Square root spiral

The governing construction rule says that a 1-unit perpendicular is drawn on the previous hypotenuse. If the previous hypotenuse is √n, then the Pythagorean theorem gives the next hypotenuse H by H² = (√n)² + 1² = n + 1. To obtain √10, we require n + 1 = 10, so n = 9. Therefore, the perpendicular must be drawn on √9, and option B is correct. Drawing it on √8 would produce √9, which is one stage too early. Drawing it on √10 would produce √11, and drawing it on √11 would produce √12. The answer is determined by identifying the immediately preceding squared value, not by choosing the target itself.

The governing concept is the Pythagorean theorem used repeatedly in the usual square root spiral. When a perpendicular side of 1 unit is added to a previous hypotenuse of length √n, the new hypotenuse satisfies h²=(√n)²+1²=n+1. Thus the squared hypotenuse increases by exactly 1, producing values such as √2, √3 and √4. If the added perpendicular is changed to 2 units, the relation becomes h²=(√n)²+2²=n+4. The construction still makes right triangles and the angle can still be 90°, but it no longer generates consecutive square roots. Therefore option A is correct; the other choices misunderstand what changes in the construction.

The governing construction principle is the Pythagorean theorem applied repeatedly. In a standard square root spiral, a perpendicular segment of length 1 unit is drawn to the existing hypotenuse. If the existing hypotenuse is √n and the next one is h, then h²=(√n)²+1²=n+1. To obtain √21, we need n+1=21, which gives n=20. Therefore the immediately preceding hypotenuse is √20, so option B is correct. Starting with √19 would produce √20, not √21. Starting with √21 would produce √22, and √22 is already beyond the required construction. The sequence advances through √19, √20, √21, and √22 in that order.

The governing concept is the Pythagorean theorem applied to the right triangle formed at the next stage. The old hypotenuse is √11, and the newly added perpendicular side has length 1 unit. Therefore the new hypotenuse h must be written as h=√((√11)²+1²). Since (√11)²=11, this later simplifies to h=√12. Option B is correct because it first uses the actual lengths of the two perpendicular sides. Option A adds 2 without justification, option C incorrectly treats the old hypotenuse as 11 rather than √11 before squaring, and option D uses subtraction instead of the required sum of squares.

Adding a (1) unit perpendicular to (\sqrt{16}) forms the new hypotenuse (\sqrt{17}). The previous hypotenuse has one less number.

Option A is correct. Since \(3^2=9\) and \(4^2=16\), we get \(9<10<16\), so \(3<\sqrt{10}<4\). Thus, the \(\sqrt{10}\) point lies between 3 and 4. Exam tip: compare with nearby perfect squares first.

In a square root spiral, one leg of the new right triangle is the hypotenuse of the previous triangle, while the other leg is 1 unit. By Pythagoras’ theorem, the new hypotenuse represents the next square root. Exam tip: identify the previous hypotenuse as the shared side.

In a square root spiral, the previous hypotenuse becomes one side and a unit segment is drawn perpendicular to it. By Pythagoras, \\(\sqrt{n}^2+1^2=n+1\\), so the new hypotenuse is \\(\sqrt{n+1}\\). Exam tip: remember “perpendicular unit side.”

The square root spiral uses right triangles to construct successive lengths. If the existing hypotenuse is (\sqrt{3}) and a perpendicular side of length 1 is added, the new hypotenuse is found by the Pythagorean theorem. It is not obtained by simply adding 1 to the old length. Therefore, the mistake to avoid is assuming that \sqrt{3}+1=\sqrt{4}.

Indeed, the square of the new hypotenuse is (\sqrt{3})^2+1^2=3+1=4, so its length is \sqrt{4}=2. Thus option A identifies the common error. Option B states the correct squared-length relation, while making a right angle and drawing the unit perpendicular are necessary construction steps.

If the new hypotenuse is (\sqrt{n+1}), the previous one is (\sqrt{n}). Therefore before (\sqrt{31}), it was (\sqrt{30}).

