{{ item.displayTitle }}

No history yet!

equalizer

rate_review

{{ r.avatar.letter }}

{{ u.avatar.letter }}

+

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }} Transformations can change the size, position, or orientation of a figure. They can also map one figure onto another. But, can a transformation map a figure onto the same figure? Throughout the lesson, the answer to this question will be developed.
### Catch-Up and Review

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

Consider a rectangle $ABCD.$ Can the rectangle be rotated such that the image exactly matches the preimage? Not so fast, there is one catch; can it be done where the angle of rotation is greater than $0_{∘}$ and less than $360_{∘}$? Try it out by rotating $ABCD$ around the movable point $P.$

If accomplished, where was point $P$ located? What was the angle of rotation?

For some geometric figures, it is possible to find a transformation that maps the figures onto themselves. In such cases, the transformation is called a symmetry of the figure.

A symmetry is a rigid motion that maps a figure onto itself. It is a non-trivial congruence of a figure with itself, that is, the sides and angles do not correspond with themselves. When a figure has a symmetry it is said to be a **symmetric figure**.

When a figure is rotated less than $360_{∘},$ the final image can look the same as the initial one — as if the rotation did nothing to the preimage. In such a case, the figure is said to have rotational symmetry.

A figure in a plane has rotational symmetry if the figure can be mapped onto itself by a rotation between $0_{∘}$ and $360_{∘}$ about the center of the figure. This point is called the **center of symmetry**. When a figure has rotational symmetry, it is said to be **rotationally symmetric**.

As shown, a square, an equilateral triangle, and the digit $0$ all have rotational symmetry, each with a particular angle measure. The order of symmetry of a figure is the number of times it maps onto itself while rotating from $0_{∘}$ to $360_{∘}.$

While walking downtown, Heichi and Paulina saw a store with the following logo. They began to discuss whether the logo has rotational symmetry. To figure it out, they went into the store and took a business card each.

At their respective homes, and using patty paper, each performed rotations on the logo. The same night, Heichi called Paulina to tell her that the logo is rotationally symmetric, but Paulina disagreed. Who is correct?{"type":"choice","form":{"alts":["Heichi","Paulina"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["3"]}}

Rotate the logo $360_{∘}$ about its center. If possible, verify where along the way the rotation matches the original logo.

To determine whether the logo has rotational symmetry, check for a rotation of less than $360_{∘}$ about the center that maps the logo onto itself. By the use of patty paper or the following applet, rotational symmetry can be checked.

As shown, a rotation of $120_{∘}$ maps the logo onto itself. Therefore, the logo is rotationally symmetric, which means Heichi was right. Additionally, two other rotations of $240_{∘}$ and $360_{∘}$ also map the logo onto itself. This means that the order of symmetry is $3.$

In the first exploratory applet, it was studied whether a rectangle has rotational symmetry. Now, a parallelogram will be studied. Consider a parallelogram $ABCD.$

Does it have rotational symmetry?{"type":"choice","form":{"alts":["Yes","No"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

a What is the center of rotation?

b What is the order of symmetry?

c What conclusions can be drawn about the sides and angles of the parallelogram?

Yes, the parallelogram has rotational symmetry.

a The point of intersection of the diagonals.

b $2$

c The opposite sides of a parallelogram are congruent as well as the opposite angles.

a How can the center of the parallelogram be found?

b Rotate the parallelogram $360_{∘}$ around its center and count the number of times it matches the original one.

c When the parallelogram is rotated $180_{∘},$ opposite vertices and sides are swapped. What does this mean in terms of the angle measures and side lengths?

To determine whether the parallelogram has rotational symmetry, it needs to be checked if a rotation of less than $360_{∘}$ about the center maps the parallelogram onto itself. In this case, the center of the parallelogram is the point of intersection of the diagonals.

As shown, a rotation of $180_{∘}$ about the intersection point of diagonals maps the parallelogram onto itself. Consequently, it is rotationally symmetric.

a The center of rotation is the point of intersection of the diagonals.

b Because rotations of $180_{∘}$ and $360_{∘}$ map the parallelogram onto itself, the order of symmetry is $2.$

c When $ABCD$ is rotated $180_{∘}$ about its center, the following events occur.

- The opposite vertices are swapped. That is, $A$ and $C$ are swapped as well as $B$ and $D.$
- The opposite sides are swapped. That is, $AB$ and $CD$ are swapped as well as $BC$ and $DA.$

The first point implies that the measures of $∠A$ and $∠C$ are equal and the measures of $∠B$ and $∠D$ as well. Therefore, in a parallelogram, the opposite angles are congruent angles. $∠A≅∠C∠B≅∠D $ Similarly, the second point implies that the lengths of $AB$ and $CD$ are equal and the lengths of $BC$ and $DA$ as well. Therefore, in a parallelogram, the opposite sides are congruent segments. $AB≅CDBC≅DA $ On the parallelogram, these congruences can be represented as follows.

The symmetries of a figure help determine the properties of that figure. For instance, since a parallelogram has $180_{∘}$ rotational symmetry, its opposite sides and angles will match when rotated $180_{∘},$ which allows for the establishment of the following property.

The opposite sides and angles of a parallelogram are congruent.

In consequence, it can be said that rectangles, rhombi, and squares are special kinds of parallelograms. Upon taking a closer glance, notice that all squares have $90_{∘}$ rotational symmetry, which implies squares belong to a particular class of parallelograms. Even more, this $90_{∘}$ rotational symmetry implies that all squares have congruent sides and angles.

Is there another type of symmetry apart from the rotational symmetry? The answer is yes. Some figures can be *folded* along a certain line in such a way that all the sides and angles will lay on top of each other. In this case, it is said that the figure has line symmetry.

