Square root spiral

Each step of the square root spiral adds 1 to the squared hypotenuse, so a target √224 must be constructed from the previous hypotenuse √223 with a new perpendicular of 1 unit: (√223)² + 1² = 224. To locate the result, compare 224 with consecutive perfect squares. Since 14² = 196 and 15² = 225, we have 196 < 224 < 225. Taking positive square roots gives 14 < √224 < 15. Therefore option B is correct. Option A has the wrong predecessor, option C reverses the correct interval, and option D would apply to √225, not √224.

Since \(4^2=16\) and \(5^2=25\), \(\sqrt{18}\) must lie between 4 and 5. Its value is about 4.24, while \(4.5^2=20.25\). Exam tip: compare the nearest perfect squares first.

The same length to be marked is taken in the compass. Drawing an arc from the origin gives the correct position.

Each new right triangle in the square root spiral has a perpendicular side of 1 unit. \(\sqrt{9^2+1^2}=\sqrt{10}\), so a 10-unit base is not needed. Exam tip: apply Pythagoras' theorem.

The direct answer is A. In a square-root spiral, the old hypotenuse becomes one side of a new right triangle, and a perpendicular side of length 1 is added. Here the old length is sqrt{4}. By Pythagoras, the square of the new hypotenuse is (sqrt{4})^2+1^2=4+1=5, so the new hypotenuse is sqrt{5}. Option A is correct because it uses squares of the perpendicular sides. Option B is wrong: lengths are not added directly, since sqrt{4}+1=3, not sqrt{5}. Option C is wrong because 5 is not equal to sqrt{5}. Option D is wrong because the Pythagorean theorem uses addition for a right triangle, not subtraction. Remember: square, add, then take the square root.

\(20^2=400\) and \(21^2=441\). Since \(400<440<441\), taking square roots gives \(20<\sqrt{440}<21\). Although it is close to \(21\), \(\sqrt{440}\) is less than \(21\) because \(440<441\). Exam tip: To locate a square root, compare the number with consecutive perfect squares.

The notation in this question directly defines the kth hypotenuse as \\(\\sqrt{k}\\). Therefore, the number under the radical identifies the position of the hypotenuse, while evaluating the radical gives its numerical length. For \\(\\sqrt{25}\\), the index is 25 and the length is 5. Hence option B contains both required parts correctly.

The calculation is \\(\\sqrt{25}=5\\), because 25 is the square of 5. The phrase “25th hypotenuse” refers to the position in the spiral, not to a length of 25. Thus option D confuses the index with the value. The 24th and 26th positions would represent \\(\\sqrt{24}\\) and \\(\\sqrt{26}\\), respectively, so options A and C do not describe the stated segment.

Apply Pythagoras to the new hypotenuse: \(h^2=14+1=15\), so \(h=\sqrt{15}\). Directly adding 1 to \(\sqrt{14}\) is invalid. Exam tip: add squares, not lengths.

The correct basis of a square root spiral is a right triangle and Pythagoras theorem. Directly adding lengths is wrong.

At each step, a unit segment is added perpendicular to the previous hypotenuse, creating a right triangle. By Pythagoras,

g(

gsqrt{n})^2+1^2=n+1, so the next hypotenuse is

gsqrt{n+1}. Exam tip: look for “perpendicular” and “unit length.”

Because (24^2=576) and (25^2=625). The number (624) lies between them, so (\sqrt{624}) lies between (24) and (25).

The square root of a positive integer is a whole number only when it is a perfect square. If it is not a perfect square, the square root is irrational.

Pythagoras theorem uses the sum of squares. Therefore (\sqrt{20}) and (1) form hypotenuse (\sqrt{21}).

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).

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.