Problem 47 - College Algebra Essentials

Question

The rectangular painting in the figure shown measures 12 inches by 16 inches and is surrounded by a frame of uniform width around the four edges. The perimeter of the rectangle formed by the painting and its frame is 72 inches. Determine the width of the frame.

Step-by-Step Solution

Answer

The width of the frame is 2 inches.

Step 1: Understand the Relationships

Knowing that the frame has a uniform width around the four edges, both the length and the width of the frame will be increased by twice the frame's width. This is essential to remember because the perimeter calculation involves both the length and the width of the rectangle. Thus, the length of the combined painting and frame is \(16 + 2x\) and the width is \(12 + 2x\), where \(x\) represents the frame's width.

Step 2: Set Up the Perimeter Equation

The perimeter of a rectangle is calculated by adding up all its sides. Therefore, the perimeter of the frame and painting is given by \(2(16 + 2x) + 2(12 + 2x)\). This value is given as 72 inches in the problem. Therefore the equation will look like this: \(2(16 + 2x) + 2(12 + 2x) = 72\)

Step 3: Solve the Equation

First simplify the left side of the equation: \(2(16+2x) + 2(12+2x) = 72\) becomes \(56 + 8x = 72\). Then, you can solve this simple linear equation for \(x\): Subtract 56 from both sides, resulting in \(8x = 16\). Now, divide both sides by 8, giving you \(x = 2\) inches.