Square root spiral

In the usual sequence of the square root spiral, (\sqrt{8}) comes after (\sqrt{7}). The new perpendicular is always (1) unit.

\(20\) is not a perfect square. Since \(20=4\times5\), \(\sqrt{20}=2\sqrt{5}\). As \(\sqrt{5}\) is irrational, \(\sqrt{20}\) is also irrational. Being even does not make \(20\) or its square root zero. Exam tip: The square root of an integer is rational only when the integer is a perfect square.

In each new triangle, the previous hypotenuse becomes one leg and a perpendicular unit-length leg is added. If the old hypotenuse is \(\sqrt{n}\), Pythagoras gives the new one as \(\sqrt{n+1}\). Two unit legs produce only \(\sqrt{2}\). Exam tip: identify the fixed unit leg first.

In the spiral, the previous hypotenuse is \(\sqrt{n}\). A perpendicular unit side gives the next hypotenuse \(\sqrt{n+1}\), since \((\sqrt{n})^2+1^2=n+1\). Exam tip: check that each new unit segment is perpendicular to the previous hypotenuse.

The new hypotenuse is \(\sqrt{6}\), since \(h^2=(\sqrt{5})^2+1^2=5+1=6\). It is not \(\sqrt{10}\), as the lengths are not added directly. Exam tip: apply Pythagoras’ theorem at every new step.

The spiral does not obtain a new length by simply adding the two side lengths. Instead, each new figure is a right triangle, so the Pythagorean theorem must be used. The square of the new hypotenuse equals the sum of the squares of the old hypotenuse and the new unit perpendicular. This explains why the radical changes from 2 to 3.

For the given construction, the two perpendicular sides have lengths
sqrt{2}
and 1. Therefore the new hypotenuse
h
satisfies
h^2=(
sqrt{2}
)^2+1^2=2+1=3
. Since a length is positive,
h=
sqrt{3}
. The expression
sqrt{2}+1
is not equal to
sqrt{3}
, and the other numerical statements do not represent the theorem. Hence option A is correct.

In a square root spiral, a perpendicular side of length 1 unit is drawn from an end of the previous hypotenuse. By Pythagoras, the new hypotenuse becomes \(\sqrt{n+1}\). Exam tip: the added side is always 1 unit.

Since \(25<27<36\), we have \(5^2<27<6^2\). Taking square roots gives \(5<\sqrt{27}<6\), so it must be placed between \(5\) and \(6\) on the number line. The interval \(4\) and \(5\) is incorrect because \(\sqrt{25}=5\) and \(27\) is greater than \(25\). Exam tip: To locate \(\sqrt{n}\), compare \(n\) with the nearest perfect squares.

\(16=4^2\), \(25=5^2\), and \(36=6^2\) are perfect squares, so their square roots are whole numbers. However, \(38\) is not a perfect square; it lies between \(36\) and \(49\). Therefore, \(\sqrt{38}\) is not a whole number. Exam tip: A square root is a whole number only when the number is a perfect square.

The claim is incorrect. The new right triangle has legs \(\sqrt{7}\) and 1, so by Pythagoras, hypotenuse² = \(7+1=8\); hence it is \(\sqrt{8}\). Exam tip: add the squares of perpendicular sides, not the hypotenuse lengths.

The governing concept is the repeated application of the Pythagorean theorem in a square-root spiral. At each stage, a perpendicular of length 1 unit is added to the existing hypotenuse. If the existing hypotenuse is √n, then the new hypotenuse h satisfies h²=(√n)²+1²=n+1, so h=√(n+1). To construct √121, we must therefore begin with n+1=121, which gives n=120. Indeed, (√120)²+1²=120+1=121, so the new hypotenuse is √121. Thus option B is correct. √119 would produce √120, while √121 is already the target rather than the preceding length. √122 would produce √123 and would occur after the required construction.

To locate a square root between two numbers, compare the number under the root with their squares. Since \(1^2=1\) and \(2^2=4\), every positive number strictly between 1 and 4 has a square root strictly between 1 and 2. Both 2 and 3 lie in this interval: \(1<2<4\) and \(1<3<4\).

Taking square roots preserves the order for positive numbers, so \(1<\sqrt{2}<2\) and \(1<\sqrt{3}<2\). Neither root equals 2, and neither 2 nor 3 is a perfect square. The statement in option A, written as \(1^2<2,3<2^2\), expresses exactly this comparison. Therefore option A is correct.

