Square root spiral

The direct answer is A, a 1-unit length. A square root spiral, also called the spiral of Theodorus, begins with a right triangle whose two perpendicular sides are 1 unit each. By Pythagoras, its hypotenuse is (\sqrt{1^2+1^2}=\sqrt{2}). Then a new 1-unit perpendicular is drawn at the end of the previous hypotenuse, producing the next right triangle and (\sqrt{3}), and the process continues. Option A is correct because the first basic segment has length 1 unit. Option B, 2 units, would not give the standard starting construction and would change the sequence. Option C, (\sqrt{2}), is the first hypotenuse obtained after starting with unit sides, not the original basic length. Option D, (\sqrt{3}), is reached later after another triangle, so it cannot be the starting length. The exact idea is that each new triangle has one side of 1 unit and the earlier hypotenuse as the other side. Memory cue: start with 1, then the hypotenuses become (\sqrt{2}), (\sqrt{3}), (\sqrt{4}) and so on.

The two perpendicular sides of the first right triangle are 1 unit each. By Pythagoras’ theorem, hypotenuse² = 1² + 1² = 2, so the hypotenuse is \(\sqrt{2}\) units. Note that 2 is the square of the hypotenuse, not the hypotenuse itself. Exam tip: To find the hypotenuse of a right triangle, add the squares of the perpendicular sides and take the square root.

The governing construction is the square-root spiral, also known as the Spiral of Theodorus. After the initial right triangle is made, each new triangle uses the preceding hypotenuse as one side. A segment of exactly 1 unit is drawn perpendicular to that hypotenuse; the new segment and the old hypotenuse then form the two perpendicular sides of the next right triangle. By the Pythagorean theorem, a previous hypotenuse √n produces a new one √(n + 1), since (√n)² + 1² = n + 1. Therefore, option A correctly describes how every new right angle is constructed. The other choices do not follow this geometric rule.

In a square root spiral, the new 1-unit segment is drawn perpendicular to the hypotenuse of the preceding triangle. The resulting hypotenuse represents the next square root. Exam tip: locate the right-angle mark; it is made with the previous hypotenuse.

Adding a (1) unit perpendicular to (\sqrt{6}) gives (\sqrt{7}). The previous hypotenuse has one less number.

The hypotenuses progress as (\sqrt{1},\sqrt{2},\sqrt{3}) and so on. Recognizing the order is useful in easy questions.

In a square root spiral, the previous hypotenuse becomes one side of the next right triangle, and a perpendicular side of 1 unit is added. By Pythagoras, the new hypotenuse gives the next square root. Exam tip: look for the 1-unit perpendicular side.

The direct answer is A: Pythagoras theorem. A square root spiral is made by constructing a sequence of right-angled triangles. In the first triangle, the two perpendicular sides can be taken as unit lengths, so the hypotenuse has length [?] No formula is needed to understand the idea: each new triangle uses the previous hypotenuse as one side and adds a new perpendicular side of one unit. The Pythagoras theorem connects the three sides of a right triangle through the relation \(c^2=a^2+b^2\), so it gives the next hypotenuse, such as \(\sqrt{2}\), \(\sqrt{3}\), and so on. Option A is correct because this repeated right-triangle calculation creates successive square-root lengths. Option B, the Remainder theorem, concerns the remainder obtained when a polynomial is divided by a linear expression; it is not the construction principle here. Option C, the Factor theorem, tests whether a value is a zero of a polynomial; it is unrelated to these triangles. Option D, the converse of Thales theorem, can help identify a right angle in some geometry situations, but it is not the theorem repeatedly used to calculate each new hypotenuse in the spiral. Exam cue: right triangle plus hypotenuse means think of Pythagoras.

In each new triangle, the previous hypotenuse becomes one side and a 1-unit perpendicular is drawn at its end. By Pythagoras’ theorem, the new hypotenuse represents the next square root. Exam tip: look for the perpendicular unit side.

A square root spiral is formed using successive right-angled triangles. The (90^\circ) angle allows the use of Pythagoras’ theorem to find each new hypotenuse, producing lengths such as \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Equilateral triangles cannot produce this construction. Exam tip: If successive hypotenuses represent square roots, look for a right triangle and Pythagoras’ theorem.

The construction follows one repeated Pythagorean pattern. Suppose the current hypotenuse has length \(\sqrt{n}\), and a new perpendicular segment of length 1 is drawn from its endpoint. If the next hypotenuse is \(L\), then \(L^2=(\sqrt{n})^2+1^2=n+1\). Taking the positive square root, because a length is positive, gives \(L=\sqrt{n+1}\).

Thus the correct formula is option A. The expression \(\sqrt{n-1}\) would represent a smaller value, \(\sqrt{2n}\) would require a different construction, and \(\sqrt{n+2}\) would add 2 rather than the square of a unit side. The formula accurately describes every successive step of the spiral.

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

The new right triangle has legs \(\sqrt{8}\) and 1. So its hypotenuse is \(\sqrt{(\sqrt{8})^2+1^2}=\sqrt{9}=3\). Adding \(\sqrt{8}+1\) is incorrect because a hypotenuse is not the ordinary sum of the legs. Exam tip: apply Pythagoras at every spiral step.

Drawing a (1) unit perpendicular on (\sqrt{11}) gives (\sqrt{12}). The previous hypotenuse has one less number.

