The triangle in Problem 1 is a 45°-45°-90° triangle. In such triangles, the legs are congruent, and the hypotenuse is 2 \sqrt{2} 2 times the length of each leg.
Given that one leg is x x x, the other leg is also x x x, and the hypotenuse is x2 x\sqrt{2} x2.
Since the hypotenuse is given as x2 x\sqrt{2} x2, we have: y=x2 y = x\sqrt{2} y=x2
For Problem 1: x=x x = x x=x y=x2 y = x\sqrt{2} y=x2
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