If the perimeter of an isosceles right triangle is 4(2+\sqrt{2}) cm, then what is its area in square cm?

Explanation

Let the legs be a, so the hypotenuse is a\sqrt{2}. Perimeter = 2a + a\sqrt{2} = a(2+\sqrt{2}). Equating this to 4(2+\sqrt{2}), we find a = 4 cm. The area is \frac{1}{2} \times a \times a = \frac{1}{2} \times 16 = 8 square cm.

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