Square root spiral

In each new right triangle of a square root spiral, one side is the previous hypotenuse and the other side is 1 unit. Hence, the new hypotenuse is \(\sqrt{(\sqrt{399})^2+1^2}=\sqrt{399+1}=\sqrt{400}\). \(\sqrt{401}\) would be obtained in the next step, not in this one. Exam tip: at each new step, add 1 to the number under the square root of the hypotenuse.

Since \(13^2=169\) and \(14^2=196\), and \(169<195<196\), we get \(13<\sqrt{195}<14\). On the square root spiral, the point for \(\sqrt{195}\) lies between \(\sqrt{169}=13\) and \(\sqrt{196}=14\). Option B is incorrect because \(\sqrt{195}\) is less than 14. Exam tip: Find the nearest smaller and larger perfect squares to determine a square root’s interval quickly.

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

Drawing a (1) unit perpendicular on (\sqrt{439}) forms (\sqrt{440}). The previous hypotenuse has one less number.

Since \(12^2=144\) and \(13^2=169\), we have \(144<150<169\). Therefore, \(12<\sqrt{150}<13\). On the other hand, \(169=13^2\), so \(\sqrt{169}=13\). Option A gives the wrong interval for \(\sqrt{150}\). Exam tip: the square root of a number between two consecutive perfect squares lies between their square roots.

The square-root spiral uses a right triangle at every step. To construct
sqrt{5}
, the previous hypotenuse is
sqrt{4}
and the newly drawn perpendicular side has length 1. The next hypotenuse is determined by adding the squares of these perpendicular sides, not by adding their lengths directly.

Let the new hypotenuse be
h
. By Pythagoras,
h^2=(
sqrt{4}
)^2+1^2=4+1=5
. Because a length is positive,
h=
sqrt{5}
. Therefore option A is correct. The expression
sqrt{4}+1
is not
sqrt{5}
, and
(
sqrt{4}
)^2-1^2
gives 3 rather than 5. The construction depends on a sum of squares.

In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on. Hence, the hypotenuse after \(\sqrt{255}\) is \(\sqrt{256}\). Since \(256=16^2\), \(\sqrt{256}=16\), so it lies at an integer value. \(\sqrt{257}\) is the following hypotenuse, while \(\sqrt{254}\) comes earlier. Exam tip: the square root of a perfect square is always an integer.

Since \(4^2=16\), \(5^2=25\), and \(6^2=36\), \(16<24<25\) gives \(4<\sqrt{24}<5\). Similarly, \(25<26<36\) gives \(5<\sqrt{26}<6\). Hence, the two square roots occur between different consecutive integers on the square root spiral. Option B is a close distractor, but \(\sqrt{26}\) is greater than 5. Exam tip: locate a square root by comparing the number with nearby perfect squares.

In a square root spiral, successive hypotenuses are formed as \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Therefore, the hypotenuse after \(\sqrt{624}\) is \(\sqrt{625}\). Since \(625=25^2\), \(\sqrt{625}=25\). \(\sqrt{626}\) is the next hypotenuse after it, while \(\sqrt{623}\) is the preceding one. Exam tip: when the number inside a square root is a perfect square, its square root is an integer.

Since \(80<81\), \(\sqrt{80}<9\), and since \(81<82<100\), \(\sqrt{82}\) lies between (9) and (10).

The square root of an integer is a whole number only when it is a perfect square. If it is not a perfect square, the square root can be irrational.

Here, \(12^2=144\), \(13^2=169\), and \(14^2=196\). Since \(144<168<169\), we get \(12<\sqrt{168}<13\). Similarly, \(169<170<196\) gives \(13<\sqrt{170}<14\). Therefore, option A is correct. Option B is incorrect because \(\sqrt{168}\) is less than 13. Exam tip: To locate a square root, compare the number with the nearest perfect squares.

In each new right triangle of a square root spiral, one leg is the previous hypotenuse and the other leg is \(1\). Therefore, by Pythagoras’ theorem, the new hypotenuse is \(\sqrt{(\sqrt{12})^2+1^2}=\sqrt{12+1}=\sqrt{13}\). Option B is incorrect because, in general, \(\sqrt{12}+1\neq\sqrt{13}\). Exam tip: square the previous hypotenuse, add \(1\), and then take the square root.

In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on. Hence, the hypotenuse after \(\sqrt{1023}\) is \(\sqrt{1024}\). Since \(1024=32^2\), \(\sqrt{1024}=32\). \(\sqrt{1025}\) is the following hypotenuse, not the immediate next one. Exam tip: remember nearby perfect squares; \(32^2=1024\).

After (\sqrt{35}), (\sqrt{36}=6) is formed. Since (6^2<37<7^2), (\sqrt{37}) lies between (6) and (7).

(36^2=1296) and (37^2=1369). The number (1368) lies between them, so (\sqrt{1368}) lies between (36) and (37).

Direct addition is not used in a square root spiral. The new hypotenuse is found using a right triangle and Pythagoras theorem.

Nearest perfect squares give the correct interval. Then the spiral length is placed on the number line using a compass.

A square root spiral is based on Pythagoras theorem. The previous hypotenuse and (1) unit perpendicular form the next square root.

The next hypotenuse is (\sqrt{728+1}=\sqrt{729}). Since (729=27^2), it will lie exactly at (27).