Perimeter of a triangle = sum of all sides ⇒ \(P = y + x + x = 2\) ⇒ \(y + 2x = 2\) ⇒ \(y= 2 - 2x\)-----(i) Using Pythagoras theorem \(y^2 = x^2 + x^2\) ⇒ \(y^2 = 2x^2\) ⇒ \(y = \sqrt2x^2\) ⇒ \(y = x\sqrt2\)-----(ii) Equate \(y\) ⇒ \(2 - 2x = x\sqrt2\) Square both sides ⇒ \((2 -2x) ^2 = (x\sqrt2)^2\) ⇒ \(4 - 8x + 4x^2 = 2x^2\) ⇒ \(4 - 8x + 4x^2 - 2x^2 = 0\) ⇒ \(2x^2 - 8x + 4 = 0\) ⇒ \(x = \frac{-(-8)\pm\sqrt(-8)^2 - 4\times2\times4}{2\times2}\) ⇒ \(x = \frac{8\pm\sqrt32}{4}\) ⇒ \(x = \frac{8\pm4\sqrt2}{4}\) ⇒ \(x = 2\pm\sqrt2\) ⇒ \(x = 2 + \sqrt2\) or \(2 - \sqrt2\) ∴ \(x = 2 - \sqrt2\) (for \(x\) has to be less than its perimeter) ∴ \(y = 2 - 2x = 2 - 2(2 - \sqrt2) = -2 + 2 \sqrt2\) ∴ The length of the longer side = -2 + 2\(\sqrt2\)m