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| 16 Theory slides |
| 13 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
PA | PA′ | ||
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PB | PB′ | ||
PC | PC′ | ||
PD | PD′ |
m∠APA′ | |
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m∠BPB′ | |
m∠CPC′ |
A rotation is a transformation in which a figure is turned about a fixed point P. The number of degrees the figure rotates α∘ is the angle of rotation. The fixed point P is called the center of rotation. Rotations map every point A in the plane to its image A′ such that one of the following statements is satisfied.
Rotations can be performed by hand with the help of a straightedge, a compass, and a protractor.
To rotate point A about point P by an angle of 130∘ measured counterclockwise, follow these five steps.
Place the center of the protractor on P and align it with PA.
The protractor is placed as illustrated above when the rotation is counterclockwise. If the rotation has to be done clockwise, the protractor needs to be placed as follows.
Locate the corresponding measure on the protractor and make a small mark. In this case, the mark will be made at 130∘.
Notice that this method of construction has also confirmed that PA is congruent to PA′.
Start by rotating the vertices of the triangle. To do so, use a protractor to draw a 180∘ angle.
Zosia's goal is to find the image of the triangle after a 180∘ clockwise rotation about its fixed vertex. Labeling the vertices of the triangle will make this rotation process simpler.
The center of rotation is O and the angle of rotation is 180∘. Perform the rotation by rotating one point at a time. To rotate N, place the center of the protractor on O and align it with NO. Use the protractor to draw a ray that starts from O and makes a 180∘ angle with NO.
Then, mark a point N′ on this ray so that ON′ is the same length as ON. This is the image of N after the rotation. Since ON is the radius of the inner circle, N′ should also be on that circle.
Repeat the same process for the vertex E to find E′. Since O is the center of rotation, O′ will be in the same position as O.
Finally connect N′, E′, and O′ to draw the image of NEO after the 180∘ rotation.
As shown, the image N′ corresponds to P and E′ corresponds to T. Therefore, the clue Vincenzo and Zosia is looking for is NEPT.
Counterclockwise Rotations Around the Origin | |
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Angle of Rotation | Rule |
90∘ | (x,y)→(-y,x) |
180∘ | (x,y)→(-x,-y) |
270∘ | (x,y)→(y,-x) |
The clue Vincenzo and Zosia found reminded them of the word Neptune.
Vincenzo had previously heard stories about this magical place. He had even heard about three ordinary students who had found a door into a mystical library. Could this be another doorway to that place? As they got closer, a puzzle appeared between the columns of the gate.
When a point with coordinates (x,y) is rotated 90∘ counterclockwise about the origin, its image becomes (-y,x).
Zosia and Vincenzo want to rotate the figure 90∘ counterclockwise around the origin. This can be achieved by rotating each point of the original figure. In this case, four points will be sufficient to determine the position of the image.
When a counterclockwise rotation is performed about the origin, the coordinates of the image can be written in relation to the coordinates of the preimage.
Counterclockwise Rotations About the Origin | ||
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90∘ Rotation | 180∘ Rotation | 270∘ Rotation |
Preimage(x,y)→Image(-y,x) |
Preimage(x,y)→Image(-x,-y) |
Preimage(x,y)→Image(y,-x) |
As they explore the magical realm, Vincenzo and Zosia come across Dilatius the Dimension Shifter, a wizard who can change the size of objects using dilations. They realize that the notebook they found belongs to the wizard and excitedly ask him to teach them about dilation. Dilatius thrilled to share his knowledge with them.
OA′=k⋅OA⇔k=OAOA′
When a point is dilated using a scale factor of k and a center of dilation at the origin, the coordinates of its image are found by multiplying the coordinates of the preimage by k.
(x,y)→(kx,ky)
The diagram shows how the image changes as the preimage and the scale factor change.
Dilatius is impressed by Vincenzo and Zosia's eagerness to learn. They seem to have picked up the dilation spell using the coordinate rule quickly, so he challenges them to dilate the following triangle.
Draw the image of the triangle after a dilation with center (0,0) and a scale factor of 3.To find the image of a vertex after a dilation with scale factor k, multiply its coordinates by k.
