Understanding Similarity Transformations: Scale Factor, Corresponding Sides, and Congruent Angles

When working with transformations, remember that the composition of rigid motions is a rigid motion. However, since a dilation is not a rigid motion, what happens if a rigid motion and a dilation are composed? The answer to that question will be developed throughout this lesson.

Catch-Up and Review

Here is some recommended reading before getting started with this lesson.

Concept and properties of dilation.
Concept and properties of rigid motion.
Concept and properties of congruence.

Explore

Investigating Similarity

The applet below shows a spiral tiling of a plane using quadrilaterals of different sizes.

Moving the left slider dilates the shaded quadrilateral using the center.
Moving the right slider rotates the shaded quadrilateral around the center.

Move the sliders to match the quadrilaterals in the tiling.

Discussion

Similarity Transformations

On the previous applet the combination of a rotation and a dilation moved the quadrilateral to match the other quadrilaterals in the tiling. This combination of transformations has its own name.

Concept

Similarity Transformation

A combination of rigid motions and dilations is called a similarity transformation. The scale factor of a similarity transformation is the product of the scale factors of the dilations.

One triangle is mapped onto the other triangle using rigid motions and dilations

Move the slider to create a similarity transformation by combining rigid motions and dilations. Similar figures are created as the result of a similarity transformation.

Two similar figures of which one is mapped onto the other after applying rigid motions

Discussion

Properties of a Similarity Transformation

Previously, it was seen that rigid motions keep the figure's size and shape. In comparison, dilations keep the figure's shape but can change its size. The next natural question is, what does a similarity transformation do to a figure?

The following is a list of a few important properties of similarity transformations.

The image of a line segment is a line segment. The image is longer or shorter than the preimage in the ratio given by the scale factor.
Similarity transformations preserve angle measures.

Proof

These are properties of both rigid motions and dilations. Since similarity transformations are combinations of rigid motions and dilations, these are also properties of similarity transformations.

Discussion

Definition of Similar Figures

Two figures are similar figures if there is a composition of similarity transformations that maps one figure onto the other. In other words, two figures are similar if they have the same shape and the ratios of their corresponding linear measures are equal. The symbol

\sim

indicates that two figures are similar.

When writing a similarity statement, the corresponding vertices must be listed in the same order as they appear. The relationship between the two given polygons has multiple similarity statements. Consider two of them.

A B C D \sim J K L M or C D A B \sim L M J K

The same definition applies to three-dimensional shapes.

Now, consider one of the possible similarity statements for the given polyhedrons.

A B C D E F G H \sim J K L M N O P R

Note that for two-dimensional figures, all squares are similar and all circles are similar. Similarly, for

3 D

figures, all cubes are similar and all spheres are similar.

Example

Analyzing Similar Figures

The figure below is put together using $39$ similar tiles.

a How many different sizes are there on the figure?

b What is the scale factor between the smallest and largest tile?

c Find a similarity transformation that maps the green tile to the blue tile. State the scale factor.

Answer

Example Answer: Translation up, followed by a rotation clockwise by $90$ degrees, followed by a dilation using scale factor $2 .$

Hint

a The tiles are smaller towards the bottom.

b Look for a triangular pattern.

c Move one vertex to the corresponding vertex first.

Solution

a On the figure below the different sizes are shaded using different colors.

In this pattern, five different sized tiles are used.

b Notice that a triangular pattern can help in finding the scale factor between the different sized tiles.

The combined width of four of the smallest tiles matches the width of the largest tile. This means that an enlargement with scale factor

4

is needed to get the largest tile from the smallest tile.

c There are several ways a similarity transformation can be put together using rigid motions and dilations. One possibility is to start with a translation to move one vertex of the preimage tile to the corresponding vertex of the image tile.

Once a vertex is at the right place, a rotation can be used to position the pre-image in the right direction.

A dilation by scale factor $2$ completes the transformation.

Discussion

Criteria for Similar Polygons

For polygons, similarity can be checked by considering angle measures and side lengths.

Two polygons are similar if and only if both of the following two properties hold.

The corresponding angles are congruent.
The corresponding sides are proportional.

