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

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.

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.

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.

Observation

Justification

P = A^{''}

The translation moves

A

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.

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 $S R .$

Observation

Justification

P = A^{'''}

The translation moves

A

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^{''}

Q,

also moves

D^{''}

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 .

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

Proportion

Solution

Expression

Substitution

\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}

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()

In the following figure, $A B C D \sim E F G H .$ Find the value of the variables.

Two similar quadrilaterals ABCD and EFGH

In the following figure, $A B C \sim D E F .$ Find the value of the variables.

Since we are given that A B C D ~ E F G H, we can identify corresponding congruent angles by matching letters following the order in which they appear in the similarity statement. ccccc A & B & C & D ↓ & ↓ & ↓ & ↓ E & F & G & H Let's highlight the corresponding vertices and angles in the diagram.

Now we can write two equations, one containing x and the other containing y. x+34^(∘)&=98^(∘) 3y-13^(∘)&= 83^(∘) Let's solve these equations one at a time.

Let's go ahead and solve the second equation in the same manner.

As in Part A, we will use the similarity statement to identify corresponding vertices. ccccc A & B & C ↓ & ↓ & ↓ D & E & F Let's add this information to the diagram.

With this information, we can write two equations — one that contains x and another that contains y. 8x-13^(∘)=75^(∘) y=m∠ F We can not yet solve for y. However, if we find the value of x, we can use this to find y.

Notice that m∠ C is not given. However, we can use that ∠ C ≅ ∠ F and apply the Interior Angles Theorem to △ DEF to find y.

For what value of $x$ is $B E F A \sim E D C B ?$

Two similar rectangles BEFA and EDCB which are attached along BE

In similar polygons, corresponding sides are proportional. Since BEFA is similar to EDCB, we can write a proportionality using the corresponding sides. To identify corresponding sides, we will separate the rectangles. Notice that the length of EDCB is the width of BEFA.

The ratio of the rectangles' longer sides is equal to the ratio of the shorter sides. AB/CD=BE/DE Let's substitute the lengths of the sides into the equation and solve for x.

When x=4, the rectangles are similar.

On Davontay's last holiday, he took some pictures that he really wants to show to his family using a digital projector. The photos are

12

inches by

9

inches when displayed on a computer screen. The projector dilates these photos from the preimage to the image at a scale factor of

1 : 8 .

How large of an area on the wall does the image display?

From the exercise, we know that the projector dilates an image such that the scale factor is 1:8. This tells us that the side lengths are 8 times greater for the image then for the preimage. To determine its length and width, we can use the following formula. Scale factor=Length of preimage/Length of image If we call the length and width of the image x and y respectively, we can use this formula to determine the dimensions of the image. Let's begin with the length followed by the width.

Length

The length is 96 inches.

Width

The width is 72 inches.

Area

Now we can determine the area that the images cover on the wall by multiplying the length and width. A=( 96)( 72)=6912 inches^2

	13 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Understanding Similarity Transformations

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

Length

Width

Area

Understanding Similarity Transformations

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