Which is the correct calculation to make (\sqrt{3}) in a square root spiral?

Correct answer: (\sqrt{(\sqrt{2})^2+1^2})

The direct answer is option A: \(\sqrt{(\sqrt{2})^2+1^2}\). In a square-root spiral, begin with a unit right triangle, whose hypotenuse is \(\sqrt{2}\). To make the next triangle, use this existing hypotenuse as one perpendicular side and add a new perpendicular side of length 1. Pythagoras gives the new hypotenuse as \(\sqrt{(\sqrt{2})^2+1^2}=\sqrt{2+1}=\sqrt{3}\). Option A works because it uses the previous length and one new unit side. Option B, \(\sqrt{2+2}=2\), adds two squared lengths of 2 and does not represent the next construction. Option C, \(\sqrt{3+1}=2\), starts with the number being sought and is circular. Option D, \(\sqrt{1^2+3^2}=\sqrt{10}\), uses the wrong side length and gives another value. Memory cue: every new spiral step adds \(1^2\) to the previous hypotenuse squared.