1. The problem states that $r(180^\circ, O)(\triangle ABC) = \triangle A'B'C'$, meaning $\triangle ABC$ is rotated 180 degrees about the origin $O$ to get $\triangle A'B'C'$. We need to find a composition of a rotation and a translation that results in the same image $\triangle A'B'C'$. 2. First, let's verify the effect of a 180° rotation about $O$ on the vertices of $\triangle ABC$: - $A(4,4)$ rotated 180° about $O$ becomes $A'(-4,-4)$. - $B(6,6)$ rotated 180° about $O$ becomes $B'(-6,-6)$. - $C(6,2)$ rotated 180° about $O$ becomes $C'(-6,-2)$. 3. However, the given $\triangle A'B'C'$ vertices are approximately $A'(-4,-4)$, $B'(-6,-2)$, and $C'(-2,-4)$, which do not match the direct 180° rotation about $O$ exactly. This suggests a translation is involved after a rotation. 4. Let's analyze each answer choice: A. Rotate $\triangle ABC$ 180° about point $C(6,2)$, then translate 4 units down and 8 units left. - Rotating 180° about $C$ sends $A(4,4)$ to $A'' = 2\times C - A = (6,2)\times 2 - (4,4) = (12,4) - (4,4) = (8,0)$. - Then translating 4 down and 8 left: $A''' = (8-8,0-4) = (0,-4)$, which does not match $A'(-4,-4)$. B. Rotate $\triangle ABC$ 90° about $O$, then translate 6 units down and 2 units left. - 90° rotation about $O$ sends $A(4,4)$ to $A''(-4,4)$. - Translate 6 down and 2 left: $A''' = (-4-2,4-6) = (-6,-2)$, which does not match $A'(-4,-4)$. C. Rotate $\triangle ABC$ 270° about $O$, then translate 2 units up and 6 units left. - 270° rotation about $O$ sends $A(4,4)$ to $A''(4,-4)$. - Translate 2 up and 6 left: $A''' = (4-6,-4+2) = (-2,-2)$, which does not match $A'(-4,-4)$. D. Rotate $\triangle ABC$ 90° about point $C(6,2)$, then translate 4 units down and 8 units left. - Rotate 90° about $C$: For $A(4,4)$, vector $\overrightarrow{CA} = (4-6,4-2) = (-2,2)$. Rotate 90° counterclockwise: $(-2,2) \to (-2,-2)$. New $A'' = C + (-2,-2) = (6-2,2-2) = (4,0)$. - Translate 4 down and 8 left: $A''' = (4-8,0-4) = (-4,-4)$, which matches $A'$. - Check $B(6,6)$: $\overrightarrow{CB} = (6-6,6-2) = (0,4)$. Rotate 90° CCW: $(0,4) \to (-4,0)$. $B'' = (6-4,2+0) = (2,2)$. Translate: $(2-8,2-4) = (-6,-2)$ matches $B'$. - Check $C(6,2)$: Rotating about itself leaves $C$ unchanged. Translate: $(6-8,2-4) = (-2,-2)$ but $C'$ is $(-2,-4)$, so slight discrepancy in $y$. Given the options, D is the closest and matches two points exactly and the problem likely expects this answer. Final answer: D. Rotate $\triangle ABC$ 90° about point $C$, then translate 4 units down and 8 units left.