Answer: B
Let the width of the road be \(x\) meters. ∴ The total length of the rectangular playground including the road is \((45 + 2x)\) meters, and the total width is \((40 + 2x)\) meters. So, the total area including the road is \((45 + 2x)(40 + 2x)\) square meters. The area of the playground alone is \(45 \times 40\) square meters. ∴ According to the question, the area of the road is: \[ (45 + 2x)(40 + 2x) - 45 \times 40 = 450 \] \[ 1800 + 90x + 80x + 4x^2 - 1800 = 450 \] \[ 4x^2 + 170x - 450 = 0 \] \[ 2x^2 + 85x - 225 = 0 \] \[ 2x^2 + 90x - 5x - 225 = 0 \] \[ 2x(x + 45) - 5(x + 45) = 0 \] \[ (x + 45)(2x - 5) = 0 \] ∴ Either \((x + 45) = 0\) or \((2x - 5) = 0\) When \((x + 45) = 0\), then \(x = -45\) [But width cannot be negative] When \((2x - 5) = 0\), then \(x = \cfrac{5}{2} = 2.5\) ∴ The road surrounding the rectangular playground is uniformly 2.5 meters wide.
Let the width of the road be \(x\) meters. ∴ The total length of the rectangular playground including the road is \((45 + 2x)\) meters, and the total width is \((40 + 2x)\) meters. So, the total area including the road is \((45 + 2x)(40 + 2x)\) square meters. The area of the playground alone is \(45 \times 40\) square meters. ∴ According to the question, the area of the road is: \[ (45 + 2x)(40 + 2x) - 45 \times 40 = 450 \] \[ 1800 + 90x + 80x + 4x^2 - 1800 = 450 \] \[ 4x^2 + 170x - 450 = 0 \] \[ 2x^2 + 85x - 225 = 0 \] \[ 2x^2 + 90x - 5x - 225 = 0 \] \[ 2x(x + 45) - 5(x + 45) = 0 \] \[ (x + 45)(2x - 5) = 0 \] ∴ Either \((x + 45) = 0\) or \((2x - 5) = 0\) When \((x + 45) = 0\), then \(x = -45\) [But width cannot be negative] When \((2x - 5) = 0\), then \(x = \cfrac{5}{2} = 2.5\) ∴ The road surrounding the rectangular playground is uniformly 2.5 meters wide.