Square root spiral

The governing concept is the Pythagorean theorem, used in the square-root spiral. If one leg of a right triangle is \(\sqrt{n}\) and the perpendicular leg is 7 units, then the square of the hypotenuse equals the sum of the squares of the two perpendicular legs: \(h^2=(\sqrt n)^2+7^2=n+49\). Taking the positive square root, because a length is positive, gives \(h=\sqrt{n+49}\). Thus option A is correct. Option B adds 7 instead of its square, C multiplies the terms incorrectly, and D uses twice 7 rather than \(7^2\).

The square root spiral uses the Pythagorean theorem to create the next length. When a perpendicular of 1 unit is drawn on a hypotenuse of length \(\sqrt{1224}\), the old hypotenuse and the new perpendicular become the two perpendicular sides of a right triangle. Their squared lengths must be added.

The new hypotenuse is \(\sqrt{(\sqrt{1224})^2+1^2}=\sqrt{1224+1}=\sqrt{1225}\). Since \(35^2=1225\), this length is exactly 35, not merely a value between 34 and 35. Thus option D is correct. Options A, B and C use the wrong arithmetic or give a nonmatching interval; in particular, \(\sqrt{1223}\) is less than 35.

Since \(22^2=484\) and \(23^2=529\), and \(484<528<529\), we get \(22<\sqrt{528}<23\). Similarly, \(530\) lies between \(23^2=529\) and \(24^2=576\), so \(23<\sqrt{530}<24\). Hence, option B is correct. Option C may seem close, but \(528<529\) means that \(\sqrt{528}\) is less than 23. Exam tip: locate a square root by comparing the number with consecutive perfect squares.

In a square root spiral, the hypotenuse after \(\sqrt{1520}\) is \(\sqrt{1521}\). Since \(1521=39^2\), we get \(\sqrt{1521}=39\). \(38\) is incorrect because \(38^2=1444\), while \(40^2=1600\). Exam tip: To check whether a square root is an integer, compare the number with nearby perfect squares.

The standard square root spiral adds a perpendicular segment of unit length to the preceding hypotenuse. If the preceding hypotenuse is √n, then the Pythagorean theorem gives the next length as √((√n)²+1²)=√(n+1). To construct √390, we must have n+1=390, so n=389. Consequently, the correct preceding hypotenuse is √389 and the new perpendicular is 1 unit, which is option C. Option A uses a 2-unit perpendicular and would give 388+2²=392. Option B treats the target as though it were the preceding hypotenuse, while option D would produce √392 because 391+1=392. Only option C satisfies the exact construction rule and reaches √390.

In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on, with the number under the root increasing by 1 each time. Therefore, the hypotenuse after \(\sqrt{1295}\) is \(\sqrt{1296}\). Since \(1296=36^2\), \(\sqrt{1296}=36\). \(\sqrt{1297}\) comes one step later, not immediately after \(\sqrt{1295}\). Exam tip: increase the radicand by 1 first, then check whether it is a perfect square.

\(25^2=625\) and \(26^2=676\). Since \(625<675<676\), the correct inequality is \(25^2<675<26^2\), so \(\sqrt{675}\) lies between 25 and 26. Option A is incorrect because 675 is greater than \(25^2=625\). Exam tip: To locate a square root, compare the number with consecutive perfect squares.

In a square root spiral, the hypotenuse after \(\sqrt{n}\) is \(\sqrt{n+1}\). Since this hypotenuse is \(52\), \(\sqrt{n+1}=52\). Squaring gives \(n+1=52^2=2704\), so \(n=2703\). Taking \(2704\) would make the next hypotenuse \(\sqrt{2705}\), so it is not correct. Exam tip: square the given hypotenuse first, then subtract 1 because the question refers to the next term.

The direct answer is C. In the standard square-root spiral, the radicand increases one unit at each construction step. Starting from sqrt{18}, the successive hypotenuses are sqrt{19}, sqrt{20}, sqrt{21} and sqrt{22}. Thus option C gives the complete increasing order. Option A skips sqrt{19} and sqrt{21}, so it is not the usual complete construction. Option B jumps from sqrt{18} to sqrt{22} and then moves backward, which is not the normal order. Option C is correct because every intermediate step is included. Option D begins at the largest value and decreases, reversing the construction and omitting sqrt{18} and sqrt{19}. Memory cue: add 1 inside each successive square root.

Since \(20^2=400\) and \(21^2=441\), and \(400<440<441\), we get \(20<\sqrt{440}<21\). Similarly, \(21^2=441\) and \(22^2=484\), and \(441<442<484\), so \(21<\sqrt{442}<22\). Therefore, option B is correct. Options C and D incorrectly place both square roots in the same interval. Exam tip: locate a square root by comparing the number with the nearest perfect squares.

After (\sqrt{624}), (\sqrt{625}=25) is formed. Since (25^2<626<26^2), (\sqrt{626}) lies between (25) and (26).

Since \(49^2=2401\) is less than \(2499\), while \(50^2=2500\) is greater than \(2499\), we get \(49<\sqrt{2499}<50\). On the other hand, \(2500=50^2\), so \(\sqrt{2500}=50\). Hence, option C is correct. Option A incorrectly states that \(\sqrt{2499}=50\); it is slightly less than \(50\). Exam tip: compare a number with nearby perfect squares to locate its square root.

