Square root spiral

The direct answer is C: Pythagoras theorem. In a right triangle, if the perpendicular sides are a and b and the hypotenuse is c, then . Thus . In a square root spiral, the old length and the newly added unit perpendicular are the two legs; Pythagoras gives the next hypotenuse. Option A, Thales theorem, concerns angles in a semicircle and is not the calculation used here. Option B, midpoint theorem, relates a segment joining midpoints to a triangle side. Option C is correct because it directly finds the hypotenuse of a right triangle. Option D, remainder theorem, belongs to polynomial division, not geometry. Memory cue: right triangle plus hypotenuse means Pythagoras.

Each new triangle is formed by adding a perpendicular side of length 1 to the previous hypotenuse. By Pythagoras, \((\sqrt{n-1})^2+1=n\), so the new hypotenuse is \(\sqrt{n}\). Exam tip: remember the order \(\sqrt2,\sqrt3,\sqrt4\).

Answer: option B, √4. A square-root spiral adds a perpendicular segment of length 1 unit at every step. If the current hypotenuse is √3 and the next one is h, apply the Pythagorean theorem: h² = (√3)² + 1² = 3 + 1 = 4. Since h is a length, h = √4, which is numerically equal to 2. Option A, √2, is the preceding stage. Option C, √5, would come after the √4 stage because the next calculation would be 4 + 1 = 5. Option D is even farther ahead. The radical form √4 is retained here because the question asks for the next member of the spiral sequence. Memory cue: the radicands increase by one at each construction step.

The defining sequence is generated by repeatedly adding a perpendicular unit and applying the Pythagorean theorem. If one stage has hypotenuse √n, the next has length √(n + 1). Beginning with the unit length represented as √1, the initial sequence is therefore √1, √2, √3, √4, and it continues in the same increasing pattern. Thus option A is correct. Option B places √2 before √1 and breaks the construction order. Option C increases the radicands by 2, whereas the spiral increases them by 1. Option D reverses the order and does not describe the forward construction. The radical notation also makes the connection with successive square roots visible, even when a term such as √4 can be simplified.

In a square root spiral, the previous hypotenuse becomes one leg and a 1-unit perpendicular leg is added at its endpoint. By Pythagoras, the next hypotenuse is \(\sqrt{n+1}\). A parallel segment will not form the required right angle. Exam tip: look for the perpendicular mark.

The direct answer is B, (\sqrt{11}). In the square root spiral, suppose the previous hypotenuse is (\sqrt{n}) and a new perpendicular side of 1 unit is added. Pythagoras gives the new hypotenuse as (\sqrt{(\sqrt{n})^2+1^2}=\sqrt{n+1}). To obtain (\sqrt{12}), we need (n+1=12), so (n=11). Therefore the previous hypotenuse is (\sqrt{11}). Option A, (\sqrt{10}), would produce (\sqrt{11}) after adding the unit perpendicular, not (\sqrt{12}). Option B is correct because it produces the required next value. Option C, (\sqrt{12}), is already the target hypotenuse, not the previous one. Option D, (\sqrt{13}), would lead to (\sqrt{14}), so it is one step too large. The important rule is: to make (\sqrt{k}), use (\sqrt{k-1}) as the previous hypotenuse and add a 1-unit perpendicular. Memory cue: previous square-root number is always one less.

A square root spiral starts with \(\sqrt{1}\). Each added right triangle increases the square of the hypotenuse by 1, so after 9 new triangles the hypotenuse is \(\sqrt{10}\). Exam tip: count from the initial \(\sqrt{1}\), not from zero.

In a square root spiral, each new right triangle is formed by adding a perpendicular side of 1 unit to the previous hypotenuse. Hence the new hypotenuse becomes \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Exam tip: the added side is always 1 unit.

A right angle measures 90°, so 90° is the correct option. In a square root spiral, each new triangle is a right-angled triangle, and the Pythagoras theorem is used to obtain the next length. A 60° angle is associated with an equilateral triangle, not a right angle. Exam tip: whenever you see ‘right angle’, recall 90° immediately.

The general rule follows directly from the Pythagorean theorem. Let the old hypotenuse be √n and let the new hypotenuse be h after a perpendicular side of length 1 is added. The two perpendicular sides of the new right triangle have lengths √n and 1, so h² = (√n)² + 1² = n + 1. Since h is a length, h = √(n + 1), which makes option B correct. Option A subtracts 1 and describes the preceding step rather than the new one. Option C incorrectly doubles n, and option D squares n without using the added unit side. This formula explains why the spiral produces successive square-root lengths.

