Square root spiral

The new hypotenuse is (\sqrt{80+1}=\sqrt{81}). Since (\sqrt{81}=9), write the exact value at a perfect square.

By Pythagoras’ theorem, the square of the new hypotenuse is \((\sqrt{17})^2+1^2=17+1=18\), so it is \(\sqrt{18}\), not \(\sqrt{19}\). To construct \(\sqrt{19}\), start with \(\sqrt{18}\). Exam tip: add 1 to the radicand at each step.

Lengths are not added directly in a square root spiral. The correct method is to add squares using Pythagoras theorem.

The new hypotenuse is (\sqrt{144}). Since (\sqrt{144}=12), it is not in an interval but exactly at (12).

In a square root spiral, each new right triangle has a perpendicular side of length 1. The new hypotenuse is obtained from the previous hypotenuse by applying the Pythagorean theorem. Thus, to obtain the length \\(\\sqrt{35}\\), we must begin with a previous hypotenuse whose square is one less than 35. This makes option A the appropriate choice.

Using the theorem, the required calculation is \\( (\\sqrt{34})^2+1^2=34+1=35 \\). Therefore, the new hypotenuse is \\(\\sqrt{35}\\). A previous length of \\(\\sqrt{33}\\) with a side of 2 would also give 37, not 35, so option B does not fit this construction. The other choices use the target or an excessive previous value.

Each new right triangle has the previous hypotenuse \(\sqrt{n-1}\) as one side and a perpendicular side of length 1. By Pythagoras, its hypotenuse is \(\sqrt{(n-1)+1}=\sqrt n\). Exam tip: track successive hypotenuses.

The previous hypotenuse \(\sqrt{n}\) and the new unit side form a right angle. By Pythagoras, the next hypotenuse is \(\sqrt{n+1}\), not \(\sqrt{n^2+1}\). Exam tip: first locate the right-angle mark.

Before (\sqrt{226}), the previous hypotenuse was (\sqrt{225}). Since (15^2<226<16^2), the new value lies between (15) and (16).

The governing concept is the Pythagorean theorem. The two perpendicular sides are √3 and 1, so their squared lengths give (√3)² + 1² = 3 + 1 = 4, and the hypotenuse is √4. However, √3 + 1 is a direct sum of side lengths and is not equal to √4; numerically it is about 2.732, whereas √4 is 2. Hence option C contains the logical error.

After \(\sqrt{168}\), \(\sqrt{169}\) is formed. Since \(169=13^2\), the hypotenuse value is (13).

Since \(15^2=225\) and \(16^2=256\), and \(225<242<256\), we get \(15<\sqrt{242}<16\). Also, \(\sqrt{256}=16\) because \(256=16^2\). Therefore, option B is correct. Option D is incorrect because \(\sqrt{242}\) is less than 16. Exam tip: To locate a square root, compare the number with the nearest perfect squares on either side.

In a square root spiral, a point’s distance from the origin is its hypotenuse label \(\sqrt{n}\). This is an integer only when \(n\) is a perfect square; for example, \(\sqrt{9}=3\). Option B fails because \(\sqrt{12}\) is not an integer. Exam tip: look for perfect-square radicands.

\(11^2=121\) and \(12^2=144\). Since \(121<125<144\), the correct inequality is \(11^2<125<12^2\), and \(\sqrt{125}\) lies between 11 and 12. \(10^2<125<11^2\) is incorrect because \(11^2=121\), which is less than 125. Exam tip: To locate a square root, compare the number with the nearest perfect squares.

Since \(49=7^2\), \(\sqrt{49}=7\), so its point is 7 units from the origin. It is not irrational. Exam tip: square roots of perfect squares are integers.

By Pythagoras theorem, the squares of sides are added. Therefore (\sqrt{7}) and (1) form hypotenuse (\sqrt{8}).

After (\sqrt{63}), (\sqrt{64}=8) is formed. Since (8^2<65<9^2), (\sqrt{65}) lies between (8) and (9).

The next hypotenuse is (\sqrt{m+1}). If (m+1) is a perfect square, its square root is a whole number.

Because (16^2=256) and (17^2=289). The number (288) lies between them, so (\sqrt{288}) lies between (16) and (17).

In the usual square root spiral, a (1) unit perpendicular is added every time. For (\sqrt{48}), the previous hypotenuse is (\sqrt{47}).

Since \(25<26<36\), we get \(5<\sqrt{26}<6\). A nearby perfect square does not make the root equal to 5. In the spiral, each new hypotenuse represents its corresponding root. Exam tip: bracket roots using consecutive perfect squares.