Zosia and Vincenzo need to dilate the triangle using a scale factor of 3 with respect to the origin. Start by identifying the vertices of the triangle.
When the center of dilation is the origin, each coordinate of the preimage is multiplied by the scale factor k to find the coordinates of the image.Dilation With Scale Factor k=3 | ||
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Preimage | Multiply by k | Image |
A(0,2) | (3⋅0,3⋅2) | A′(0,6) |
B(3,1) | (3⋅3,3⋅1) | B′(9,3) |
C(2,-1) | (3⋅2,3⋅(-1)) | C′(6,-3) |
Dilatius teaches Vincenzo and Zosia to use reducio
to make things smaller and enlargio
to make them bigger. These phrases produce a reduction and an enlargement, respectively. He then quizzes them about the magic behind the drawings in his notebook.
The scale factor is the ratio of the sides lengths of the image to the corresponding side lengths of the original figure.
There are two types of dilations.
In the given coordinate plane, it can be seen that the green square is smaller than the blue square. This means that the dilation is a reduction.
Next, remember that the scale factor is the ratio of the sides lengths of the image to the corresponding side lengths of the preimage.The vertices of quadrilateral ABCD are A(-3,2), B(3,2), C(2,5), and D(-1,4). Which is the image of the quadrilateral after a 90∘ counterclockwise rotation about vertex B?
Let's start by plotting the given vertices A(- 3,2), B(3,2), C(2,5), and D(- 1,4). Then we can connect them with line segments to draw the quadrilateral.
We will now draw the image of ABCD after a 90^(∘) counterclockwise rotation about vertex B. For simplicity, we start by rotating only one point. Let's start with vertex A. To rotate it, we will use a protractor to draw a ray that makes a 90^(∘) angle with BA at B.
Now we mark a point A' on this ray such that BA' is the same length as BA. This point is the image of A after the rotation. We can see that BA is 6 units long, so BA' must also be 6 units long.
Next, we repeat the same process for vertices C and D to plot C' and D'. Since B is the center of rotation, B' will be in the same position as B.
The coordinates of the images of the vertices are A'(3,- 4), B'(3,2), C'(0,1), and D'(1,- 2). Finally, we will connect A', B', C', and D' to draw the image of ABCD.
This corresponds to B.
Let's rotate ABCD 90^(∘) counterclockwise about B so that we can see how it is mapped onto A'B'C'D'.
Triangle JKL has vertices J(-4,1), K(-6,-5), and L(-2,-3). Which graph shows the image of the triangle after a 180∘ counterclockwise rotation about the origin?
We want to rotate a triangle 180^(∘) counterclockwise about the origin. Vertices of Triangle J(- 4,1), K(- 6,- 5),andL(-2,- 3) The center of rotation is the origin. When a counterclockwise rotation is performed about the origin, the coordinates of the image can be written in relation to the coordinates of the preimage.
Counterclockwise Rotations About the Origin | ||
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90^(∘) Rotation | 180^(∘) Rotation | 270^(∘) Rotation |
ccc Preimage & & Image [0.5em] (x,y) & → & (- y,x) |
ccc Preimage & & Image [0.5em] (x,y) & → & (- x,- y) |
ccc Preimage & & Image [0.5em] (x,y) & → & (y,- x) |
We can use the rule shown in the table that corresponds to a 180^(∘) counterclockwise rotation to determine the coordinates of the image of each vertex. ccc Preimage & & Image [0.3em] (x,y) & → & (- x,- y) [0.5em] [-0.5em] J(- 4,1) & & J'(4,- 1) [0.5em] K(- 6,- 5) & & K'(6,5) [0.5em] L(- 2,- 3) & & L'(2,3) Let's draw triangle JKL first. Then, we will plot the points we found and connect them to see the image of the triangle after rotation.
This corresponds to A.
Let's rotate JKL 180^(∘) counterclockwise about the origin so that we can see how it is mapped onto J'K'L'.
Square EFGH is rotated about the origin.
Which angle of rotation maps square EFGH to square E′F′G′H′?We know that square EFGH is rotated about the origin. We want to identify the angle of rotation.
