Questions: A club swimming pool is 25 ft wide and 41 ft long. The club members want an exposed aggregate border in a strip of uniform width around the pool. They have enough material for 280 ft². How wide can the strip be? The width of the border will be

Solution Steps

Let \( x \) be the width of the strip around the pool.

The dimensions of the larger rectangle (pool + strip) will be:

• Width: \( 25 + 2x \)
• Length: \( 41 + 2x \)

The area of the larger rectangle is: \[ (25 + 2x)(41 + 2x) \]

The area of the pool is: \[ 25 \times 41 = 1025 \, \text{ft}^2 \]

The area of the strip is the area of the larger rectangle minus the area of the pool: \[ (25 + 2x)(41 + 2x) - 1025 = 280 \]

\[ (25 + 2x)(41 + 2x) - 1025 = 280 \] \[ (25 + 2x)(41 + 2x) = 1305 \]

\[ 25 \times 41 + 25 \times 2x + 41 \times 2x + 4x^2 = 1305 \] \[ 1025 + 50x + 82x + 4x^2 = 1305 \] \[ 4x^2 + 132x + 1025 = 1305 \]

\[ 4x^2 + 132x + 1025 - 1305 = 0 \] \[ 4x^2 + 132x - 280 = 0 \] \[ x^2 + 33x - 70 = 0 \]

\[ (x + 35)(x - 2) = 0 \]

\[ x + 35 = 0 \quad \text{or} \quad x - 2 = 0 \] \[ x = -35 \quad \text{or} \quad x = 2 \]

Since \( x \) represents a width, it must be positive: \[ x = 2 \]

Final Answer