Square root spiral

In a square root spiral, a unit segment is drawn perpendicular to the previous hypotenuse. By Pythagoras, the square of the new hypotenuse equals the previous square plus 1. A parallel segment would not form a right angle. Exam tip: remember “perpendicular unit segment.”

The two perpendicular sides of the first right triangle are 1 unit each. By Pythagoras’ theorem, the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). \(2\) is the value of the square of the hypotenuse, not the hypotenuse itself. Exam tip: the first hypotenuse in a square root spiral is always \(\sqrt{2}\).

The governing concept is the construction of a square-root spiral using the Pythagorean theorem. At this stage, the existing hypotenuse of length √2 becomes one leg of a new right triangle, and the newly drawn perpendicular is the other leg of length 1. If H is the next hypotenuse, then H² = (√2)² + 1² = 2 + 1 = 3. Since a length is positive, H = √3. Therefore option B is correct. Option A is only the previous hypotenuse. Option C would incorrectly add 4 or treat the lengths without squaring, while option D would add an extra unit to the squared result. The theorem must be applied to the squares of the perpendicular sides.

With (\sqrt{3}) and (1) unit, the new hypotenuse is (\sqrt{3+1}=\sqrt{4}). The next square root appears in order.

The governing construction in a square root spiral is the repeated addition of a unit-length perpendicular segment. At each stage, the previously obtained radius and the new perpendicular side form a triangle with a 90-degree angle. By the Pythagorean theorem, if the existing length is √n and the added side has length 1, the new hypotenuse is √(n+1), since (√n)² + 1² = n + 1. Thus every newly constructed triangle is a right triangle, so option A is correct. An equilateral triangle has three equal sides and is not produced by this perpendicular construction; a circle is not a triangle.

In a square root spiral, each new right triangle uses the hypotenuse of the previous triangle as one side, and its new perpendicular side is kept 1 unit long. By Pythagoras’ theorem, the successive hypotenuses become \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. \(\sqrt{2}\) is not the new perpendicular side; it is the hypotenuse of the first new triangle. Exam tip: remember that the side kept constant at every step is 1 unit.

The governing concept is the Pythagorean relationship in a right triangle. If the perpendicular sides have lengths a and b and the hypotenuse is c, then c² = a² + b², so c = √(a² + b²). A square root spiral repeatedly creates right triangles; for example, legs √n and 1 give a hypotenuse √(n+1). Therefore option B, Pythagoras theorem, is correct. Thales theorem can help identify a right angle in some circle constructions, but it is not the calculation used here. The midpoint theorem concerns a segment joining midpoints, and the remainder theorem belongs to polynomial division, so neither finds the spiral’s hypotenuse.

Adding a perpendicular 1-unit side forms a right triangle. By Pythagoras, the square of the new hypotenuse equals the previous square plus 1, producing successive square roots. Exam tip: identify the perpendicular unit side first.

In a square root spiral, each new right triangle uses the previous hypotenuse and a new perpendicular side of length 1. By the Pythagorean theorem, the squared hypotenuse increases by 1 at every step. Starting with a hypotenuse of \sqrt{1}, the next one is \sqrt{2}, and the following one is \sqrt{3}.

Counting the hypotenuses in order gives first: \sqrt{1}; second: \sqrt{2}; third: \sqrt{3}. Therefore the third hypotenuse is \sqrt{3}, which is option B. It is important to count the initial \sqrt{1} as the first member of the sequence. If the counting began after the initial triangle, a different label might be obtained, but that is not the convention stated here.

Since \(2^2=4\) and \(3^2=9\), and 7 lies between 4 and 9, \(\sqrt{7}\) lies between 2 and 3. Do not choose 1 and 2: their squares bracket numbers only up to 4. Exam tip: compare with nearby perfect squares.

A square root spiral helps construct lengths like (\sqrt{2},\sqrt{3},\sqrt{5}). These can be shown on the number line.

In a square root spiral, the previous hypotenuse becomes a side of the next triangle, and a new perpendicular side of 1 unit is added. The new hypotenuse follows from Pythagoras’ theorem. Exam tip: identify the newly added unit side at every step.

The square-root spiral uses the Pythagorean theorem to create successive lengths. If the existing hypotenuse is \(\sqrt{8}\) and a perpendicular side of length 1 is added, the new hypotenuse \(L\) satisfies \(L^2=(\sqrt{8})^2+1^2\). Therefore \(L^2=8+1=9\), so \(L=\sqrt{9}\).

Thus the next hypotenuse is \(\sqrt{9}\), which is also 3. The construction does not subtract 1, keep the same length, or produce \(\sqrt{10}\), because the added perpendicular side contributes the square of its length, namely 1. Hence option C follows directly from the Pythagorean theorem.