In a square root spiral, the new unit segment is drawn perpendicular to the previous hypotenuse. Thus, \((\sqrt{10})^2+1^2=11\), giving a hypotenuse of \(\sqrt{11}\). A parallel segment does not form the required right triangle. Exam tip: apply Pythagoras at each step.

In a square root spiral, the obtained hypotenuse represents a square-root length. This length is taken in a compass and transferred from 0 onto the number line by drawing an arc. A ruler helps draw or measure lines, but a compass is the appropriate instrument for transferring an exact length. Exam tip: use a compass whenever a construction asks you to transfer a length.

In a square root spiral, a perpendicular side of length 1 unit is drawn on the previous hypotenuse. By Pythagoras’ theorem, the square of the new hypotenuse equals the previous square plus 1. Exam tip: do not confuse the added perpendicular with the hypotenuse.

At each step, a 1-unit side is drawn perpendicular to the previous radius. Thus, hypotenuse² = \(6+1=7\), so the hypotenuse is \(\sqrt{7}\). In exams, always check the right angle.

In a square root spiral the addition is of squares, not direct lengths. The correct reasoning is Pythagoras theorem.

A square-root spiral is formed by attaching successive right triangles, usually with one unit-length perpendicular side, to the preceding hypotenuse. Each new hypotenuse becomes longer and changes direction, so the connected construction winds outward. Consequently, the overall figure resembles a spiral. It is not a straight line, a single square, or merely a circle, making option A correct.

A square root spiral constructs lengths like (\sqrt{2},\sqrt{3},\sqrt{5}). It is useful on the number line.

The direct answer is B: \(\sqrt{17}\). In the spiral, the construction proceeds one step at a time. If a hypotenuse of length \(\sqrt{17}\) has already been made, a new perpendicular side of length 1 is attached to it. By Pythagoras theorem, the new hypotenuse satisfies \(c^2=(\sqrt{17})^2+1^2=17+1=18\), so its length is \(\sqrt{18}\). Therefore the immediately preceding hypotenuse is \(\sqrt{17}\). Option A, \(\sqrt{16}\), is one more step earlier and is not the hypotenuse immediately before \(\sqrt{18}\). Option B is correct because adding the next unit perpendicular side to \(\sqrt{17}\) produces \(\sqrt{18}\). Option C, \(\sqrt{18}\), is the length being constructed, not the one already constructed before it. Option D, \(\sqrt{19}\), comes after \(\sqrt{18}\), so it cannot be the previous hypotenuse. The sequence is \(\sqrt{16},\sqrt{17},\sqrt{18},\sqrt{19}\); “before” means move one place left.

In a square root spiral, one leg of every new right triangle is 1 unit, while the other leg is the previous hypotenuse. By Pythagoras’ theorem, the hypotenuses become \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Exam tip: remember that the fixed added side is always 1 unit.

By Pythagoras’ theorem, the square of the new hypotenuse is \((\sqrt{n})^2+1^2=n+1\), so its length is \(\sqrt{n+1}\). The triangles need not be isosceles. Exam tip: follow the sequence of squared hypotenuse lengths.

If the previous hypotenuse is \(\sqrt{n}\) and a unit segment is drawn perpendicular to it, the square of the new hypotenuse is \(n+1\). Hence its length is \(\sqrt{n+1}\). In exams, first identify the right angle.

At each step of a square root spiral, a new side of length 1 unit is drawn at 90° to the previous side, forming a right-angled triangle. Pythagoras’ theorem then gives successive hypotenuse lengths such as \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Option B is incorrect because the new hypotenuse is generally longer than the previous side. Exam tip: A 90° angle in such a construction signals the use of a right triangle and Pythagoras’ theorem.

The governing idea is the repeated use of the Pythagorean theorem in the square root spiral. At each stage, a perpendicular segment of length 1 unit is added to the preceding hypotenuse. If the preceding hypotenuse is √n, the new hypotenuse h satisfies h²=(√n)²+1²=n+1. To construct √50, we need n+1=50, so n=49. Hence the immediately previous hypotenuse is √49, and option B is correct. √48 is two stages earlier in the sequence. √50 is the target length being constructed, not its predecessor. √51 would occur at the following stage after √50. The relationship between consecutive squared lengths therefore identifies √49 uniquely as the required previous hypotenuse.

In a square root spiral, each new right triangle has one leg of length 1, so its hypotenuse becomes \(\sqrt{2}, \sqrt{3}\), and so on. Thus, the distance from the starting point to the relevant point represents \(\sqrt{n}\). Exam tip: identify the hypotenuse, not the unit side.

In a square root spiral, the previous hypotenuse becomes one side and a perpendicular side of 1 unit is added. The new hypotenuse then represents √2, √3 and so on. Exam tip: remember that the added perpendicular side is always 1 unit.