Proof

A biconditional statement can be proven by separately proving the corresponding conditional statement and its converse.

Conditional Statement	Two polygons are similar if the corresponding angles are congruent and the corresponding sides are proportional.
Converse	If the corresponding angles in two polygons are congruent and the corresponding sides are proportional, then the polygons are similar.

Consider and prove each statement one at a time.

Proving the Conditional Statement

If two polygons are similar, then a similarity transformation that maps one polygon to the other exists. Consider how that relationship affects the corresponding angles and sides of the similar polygons.

Angles: Similarity transformations preserve angle measures. That means the corresponding angles of the two polygons are congruent. $✓$
Sides: Similarity transformations map line segments to other line segments. The length of all line segments change according to the scale factor. That means the corresponding sides of the two polygons change proportionally. $✓$

These observations conclude the proof of the conditional statement.

Proving the Converse

Consider two polygons with congruent corresponding angles and proportional corresponding sides. The proof here will be carried out for quadrilaterals $A B C D$ and $P Q R S,$ but it can be generalized to any polygon.

Since the corresponding angles are congruent and the corresponding sides are proportional, the following statements are true.

∠ A ∠ B ∠ C ∠ D \frac{P Q}{A B} = \frac{Q R}{B C} ≅ ∠ P ≅ ∠ Q ≅ ∠ R ≅ ∠ S = \frac{R S}{C D} = \frac{S P}{D A}

To show that the polygons

A B C D

and

P Q R S

are similar, a similarity transformation can be built to map

A B C D

P Q R S .

This can be done in several ways, so here is just an example of one possibility.

Translate One Vertex to the Corresponding Vertex

Use a translation that moves

A

P .

The image of this translation of

A B C D

A^{'} B^{'} C^{'} D^{'} .

Translation of ABCD to A'B'C'D' that maps A onto P

Rotate to Get the Correct Direction of One Pair of Segments

Use a rotation around the common vertex that moves

B^{'}

P Q .

The image of this rotation of

A^{'} B^{'} C^{'} D^{'}

A^{''} B^{''} C^{''} D^{''} .

The following table contains some observations about the position of points

A^{''},

B^{''},

C^{''},

and

D^{''}

relative to

P Q R S .

Observation	Justification
$P = A^{''}$	The translation moves $A$ to $P$ and, since this is the center of rotation, it stays there.
$B^{''}$ is on $P Q$	This is how the angle of rotation was chosen.
$D^{''}$ is on $P S$	This is true, because by assumption $∠ A$ is congruent to $∠ P$ and because rigid motions preserve angle measures. Note that, in this case, the orientation of $A B C D$ and $P Q R S$ is the same. If the orientations are different, then a reflection of $A^{''} B^{''} C^{''} D^{''}$ in line $P Q$ is also needed to match the orientations of the polygons.

Dilate to Get a Second Vertex to the Corresponding Vertex

Use a dilation from the common vertex to move

B^{''}

Q .

This dilation maps

A^{''} B^{''} C^{''} D^{''}

A^{'''} B^{'''} C^{'''} D^{'''} .

The following table contains some observations about the position of points

A^{'''},

B^{'''},

C^{'''},

and

D^{'''}

relative to

P Q R S .

Observation	Justification
$P = A^{'''}$	The translation moves $A$ to $P$ and, since this is the center of rotation and also the dilation, it stays there.
$Q = B^{'''}$	This is how the scale factor of the dilation was chosen.
$S = D^{'''}$	Since translations and rotations are rigid motions, $A B = A^{''} B^{''}$ and $A D = A^{''} D^{''} .$ It is assumed that $P Q / A B = P S / A D,$ so the dilation that moves $B^{''}$ to $Q,$ also moves $D^{''}$ to $S .$
$C^{'''}$ is on $Q R$	It is assumed that $∠ B ≅ ∠ Q .$ Since rigid motions and dilations preserve angles, this means that $∠ B^{'''} ≅ ∠ Q .$
$C^{'''}$ is on $S R$	It is assumed that $∠ D ≅ ∠ S .$ Since rigid motions and dilations preserve angles, this means that $∠ D^{'''} ≅ ∠ S .$
$R = C^{'''}$	Both $R$ and $C^{'''}$ is the intersection of $Q R$ and $S R .$

The steps above give a similarity transformation that maps

A B C D

P Q R S,

so these two quadrilaterals are similar. This proves the converse statement.

