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| 15 Theory slides |
| 10 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.
Identifying whether one figure is the image of another figure under rotation can be difficult. A key aspect to observe is whether the center of rotation is the same distance from an image as it is from its preimage.
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.
Remember, after performing a rotation, the preimage and the image of a point are the same distance from the center of rotation. The angle of rotation is formed by a preimage, the center of rotation, and the corresponding image.
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′.
On a geometry test, Ignacio was asked to perform a 70∘ counterclockwise rotation to △ABC about point P.
Draw △ABC and its image under this rotation.
Finally, the image of △ABC under the given rotation is the triangle formed by A′, B′, and C′.
Given a figure and its image under a rotation, the following theorem can be used to find the center of rotation.
If a point is equidistant from the endpoints of a line segment, then it lies on the perpendicular bisector of the segment. |
With this theorem in mind, consider the following example. In the diagram, quadrilateral A′B′C′D′ is the image of ABCD under a certain rotation.
Find the center and angle of rotation.
Angle of Rotation: 120∘ clockwise or 240∘ counterclockwise.
Graph:
Remember that the center of rotation is equidistant from the preimage and the image of each vertex. Use the Converse Perpendicular Bisector Theorem. The center is the intersection point between two perpendicular bisectors.
The first step is to find the center of rotation. Remember, by definition, a point and its image under a rotation are the same distance from the center.
The center of rotation is equidistant from a point and its image.
Therefore, by the Converse of the Perpendicular Bisector Theorem, the center lies on the perpendicular bisector of AA′, for instance. Then, with the aid of a compass and a straightedge, start by constructing the perpendicular bisector of this segment.
To determine the center's exact position, draw a second segment joining a vertex and its image, for example, DD′. Then, draw the perpendicular bisector of this segment. The intersection between both perpendicular bisectors is the center of rotation.
Notice that drawing only two perpendicular bisectors is enough to find the center of rotation because all will intersect at the same point. Since the sense of rotation was not specified, both measures will be found using a protractor.
The angle of rotation is either 240∘ counterclockwise or 120∘ clockwise.
If the point is not placed close enough to the center of rotation, when the Check Answer
button is pushed, a red area is highlighted indicating the region where the center of rotation is located.
Recall that rotations are transformations and that transformations can be composed. Therefore, it is possible to have a composition of two or more rotations. On a geometry exercise, the following two rotations are given.
LaShay has to perform both rotations to △ABC, one after the other, but the book does not indicate the composition's order.
In real life, there are plenty of situations where rotations can be appreciated. For instance, take a look at a door.
Consider a triangle ABC, a triangle A′B′C′, and a point P. Determine whether the following statement is true or false.
The statement tells us that as long as the distances to a given point and what would be corresponding vertices in two shapes are congruent, this must describe a rotation about that point.
Let's show an example of a general triangle ABC that has undergone a rotation about a point P.
As we can see, the rotation fulfills the given criterion regarding segment length between corresponding vertices and the center of rotation. PA&=PA' && ✓ PB&=PB' && ✓ PC&=PC' && ✓
Remember, a rotation is a rigid motion. Therefore, if it is possible to change at least one of the triangle's vertices, while keeping these three equations true, then the given statement that this describes a rotation is false. Below we have moved A' in a way that keeps PA=PA' true.
Since we can draw two triangles that fulfills the criterion without △ A'B'C' being a rotation of △ ABC, the statement must be false.
This time, we are given that the angle of rotation between what would be corresponding vertices on a rotated triangle, its preimage, and a point P, are all congruent. This is indeed a criterion when performing a rotation.
As we can see, the given angle criterion is fulfilled. m∠ APA'= m∠ BPB'=m∠ CPC' ✓ However, there is another criterion for a rotation. In addition to every angle between corresponding vertices and the center of rotation being congruent, the segments connecting P with corresponding vertices must also be congruent. Below we have moved one of the image's vertices without changing the angle of rotation.
As we can see, the given angle criterion is still fulfilled. m∠ APA'= m∠ BPB'=m∠ CPC' ✓ However, this shows a different transformation than △ ABC. Therefore, the statement is false.
Point A has been rotated 120∘ counterclockwise about a center of rotation P until it ends up at B.
When a polygon is rotated, the coordinates of the center of rotation P can be found by following these steps.
Below we see an example of this.
However, in this case, we are rotating a point. This means that we can only determine the perpendicular bisector of one segment.
The center of rotation must fall somewhere on this perpendicular bisector. But where? Remember that the center of rotation is equidistant to the image and preimage. This means the center of rotation, along with the points A and B, form an isosceles triangle ABP. Recall the Isosceles Triangle Theorem.
Isosceles Triangle Theorem |- If two sides of a triangle are congruent, then the angles opposite them are congruent.
Additionally, because the angle of rotation is the vertex angle of this isosceles triangle, we can calculate its base angles by using the Interior Angles Theorem. Let's label these base angles x.
If we use a protractor, we can find the segments AP and BP. We have two options here. The center of rotation could be above or below AB.
Since A is rotated counterclockwise, the only viable option is P_1. We see that this point falls on (4,1).