Square root spiral

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.

Hypotenuses increase one by one in the spiral. In the usual construction, intermediate steps are not skipped.

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.

The next hypotenuse is (\sqrt{256}), and (\sqrt{256}=16). At a perfect square, the hypotenuse becomes a whole number.

\(48^2=2304\) 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.