JAMB Mathematics 2023 Paper

Perimeter of a triangle = sum of all sides

⇒ $P = y + x + x = 2$

⇒ $y + 2x = 2$

⇒ $y = 2 - 2x$-----(i)

Using Pythagoras theorem

$y^2 = x^2 + x^2$

⇒ $y^2 = 2x^2$

⇒ $y = \sqrt{2} x^2$

⇒ $y = x\sqrt{2}$-----(ii)

Equate $y$

⇒ $2 - 2x = x\sqrt{2}$

Square both sides

⇒ $(2 - 2x)^2 = (x\sqrt{2})^2$

⇒ $4 - 8x + 4x^2 = 2x^2$

⇒ $4 - 8x + 4x^2 - 2x^2 = 0$

⇒ $2x^2 - 8x + 4 = 0$

⇒ $x = \frac{-(-8)\pm\sqrt{(-8)^2 - 4\times2\times4}}{2\times2}$

⇒ $x = \frac{8\pm\sqrt[3]{2}}{4}$

⇒ $x = \frac{8\pm4\sqrt{2}}{4}$

⇒ $x = 2\pm\sqrt{2}$

⇒ $x = 2 + \sqrt{2}$ or $2 - \sqrt{2}$

∴ $x = 2 - \sqrt{2}$ (for $x$ has to be less than its perimeter)

∴ $y = 2 - 2x = 2 - 2(2 - \sqrt{2}) = -2 + 2\sqrt{2}$

∴ The length of the longer side = -2 + 2$\sqrt{2}$m