A rectangular field is 40 m long and 30 m wide. A uniform path is constructed inside the field along its boundary. The area of the path is 264 m². What is the width of the path?

Let width of path = x m Outer rectangle (field): 40 m × 30 m Area = 40 × 30 = 1200 m² Inner rectangle (after path on all sides): Length = 40 − 2x Breadth = 30 − 2x Area of inner rectangle = (40 − 2x)(30 − 2x) Area of path = outer area − inner area ⇒ 1200 − (40 − 2x)(30 − 2x) = 264a (40 − 2x)(30 − 2x) = 1200 − 264 = 936 Expand: (40 − 2x)(30 − 2x) = 40 × 30 − 40 × 2x − 30 × 2x + 4x² = 1200 − 80x − 60x + 4x² = 1200 − 140x + 4x² So: 1200 − 140x + 4x² = 936 4x² − 140x + 1200 − 936 = 0 4x² − 140x + 264 = 0 Divide by 4: x² − 35x + 66 = 0 Solve the quadratic: x² − 35x + 66 = 0 Discriminant D = 35² − 4 × 66 = 1225 − 264 = 961 = 31² x = [35 ± 31] / 2 x₁ = (35 + 31) / 2 = 66 / 2 = 33 x₂ = (35 − 31) / 2 = 4 / 2 = 2 Width must be less than half of the smaller side (15 m), so x = 2 m is valid.