A labelled graph showing the original triangle and all images

Here are the coordinates for all the triangles to help you draw the labeled graph:

• Original Triangle $\triangle ABC$: $A(1, 2)$ $B(3, 4)$ $C(2, 0)$

• First Image $\triangle A'B'C'$ (after translation): $A'(-2, 3)$ $B'(0, 5)$ $C'(-1, 1)$

• Second Image $\triangle A''B''C''$ (after reflection): $A''(2, 3)$ $B''(0, 5)$ $C''(1, 1)$

• Third Image $\triangle A'''B'''C'''$ (after rotation): $A'''(-3, 2)$ $B'''(-5, 0)$ $C'''(-1, 1)$

• Final Image $\triangle A''''B''''C''''$ (after enlargement): $A''''(-6, 4)$ $B''''(-10, 0)$ $C''''(-2, 2)$

To create the labeled graph, plot these points on a coordinate plane and connect the vertices for each triangle. Label each triangle clearly as $\triangle ABC$, $\triangle A'B'C'$, $\triangle A''B''C''$, $\triangle A'''B'''C'''$, and $\triangle A''''B''''C''''$.

Regarding the comparison of perimeters from the previous step:

Step 5: Compare the perimeter of $\triangle ABC$ and $\triangle A''''B''''C''''$.
The perimeter of $\triangle ABC$ is $P_{ABC} = \sqrt{8} + \sqrt{17} + \sqrt{5}$.
The final transformation from $\triangle A'''B'''C'''$ to $\triangle A''''B''''C''''$ is an enlargement with a scale factor of $k=2$. All previous transformations (translation, reflection, rotation) are rigid transformations, meaning they preserve length and shape. Therefore, $\triangle A'''B'''C'''$ has the same perimeter as $\triangle ABC$.
When a shape is enlarged by a scale factor $k$, its perimeter is multiplied by $k$.
So, the perimeter of $\triangle A''''B''''C''''$ is $k$ times the perimeter of $\triangle ABC$.

$$P_{A''''B''''C''''} = 2 \times P_{ABC}$$ $$P_{A''''B''''C''''} = 2 \times (\sqrt{8} + \sqrt{17} + \sqrt{5})$$

Numerically:

$$P_{ABC} \approx 2.828 + 4.123 + 2.236 \approx 9.187 \text{ units}$$ $$P_{A''''B''''C''''} \approx 2 \times 9.187 \approx 18.374 \text{ units}$$

The perimeter of $\triangle A''''B''''C''''$ is twice the perimeter of $\triangle ABC$.