is similar to y = ax^2 for every a > 0.

Solution

A point of y = x^2 has the form (t, t^2). The dilation multiplies both coordinates by \tfrac13:

(t,\ t^2) \longmapsto \left(\frac{t}{3},\ \frac{t^2}{3}\right)

The image is on y = 3x^2, because 3\left(\tfrac{t}{3}\right)^2 = 3 \cdot \tfrac{t^2}{9} = \tfrac{t^2}{3}. For instance (3, 9) \longmapsto (1, 3), and 3 \cdot 1^2 = 3.

Every point of y = 3x^2 is reached as well. A point (u, 3u^2) on it is the image of (3u, 9u^2), and that point is on y = x^2 because (3u)^2 = 9u^2. So the dilation carries the whole of one parabola onto the whole of the other, and they are similar.

The same works for any a > 0 with the scale factor \tfrac1a: the point (t, t^2) goes to \left(\tfrac{t}{a}, \tfrac{t^2}{a}\right), and a\left(\tfrac{t}{a}\right)^2 = \tfrac{t^2}{a}. When a < 0, follow the dilation by \tfrac{1}{|a|} with the reflection in the x-axis. Every parabola y = ax^2 with a \neq 0 is similar to y = x^2, and by chaining they are all similar to one another. A narrow parabola is a wide one seen from further away.

Boundaries

Scale factors are positive. A dilation with a negative factor, which some books allow, is a dilation with a positive factor followed by a half turn (dilations). Allowing it adds no new similar figures.

One factor in every direction. A similarity transformation multiplies every length by the same k. A move that stretches one direction more than another, such as (x, y) \longmapsto (x, 3y), fails that: it turns a 1 by 1 square into a 1 by 3 rectangle, whose sides are no longer in the ratio 1 : 1, so the image is not similar to the original. Such a move is a stretch, studied with graphs.

Onto the whole figure. Carrying F onto part of G shows nothing. For a polygon it is enough to carry the vertices onto the vertices in order, since each side then goes onto the side between the images. For a curve, every point of each must be accounted for, both ways round (Examples 3 and 10).

The matching is part of the claim. \triangle ABC \sim \triangle DEF says which vertex goes to which. \triangle ABC \sim \triangle EFD is a different statement, and it can be false while the first is true.

Polygons need both conditions. Equal angles alone fail for rectangles (Example 2) and proportional sides alone fail for rhombuses (Example 4). Triangles are special, and either half is enough there.

A sequence is not unique, and its order matters. Many different sequences carry one figure onto another, and any correct one shows similarity. Reordering the moves of a given sequence usually changes where the image lands (Example 8).

Looking similar is not being similar. A drawing can be off by a little. Similarity is shown by moves that land exactly, or by measurements and coordinates, never by eye.

In space. The definition works unchanged in three dimensions, with rigid motions and dilations of space. Any two spheres are similar, and so are any two cubes.

Where this goes next

The result for polygons is the working test for similar figures: matching angles equal and matching sides in one ratio. Once two figures are known to be similar, a missing length comes from that ratio, which is proportions in similar triangles, and perimeters and areas follow the rules in scale factor and perimeter, area and volume.

For triangles the test shrinks. Two pairs of equal angles already force a similarity transformation to exist, which is AA similarity, and the argument in “Why it is true” above, where a dilation is followed by a congruence, is the heart of SSS and SAS similarity too. Finding those angle pairs in a figure is proving two triangles similar, and writing the moves out as the proof itself is transformational proof.

Similarity is behind much of what comes later. A line parallel to one side of a triangle cuts off a smaller triangle similar to the whole, which gives the triangle proportionality theorem. All right triangles with the same acute angle are similar, so the ratio of two of their sides depends on the angle alone, and that is where sine, cosine and tangent come from.

Exercises

Part 1: what the image looks like

A triangle has angles of 35^\circ, 65^\circ and 80^\circ, and its longest side is 12 cm. It is rotated 90^\circ, dilated with scale factor 1.5, and then reflected in a line. Find the angles and the longest side of the image.
A sequence is a dilation with scale factor 4, a translation, a dilation with scale factor \tfrac16 about a different center, and a reflection. Find the scale factor of the whole sequence, and the length of the image of a segment 9 cm long.
A similarity transformation with scale factor \tfrac52 carries figure F onto figure G. A segment of G is 15 cm long. How long is the matching segment of F?
A similarity transformation has scale factor 3. Say what it does to each of these: (a) the measure of an angle, (b) the length of a segment, (c) the ratio of two sides of the figure, (d) a pair of parallel lines, (e) the area of the figure.

Part 1: similar or not

Is a rectangle 8 by 12 similar to a rectangle 14 by 21? If so, give the scale factor from the first to the second.
Is a rectangle 8 by 12 similar to a rectangle 12 by 20? Explain.
One rhombus has sides of 5 and an angle of 70^\circ. Another has sides of 8 and an angle of 110^\circ. Are they similar? If so, give the scale factor from the first to the second.
Two isosceles triangles have apex angles of 30^\circ and 50^\circ. Are they similar? Explain.
Is an equilateral triangle with sides of 4 similar to one with sides of 10? If so, give the scale factor from the first to the second.