The governing pattern is that each new hypotenuse is obtained by adding 1 to the square of the preceding hypotenuse. If the current hypotenuse is √n and a perpendicular of length 1 unit is added, then the new hypotenuse h satisfies h² = (√n)² + 1² = n + 1. Taking n = 15 gives h² = 16, so h = √16, the positive length equal to 4 units. Therefore, option C is correct. √14 is the preceding hypotenuse and √15 is the current one, so neither is the next result. √30 is also incorrect because the radicand is increased by 1, not doubled. The construction depends on the Pythagorean theorem, not on ordinary addition of radical expressions.

Since \(3^2=9\) and \(4^2=16\), any square root of a number between 9 and 16 lies between 3 and 4. Therefore, \(\sqrt{10}\) fits. \(\sqrt{17}\) is greater than 4. Exam tip: compare with nearby perfect squares first.

The governing concept is the Pythagorean theorem applied to the first right triangle of the square-root spiral. The two perpendicular sides of that triangle each have length 1 unit. If h denotes the hypotenuse, then h² = 1² + 1² = 1 + 1 = 2. Taking the positive square root gives h = √2, so option A is the correct calculation. Option B uses two sides of length 2 and therefore gives √(4 + 4) = √8, not √2. Option C omits the required squaring of the perpendicular sides and is not the Pythagorean expression. Option D subtracts instead of adding and gives √1 = 1. Thus only option A correctly represents the first construction step.

The direct answer is option A: \(\sqrt{(\sqrt{2})^2+1^2}\). In a square-root spiral, begin with a unit right triangle, whose hypotenuse is \(\sqrt{2}\). To make the next triangle, use this existing hypotenuse as one perpendicular side and add a new perpendicular side of length 1. Pythagoras gives the new hypotenuse as \(\sqrt{(\sqrt{2})^2+1^2}=\sqrt{2+1}=\sqrt{3}\). Option A works because it uses the previous length and one new unit side. Option B, \(\sqrt{2+2}=2\), adds two squared lengths of 2 and does not represent the next construction. Option C, \(\sqrt{3+1}=2\), starts with the number being sought and is circular. Option D, \(\sqrt{1^2+3^2}=\sqrt{10}\), uses the wrong side length and gives another value. Memory cue: every new spiral step adds \(1^2\) to the previous hypotenuse squared.

By Pythagoras’ theorem, \(h_{new}^2=h_{old}^2+1^2\). Hence, each step adds 1 inside the square, producing successive square roots. The 1-unit segment is a leg, not the hypotenuse. Exam tip: identify the perpendicular legs before applying the theorem.

Each new right triangle uses a fresh side of length 1 unit and the previous hypotenuse as its other leg. By Pythagoras, the new hypotenuse becomes \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Two unit legs would not describe later triangles. Exam tip: identify the previous hypotenuse first.

\(\sqrt{8}=2\sqrt{2}\), and \(\sqrt{2}\) is irrational; therefore, \(\sqrt{8}\) is irrational. In contrast, \(\sqrt{1}=1\), \(\sqrt{4}=2\), and \(\sqrt{9}=3\) are rational because 1, 4, and 9 are perfect squares. Exam tip: simplify the number under the root using a perfect-square factor, for example \(8=4\times2\), before deciding.

Each new triangle in a square root spiral is constructed by drawing a perpendicular side of 1 unit on the previous hypotenuse. By Pythagoras, the new hypotenuse becomes \(\sqrt{n+1}\). Using 2 units would not produce the standard spiral. Exam tip: remember the repeated added side is 1 unit.

A unit segment perpendicular to the previous hypotenuse forms the next right triangle. By Pythagoras,

ot valid?

A square root spiral is built by repeatedly adding a right triangle with one new perpendicular side of length 1. If the existing hypotenuse has length \(\sqrt{n}\), then the next hypotenuse has length \(\sqrt{n+1}\), because the Pythagorean theorem gives \((\sqrt{n})^2+1^2=n+1\). The construction therefore proceeds in order: \(\sqrt{28}\), \(\sqrt{29}\), \(\sqrt{30}\), and so on.

Just before constructing \(\sqrt{30}\), the existing hypotenuse must be \(\sqrt{29}\). A perpendicular segment of length 1 is then drawn, giving \((\sqrt{29})^2+1^2=29+1=30\), so the new hypotenuse is \(\sqrt{30}\). Hence option B is correct. \(\sqrt{28}\) is one stage earlier, while \(\sqrt{30}\) is the hypotenuse being made, not the one already present.

The direct answer is A: compass. In a square root spiral, a right triangle produces a hypotenuse whose length represents a square root. To place that length on a number line, the length must be transferred accurately from the geometric figure. A compass can keep the distance between its two points fixed. One point is placed at the starting point of the number line, and an arc is drawn through the required length; the point where the arc meets the number line represents that length. Option A is correct because a compass transfers a length and draws the necessary arc. Option B, a clock, measures time, not a geometrical distance, so it cannot construct or transfer the length. Option C, a balance, compares masses or weights, not line lengths. Option D, a calculator, may approximate a numerical value such as \(\sqrt{2}\), but it does not physically transfer the constructed length to the number line; the question asks about the geometric construction. Therefore the instrument used in the construction is the compass. Memory cue: compass carries a distance; ruler mainly draws a straight line.