We are tasked with finding the length of segment $EF$ using the altitude and similarity properties of triangles. Here's how we solve the problem step by step:

Step 1: Identify triangle similarity

The triangle $\triangle EDG$ has an altitude $EF$ drawn from vertex $E$ to the base $DG$. This creates two smaller right triangles: $\triangle DEF$ and $\triangle EFG$, which are similar to the larger triangle $\triangle EDG$ (by AA similarity).

Step 2: Use proportions

The similarity of the triangles gives us the following relationships: $\frac{EF}{DE} = \frac{FG}{EF} \quad \text{or} \quad EF^2 = DE \cdot FG$ where:

• $DE = 6$
• $FG = DG - DF = 8 - 4.8 = 3.2$

Step 3: Calculate $EF$

Substitute the known values into the equation: $EF^2 = 6 \cdot 3.2$ $EF^2 = 19.2$ Take the square root: $EF = \sqrt{19.2} \approx 4.4$

Final Answer:

The closest answer is 4.3.

Would you like further clarification or additional details?

Here are 5 related questions for deeper understanding:

1. How can we verify the similarity of the triangles in this problem?
2. What are the conditions for two triangles to be similar?
3. How is the geometric mean property applied in triangle altitude problems?
4. Can we use the Pythagorean theorem here to check our results?
5. How would the problem change if the base $DG$ or $DF$ were unknown?

Tip:

For problems involving triangle similarity and altitudes, always start by identifying the similar triangles and setting up proportions between their corresponding sides.