Correct Answer: Option D

Explanation

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

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