Square root spiral

The governing concept is the Pythagorean theorem applied to the successive right triangles of a square root spiral. The old hypotenuse has length √80, and the newly drawn perpendicular has length 1. Thus, if h is the new hypotenuse, h² = (√80)² + 1² = 80 + 1 = 81. Since a length is positive, h = √81 = 9. Therefore option A is correct. Option B incorrectly subtracts 1 from the radicand. Option C doubles 80 without any geometric justification, while option D adds 2 rather than adding the square of the new perpendicular. The next point consequently represents the exact length 9.

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.

The direct answer is B. In the usual square-root spiral, each new squared hypotenuse increases by 1. Therefore, immediately before sqrt{226} comes sqrt{225}. To locate the value, compare 226 with neighbouring perfect squares: 15^2=225 and 16^2=256, so 15^2<226<16^2 and hence 15<sqrt{226}<16. Option A is wrong because sqrt{224} is not the immediately preceding term and its interval is also not the stated one. Option B correctly gives both the previous hypotenuse and the interval. Option C repeats the new hypotenuse instead of the previous one. Option D gives a later value and an incorrect interval. Exam cue: compare the radicand with consecutive squares.

The relevant rule is that adding a perpendicular of length 1 at a right angle increases the square of the current hypotenuse by 1. Starting with a hypotenuse √168, the next one is therefore √((√168)² + 1²) = √(168 + 1) = √169. Since 169 = 13 × 13 = 13², the radical simplifies exactly to the whole number 13. Hence option A is correct. Option B results from subtracting 1, which is not the rule of the construction. Option C doubles the radicand without using the Pythagorean theorem, and option D wrongly assumes that the added perpendicular causes no change. The special simplification occurs because 169 is a perfect square.

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 governing construction principle is that a unit perpendicular is added at a right angle, changing a hypotenuse √m into √(m+1). The condition says that m is exactly one less than a perfect square. Thus there is some whole number r such that m+1 = r². The next hypotenuse is therefore √(m+1) = √(r²) = r, which is a whole number. Option B is correct. It is not always irrational, as option A claims, because the condition deliberately makes the new radicand a perfect square. Option C has no relation to the triangle, and option D ignores the increase caused by the new perpendicular.

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.

The hypotenuses increase in order in the spiral. Just before \(\sqrt{10}\) comes \(\sqrt{9}\), and before that \(\sqrt{8}\).

Actually (13^2=169) and (14^2=196). Therefore (\sqrt{195}) lies between (13) and (14), less than (14).

In Pythagoras theorem, the square of the hypotenuse equals the sum of squares of the sides. This is the general rule.

Since \(196=14^2\), \(\sqrt{196}=14\), which is rational. The numbers 198, 200 and 202 are not perfect squares, so their square roots are irrational. Exam tip: check for a perfect square first.

\(\sqrt{n}\) is rational only when \(n\) is a perfect square. For example, \(\sqrt{49}=7\), so that spiral segment has rational length. Being prime or odd is not sufficient. Exam tip: check for perfect squares first.

Let the previous hypotenuse be \(\sqrt{n}\). With a new perpendicular side of 1, Pythagoras gives \(\sqrt{(\sqrt{n})^2+1^2}=\sqrt{n+1}\). It is not obtained by simply adding 1 to the length. Exam tip: add the squares of perpendicular sides.