Let the length of the park be \(x\) meters. ∴ Breadth = \(\cfrac{600}{x}\) meters ∴ Perimeter of the park = \(2\left(x + \cfrac{600}{x}\right)\) meters According to the question, \(2\left(x + \cfrac{600}{x}\right) = 100\) Or, \(\cfrac{x^2 + 600}{x} = 50\) Or, \(x^2 + 600 = 50x\) Or, \(x^2 - 50x + 600 = 0\) Or, \(x^2 - (30 + 20)x + 600 = 0\) Or, \(x^2 - 30x - 20x + 600 = 0\) Or, \(x(x - 30) - 20(x - 30) = 0\) Or, \((x - 30)(x - 20) = 0\) ∴ Either \(x - 30 = 0\) ⇒ \(x = 30\) Or, \(x - 20 = 0\) ⇒ \(x = 20\) Since length is greater than breadth, Therefore, the length of the park is 30 meters and the breadth is \(\cfrac{600}{30} = 20\) meters.