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How is each new right triangle constructed in a square root spiral to obtain successive lengths \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on?

Answer and explanation

Correct answer: By drawing a 1-unit perpendicular side at an end of the previous hypotenuse

With previous hypotenuse \(\sqrt{n}\) and a new perpendicular side of 1 unit, Pythagoras gives \(\sqrt{n+1}\). A parallel side does not form the required right triangle. Exam tip: identify the fixed 1-unit leg.

Related tags

Number SystemsSquare Root SpiralPythagoras TheoremRight TrianglesIrrational NumbersGeometryMathematicsClass 9 Mcq

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