To construct (\sqrt{80}) in a square root spiral, which previous hypotenuse will be used?

Correct answer: (\sqrt{79})

The direct answer is B, (\sqrt{79}). In each step of the square root spiral, a 1-unit segment is drawn perpendicular to the previous hypotenuse. If that previous hypotenuse is (\sqrt{n}), then the new hypotenuse is (\sqrt{n+1}) because (\sqrt{n})^2+1^2=n+1. For the target (\sqrt{80}), solve (n+1=80); hence (n=79) and the previous hypotenuse is (\sqrt{79}). Option A, (\sqrt{78}), would give (\sqrt{79}) next, so it is one step short. Option B is correct because adding the unit perpendicular gives exactly (\sqrt{80}). Option C, (\sqrt{80}), is the required new hypotenuse, not the previous one. Option D, (\sqrt{81}), would produce (\sqrt{82}), one step beyond the target. The safe exam method is to subtract 1 from the number inside the target square root: (80-1=79).