In a square root spiral, consecutive hypotenuses have lengths √n and √(n+1). Their squares differ by 1, not their lengths; for example, √5 − 2 is about 0.24. Exam tip: compare squares when checking successive hypotenuses.

The first step of a square root spiral begins with two perpendicular sides, each of length 1. The hypotenuse of this right triangle is the distance from the starting point to the new point. By the Pythagorean theorem, the square of this distance is the sum of the squares of the two perpendicular sides.

Thus, (\sqrt{1})^2+1^2=1+1=2, and the hypotenuse is \sqrt{2}. Therefore option A gives the correct relation. Adding lengths directly, using 2 instead of 1, or subtracting squares would not describe this construction. The right angle is essential because the Pythagorean theorem applies to a right triangle.

(\sqrt{75}) is formed from (\sqrt{74}) and a (1) unit perpendicular. The previous hypotenuse always has one less number.

A square root spiral is formed by adding a unit perpendicular side to each right triangle. By Pythagoras’ theorem, successive hypotenuses are \(\sqrt{2}, \sqrt{3}, \ldots, \sqrt{n}\); hence the point is \(\sqrt{n}\) units from the origin. Exam tip: identify the hypotenuse as the radius.

The governing ideas are perfect squares and the one-unit construction rule of the square-root spiral. Since 64=8², its principal square root is √64=8 exactly. Thus √64 is a whole number and a rational number, not an irrational number. When a perpendicular of length 1 is erected on the segment of length 8, the Pythagorean theorem gives h²=8²+1²=64+1=65. Therefore the new hypotenuse is h=√65. Option A correctly states both facts. Option B misclassifies √64 and reverses the role of the old and new lengths. Option C is false because √65 is greater than 8, and option D wrongly treats the two successive hypotenuses as equal.

The hypotenuse length of the required square root is taken in the compass. Then an arc is drawn from the origin.

In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on. Therefore, the hypotenuse after \(\sqrt{45}\) is \(\sqrt{46}\). Since \(6^2=36\), \(7^2=49\), and \(36<46<49\), \(\sqrt{46}\) lies between \(6\) and \(7\). Although \(\sqrt{47}\) also lies in this interval, it is not the immediate next hypotenuse. Exam tip: compare the number under the root with consecutive perfect squares to find its interval.

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

At each step, a unit segment is drawn perpendicular to the previous hypotenuse, making a new right triangle. If the old hypotenuse is \(\sqrt{n}\), Pythagoras gives the next one as \(\sqrt{n+1}\). Exam tip: look for “perpendicular.”

In a square root spiral, \(\sqrt{2}\) is the hypotenuse of a right-angled triangle with two sides of 1 unit. To construct \(\sqrt{3}\), a right-angled triangle is formed using \(\sqrt{2}\) as one side and a perpendicular side of 1 unit; its hypotenuse is \(\sqrt{3}\). Thus, both constructions use Pythagoras’ theorem to obtain the hypotenuse. Option B applies only to \(\sqrt{2}\), not to \(\sqrt{3}\). Exam tip: each new step of the spiral uses the previous hypotenuse and a perpendicular side of 1 unit.

After (\sqrt{15}), (\sqrt{16}) is formed and (\sqrt{16}=4). Therefore the next hypotenuse is the whole number (4).

The governing idea of a square root spiral is the Pythagorean theorem. If the existing hypotenuse is √6 and a perpendicular segment of length 1 is drawn, the new hypotenuse has length √((√6)² + 1²) = √(6 + 1) = √7. Thus option A gives the correct construction. The other choices either add lengths directly or use unsuitable values.

The direct answer is A, because that statement is wrong. In a square root spiral, each new figure is a right triangle and its hypotenuse represents a square-root length. These lengths are not all whole numbers. For example, \(\sqrt{2}\) and \(\sqrt{3}\) are irrational numbers, so they cannot be written as whole numbers. Option A is therefore wrong: the fact that every hypotenuse is a square root does not mean every square root is a whole number. Option B is correct because each new triangle is constructed with a right angle, which is the basis of the spiral. Option C is correct because the usual construction adds a new perpendicular side of length 1 unit at each step. Option D is correct because the early hypotenuses include \(\sqrt{2}\) and \(\sqrt{3}\). The statement asks which one is wrong, so choose A, not one of the true construction facts. A useful caution is that “square root” is a type of number expression, not a guarantee that the result is whole; only square roots of perfect squares, such as \(\sqrt{49}=7\), are whole numbers.