\(48^2=2304\) and \(49^2=2401\). Since \(2304<2400<2401\), taking positive square roots gives \(48<\sqrt{2400}<49\). It is less than \(49\) because \(2400<49^2\). Exam tip: To locate a square root, compare the number with consecutive perfect squares.

In a square root spiral, the hypotenuse after \(\sqrt{3480}\) is \(\sqrt{3481}\). Since \(3481=59^2\), \(\sqrt{3481}=59\). Also, \(58^2=3364\) and \(60^2=3600\), so neither is correct. Exam tip: check the squares of nearby integers to identify a perfect square quickly.

Since \(11^2=121\) and \(12^2=144\), we get \(121<143<144\), so \(11<\sqrt{143}<12\). Similarly, \(13^2=169\) and \(14^2=196\) give \(169<170<196\), so \(13<\sqrt{170}<14\). Hence, option A is correct. Option D incorrectly places \(\sqrt{170}\) between 12 and 13. Exam tip: To locate a square root, compare the number with the nearest perfect squares.

A square root spiral adds one new perpendicular side of length 1 at each stage. Starting with a hypotenuse \(\sqrt{84}\), the next right triangle has side lengths \(\sqrt{84}\) and 1. The new hypotenuse must be calculated from the squares of these lengths, as required by the Pythagorean theorem.

The calculation is \((\sqrt{84})^2+1^2=84+1=85\). Taking the positive square root gives the new hypotenuse \(\sqrt{85}\). Therefore option D is correct. The statement \(\sqrt{84}+1=\sqrt{85}\) in option A is not valid, because the square root of a sum is generally not the sum of square roots or lengths. Multiplication and adding a squared length of 2 are also incorrect.

We have \(64^2=4096\) and \(65^2=4225\). Since \(4096<4224<4225\), taking positive square roots gives \(64<\sqrt{4224}<65\). Although 4224 is very close to \(4225\), it is still less than \(65^2\), so its square root is less than 65. Exam tip: To locate a square root, compare the number with consecutive perfect squares.

In a square root spiral, each new right triangle is formed by adding a unit side to the previous hypotenuse. Hence, if the previous hypotenuse is \(\sqrt{1935}\), the square of the new hypotenuse is \(1935+1=1936\). Therefore, the new hypotenuse is \(\sqrt{1936}\). Since \(44^2=1936\), its value is \(44\). Option B has the correct radicand but an incorrect value. Exam tip: first add \(1\) to the radicand, then check whether the result is a perfect square.

We have \(59^2=3481\) and \(60^2=3600\). Since \(3481<3599<3600\), it follows that \(59<\sqrt{3599}<60\). Therefore, its position on the square root spiral is between 59 and 60. The close distractor \(60<\sqrt{3599}<61\) is incorrect because 3599 is less than \(60^2\). Exam tip: locate a square root by comparing the number with consecutive perfect squares.

We have \(38^2=1444\), \(39^2=1521\), and \(40^2=1600\). Since \(1444<1520<1521\), \(38<\sqrt{1520}<39\). Similarly, \(1521<1522<1600\), so \(39<\sqrt{1522}<40\). Therefore, option A is correct. Options B and C incorrectly place both square roots in the same interval. Exam tip: To locate a square root, compare the number with consecutive perfect squares.

The spiral moves from one square-root length to the next by adding a perpendicular side of length 1. If the old hypotenuse is
sqrt{n}
, then the new hypotenuse has square length
n+1
. This follows because
(
sqrt{n}
)^2+1^2=n+1
. The wording identifies
sqrt{1368}
as the previous hypotenuse, so here
n=1368
.

Substituting this value gives the new length
sqrt{1368+1}=
sqrt{1369}
. Hence option B is correct. The answer is not
sqrt{1367}
, which would be a preceding value, and it is not the unchanged
sqrt{1368}
. The value
sqrt{2736}
would incorrectly double the radicand rather than add one. No decimal approximation is needed.

In a square root spiral, each new hypotenuse represents the square root of the corresponding number. Since \(1369=37\times37=37^2\), \(\sqrt{1369}=37\). Option 36 is incorrect because \(36^2=1296\), while \(38^2=1444\). Exam tip: check nearby perfect squares to identify a square root quickly.

In a square root spiral, when one leg is \(\sqrt{n}\) and the perpendicular leg is 1 unit, the new hypotenuse is \(\sqrt{n+1}\). Therefore, drawing a 1-unit perpendicular on \(\sqrt{3024}\) gives \(\sqrt{3024+1}=\sqrt{3025}\). \(\sqrt{3023}\) represents the previous stage, while \(\sqrt{3026}\) would be the next stage. Exam tip: add exactly 1 to the radicand at each new step.

\(50^2=2500\) and \(51^2=2601\). Since \(2500<2600<2601\), taking square roots gives \(50<\sqrt{2600}<51\). \(\sqrt{2600}=51\) is incorrect because \(51^2=2601\), not 2600. Exam tip: To find the interval of a square root, compare the number with the nearest perfect squares.

The hypotenuse length of the square root to be marked is taken in the compass. An arc from the origin gives the correct location.