In a square root spiral, each new triangle must be a right-angled triangle. Therefore, the 1-unit perpendicular segment is drawn at \(90^\circ\) to the previous side. A line at \(60^\circ\) or \(45^\circ\) would not be perpendicular and would not form the required right triangle. Exam tip: the word “perpendicular” always indicates an angle of \(90^\circ\).

The governing concept is the Pythagorean theorem used successively in a square-root spiral. At every stage, a perpendicular segment of length 1 unit is erected on the preceding hypotenuse. If that preceding hypotenuse is √n, then the new hypotenuse h satisfies h² = (√n)² + 1² = n + 1, so h = √(n + 1). For the desired length √10, we require n + 1 = 10, which gives n = 9. Hence the perpendicular must be drawn on the previous hypotenuse √9, making option B correct. Using √8 would produce √9; using √10 would produce √11; and using √11 would produce √12. Thus the other choices represent different stages.

At each step, the previous hypotenuse becomes one side and (1) unit perpendicular is the other side. The new hypotenuse gives the next square root.

The relevant construction principle is that a square-root spiral produces successive hypotenuses of lengths √1, √2, √3, and so forth. To construct √2, begin with two perpendicular unit segments. By the Pythagorean theorem, the resulting hypotenuse has square length 1²+1²=2, so its length is √2. The compass is opened to this constructed hypotenuse and an arc is drawn to transfer that length to the number line. Therefore option A is correct. The √3 hypotenuse belongs to the next stage of the spiral, while a 2-unit perpendicular is not the radius required for this construction and zero length cannot locate √2.

The governing concept is the repeated use of the Pythagorean theorem in a square-root spiral. At each stage, a perpendicular segment of length 1 unit is added to the previous hypotenuse. If one hypotenuse is √n, the next hypotenuse has square (√n)² + 1² = n + 1, and therefore has length √(n + 1). The sequence is consequently √1, √2, √3, √4, √5, and so on. Since the second hypotenuse is given as √2, the third is √3 and the fourth is √4. Therefore, option B is correct. Option A is one stage earlier, while √5 and √6 occur at later stages. The answer follows from the construction sequence, not from simply adding 1 to the radical itself.

In a square root spiral, each new right triangle uses the previous hypotenuse and a unit side. \((\sqrt{12})^2+1^2=12+1=13\), so its hypotenuse is \(\sqrt{13}\). Exam tip: apply Pythagoras’ theorem.

In a square root spiral, each new right triangle uses the hypotenuse of the previous triangle as one of its sides, while the other side is 1 unit. By Pythagoras, the new hypotenuse becomes \(\sqrt{2},\sqrt{3},\sqrt{4}\), and so on. Exam tip: the shared side is always the previous hypotenuse.

A square root spiral is made from successive right triangles, each with a new perpendicular side of 1 unit. By Pythagoras, the hypotenuses become
1,
2,
3, and so on. Exam tip: identify the constant 1-unit added side.

To construct \(\sqrt{7}\), draw a perpendicular side of length 1 at the previous hypotenuse \(\sqrt{6}\). Since \((\sqrt{6})^2+1^2=7\), the new hypotenuse is \(\sqrt{7}\). \(\sqrt{5}\) belongs to an earlier step. Exam tip: add 1 at each new step.

The governing concept is the Pythagorean construction used in a square-root spiral. A segment of length 1 unit is drawn perpendicular to the previous hypotenuse. If that previous hypotenuse is √n, then the new hypotenuse H satisfies H² = (√n)² + 1² = n + 1. To construct √11, we set n + 1 = 11, giving n = 10. Therefore, the previous hypotenuse must be √10, so option B is correct. If √9 were used, the next hypotenuse would be √10. If √11 were used, the next one would be √12, and √12 would lead to √13. These alternatives therefore represent the wrong stage of the construction.

The governing geometric idea is transferring a previously constructed length without changing it. A compass is designed for this purpose: its two points can be placed on the endpoints of the hypotenuse, preserving that opening, and an arc can then be drawn from the required point on the number line. The intersection of the arc with the line marks the corresponding irrational length. Therefore, option A, compass, is correct. A balance measures mass, not geometric distance; a clock measures time; and a calculator can evaluate or approximate a number but cannot transfer a segment in a geometric construction. A ruler may help draw or measure, but it is not the listed correct tool for copying the hypotenuse as an arc.

By Pythagoras theorem, the new hypotenuse becomes (\sqrt{2+1}=\sqrt{3}). Do not treat (\sqrt{2}+1) as (\sqrt{3}).