Part 2: following a sequence

The triangle with vertices (2, 0), (4, 0) and (2, 6) is dilated with center the origin and scale factor \tfrac12, then translated by \langle -3, 2 \rangle. Find the image.
The triangle with vertices (-1, 2), (1, 2) and (0, 4) is reflected in the x-axis, then dilated with center the origin and scale factor 3. Find the image.
The point (2, -1) is rotated 90^\circ counterclockwise about the origin, then dilated with center the origin and scale factor 2, then translated by \langle 0, 5 \rangle. Find its image.
A sequence dilates with center the origin and scale factor 3, rotates 180^\circ about the origin, then translates by \langle 4, 1 \rangle. Write a single rule for it, and find the image of (2, 2).

Part 2: finding a sequence

Find a similarity transformation that carries the square with vertices A(0, 0), B(2, 0), C(2, 2) and D(0, 2) onto the square with vertices (3, 1), (9, 1), (9, 7) and (3, 7), in that order.
Find a similarity transformation that carries \triangle ABC, with A(1, 2), B(3, 2) and C(1, 5), onto \triangle DEF, with D(-2, -4), E(-6, -4) and F(-2, -10), taking A to D, B to E and C to F.
Find a similarity transformation that carries \triangle PQR, with P(0, 0), Q(4, 0) and R(0, 2), onto \triangle STU, with S(1, 1), T(1, 3) and U(2, 1), taking P to S, Q to T and R to U.

Part 2: undoing, order and single moves

A similarity transformation dilates with center the origin and scale factor 2, then translates by \langle -3, 4 \rangle. Describe a sequence that undoes it, and find the point it sends to (5, 0).
Find the rule for “translate by \langle 2, 2 \rangle, then dilate with center the origin and scale factor 3”, and the rule for the same two moves in the other order. What translation, done after the dilation, matches the first order?
The rule (x, y) \longmapsto (3x - 4,\ 3y + 2) dilates with center the origin and scale factor 3, then translates by \langle -4, 2 \rangle. Show that it is a single dilation, and find its center.
Find the scale factor of the dilation with center the origin that carries the parabola y = x^2 onto the parabola y = \tfrac14 x^2, and show that it does.

Reasoning

Explain why any two squares are similar, by describing the moves that carry one onto the other.
Explain why every figure is similar to itself, and why F \sim G means that G \sim F as well.
F \sim G with scale factor 2, and G \sim H with scale factor 3. Explain why F \sim H, and find the scale factor from F to H.
The vertices of \triangle ABC run counterclockwise and those of its image \triangle A'B'C' run clockwise. Explain why every similarity transformation that carries one onto the other includes a reflection.
Two figures are similar and have equal areas. Explain why they are congruent.