Let's consider point G and its image G'. First, we will draw a segment that connects G and the origin. We will then connect the origin with G' and measure the angle between the segments.
We can see that the segments create a 90^(∘) angle. Therefore, square EFGH is rotated either 90^(∘) clockwise or 270^(∘) counterclockwise. Of these, only 90^(∘) clockwise is an available option, so the graph describes a 90^(∘) clockwise rotation.
Let's rotate EFGH 270^(∘) counterclockwise about the origin so that we can see how it is mapped onto E'F'G'H'. Then we will rotate it 90^(∘) clockwise to see it mapped onto E'F'G'H' to see it mapped that way.
The vertices of a trapezoid are P(2,-4), Q(4,-4), R(6,-8), and S(-2,-8). Which is the image of the trapezoid after a dilation with center (0,0) and a scale factor of 21?
When the center of dilation is the origin, each coordinate of the preimage is multiplied by the scale factor k to find the coordinates of the image. ccc Preimage & & Image [0.5em] (x,y)& ⇒ & ( kx, ky) Let's find the coordinates of the vertices of the trapezoid after a dilation with a scale factor k= 12.
Dilation With Scale Factor k= 12 | ||
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Preimage | Multiply by k | Image |
P(2,- 4) | ( 1/2 (2), 1/2(- 4)) | P'(1,- 2) |
Q(4,- 4) | ( 1/2 (4), 1/2(- 4)) | Q'(2,- 2) |
R(6,- 8) | ( 1/2 (- 6), 1/2(- 8)) | A'(- 3,- 4) |
S(- 2,- 8) | ( 1/2 (- 2), 1/2(- 8)) | S'(- 1,- 4) |
Let's draw the preimage trapezoid first. Then we will plot the points we found and connect them to see its image after dilation.
The answer is D.
The green figure PQR is a dilation of the blue figure P′Q′R′.
Is the dilation is an enlargement or a reduction?We have a graph showing triangles PQR and P'Q'R'. We know that △ P'Q'R' is the image of △ PQR after a dilation. We want to decide if the dilation is an enlargement or a reduction.
To decide, recall that there are two types of dilation.
In the graph we see that the image, triangle P'Q'R', is larger than the preimage, △ PQR. This tells us that the dilation is an enlargement.
The green triangle K′L′M′ is a dilation of the blue triangle KLM.
What type of dilation is applied, and what is the scale factor?We want to identify the type of dilation and the scale factor of the given triangles.
We know that there are two types of dilation.
In the given coordinate plane, we can see that the green triangle is larger than the blue triangle. This means that the dilation is an enlargement. We can find the ratio of the sides lengths of the image to the corresponding side lengths of the original figure to determine the scale factor. Scale Factor= Image Side Length/Preimage Side Length In the graph we can see that KL and K'L' are bounded by lattice points. Since the length of a diagonal of a unit square is sqrt(2) by the Pythagorean Theorem, we have KL=4sqrt(2) and K'L' = 20sqrt(2).
Let's substitute these values into the formula to find the scale factor. Scale Factor = 20sqrt(2)/4sqrt(2) = 5 The scale factor is 5, so the green triangle is the image of the blue triangle after a dilation with scale factor 5. The answer is D.
Recall that when a dilation by a scale factor of k is applied to a point, both of its coordinates are multiplied by the scale factor, k. Preimage & & Image (x,y) & → & ( kx, ky) Let's find the coordinates of all the vertices and check if there is a scale factor k.
We will divide the x- and y-coordinate of each preimage point by its corresponding coordinate in the image. Then we can check if the scale factors from each coordinate are equal. Let's do it!
Scale Factor | ||
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Ratio of x-coordinates | Ratio of y-coordinates | |
K(- 2,2) and K'(- 10,10) | -10/-2=5 | 10/2=5 |
L(2,- 2) and L'(10,- 10) | 10/2=5 | - 10/- 2=5 |
M(- 3,- 3) and M'(- 15,- 15) | -15/-3=5 | - 15/- 3=5 |
The scale factor is the same for every coordinate in the points, which means that the scale factor is 5. Since it is greater than 1, the dilation is an enlargement.