Since \(49=7^2\), \(\sqrt{49}=7\), which is a whole number. However, \(50\) is not a perfect square; it lies between \(49\) and \(64\), so \(\sqrt{50}\) is not a whole number and is irrational. Option C is incorrect because only \(\sqrt{49}\) is a whole number. Exam tip: A square root is a whole number only when the number is a perfect square.

In a square root spiral, each new triangle must be right-angled. Drawing the 1-unit segment at 90° to the previous hypotenuse allows Pythagoras’ theorem to be used, giving successive hypotenuse lengths such as \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. A 1-unit side alone does not make the triangle equilateral; the right angle is essential. Exam tip: For successive square-root lengths obtained as hypotenuses, look for a right triangle and use Pythagoras’ theorem.

The previous hypotenuse \(\sqrt{n-1}\) with a unit perpendicular side gives \(\sqrt{(n-1)+1}=\sqrt n\) by Pythagoras. The sides are not parallel. Tip: mark the right angle.

The governing principle is the Pythagorean theorem applied successively to the right triangles of the spiral. If the previous hypotenuse is √n and a perpendicular segment of length 1 unit is added, then the new hypotenuse h satisfies h²=(√n)²+1²=n+1. Taking the positive square root gives h=√(n+1). Therefore option C states the correct mathematical reason. Option A incorrectly moves an addition through a radical, option B confuses n with n² and is not generally true, and option D has no role in the construction. The sequence advances because the square of the hypotenuse increases by exactly one at every step.

Taking the hypotenuse length in the compass, an arc is drawn from the origin. This gives the correct point on the number line.

In a square root spiral, each new right triangle uses the previous hypotenuse and a new perpendicular side of 1 unit. By Pythagoras, the new hypotenuse satisfies \(h^2=a^2+1^2\). Exam tip: identify the added 1-unit perpendicular.

Since \(36=6^2\), \(\sqrt{36}=6\), which is a whole number. However, \(37\) is not a perfect square, so \(\sqrt{37}\) is irrational. Option D is incorrect because only \(36\), not \(37\), is a perfect square. Exam tip: The square root of a number is a whole number only when the number is a perfect square.

In a square root spiral, a unit-length side is drawn perpendicular to the previous hypotenuse. By Pythagoras’ theorem, the new hypotenuse becomes \(\sqrt{2},\sqrt{3},\sqrt{4}\), and so on. Exam tip: remember “perpendicular unit side.”

In the usual spiral (1^2) is added each time to give the next square root. Taking (2) units will add (4).

In a square root spiral, the successive hypotenuses are \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on; the number under the square root increases by 1 at each step. Therefore, the hypotenuse after \(\sqrt{99}\) is \(\sqrt{100}\). \(\sqrt{98}\) is the previous hypotenuse, while \(\sqrt{198}\) is not the next term in this sequence. Exam tip: the squares of successive spiral hypotenuses are consecutive natural numbers.

A square root spiral is a chain of right triangles. The previous hypotenuse and a (1) unit perpendicular form the new triangle.

Each new triangle is formed by drawing a perpendicular side of length 1 on the previous hypotenuse \(\sqrt{n-1}\). By Pythagoras, \((\sqrt{n-1})^2+1^2=n\), so its hypotenuse is \(\sqrt{n}\). The 1-unit segment is a leg, not the hypotenuse. Exam tip: every new hypotenuse represents the next square root.

The previous hypotenuse (\sqrt{2}) becomes one side and a (1) unit perpendicular is added. The new hypotenuse is (\sqrt{3}).

Since \(5^2=25\) and \(6^2=36\), and \(25<32<36\), we get \(5<\sqrt{32}<6\). Therefore, \(\sqrt{32}\) lies between 5 and 6 on the number line. It cannot lie between 4 and 5, because that interval corresponds to numbers between 16 and 25. Exam tip: compare the number with the nearest perfect squares to locate its square root.

Adding a (1) unit perpendicular to (\sqrt{47}) forms (\sqrt{48}). In the previous step the number is (1) less.