Square root spiral

The new hypotenuse is (\sqrt{120+1}=\sqrt{121}), and (\sqrt{121}=11). When a perfect square appears, write its exact value.

The correct hypotenuse is (\sqrt{(\sqrt{18})^2+1^2}=\sqrt{19}). In a square root spiral, the number increases by (1) at each step.

The statement is incorrect. Adding a perpendicular side of length 1 gives successive hypotenuses \(\sqrt{n}\); for example, \(\sqrt{3}\) follows \(\sqrt{2}\). Thus, non-perfect-square roots also occur. Exam tip: track the radicand step by step.

If the new hypotenuse is (\sqrt{n+1}), the previous hypotenuse is (\sqrt{n}). Therefore before (\sqrt{35}), it is (\sqrt{34}).

\(6^2=36\) and \(7^2=49\). Since \(36<45<49\), we get \(6<\sqrt{45}<7\). Therefore, the length \(\sqrt{45}\) in the square root spiral lies between \(6\) and \(7\) on the number line. It cannot lie between \(5\) and \(6\), because numbers in that interval have squares between \(25\) and \(36\). Exam tip: compare the radicand with the nearest perfect squares to locate a square root.

The previous hypotenuse is \(\sqrt{12}\) and the new perpendicular side is 1 unit. By Pythagoras, its square is \((\sqrt{12})^2+1^2=12+1=13\), so the new hypotenuse is \(\sqrt{13}\). Exam tip: add squares of perpendicular sides.

A 1-unit side is drawn perpendicular to the previous hypotenuse, forming a right triangle. By Pythagoras, new hypotenuse² = previous hypotenuse² + 1, giving \(\sqrt{2},\sqrt{3}\), etc. Exam tip: look for “perpendicular,” not parallel.

Since \(8^2=64<72<81=9^2\), \(\sqrt{72}\) lies between \(8\) and \(9\). Also, \(81=9^2\), so \(\sqrt{81}=9\), which is a whole number. Therefore, option B is correct. Option D is incorrect because \(\sqrt{81}\) is not irrational. Exam tip: To locate a square root, compare the number with the squares of nearby whole numbers.

(\sqrt{120}) is formed from (\sqrt{119}) and a (1) unit perpendicular. The previous hypotenuse number is always (1) less.

Since \(10^2=100\) and \(11^2=121\), and \(100<120<121\), we get \(10<\sqrt{120}<11\). Therefore, \(\sqrt{120}\) lies between 10 and 11 on the number line. For it to lie between 9 and 10, its radicand would need to be between 81 and 100. Exam tip: compare the number with the nearest perfect squares to locate a square root.

In the usual spiral (1^2) is added every time. Taking (4) units adds (16), so the sequence changes.

In the next right triangle of the square root spiral, the previous hypotenuse \(\sqrt{14}\) becomes one side and the new perpendicular side has length 1. By Pythagoras’ theorem, the square of the new hypotenuse is \((\sqrt{14})^2+1^2=14+1=15\), so the new hypotenuse is \(\sqrt{15}\). Note that \(\sqrt{14}+1\) is not equal to \(\sqrt{15}\). Exam tip: at each new step, add \(1^2\) to the square of the previous hypotenuse.

Since \(11^2=121\) and \(12^2=144\), and \(121<130<144\), we get \(11<\sqrt{130}<12\). Hence, on the square root spiral, its number-line position is between 11 and 12. Option A is incorrect because \(130\) is greater than \(11^2\). Exam tip: To locate a square root, compare the number with the nearest perfect squares.

At every step, a perpendicular of length 1 unit is drawn at an endpoint of the previous hypotenuse. By Pythagoras, the square of the new hypotenuse is the previous square plus 1, producing √2, √3, √4, and so on. Exam tip: identify the fixed 1-unit side first.

In the square-root spiral, a unit perpendicular is added to the existing hypotenuse. If the existing hypotenuse is
sqrt{48}
, the Pythagorean theorem says that the square of the new hypotenuse is
(
sqrt{48}
)^2+1^2=48+1=49
. Thus the new hypotenuse is
sqrt{49}
.

Since 49 is a perfect square, its positive square root is exactly 7. Therefore the answer is option B, not the less precise expression “no whole number.” The values 6 and 8 would correspond to
sqrt{36}
and
sqrt{64}
, so they do not result from this single step. The important point is that the spiral normally gives
sqrt{49}
here, which simplifies to the whole number 7.

Since \(5^2=25\) and \(6^2=36\), we have \(25<26<36\) and \(25<27<36\). Hence, \(5<\sqrt{26}<6\) and \(5<\sqrt{27}<6\). Option B is incorrect because \(\sqrt{27}\) is also less than \(6\). Exam tip: To locate a square root, compare the number with the nearest perfect squares on either side.

The governing rule of the standard square-root spiral is that each one-unit perpendicular changes the current hypotenuse √n into √(n+1). This follows from h²=(√n)²+1²=n+1. Applying the rule to √143 gives the next hypotenuse √144. Since 144 is a perfect square, 144=12², and therefore √144=12 exactly. Option A is correct because it gives both the correct next radical and its exact value. Option B moves backward to √142. Option C names the correct radical but gives an incorrect interval: 12 is not strictly between 11 and 12; it is exactly the endpoint 12. Option D changes the radicand without any construction rule and is unrelated to the next step.

A compass is used to take the hypotenuse length and draw an arc on the number line. This is the correct geometric transfer method.

(55) is not a perfect square and (7^2<55<8^2). Therefore (\sqrt{55}) is irrational and lies between (7) and (8).

In a square root spiral, hypotenuses are formed in order. In the usual construction, the \(\sqrt{3}\) step is not skipped.

Since \(\sqrt{9}=3\), this segment has a rational length. As 2, 3 and 5 are not perfect squares, their square roots are irrational. Exam tip: first check whether the radicand is a perfect square.

In a square root spiral, the previous hypotenuse √n and a unit side form the next right triangle. By Pythagoras, the new hypotenuse is √(n+1). Thus A is correct; the hypotenuses are not equal. Exam tip: add 1 at each step.

In a square root spiral, the hypotenuse of one right triangle becomes a side of the next triangle, while the other new side is 1 unit. Thus the hypotenuses progress as \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Exam tip: identify the newly added 1-unit side.

In a square root spiral, the distance of a point from the origin is \(\sqrt{n}\). Since \(36=6^2\), \(\sqrt{36}=6\) is an integer. The others are not perfect squares. Exam tip: check for a perfect square first.

The square-root spiral uses the previous hypotenuse and a perpendicular of 1 unit to construct the next one. Immediately before √60 comes √59. Applying the Pythagorean theorem gives the square of the new hypotenuse as (√59)²+1²=59+1=60, so the new hypotenuse is √60. Thus option A is correct. Option B incorrectly treats √60 as the previous side; adding another unit would produce a square of 61. Option C uses a 2-unit perpendicular, giving 58+4=62 rather than 60. Option D uses ordinary addition instead of the relation between the squares of the sides of a right triangle.