The problem involves a rhombus $DEFG$, and we are tasked with finding the length of $\overline{GE}$.

Given the information in the image:

• The rhombus has diagonals that intersect at point $H$.
• $DH = d + 154$ and $GH = 4d + 4$.
• We know that in a rhombus, diagonals bisect each other, meaning that $DH = HE$ and $GH = HF$.

Since the diagonals bisect each other, we can set up an equation based on the total diagonal lengths.

1. The length of diagonal $DF$ is: $DF = 2 \times DH = 2 \times (d + 154) = 2d + 308$

2. The length of diagonal $GE$ is: $GE = 2 \times GH = 2 \times (4d + 4) = 8d + 8$

Now, we will use the properties of the rhombus to find the length of $GE$. From the options given (408, 432, 50, 204), we can try to solve for $d$.

Solving for $d$ based on diagonal lengths:

We’ll substitute values for $GE = 204$ and solve for $d$: $8d + 8 = 204$ $8d = 196$ $d = 24.5$

Now, let’s verify:

• $DH = d + 154 = 24.5 + 154 = 178.5$
• $GE = 8d + 8 = 8 \times 24.5 + 8 = 204$

This confirms that the length of $\overline{GE}$ is 204, which corresponds to option D.

Would you like further details or explanations?

Here are some related questions:

1. What are the properties of a rhombus, and how do they help solve problems involving diagonals?
2. How does the relationship between diagonals in a rhombus differ from other quadrilaterals like rectangles?
3. What role does symmetry play in solving geometry problems involving bisected diagonals?
4. How can you generalize the approach used here to solve for the diagonals in other types of polygons?
5. What would the outcome be if we changed the given lengths or formulas?

Tip: Remember that diagonals of a rhombus not only bisect each other but also intersect at right angles. This is a key property when solving problems like this one!