Example

Investigating Similarity of Different Polygons

Not all rectangles are similar.

f All squares are similar.

Hint

Do all squares have the same shape?

Solution

Consider the angles and the sides of a square with side length $m$ and a square with side length $n .$

All angles of both squares are right angles, so the corresponding angles of the two squares are congruent.
The ratio between any sides of the two squares is $m : n,$ so the corresponding sides are proportional.

These two properties guarantee that the squares are similar.

Any two squares are similar.

Example

Solving Problems Using Similarity of Polygons

The triangles on the diagram are similar. It is given that the length of

A B

4

centimeters, the length of

B C

3

centimeters, and the length of

A C

5

centimeters.

△ A B C \sim △ A C D \sim △ A D E \sim △ A E F \sim △ A F G

Find

A G .

Give your answer rounded to the nearest millimeter.

There are five triangles joined by their sides: ABC, ACD, ADE, AEF, and AFG.

Hint

Find the length of $\overline{A D}$ first.

Solution

Focus on the first two triangles to find the length of $\overline{A D}$ first.

It is given that

△ A B C

is similar to

△ A C D,

so the corresponding sides are proportional.

\frac{D A}{C A} = \frac{C A}{B A}

The lengths of

\overline{C A}

and

\overline{B A}

are given in the question. Substituting these values in the equation gives the length of

\overline{D A} .

\frac{D A}{C A} = \frac{C A}{B A}

SubstituteII

$C A = 5$ , $B A = 4$

\frac{D A}{5} = \frac{5}{4}

MultEqn

$LHS \cdot 5 = RHS \cdot 5$

D A = \frac{5}{4} \cdot 5

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

D A = \frac{2 5}{4}

Similar argument gives the length of

\overline{E A},

\overline{F A},

and

\overline{G A} .

The other three triangles are also similar to $△ A B C,$ so the corresponding sides are proportional.

Proportion		Solution
Expression	Substitution	Solution
$\frac{E A}{C A} = \frac{D A}{B A}$	$\frac{E A}{5} = \frac{2 5 / 4}{4}$	$E A = \frac{2 5 / 4}{4} \cdot 5 = \frac{1 2 5}{1 6}$
$\frac{F A}{C A} = \frac{E A}{B A}$	$\frac{F A}{5} = \frac{1 2 5 / 1 6}{4}$	$F A = \frac{1 2 5 / 1 6}{4} \cdot 5 = \frac{6 2 5}{6 4}$
$\frac{G A}{C A} = \frac{F A}{B A}$	$\frac{G A}{5} = \frac{6 2 5 / 6 4}{4}$	$G A = \frac{6 2 5 / 6 4}{4} \cdot 5 = \frac{3 1 2 5}{2 5 6}$

The length of $\overline{G A}$ is $\frac{3 1 2 5}{2 5 6},$ or approximately $12.2$ centimeters.

Example

Solving Problems Using Similarity and Congruence of Polygons

In the diagram all quadrilaterals are similar, and the two shaded quadrilaterals are congruent. The length of three sides of the shaded quadrilaterals are $1,$ $w,$ and $w^{2} .$

Find the value of

w .

Write your answer rounded to two decimal places.

Hint

Write the horizontal base of the top left corner's quadrilateral in two different ways.

Solution

The solution has several steps. First, look at the quadrilateral in the top left corner and compare it to the quadrilateral next to it.

These two quadrilaterals are similar, so the corresponding sides are proportional.

\frac{x}{w ^{2}} = \frac{1}{w}

This gives the length of the bottom side of the quadrilateral in the top left.

x = w

Next, consider the quadrilaterals in the bottom left corner.

These are also similar quadrilaterals, so the corresponding sides are proportional.

