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Let's work through this problem step by step:

1. **Rectangular dimensions**: We are told that the length of the rectangle is twice the width and that the area is 92 square units.

2. **Dividing the rectangle into two squares**: Since the rectangle is divided into two squares, the length of the rectangle is equal to twice the side length of each square, and the width of the rectangle is equal to the side length of each square.

Let's say:
- The side length of each square is \( s \).
- Then, the length of the rectangle is \( 2s \) and the width of the rectangle is \( s \).

3. **Calculating the area**: 
   The area of the rectangle can be expressed as:
   \[
   \text{Area} = \text{Length} \times \text{Width} = 2s \times s = 2s^2
   \]
   We're told the area of the rectangle is 92 square units:
   \[
   2s^2 = 92
   \]
   Solving for \( s^2 \):
   \[
   s^2 = \frac{92}{2} = 46
   \]
   So:
   \[
   s = \sqrt{46}
   \]

Thus, the correct estimate for the side length of each square is \( \sqrt{46} \), which corresponds to **option A**.

Would you like further clarification or details? Here are some related questions to deepen understanding:

1. How can we verify the side length calculation?
2. If the area were 184 square units, what would the side length of each square be?
3. How does the concept of squaring help in solving problems involving areas of geometric shapes?
4. What would be the new length and width if the rectangle's area was doubled while maintaining the same proportions?
5. How can this problem be extended to a 3D scenario with a rectangular prism?

**Tip**: Always double-check the relationship between dimensions and area to ensure that calculations align with the given geometric properties.

The length of a rectangle is twice the width. The area of the rectangle is 92 square units. Notice that you can divide the rectangle into two squares with equal area. How can you estimate the side length of each square?

Learn how to estimate the side length of squares derived from a rectangle with an area of 92 square units. The length is twice the width. Solve step-by-step using area formulas and square roots. Suitable for Grades 6-8.

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