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In the study of geometry, understanding similarity transformations is vital. This subject focuses on how one shape can be mapped onto another through a series of steps like scaling, rotating, and translating. The key elements to consider are the scale factor, which determines how much a shape is enlarged or reduced, and the corresponding sides and angles, which must be proportional and congruent, respectively. These transformations are not just theoretical concepts; they have practical applications in various fields like design, architecture, and even in understanding natural phenomena like fractals. By mastering similarity transformations, one gains the tools to analyze and replicate complex shapes and patterns.
Show less Show more expand_more| Student Learning Objectives: |
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| | 13 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
The applet below shows a spiral tiling of a plane using quadrilaterals of different sizes.
Move the sliders to match the quadrilaterals in the tiling.
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.
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.
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.
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 ~
indicates that two figures are similar.
ABCD~ JKLM or CDAB~ LMJK
The same definition applies to three-dimensional shapes.
ABCDEFGH~ JKLMNOPR
The figure below is put together using 39 similar tiles.
How many different sizes are there on the figure?
What is the scale factor between the smallest and largest tile?
Find a similarity transformation that maps the green tile to the blue tile. State the scale factor.
The tiles are smaller towards the bottom.
Look for a triangular pattern.
Move one vertex to the corresponding vertex first.
On the figure below the different sizes are shaded using different colors.
Notice that a triangular pattern can help in finding the scale factor between the different sized tiles.
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.
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.
| Conditional Statement | Two polygons are similar if the corresponding angles are congruent and the corresponding sides are proportional. |
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| 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.
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.
These observations conclude the proof of the conditional statement.
Consider two polygons with congruent corresponding angles and proportional corresponding sides. The proof here will be carried out for quadrilaterals ABCD and PQRS, 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&≅∠ P ∠ B&≅∠ Q ∠ C&≅∠ R ∠ D&≅∠ S PQ/AB=QR/BC&=RS/CD=SP/DA To show that the polygons ABCD and PQRS are similar, a similarity transformation can be built to map ABCD to PQRS. This can be done in several ways, so here is just an example of one possibility.
The following table contains some observations about the position of points A'', B'', C'', and D'' relative to PQRS.
| Observation | Justification |
|---|---|
| P=A'' | The translation moves A to P and, since this is the center of rotation, it stays there. |
| B'' is on PQ | This is how the angle of rotation was chosen. |
| D'' is on PS | 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 ABCD and PQRS is the same. If the orientations are different, then a reflection of A''B''C''D'' in line PQ is also needed to match the orientations of the polygons. |
The following table contains some observations about the position of points A''', B''', C''', and D''' relative to PQRS.
| 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, AB=A''B'' and AD=A''D''. It is assumed that PQ/AB=PS/AD, so the dilation that moves B'' to Q, also moves D'' to S. |
| C''' is on QR | It is assumed that ∠ B≅ ∠ Q. Since rigid motions and dilations preserve angles, this means that ∠ B'''≅ ∠ Q. |
| C''' is on SR | 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 QR and SR. |
The steps above give a similarity transformation that maps ABCD to PQRS, so these two quadrilaterals are similar. This proves the converse statement.
Determine whether the following statements are true or false.
All quadrilaterals are similar.
All trapezoids are similar.
All parallelograms are similar.
All rhombi are similar.
All rectangles are similar.
Not all rectangles are similar.
All squares are similar.
These two properties guarantee that the squares are similar.
Any two squares are similar.
The triangles on the diagram are similar. It is given that the length of AB is 4 centimeters, the length of BC is 3 centimeters, and the length of AC is 5 centimeters. △ ABC ~ △ ACD ~ △ ADE ~ △ AEF ~ △ AFG Find AG. Give your answer rounded to the nearest millimeter.
It is given that △ ABC is similar to △ ACD, so the corresponding sides are proportional. DA/CA=CA/BA The lengths of CA and BA are given in the question. Substituting these values in the equation gives the length of DA.
CA= 5, BA= 4
LHS * 5=RHS* 5
a/c* b = a* b/c
Similar argument gives the length of EA, FA, and GA.
The other three triangles are also similar to △ ABC, so the corresponding sides are proportional.
| Proportion | Solution | |
|---|---|---|
| Expression | Substitution | |
| EA/CA=DA/BA | EA/5=25/4/4 | EA=25/4/4* 5=125/16 |
| FA/CA=EA/BA | FA/5=125/16/4 | FA=125/16/4* 5=625/64 |
| GA/CA=FA/BA | GA/5=625/64/4 | GA=625/64/4* 5=3125/256 |
The length of GA is 3125256, or approximately 12.2 centimeters.
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.
These two quadrilaterals are similar, so the corresponding sides are proportional. x/w^2=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. y/w=w/w^2 ⟹ y=1 Consider one more quadrilateral.
This time, proportionality gives an expression for z. z/1=1/w^2 ⟹ z=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=1/w^2+1 This equation is not easy to solve algebraically. Graphical calculators have applications that can solve an equation like this. w≈ 1.47
This is a cubic equation, the exact solution is the following. w=1/3(1+sqrt(29/2-3sqrt(93)/2)+sqrt(29/2+3sqrt(93)/2)) This solution can be calculated manually using Cardano's formula or it can be obtained using a computer algebra system.
On the diagram all quadrilaterals are similar.
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.
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 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.
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!
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 BEFA ~ EDCB?
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.
The length is 96 inches.
The width is 72 inches.
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