\frac{y}{w} = \frac{w}{w ^{2}} ⟹ y = 1

Consider one more quadrilateral.

This time, proportionality gives an expression for

z .

\frac{z}{1} = \frac{1}{w ^{2}} ⟹ z = \frac{1}{w ^{2}}

Putting these together gives the following diagram.

Comparing the two ways the length of the bottom side of the top left quadrilateral can be expressed gives an equation for

w .

w = \frac{1}{w ^{2}} + 1

This equation is not easy to solve algebraically. Graphical calculators have applications that can solve an equation like this.

w \approx 1.47

Extra

Exact solution

At the end of the solution the following equation was presented.

w = \frac{1}{w ^{2}} + 1

The approach was to use some form of technology to find the solution. It is a natural question to wonder how to find the solution algebraically. One way of doing this is to rearrange the equation.

w = \frac{1}{w ^{2}} + 1

MultEqn

$LHS \cdot w^{2} = RHS \cdot w^{2}$

w^{3} = 1 + w^{2}

SubEqn

$LHS - (1 + w^{2}) = RHS - (1 + w^{2})$

w^{3} - w^{2} - 1 = 0

This is a cubic equation, the exact solution is the following.

w = \frac{1}{3} ⎝ ⎜ ⎛ 1 + 3 \frac{2 9}{2} - \frac{3 9 3}{2} + 3 \frac{2 9}{2} + \frac{3 9 3}{2} ⎠ ⎟ ⎞

This solution can be calculated manually using Cardano's formula or it can be obtained using a computer algebra system.

Pop Quiz

Practice Using Similarity to Solve Problems

On the diagram all quadrilaterals are similar.

Illustration

Similarity in the Natural World

There are examples where similar shapes appear in nature.

It is interesting to investigate the three-dimensional self-similar nature of a romanesco broccoli. It is built up of parts that are similar to the whole.

Self-similarity is used as an inspiration in fractals. The image below is not a living plant, it is a computer generated image using a construction that uses similarity.

Digital Tools

Python code to draw the image

The following is the Python source code the author used to draw the image.

import random
import tkinter as tk
width, height = 1024, 1024
pixels = [0] * (width * height) x, y = 0, 1
for n in range(60 * width * height): r = random.random() * 100 xn, yn = x, y if r < 1: x = 0 y = 0.16 * yn elif r < 86: x = 0.85 * xn + 0.04 * yn y = -0.04 * xn + 0.85 * yn + 1.6 elif r < 93: x = 0.20 * xn - 0.26 * yn y = 0.23 * xn + 0.22 * yn + 1.6 else: x = -0.15 * xn + 0.28 * yn y = 0.26 * xn + 0.24 * yn + 0.44 x_pix = int(width * (0.45 + 0.195 * x)) y_pix = int(height * (1 - 0.099 * y )) pixels[x_pix + y_pix * width] += 1 greys = [ max(0, (256 - p) / 256) for p in pixels]
colors = [int(c * 255) for g in greys for c in [g ** 6, g, g ** 6]]
root = tk.Tk()
p6header = bytes("P6\n{} {}\n255\n".format(width, height), "ascii")
img = tk.PhotoImage(data=p6header + bytes(colors))
tk.Label(root, image=img).pack()
img.write("barnsley-fern.png", format='png')
tk.mainloop()

Closure

Similarity and the Golden Ratio

In the diagram below move the point to adjust the dimensions of the rectangle. Different shapes will be created. Some rectangles are narrow and tall, some are wide and flat. Move the point to create rectangle of any preferred combination.

Well, personal preference is very subjective. Nevertheless, there is a specific ratio that is used often in design. If a rectangle is such that cutting off a square gives a similar rectangle, then the ratio of the sides is called the golden ratio. Such a rectangle is called a golden rectangle.

The image above is a reconstruction of Composition with Yellow, Blue and Red, a painting of Piet Mondrian, which he painted between

1937

and

1942 .

Move the slider points to search for golden rectangles in this famous painting!

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Catch-Up and Review

Proof

Answer

Hint

Solution

Proof

Proving the Conditional Statement

Proving the Converse

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Extra

Digital Tools