The perimeter of an isosceles right-angled triangle is 2 meters. Find the length of its longer side.

2 - $\sqrt2$

-4 + 3 $\sqrt2$

It cannot be determined

-2 + 2 $\sqrt2$ m

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Correct answer is D

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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The perimeter of an isosceles right-angled triangle is 2 meters. Find the length of its longer side.

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