In this case, you have two rectangles: one larger (outer) rectangle and one smaller (inner) rectangle, with a "frame" between them that is 1 unit wide. To find the area of the shaded region (the frame), we can follow these steps:

Steps:

1. Define Dimensions of the Outer and Inner Rectangles:

   • Let the dimensions of the outer rectangle be $L \times W$, where $L$ is the length and $W$ is the width.
   • Since the frame is 1 unit wide, the dimensions of the inner rectangle will be reduced by 2 units for both the length and width. Hence, the dimensions of the inner rectangle are $(L-2) \times (W-2)$.

2. Calculate the Area of Both Rectangles:

   • The area of the outer rectangle is: $A_{\text{outer}} = L \times W$
   • The area of the inner rectangle is: $A_{\text{inner}} = (L - 2) \times (W - 2)$

3. Find the Area of the Shaded Region: The area of the shaded region (the frame) is the difference between the area of the outer rectangle and the inner rectangle: $A_{\text{shaded}} = A_{\text{outer}} - A_{\text{inner}} = (L \times W) - ((L-2) \times (W-2))$

4. Expand the Expression for the Inner Rectangle: $(L-2) \times (W-2) = LW - 2L - 2W + 4$

   So, the area of the shaded region becomes: $A_{\text{shaded}} = LW - (LW - 2L - 2W + 4) = 2L + 2W - 4$

Thus, the area of the shaded region (frame) is: $A_{\text{shaded}} = 2L + 2W - 4$

If you know the specific dimensions of the outer rectangle, you can substitute them into the formula to find the exact area.

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Do you have the dimensions of the outer rectangle or any other details?

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Related Questions:

1. How do you calculate the perimeter of a rectangle?
2. How would the formula change if the frame width was not uniform?
3. What would be the area of the inner rectangle alone?
4. How does the formula for the shaded area change if the frame is 2 units wide instead of 1?
5. Can the shaded area ever be larger than the inner rectangle?

Tip: Always subtract the areas of shapes inside one another to find the area of regions like frames.