The inner dimensions of a rectangular picture frame are 4 inches by 9 inches. The frame has a uniform width...

1. VISUALIZE the frame structure

• Given information:
  • Inner rectangle: 4 inches by 9 inches
  • Frame width: 1 inch uniform around all sides
• What this tells us: The frame completely surrounds the inner rectangle, adding material on all four sides

2. INFER how frame width affects outer dimensions

• Key insight: "Uniform width of 1 inch" means the frame extends 1 inch beyond the inner rectangle on every side
• This means each outer dimension = inner dimension + 1 inch (left side) + 1 inch (right side)
• Strategy: Calculate outer length and width first, then find perimeter

3. Calculate outer dimensions

• Outer length = \(4 + 2(1) = 6\) inches
• Outer width = \(9 + 2(1) = 11\) inches

4. TRANSLATE to perimeter formula

• Use rectangle perimeter formula: \(\mathrm{P} = 2(\mathrm{length} + \mathrm{width})\)
• Outer perimeter = \(2(6 + 11) = 2(17) = 34\) inches

Answer: D. 34

Why Students Usually Falter on This Problem

Most Common Error Path:

Poor TRANSLATE reasoning: Students misunderstand "uniform width of 1 inch" and think it means adding only 1 inch total to each dimension instead of 1 inch on each side.

This leads them to calculate:

• Outer length = \(4 + 1 = 5\) inches
• Outer width = \(9 + 1 = 10\) inches
• Perimeter = \(2(5 + 10) = 30\) inches

This may lead them to select Choice B (30).

Second Most Common Error:

Weak VISUALIZE skill: Students correctly understand the frame adds to both sides but get confused about which dimension is which, accidentally switching length and width in their calculations.

This typically doesn't lead to a wrong answer choice since the perimeter formula treats both dimensions equally, but it can cause calculation confusion and time waste.

The Bottom Line:

This problem tests whether students can visualize how a frame surrounds a rectangle and correctly interpret "uniform width" as extending equally in all directions from the inner boundary.