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{{ printedBook.courseTrack.name }} {{ printedBook.name }} A transformation that preserves the shape, but not necessarily the size, of a figure is called a similarity transformation. This includes:

Two figures that can be mapped onto one another using one or several similarity transformations are said to be similar.
Dilations, rotations, reflections, and translations are all similarity transformations. Since rotation, reflection, and translation are rigid motions, they preserve both size and shape, whereas dilation only ensures that the shape is preserved. Since rigid motions are congruence transformations, all congruent figures are also similar.

The quadrilateral $ABCD$ is defined by its vertices $A(-3,-2),$ $B(-2,-1),$ $C(-1,-2),$ and $D(-2,0.5).$ It is then translated by the mapping $(x,y)→(x+2,y+1),$ and dilated by the scale factor $2.$ Lastly, it is reflected in the $x$-axis. Express the composite transformation as a mapping. Then, draw the image in a coordinate plane. Are the image and preimage similar? Are they congruent?

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To begin, we can start with the first transformation. The mapping is given as $(x,y)→(x+2,y+1).$ This is a translation $2$ units to the right and $1$ unit up. The next transformation is a dilation with the center of dilation being the origin. Thus, the mapping is found by multiplying both the $x$- and $y$-coordinates by the scale factor $2.$ This gives us $(x+2,y+1)→(2(x+2),2(y+1)).$ The last transformation is a reflection in the $x$-axis. This is equivalent to changing the sign of the $y$-coordinate for every point. Thus, we can get the final mapping by making the $y$-coordinate negative. $(2(x+2),2(y+1))→(2(x+2),-2(y+1))$ We have now performed each transformation algebraically in order. By skipping the intermediate steps, we can express the composite transformation as $(x,y)→(2(x+2),-2(y+1)).$ To draw the image, we first have to calculate the coordinates for the vertices in the image.

$(x,y)$ | $(2(x+2),-2(y+1))$ | $(x_{′},y_{′})$ |
---|---|---|

$A(-3,-2)$ | $(2(-3+2),-2(-2+1))$ | $A_{′}(-2,2)$ |

$B(-2,-1)$ | $(2(-2+2),-2(-1+1))$ | $B_{′}(0,0)$ |

$C(-1,-2)$ | $(2(-1+2),-2(-2+1))$ | $C_{′}(2,2)$ |

$D(-2,0.5)$ | $(2(-2+2),-2(0.5+1))$ | $D_{′}(0,-3)$ |

We can draw the preimage and the image in a coordinate plane by plotting their points.

Are these two quadrilaterals similar? We know there is a composite transformation made up of a translation, dilation, and then reflection that maps the preimage onto the image. These are all similarity transformations. Thus, the quadrilaterals are similar. As the side lengths have **not** been preserved, due to the dilation, the quadrilaterals are **not** congruent.

Similar objects have the same shape. But what does it actually mean for two geometric objects to have the same shape? Polygons are defined by their angles and side lengths. Thus, these can be used to define shape. Angles are preserved by similarity transformations. Thus, similar polygons have congruent angles. Dilations scale all sides by the same factor, resulting in the ratio between corresponding sides' length being constant.

For a polygon with side lengths $a,b,c,…,$ the corresponding sides must have the lengths $ka,kb,kc,…,$ where $k$ is some scale factor. This also means that all regular polygons with the same number of vertices are similar. Also, all circles are similar.Determine if the rectangles are similar.

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The rectangles are similar if corresponding angles are congruent and the ratio between corresponding sides is constant. Since all angles are right, they're congruent. To determine if the ratio is constant, we need to identify the corresponding sides. The longest side in the rectangle to the left is $12$ units and the longest side in the rectangle to the right is $6.$ Therefore, the ratio is $612 =2.$ If the rectangles are similar, the ratio between the short sides must also be $2.$ They are $6$ and $3,$ respectively, so the ratio is $36 =2.$ In summary, corresponding angles are congruent and the ratio of the corresponding side lengths is constant. Thus, the rectangles are similar.

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