The diagram in the Unit 1 Lesson 11, the description of the rotation of the image ∆P'O'G' is as follows;
1. C. Rotate 60° counterclockwise around <em>O</em>
<h3>What is the description of the transformation of ∆POG?</h3>
The diagram of the triangle ∆POG consists of a triangle placed in a grid consisting of equilateral triangles.
The location of point O<em> </em>in both the pre-image and the image are (approximately) the same.
Therefore;
The point about which ∆POG is rotated is the point <em>O</em>.
The side OG in the pre-image is vertical, while O'G' in the image 1 is rotated to the right, through to the next adjacent leg of one of the triangles in the grid with vertex at <em>O</em>.
Given that the triangles in the grid are equilateral triangles, that have an 60° interior angles, we have;
The rotation transformation is a 60° counterclockwise rotation about the point <em>O</em>.
The correct option is therefore;
- C. Rotate 60° counterclockwise around <em>O</em>.
Learn more about rotation transformation in geometry here:
brainly.com/question/28084964
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