Square root spiral

In each new right triangle of a square root spiral, one perpendicular side has length 1. By the Pythagorean theorem, the square of the new hypotenuse is \(483+1=484\). Hence, the new hypotenuse is \(\sqrt{484}=22\). \(\sqrt{485}\) is not correct because adding a unit side increases the square of the hypotenuse by 1, not the hypotenuse itself. Exam tip: add 1 to the number under the previous radical, then take the square root.

In a square root spiral, successive hypotenuses are of the form \(\sqrt{n}\), with the number inside the square root increasing by 1 at each step. The given new hypotenuse is \(31=\sqrt{961}\). Therefore, the immediately preceding hypotenuse is \(\sqrt{960}\). \(\sqrt{961}\) is the given current hypotenuse, not the previous one. Exam tip: Express an integral hypotenuse as its square root form first, then subtract 1 from the radicand for the preceding step.

We have \(33^2=1089\) and \(34^2=1156\). Since \(1089<1155<1156\), taking square roots gives \(33<\sqrt{1155}<34\). Although it is very close to \(34\), \(\sqrt{1155}\) is less than \(34\) because \(1155<34^2\). Exam tip: To locate a square root, find the nearest perfect squares on either side of the given number.

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

By Pythagoras, \((\sqrt{n})^2+6^2=n+36\). So the usual \(\sqrt{n+1}\) sequence will change.

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.

\((\sqrt{209})^2+1^2=210\). So the previous hypotenuse for \(\sqrt{210}\) is \(\sqrt{209}\).

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 \(n\).

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

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

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 next hypotenuse is (\sqrt{81}), and (\sqrt{81}=9). At a perfect square, the hypotenuse becomes a whole number.

\(38^2=1444\) 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.