To make (\sqrt{12}) in a square root spiral, what should be the previous hypotenuse?

Correct answer: (\sqrt{11})

The direct answer is B, (\sqrt{11}). In the square root spiral, suppose the previous hypotenuse is (\sqrt{n}) and a new perpendicular side of 1 unit is added. Pythagoras gives the new hypotenuse as (\sqrt{(\sqrt{n})^2+1^2}=\sqrt{n+1}). To obtain (\sqrt{12}), we need (n+1=12), so (n=11). Therefore the previous hypotenuse is (\sqrt{11}). Option A, (\sqrt{10}), would produce (\sqrt{11}) after adding the unit perpendicular, not (\sqrt{12}). Option B is correct because it produces the required next value. Option C, (\sqrt{12}), is already the target hypotenuse, not the previous one. Option D, (\sqrt{13}), would lead to (\sqrt{14}), so it is one step too large. The important rule is: to make (\sqrt{k}), use (\sqrt{k-1}) as the previous hypotenuse and add a 1-unit perpendicular. Memory cue: previous square-root number is always one less.