To show that the rectangle in Quadrant III is congruent to the rectangle in Quadrant I Darnell rotates the first rectangle 90 degrees clockwise into Quadrant II. What is his next step?
rectangles vertex's
(2, 3) (2, 5) (7, 5) (7, 3)
Darnell's next step is to translate the rectangle in Quadrant II vertically downwards by the same distance as the side length of the rectangle, while keeping the orientation of the rectangle unchanged. This will place the rectangle in Quadrant III.
do the math bot
Apologies for the confusion. Let's calculate the translation needed to move the rectangle in Quadrant II to Quadrant III.
The rectangle in Quadrant II has vertices:
(2, 3), (2, 5), (7, 5), (7, 3)
To move it to Quadrant III, we need to shift it vertically downwards by the same distance as the side length of the rectangle.
The side length of the rectangle is 5 - 3 = 2 units.
To translate the rectangle downwards, we subtract 2 from the y-coordinate of each vertex:
(2, 3) -> (2, 3 - 2) = (2, 1)
(2, 5) -> (2, 5 - 2) = (2, 3)
(7, 5) -> (7, 5 - 2) = (7, 3)
(7, 3) -> (7, 3 - 2) = (7, 1)
Therefore, the new vertices of the rectangle in Quadrant III are:
(2, 1), (2, 3), (7, 3), (7, 1)