A figure in the plane has line symmetry if the figure can be mapped onto itself by a reflection in a line. This line of reflection is called the line of symmetry. When a figure has line symmetry, it is said to be reflectionally symmetric or line symmetric.

Some figures can have more than one line of symmetry.

After learning about line symmetry, Paulina and Heichi decided to study whether some quadrilaterals are line symmetric. To start, they picked a parallelogram.

Without thinking too much, Paulina said it has no line symmetry. However, Heichi said that the parallelogram is symmetric along the line connecting the midpoints of $AB$ and $CD.$ Who is correct?{"type":"choice","form":{"alts":["Paulina","Heichi"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

Try to find a line along which the parallelogram can be bent so that all the sides and angles are on top of each other. The lines containing the diagonals or the lines connecting the midpoints of opposite sides are always good options to start.

To determine whether the parallelogram $ABCD$ is line symmetric, it needs to be checked if there is a line such that when $ABCD$ is reflected on it, the image lies on top of the preimage. Before start testing lines, mark the midpoints of each side.

In the event that $ABCD$ is line symmetric, the lines connecting the midpoints of opposite sides are good candidates of being lines of symmetry. Also, the lines containing the diagonals could work. All possible lines can be tried in the following applet.
As can be seen, no reflection worked! In fact, there is

Despite the previous example showing a parallelogram with no line symmetry, other types of parallelograms should be studied first before making a general conclusion. Consider a rectangle and a rhombus. Study whether or not they are line symmetric.

If both polygons are line symmetric, compare their lines of symmetry.

Both rectangles and rhombi have line symmetry. Rectangles have lines of symmetry connecting midpoints of opposite sides, while rhombi have lines of symmetry along its diagonals.

For each polygon, consider the lines along the diagonals and the lines connecting midpoints of opposite sides.

When studying a polygon's line symmetries, it is always good to consider the lines along the diagonals and the lines connecting midpoints of opposite sides. For simplicity, consider one polygon at a time, starting with the rectangle.

From the diagram, it can be seen that the rectangle is line symmetric. It has two lines of symmetries, each connecting the midpoints of opposite sides. Next, repeat the same steps with the rhombus.

As can be seen, the rhombus is line symmetric and also has two lines of symmetries. In contrast to the rectangle, however, the lines of symmetries are the lines containing its diagonals.

Before, it was said that rectangles, rhombi, and squares are special kinds of parallelograms because they all have $180_{∘}$ rotational symmetry. In addition, rectangles and rhombi have line symmetry.

Polygon | Line Symmetry |
---|---|

Rectangles | Along the lines connecting midpoints of opposite sides |

Rhombi | Along the lines containing the diagonals |

Does a square have line symmetry?

If so, how many different lines of symmetry does it have?{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["4"]}}

Yes, squares are line symmetric. They have four lines of symmetries — two of them connect the midpoints of opposite sides and the other two contain its diagonals.

The lines containing the diagonals or the lines connecting the midpoints of opposite sides are always good options.

To determine whether the square is line symmetric, it is always good to consider the lines along the diagonals and the lines connecting midpoints of opposite sides. All possible lines can be tried in the following applet. Give it a go!

As might be verified, the square is indeed line symmetric. It has four lines of symmetry — two connect the midpoints of opposite sides and two others contain its diagonals.

The line symmetries of a square confirm the claim made previously that squares belong to a particular class of parallelograms.

Polygon | Number of Line Symmetries | Line Symmetry |
---|---|---|

Rectangles | $2$ | Along the lines connecting midpoints of opposite sides |

Rhombi | $2$ | Along the lines containing the diagonals |

Squares | $4$ | Two along the lines connecting midpoints of opposite sides and two along the lines containing the diagonals |

Notice that two symmetries of the square correspond to the rectangle's symmetries and the other two correspond to the rhombus symmetries. This suggests that squares are a particular case of rectangles and rhombi.

Paulina and Heichi have the task of investigating whether trapezoids have some symmetry. The teacher gave them the diagram below, but she told them to try any trapezoid they could imagine.

What conclusion should Paulina and Heichi reach?

Trapezoids have no rotational symmetry and have line symmetry only when the trapezoid is isosceles. In this case, the line of symmetry is the line passing through the midpoints of the bases.

A trapezoid is a quadrilateral with exactly one pair of parallel sides. Then, parallelograms are not trapezoids. Do not forget to try with isosceles trapezoids.

Consider any trapezoid $JKLM$ where $JK$ and $LM$ are parallel. First, start investigating whether $JKLM$ has rotational symmetry. To figure it out, set different dimensions for the trapezoid using the following applet. But remember, parallelograms are not trapezoids.

As investigated, regardless of the dimensions, no trapezoid has rotational symmetry. Next, repeat the same investigation but now checking for line symmetry.

From this last investigation, it may be seen that some trapezoids have line symmetry. Going a bit deeper, it can be concluded that these trapezoids are isosceles trapezoids.

A trapezoid has line symmetry **only** when it is isosceles trapezoid.

In this case, the line of symmetry is the line passing through the midpoints of each base.

Symmetries are not defined only for two-dimensional figures. The definition can also be extended to three-dimensional figures. In the real world, there are plenty of three-dimensional figures that have some symmetry. For example, sunflowers are rotationally symmetric while butterflies are line symmetric.

A three-dimensional figure is said to have axis symmetry if it can be rotated about a line onto itself by a rotation between $0_{∘}$ and $360_{∘}.$ This is the equivalent of rotational symmetry.
A three-dimensional figure has plane symmetry if there is a plane that divides the figure into two halves and each half is a reflection of the other across the plane. This is the equivalent of line symmetry.

{{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

{{ exercise.headTitle }}

{{ 'ml-heading-exercise' | message }} {{ focusmode.exercise.exerciseName }}