Square root spiral

The governing concept is the Pythagorean theorem. The existing hypotenuse has length √n, and the modified perpendicular has length 6. If H denotes the new hypotenuse, then H² = (√n)² + 6² = n + 36. Taking the positive square root, because H represents a length, gives H = √(n+36). Thus option C is correct. The familiar expression √(n+1) applies only when the added perpendicular is 1 unit. Option B adds the length 6 rather than its square, and option D multiplies n by 6, which is not how the sides of a right triangle determine its hypotenuse. The essential operation is squaring the perpendicular.

In a square root spiral, constructing a 1-unit perpendicular on a hypotenuse of length \(\sqrt{n}\) produces a new hypotenuse of length \(\sqrt{n+1}\). Therefore, after \(\sqrt{624}\), the new hypotenuse is \(\sqrt{625}\). Since \(625=25^2\), \(\sqrt{625}=25\), so it lies exactly at \(25\). \(\sqrt{626}\) would be the next value and would lie between \(25\) and \(26\). Exam tip: check nearby perfect squares to locate a square root quickly.

Since \(10^2=100<120<121=11^2\), we get \(10<\sqrt{120}<11\). Similarly, \(11^2=121<122<144=12^2\), so \(11<\sqrt{122}<12\). Therefore, option B is correct. Option A places \(\sqrt{122}\) in the wrong interval because \(122>121\). Exam tip: Compare the number with nearby perfect squares to locate its square root.

In a square root spiral, the hypotenuse following \(\sqrt{675}\) is \(\sqrt{676}\). Since \(676=26^2\), \(\sqrt{676}=26\). The values 25 and 27 correspond to \(\sqrt{625}\) and \(\sqrt{729}\), respectively, so they are not correct here. Exam tip: Add 1 to the number inside the radical, then check whether it is a perfect square.

Under the rule given in the question, the kth hypotenuse has length \\(\\sqrt{k}\\). Consequently, the expression \\(\\sqrt{81}\\) represents the hypotenuse at position 81. Its actual numerical length is found by simplifying the square root. Since 81 is a perfect square, the length is 9. These two facts together make option B correct.

The calculation is \\(\\sqrt{81}=9\\), because \\(9\\times9=81\\). The position number remains 81; it is not replaced by the value 9. Option D therefore confuses the location of the hypotenuse with its length. Positions 80 and 82 would correspond to \\(\\sqrt{80}\\) and \\(\\sqrt{82}\\), not to the stated segment, so A and C are also unsuitable.

In the standard square root spiral, each new right triangle is formed by adding a perpendicular of length 1 unit to the previous hypotenuse. If the previous hypotenuse is √n and the new one is h, then h²=(√n)²+1²=n+1. To obtain √210, we require n+1=210, so n=209. The correct pair is therefore a previous hypotenuse of √209 and a new perpendicular of 1 unit, making option B correct. Option A gives 208+2²=212, not 210. Option C incorrectly uses the desired final hypotenuse as the previous one, and option D would produce √212. Only option B satisfies both the unit-perpendicular rule and the required final square.

In a square root spiral, each successive hypotenuse represents the square root of the next natural number. Hence, the hypotenuse after \(\sqrt{1088}\) is \(\sqrt{1089}\). Since \(1089=33^2\), \(\sqrt{1089}=33\). Option B has the correct radicand but an incorrect value, because \(32^2=1024\). Exam tip: Compare with nearby perfect squares to check whether a square root is an integer.

\(29^2=841\) and \(30^2=900\). Since \(841<899<900\), the correct inequality is \(29^2<899<30^2\), so \(\sqrt{899}\) lies between 29 and 30. Option A is incorrect because \(899>29^2\). Exam tip: Compare the number with the nearest perfect squares to locate its square root.

In a square root spiral, the hypotenuse after \(\sqrt{n}\) is \(\sqrt{n+1}\). Since this hypotenuse is \(41\), \(\sqrt{n+1}=41\). Squaring both sides gives \(n+1=41^2=1681\), so \(n=1680\). Option 1681 is the value of \(n+1\), not of \(n\). Exam tip: first express the next hypotenuse as \(\sqrt{n+1}\), then square to find \(n\).

The direct answer is B. In the usual square-root spiral, every new right triangle adds one unit to the squared hypotenuse. Starting at sqrt{8}, the next successive hypotenuses are sqrt{9}, sqrt{10}, sqrt{11} and sqrt{12}. Therefore option B gives the complete order. Option A skips sqrt{9} and sqrt{11}, so it is not the usual step-by-step construction. Option B is correct because no intermediate stage is omitted. Option C jumps to sqrt{12} and then goes backward, which is not the increasing construction order. Option D starts at the end and decreases, so it reverses the required direction and also omits a stage. Exam cue: the radicands rise one at a time.

Since \(15^2=225<255<256=16^2\), we get \(15<\sqrt{255}<16\). Similarly, \(16^2=256<257<289=17^2\), so \(16<\sqrt{257}<17\). Therefore, option A is correct. Options B and C are incorrect because the two square roots lie on opposite sides of \(16\). Exam tip: compare a number with consecutive perfect squares to locate its square root quickly.

After (\sqrt{323}), (\sqrt{324}=18) is formed. Since (18^2<325<19^2), (\sqrt{325}) lies between (18) and (19).