Answers

The angles are 35^\circ, 65^\circ and 80^\circ, and the longest side is 18 cm. Angles are kept, and the only dilation multiplies lengths by 1.5: 1.5 \cdot 12 = 18. Multiplying by a positive number keeps the order of sizes, so the longest side stays the longest.
The scale factor is \tfrac23 and the segment becomes 6 cm. The rigid motions contribute nothing: k = 4 \cdot \tfrac16 = \tfrac23, and \tfrac23 \cdot 9 = 6.
6 cm. Lengths of G are \tfrac52 times those of F, so the segment of F is 15 \div \tfrac52 = 6.
(a) Unchanged. (b) Multiplied by 3. (c) Unchanged, since both sides are multiplied by 3. (d) They stay parallel. (e) Multiplied by 3^2 = 9.
Yes, with scale factor \tfrac74. All angles are right angles, and \tfrac{14}{8} = \tfrac{21}{12} = \tfrac74.
No. Short sides must go to short sides, which needs k = \tfrac{12}{8} = \tfrac32, and long sides to long sides, which needs k = \tfrac{20}{12} = \tfrac53. One transformation has one scale factor. In ratio terms, \tfrac{8}{12} = \tfrac23 but \tfrac{12}{20} = \tfrac35.
Yes, with scale factor \tfrac85. In a rhombus neighboring angles add to 180^\circ and opposite angles are equal, so the first has angles 70^\circ, 110^\circ, 70^\circ, 110^\circ, and the second, with 180^\circ - 110^\circ = 70^\circ, has the same four. Match the 70^\circ angles; every side is multiplied by \tfrac85.
No. The base angles are \tfrac{180^\circ - 30^\circ}{2} = 75^\circ in the first and \tfrac{180^\circ - 50^\circ}{2} = 65^\circ in the second, so the angles are 30^\circ, 75^\circ, 75^\circ and 50^\circ, 65^\circ, 65^\circ. A similarity transformation keeps every angle.
Yes, with scale factor \tfrac52. Every angle of each is 60^\circ, and every side is multiplied by \tfrac{10}{4} = \tfrac52.
(-2, 2), (-1, 2) and (-2, 5). Halving gives (1, 0), (2, 0) and (1, 3), and the translation adds -3 across and 2 up.
(-3, -6), (3, -6) and (0, -12). The reflection gives (-1, -2), (1, -2) and (0, -4), and the dilation triples both coordinates.
(2, 9). The rotation (x, y) \longmapsto (-y, x) gives (1, 2), the dilation gives (2, 4), and the translation gives (2, 9).
(x, y) \longmapsto (-3x + 4,\ -3y + 1), and (2, 2) \longmapsto (-2, -5). The moves give (3x, 3y), then (-3x, -3y), then add \langle 4, 1 \rangle. For (2, 2): -6 + 4 = -2 and -6 + 1 = -5.
Dilate with center the origin and scale factor 3, then translate by \langle 3, 1 \rangle. The sides go from 2 to 6, so k = 3. The dilation gives (0, 0), (6, 0), (6, 6) and (0, 6), and the translation gives (3, 1), (9, 1), (9, 7) and (3, 7). This is in fact the single dilation with center \left(-\tfrac32, -\tfrac12\right) and factor 3, since \tfrac{3}{1 - 3} = -\tfrac32 and \tfrac{1}{1 - 3} = -\tfrac12.
Dilate with center the origin and scale factor 2, then rotate 180^\circ about the origin. AB = 2 and DE = 4, so k = 2. \overline{AB} runs right and \overline{DE} runs left; \overline{AC} runs up and \overline{DF} runs down. Both directions are reversed and the turning direction is not, which is a half turn. The dilation gives (2, 4), (6, 4) and (2, 10), and the rotation (x, y) \longmapsto (-x, -y) gives (-2, -4), (-6, -4) and (-2, -10). No translation is needed.
Dilate with center the origin and scale factor \tfrac12, reflect in the line y = x, then translate by \langle 1, 1 \rangle. PQ = 4 and ST = 2, so k = \tfrac12. P \to Q \to R runs counterclockwise and S \to T \to U runs clockwise, so a reflection is needed; \overline{PQ} runs along the x direction and \overline{ST} along the y direction, which the reflection in y = x swaps. The moves give (0, 0), (2, 0), (0, 1), then (0, 0), (0, 2), (1, 0), then (1, 1), (1, 3) and (2, 1).
Translate by \langle 3, -4 \rangle, then dilate with center the origin and scale factor \tfrac12. The point is (4, -2). Undoing (5, 0) gives (8, -4) and then (4, -2). Check: the transformation sends (4, -2) to (8, -4) and then to (5, 0).
Translating first gives (x, y) \longmapsto (3x + 6,\ 3y + 6); dilating first gives (x, y) \longmapsto (3x + 2,\ 3y + 2). After the dilation, the translation by \langle 6, 6 \rangle matches the first order, since the dilation triples the slide.
The center is (2, -1) and the factor is 3. A fixed point needs 3x - 4 = x, so x = 2, and 3y + 2 = y, so y = -1. The dilation with that center and factor 3 sends x to 2 + 3(x - 2) = 3x - 4 and y to -1 + 3(y + 1) = 3y + 2, which is the rule.
Scale factor 4. The point (t, t^2) goes to (4t, 4t^2), and \tfrac14 (4t)^2 = 4t^2, so the image is on y = \tfrac14 x^2. Every point \left(u, \tfrac14 u^2\right) of that parabola is the image of \left(\tfrac{u}{4}, \tfrac{u^2}{16}\right), which is on y = x^2.
Take squares with sides a and b. Dilate the first with scale factor \tfrac{b}{a} about one of its corners. The image is a square with sides b, so it is congruent to the second square, and a rigid motion carries it onto the second. A dilation followed by a rigid motion is a similarity transformation.
The rigid motion that moves nothing (a translation by \langle 0, 0 \rangle, say) carries F onto itself, so F \sim F with scale factor 1. If a sequence with scale factor k carries F onto G, undoing its moves in reverse order, each by its own reverse, carries G back onto F, and each of those reverses is again a rigid motion or a dilation. So G \sim F, with scale factor \tfrac1k.
Do the sequence that carries F onto G, then the one that carries G onto H. Together they are one sequence of rigid motions and dilations carrying F onto H, so F \sim H. Lengths are multiplied by 2 and then by 3, so the scale factor is 2 \cdot 3 = 6.
Translations, rotations and dilations with positive factors all keep the direction in which the vertices run, and each reflection reverses it. A sequence with no reflection would leave \triangle A'B'C' running counterclockwise. To end up clockwise the sequence needs a reflection, and in fact an odd number of them.
A similarity transformation with scale factor k multiplies area by k^2. Equal areas give k^2 = 1, and since k > 0, k = 1. A similarity transformation with scale factor 1 keeps every distance, so it is a rigid motion, and figures carried onto each other by a rigid motion are congruent.