The problem involves the triangle $\triangle CDE$ where $F$ and $G$ are the midpoints of $\overline{CE}$ and $\overline{DE}$, respectively. The segment $\overline{FG}$ is a midsegment of the triangle, meaning it is parallel to the base $\overline{CD}$ and its length is half of $\overline{CD}$.

We are given:

• $FG = 5x - 25$
• $CD = 9x - 41$

Using the midsegment property: $FG = \frac{1}{2}CD$

Step-by-step solution:

1. Substitute the given values into the equation: $5x - 25 = \frac{1}{2}(9x - 41)$

2. Eliminate the fraction by multiplying through by 2: $2(5x - 25) = 9x - 41$ $10x - 50 = 9x - 41$

3. Simplify and solve for $x$: $10x - 9x = -41 + 50$ $x = 9$

4. Substitute $x = 9$ back into the expression for $CD$: $CD = 9x - 41$ $CD = 9(9) - 41$ $CD = 81 - 41 = 40$

Final Answer:

The measure of $CD$ is 40 units.

Would you like a detailed explanation of the midsegment theorem or have further questions? Here are some related questions to consider:

1. What is the midsegment theorem, and why is it applicable here?
2. How can you verify that $FG$ is half of $CD$ after finding $x$?
3. Are there other properties of midsegments that could apply to $\triangle CDE$?
4. What if $FG$ and $CD$ were expressed in different units? How would that affect the calculations?
5. Can this method generalize to other polygons with midsegments?

Tip: Always double-check calculations by substituting back to confirm consistency