Square root spiral

A square-root spiral is built by repeatedly forming a right triangle. At each stage, the existing hypotenuse becomes one leg of the next right triangle, and a segment of length 1 is drawn perpendicular to it. This perpendicularity is essential because it allows the Pythagorean theorem to produce the next length: if the current hypotenuse is √n, the next one has square equal to n + 1², or n + 1. Thus the direction is determined by the previous hypotenuse, not by the page or the origin. Option A is correct. A segment need not remain horizontal, so B is false; it need not point toward the origin, so C is false; and a parallel segment would not create the required right triangle, so D is false.

((\sqrt{7})^2+1^2=8), so the new hypotenuse becomes (\sqrt{8}). The process always starts from the previous square root.

In the square root spiral, (PQ=1) unit is used to form the next triangle. This makes (OQ^2=m+1).

Because the new (1) unit segment is perpendicular to the previous hypotenuse, (OQ^2=OP^2+1). Keep squares and lengths separate while writing the formula.

In a square root spiral, a 1-unit side is drawn perpendicular to the previous hypotenuse. By Pythagoras, if the previous hypotenuse has square n, the new hypotenuse has square n + 1. Exam tip: look for the right-angle mark.

The first triangle gives (\sqrt{2}), so for (\sqrt{12}), the order is (12-1=11). Subtract (1) from the inner number.

In the standard square-root spiral, the first initial radius is usually taken as 1, representing √1. Each newly constructed right triangle has one side of length 1 perpendicular to the current hypotenuse. If the current hypotenuse is √n, the next hypotenuse is √(n+1), because its square is n + 1² = n + 1. Consequently, the first triangle after the initial radius represents √2, the second represents √3, and in general the rth constructed triangle represents √(r+1). Substituting r = 10 gives √(10+1) = √11. Therefore option C is correct. √10 corresponds to the ninth constructed triangle under this indexing, while √9 and √12 are one step before and one step after the required value.

In the square root spiral, the labels represent lengths of hypotenuses. The question asks for the difference between the squares of two lengths, not the difference between the lengths themselves. Squaring removes the radical in each case: \\(\sqrt{2})^2=2\\) and \\(\sqrt{3})^2=3\\). Thus the requested difference is obtained by subtracting 2 from 3.

The calculation is \\(3-2=1\\), so option A is correct. Option B, \\(\sqrt{1}\\), has the same numerical value as 1, but it is not the usual listed form of the difference of the squares; option C gives the difference of the two hypotenuse lengths, not their squares. Option D is unrelated. Hence A gives the intended answer.

The start is (OA=1), and then the hypotenuses (\sqrt{2},\sqrt{3},\sqrt{4}) are formed. The number under the root increases continuously.

(\sqrt{27}) is formed by the (26)th triangle, so (26) new perpendicular segments of (1) unit are used. The first (1) unit perpendicular forms (\sqrt{2}).

A square root asks for the non-negative number that, when multiplied by itself, gives the number inside the radical. Since 16 is a perfect square, its square root can be simplified exactly. The number 4 satisfies \\(4^2=16\\), so the constructed length \\(\\sqrt{16}\\) has actual numerical value 4. Therefore option A is correct.

The result should not be confused with the radicand, 16, which is the number under the root sign. Likewise, 8 is not correct because \\(8^2=64\\), and \\(\\sqrt{8}\\) is a different, non-integer value. The useful method is to recognize perfect squares immediately: \\(\\sqrt{16}=\\sqrt{4^2}=4\\).

The spiral labels its successive hypotenuses by square roots. To find the step whose hypotenuse has actual length 3, express 3 as a square root: 3 = √9, because 3² = 9. Therefore the required hypotenuse appears at the √9 step. Option D is correct. The values √6, √7 and √8 are all less than 3, since their radicands are less than 9; hence none of them can represent a length of exactly 3 units. This question tests the distinction between the number under the radical and the value of the radical itself. One should square the desired length, not select the radicand that is numerically closest by appearance. In the spiral, the √9 stage is also the stage at which the hypotenuse is exactly 3 units long.

The new perpendicular side between consecutive hypotenuses is (1) unit. This can also be seen from (PQ^2=31-30=1).

In one step, the number under the root increases only by (1), so (\sqrt{15}) should come after (\sqrt{14}). Skipping a step is incorrect.

In a square root spiral, the new (1) unit segment is drawn perpendicular to the previous hypotenuse. A parallel segment will not allow Pythagoras theorem to apply.

\(\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}\). Since \(\sqrt{2}\) is irrational, \(5\sqrt{2}\) is also irrational. In contrast, \(\sqrt{25}=5\), \(\sqrt{36}=6\), and \(\sqrt{49}=7\) are rational because 25, 36, and 49 are perfect squares. Exam tip: Check whether the number inside the square root is a perfect square; if it is not, its square root is generally irrational.

This relation matches Pythagoras theorem, so (OP) and (PQ) are perpendicular. This right-angle construction is essential in the square root spiral.

Taking (1=\sqrt{1}) makes the whole sequence clear. Then the inner number increases by (1) at each step.

The squares of the hypotenuses are (2,3,4), which increase by (1). The actual lengths do not have equal differences.

The exact value is (\sqrt{5}), while (2.236) is only an approximation. In exams, prefer the exact square root form.

(\sqrt{6}) is formed from the previous hypotenuse (\sqrt{5}) and the new perpendicular side (1). Therefore, ((\sqrt{5})^2+1^2=6) is correct.

(\sqrt{2}) is given by the first triangle, so up to (\sqrt{n}), the number of triangles is (n-1). Remember the starting term while writing a general formula.

The first hypotenuse is formed from (1) and (1), giving (\sqrt{2}). (\sqrt{4}) appears in a later step, not as the first hypotenuse.

The square-root spiral proceeds in consecutive stages: √1, √2, √3, and so on, with each new stage increasing the radicand by 1. Therefore, if the final stated hypotenuse is √32, the immediately preceding stage must have radicand 32 − 1 = 31. Its hypotenuse is consequently √31, so option B is correct. √30 would be two stages earlier, √32 is the final stage itself rather than the stage before it, and √33 would be the next stage, which has not yet been constructed. The important governing pattern is that the construction does not skip an integer radicand: each perpendicular unit segment changes the squared hypotenuse from n to n+1. Hence reading the sequence backward by one step gives √31 without needing to calculate decimal approximations.

The square root spiral represents square roots like (\sqrt{2},\sqrt{3},\sqrt{5}) geometrically. It helps in understanding the number line and irrational numbers.