Square root spiral

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 next hypotenuse is (\sqrt{144}), and (\sqrt{144}=12). It is not in an interval, but exactly at (12).

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.

(\sqrt{59}) is the previous hypotenuse and the new perpendicular is (1) unit. Therefore the new hypotenuse is (\sqrt{60}).

By Pythagoras’ theorem, \(h_{new}^2=h_{old}^2+1^2\). Thus, the square of the new hypotenuse increases by 1, giving \(\sqrt2,\sqrt3,\sqrt4\) in order. Exam tip: write the squares first to avoid adding lengths directly.

In a square root spiral, the previous hypotenuse becomes one leg of the next right triangle, and a unit segment is drawn perpendicular to it. Hence the new hypotenuse represents the next square root. Exam tip: look for the perpendicular unit side.

In a square root spiral, the sum of squares is taken by Pythagoras theorem. Directly adding lengths is wrong.

In the square root spiral, each new right triangle has the previous hypotenuse as one side and a new perpendicular side of length 1. If the current hypotenuse is \(\sqrt{224}\), then the next one has square equal to \((\sqrt{224})^2+1^2=224+1=225\). Therefore the next hypotenuse is \(\sqrt{225}\). This follows directly from the Pythagorean theorem and the fixed unit segment used in the spiral.

Since 225 is a perfect square, \(\sqrt{225}=15\), because \(15^2=225\). Thus option A gives both the correct next hypotenuse and its exact value. Option D has the right radical but the wrong value, since 14 squared is 196, not 225. Options B and C do not follow the successive construction rule. The supplied answer is therefore correct.

Neither \(2\) nor \(8\) is a perfect square. The square root of a positive integer is rational only when the integer is a perfect square. Hence \(\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\) are both irrational. Option C is incorrect because \(\sqrt{2}\) is not a whole number. Exam tip: First check whether the number under the square root is a perfect square.

By Pythagoras' theorem, if the square of the previous hypotenuse is \(n\) and the new perpendicular side is 1, the new hypotenuse has square \(n+1\). Hence it represents \(\sqrt{n+1}\). Thales' theorem is not used here. Exam tip: identify the right angle first.