Square root spiral

Correct answer: B. √9

Explanation: 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.

Correct answer: A. Because the rule of adding 1 at each step will break

Explanation: 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.

Correct answer: B. √20

Explanation: 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.

Correct answer: B. √((√11)²+1²)

Explanation: 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.

Correct answer: A. (\sqrt{16}) and (1)

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

Correct answer: A. The statement is correct, because \(3^2<10<4^2\).

Explanation: 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.

Correct answer: A. पिछले त्रिभुज का कर्ण

Explanation: 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.

Correct answer: B. A new unit-length side is drawn perpendicular to the previous hypotenuse

Explanation: 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.”

Correct answer: A. Assuming (\sqrt{3}+1=\sqrt{4})

Explanation: 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.

Correct answer: B. (\sqrt{30})

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

Correct answer: B. \(\sqrt{49}\) is a whole number and \(\sqrt{50}\) is irrational

Explanation: 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.

Correct answer: A. To form a right-angled triangle and apply Pythagoras’ theorem

Explanation: 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.

Correct answer: B. In each new right triangle, one side is 1 unit and the other side is the hypotenuse of the previous triangle.

Explanation: 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.

Correct answer: C. 1² is added to the square of the previous hypotenuse

Explanation: 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.

Correct answer: A. At the origin

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

Correct answer: A. At each new step, a perpendicular of length 1 unit is drawn at the end of the previous hypotenuse.

Explanation: 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.

Correct answer: B. \(\sqrt{36}\) is a whole number and \(\sqrt{37}\) is irrational

Explanation: 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.

Correct answer: A. By adding a unit-length side perpendicular to the previous hypotenuse

Explanation: 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.”

Correct answer: A. Because (2^2) will be added instead of (1^2)

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

Correct answer: C. \(\sqrt{100}\)

Explanation: 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.

Correct answer: A. Each new triangle is made from the previous hypotenuse and a (1) unit perpendicular

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

Correct answer: A. It is the hypotenuse of the new right triangle whose other sides are \(\sqrt{n-1}\) and 1.

Explanation: 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.

Correct answer: A. (\sqrt{2}) and (1)

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

Correct answer: B. Between 5 and 6

Explanation: 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.

Correct answer: B. (\sqrt{47})

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