Exercise 4 Page 513

We want to find the lengths of the sides of $△CDE.$ We can see from the graph that the length of $\overline{CD}$ is $3$ units and the length of $\overline{DE}$ is $4$ units. We also know that angle $CDE$ is a right angle.

Since this is a right triangle, let's apply the Pythagorean Theorem to find the side length of $\overline{EC}.$ We will substitute the values into the formula and solve for $\overline{EC}.$ Since $\overline{EC}$ is the side length of the triangle and lengths cannot be negative, we know that the length of the side must be positive. We can write the side lengths of $△CDE$ as follows.

Let's reflect $△CDE$ over the $y\text{-}$axis. The image of the reflection will be $△C^{\prime \prime }{D}^{\prime \prime }{E}^{\prime \prime }.$ The side lengths of the triangles are the same but the orientation of the triangle is reserved after the reflection. Next, we can translate $△C^{\prime \prime }{D}^{\prime \prime }{E}^{\prime \prime }$ two units to the left. Now we can write the coordinates of the vertices of the image of the series of transformations.

Now we want to find the lengths of the sides of $△C^{\prime }{D}^{\prime }{E}^{\prime }.$ Let's remember the properties of the reflection and translation.

Translation	Reflection
Lengths are the same.	Lengths are the same.
Orientations are the same.	Orientations are reserved.

We know that the side lengths of figures after a reflection and a translation do not change. We found the side lengths of the preimage $△CDE$ in Part B. We can use these lengths to write the lengths of the sides of $△C^{\prime }{D}^{\prime }{E}^{\prime }.$

We are asked to determine whether $△CDE$ is congruent to $△C^{\prime }{D}^{\prime }{E}^{\prime }.$ We found a series of the transformations that maps $△CDE$ onto $△C^{\prime }{D}^{\prime }{E}^{\prime }$ in Part C. We also know that the sizes and shapes of the triangles are the same, so these two triangles — $△CDE$ and $△C^{\prime }{D}^{\prime }{E}^{\prime }$ — are congruent.