The correct answer is A. If m is exactly one less than the next perfect square, there is an integer k such that m + 1 = k². The square root spiral adds a unit perpendicular, so the next hypotenuse is √(m + 1) = √(k²) = k, a whole number. For example, if m = 80, then the next perfect square is 81 and the next hypotenuse is √81 = 9. It is not necessarily irrational, because the radicand is a perfect square; it is not zero unless the next square is zero; and it changes from √m to √(m + 1). Therefore option A follows directly from the construction rule.

Since \(40^2=1600\) and \(41^2=1681\), we have \(1600<1680<1681\). Hence, \(40<\sqrt{1680}<41\), whereas \(\sqrt{1681}=\sqrt{41^2}=41\). Option B is incorrect because 1680 is smaller than \(41^2\), so its square root cannot be 41. Exam tip: Find the two consecutive perfect squares around a number to locate its square root quickly.

The correct answer is B. In the standard square root spiral, the next right triangle is made by adding a perpendicular of length 1 to the existing hypotenuse √80. Therefore the new hypotenuse is √[(√80)² + 1²] = √(80 + 1) = √81. Because 81 is the perfect square 9², √81 = 9. Option A moves backward to 79, option C incorrectly doubles the radicand, and option D adds 2 rather than the square of the unit perpendicular. The important rule is that each unit step changes the radicand from n to n + 1, and a perfect-square radicand gives an exact whole-number length.

\(38^2=1444\) and \(39^2=1521\). Since \(1444<1520<1521\), taking positive square roots gives \(38<\sqrt{1520}<39\). Option C is incorrect because \(39^2=1521\), which is greater than 1520. Exam tip: To locate a square root, compare the number with consecutive perfect squares around it.

In a square root spiral, the hypotenuse following \(\sqrt{n}\) is \(\sqrt{n+1}\). Therefore, the hypotenuse after \(\sqrt{2024}\) is \(\sqrt{2025}\). Since \(2025=45^2\), \(\sqrt{2025}=45\). Option 44 is not correct because \(44^2=1936\), not 2025. Exam tip: Before simplifying a square root, check whether the number inside it is a perfect square.

Since \(7^2=49\) and \(8^2=64\), we have \(49<50<64\); hence \(7<\sqrt{50}<8\). Similarly, \(80\) lies between \(8^2=64\) and \(9^2=81\), so \(8<\sqrt{80}<9\). Option B may seem close, but \(\sqrt{80}>8\). Exam tip: locate a square root by comparing the number with nearby perfect squares.

A square-root spiral repeatedly uses the Pythagorean theorem. If the existing radius or hypotenuse has length sqrt{41} and a perpendicular segment of length 1 is added, the new hypotenuse has square equal to the sum of the two squared lengths. Therefore, (sqrt{41})^2+1^2=41+1=42, so the new length is sqrt{42}. The essential point is that lengths are combined through their squares, not by simply adding the lengths.

Thus option A gives the correct construction and equation. Option B is a common error because sqrt{41}+1 is not generally equal to sqrt{42}. Option C uses a perpendicular of length 2, which would give 45 under the square-root sign, and option D uses multiplication rather than the Pythagorean relation. Hence A follows directly from the theorem.

\(43^2=1849\) and \(44^2=1936\). Since \(1849<1935<1936\), taking square roots gives \(43<\sqrt{1935}<44\). Although it is very close to \(44\), \(1935<1936\), so its square root must be less than \(44\). Exam tip: To find the interval of a square root, compare the number with consecutive perfect squares.

In a square root spiral, each new right triangle has the other leg equal to \(1\). Therefore, starting from the hypotenuse \(\sqrt{288}\), the square of the next hypotenuse is \(288+1=289\). Hence, the new hypotenuse is \(\sqrt{289}=17\). The option with \(\sqrt{290}\) is incorrect because only \(1\) is added to the radicand. Exam tip: for the next hypotenuse, add \(1\) to the radicand and then check whether the result is a perfect square.

We have \(49^2=2401\) and \(50^2=2500\). Since \(2401<2499<2500\), taking square roots gives \(49<\sqrt{2499}<50\). Therefore, its position on the square root spiral is between 49 and 50. Although it is very close to 50, it cannot equal 50 because \(2499<2500\). Exam tip: locate a square root by comparing the number with consecutive perfect squares.

Since \(19^2=361\), \(20^2=400\), and \(21^2=441\), \(361<399<400\) gives \(19<\sqrt{399}<20\). Similarly, \(400<401<441\) gives \(20<\sqrt{401}<21\). Options B and C are incorrect because the numbers lie on opposite sides of 400, so their square roots lie on opposite sides of 20. Exam tip: Compare a number with the nearest perfect squares to locate its square root.

The previous hypotenuse is \(\sqrt{728}\), so \(n=728\). In a square root spiral, the next hypotenuse is \(\sqrt{n+1}\). Therefore, it is \(\sqrt{728+1}=\sqrt{729}\). \(\sqrt{728}\) is the previous hypotenuse, while \(\sqrt{727}\) represents an earlier step. Exam tip: add 1 to the number inside the square root to obtain the next hypotenuse.

In a square root spiral, each new hypotenuse represents \(\sqrt{n}\). Here the new hypotenuse is \(\sqrt{729}\), and \(729=27^2\), so its value is \(27\). Option \(26\) is incorrect because \(26^2=676\), while \(28^2=784\). Exam tip: Remember squares of nearby whole numbers to identify perfect squares quickly.