The square constructed on the diagonal of a rectangle has an area that is 108 1/3% greater than the area of the rectangle itself. If the rectangle's perimeter measures 28 units, what is the difference between its length and width?

Explanation

Let the rectangle's length and width be l and b. The diagonal length is √(l² + b²). The problem states that the area of the square on the diagonal is 108 1/3% (which is 13/12 in fraction) more than the rectangle's area, so: l² + b² = (1 + 13/12) × lb = (25/12) × lb Rearranging: 12(l² + b²) = 25 lb Adding 24 lb to both sides: 12l² + 12b² + 24 lb = 49 lb Notice that 12(l² + b² + 2lb) = 49 lb Since (l + b)² = l² + b² + 2lb, we get: 12 (l + b)² = 49 lb Given the perimeter 2(l + b) = 28, so l + b = 14: 12 × 14² = 49 lb Calculating: 12 × 196 = 49 lb 2352 = 49 lb Therefore, lb = 48 With l + b = 14 and lb = 48, solving the quadratic gives l = 8 and b = 6. The difference between the sides is 8 - 6 = 2 units.