If the perimeter of a right-angled isosceles | vedclass

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(C) Let the two equal sides of the right-angled isosceles triangle be $x \, m$.By the Pythagorean theorem,the hypotenuse $h$ is given by $h = \sqrt{x^

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$4$

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(C) Let the two equal sides of the right-angled isosceles triangle be $x \, m$.By the Pythagorean theorem,the hypotenuse $h$ is given by $h = \sqrt{x^2 + x^2} = x\sqrt{2}$.The perimeter $P$ is the sum of all sides: $P = x + x + x\sqrt{2} = 2x + x\sqrt{2} = x(2 + \sqrt{2})$.We can rewrite this as $P = x\sqrt{2}(\sqrt{2} + 1) = h(\sqrt{2} + 1)$.Given the perimeter $P = 4\sqrt{2} + 4 = 4(\sqrt{2} + 1) \, m$.Equating the two expressions: $h(\sqrt{2} + 1) = 4(\sqrt{2} + 1)$.Therefore,the hypotenuse $h = 4 \, m$.

If the perimeter of a right-angled isosceles triangle is $4 \sqrt{2}+4 \,m ,$ then the hypotenuse is.........$m$

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