Square root spiral

Each new right triangle uses the previous hypotenuse as one leg and a unit length as the other, producing \(\sqrt{2}, \sqrt{3}\), and so on. Exam tip: identify the newly added 1-unit leg first.

Because (13^2=169). The number (168) is slightly less, so it is between (12) and (13), while (170) is between (13) and (14).

In a square root spiral, the newly added perpendicular side is always 1 unit. After
√7
, the hypotenuse is
√(7+1)=√8
. Using
√2
would produce
√9
instead. Exam tip: apply the fixed 1-unit side rule at every step.

Because (28^2=784) and (29^2=841). The number (840) lies between them, so (\sqrt{840}) lies between (28) and (29).

After (\sqrt{24}), (\sqrt{25}=5) is formed. Also (4^2<24<5^2), so (\sqrt{24}) lies between (4) and (5).

Nearest perfect squares give the correct interval. Then mark the spiral length using a compass.

In a square root spiral, the distance OP is \(\sqrt{18}\). Since \(4^2=16\) and \(5^2=25\), \(4<\sqrt{18}<5\). Option B is wrong because \(3^2=9\). Exam tip: compare with nearby perfect squares.

Every radial segment in a square root spiral has length \(\sqrt{n}\), where \(n\) is a positive integer. \(\sqrt{2}\), \(\sqrt{12}\), and \(\sqrt{81}\) fit this form, but \(\frac{3}{2}\) does not. Exam tip: test whether the length can be written as \(\sqrt{n}\).

Because (30^2=900) and (31^2=961). The number (960) lies between them, so (\sqrt{960}) lies between (30) and (31).

A square root spiral is based on Pythagoras theorem. The previous hypotenuse and (1) unit perpendicular form the next square root.

Since \(7^2=49\) and \(8^2=64\), \(49<50<64\) gives \(7<\sqrt{50}<8\). Hence A is correct; \(\sqrt{50}\) is not exactly 7. In exams, compare the nearest perfect squares.

In a square root spiral, the successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}, \ldots\). The given new hypotenuse is \(14\), and \(14=\sqrt{196}\). Therefore, the hypotenuse immediately before it is \(\sqrt{195}\). \(\sqrt{196}\) is the current hypotenuse, not the preceding one. Exam tip: Express the whole number as \(\sqrt{n}\), then subtract 1 from the radicand to find the previous hypotenuse.

In each new right triangle, the previous hypotenuse is one leg and 1 is the other. Thus, new hypotenuse² = 12 + 1 = 13, so its length is \(\sqrt{13}\). Adding \(\sqrt{12}+1\) is incorrect. Exam tip: apply Pythagoras’ theorem.

In a square root spiral, the hypotenuse is formed by the sum of squares. Do not treat (\sqrt{18}+1) as (\sqrt{19}).

The existing hypotenuse is \(\sqrt{n}\), and the new perpendicular side is \(3\) units. By the Pythagorean theorem, the square of the new hypotenuse is \((\sqrt{n})^2+3^2=n+9\). Therefore, the new hypotenuse is \(\sqrt{n+9}\). \(\sqrt{n+3}\) is incorrect because the perpendicular contributes \(3^2=9\), not \(3\). Exam tip: while finding a hypotenuse, add the squares of the two perpendicular sides.

In a square root spiral, the next 1-unit side is drawn perpendicular to the previous hypotenuse, making a right triangle. Hence the new hypotenuse satisfies \(h^2=a^2+1\). Exam tip: always check the perpendicular condition.

The \(\sqrt{n}\) segment and a perpendicular unit segment are the legs. Pythagoras gives hypotenuse squared \(=n+1\), hence \(\sqrt{n+1}\). Exam tip: verify the right angle.

The next hypotenuse is \(\sqrt{144}\), and \(\sqrt{144}=12\). A perfect square gives an exact value.

The stated rule says that the kth hypotenuse has length \\(\\sqrt{k}\\). To identify the hypotenuse represented by \\(\\sqrt{36}\\), read 36 as the index and then evaluate the radical to find its length. Thus it is the 36th hypotenuse, and its numerical length is 6. Option B is therefore correct.

Since \\(36=6^2\\), we have \\(\\sqrt{36}=6\\). The number 36 must be retained as the position, while 6 is the measured value of the hypotenuse. Option D incorrectly gives 36 as the length. The 35th and 37th hypotenuses would instead have lengths \\(\\sqrt{35}\\) and \\(\\sqrt{37}\\), so neither option A nor C matches the required radical.

Only the newly added perpendicular has length 1; the other leg is the previous hypotenuse. For example, after a hypotenuse of \(\sqrt{2}\), the next one is \(\sqrt{3}\), so the triangles are not congruent. Exam tip: track the changing hypotenuse.