Square root spiral

Correct answer: B. (\sqrt{41})

Explanation: In the previous step, the number under the root is (1) less. Therefore, (\sqrt{41}) comes before (\sqrt{42}).

Correct answer: B. (47)

Explanation: (\sqrt{2}) is given by the first triangle, so up to (\sqrt{48}), the number of triangles is (48-1=47). Keep the initial (\sqrt{1}) separate in counting.

Correct answer: A. (\sqrt{22},\sqrt{23},\sqrt{24})

Explanation: In consecutive hypotenuses, the numbers under the roots increase by (1). Therefore, (\sqrt{22},\sqrt{23},\sqrt{24}) is the correct order.

Correct answer: A. The newly added side has length 1 and is perpendicular to the previous hypotenuse.

Explanation: In a square root spiral, a unit side is drawn perpendicular to the preceding hypotenuse. By Pythagoras, \((\sqrt{n-1})^2+1^2=n\), so the new hypotenuse is \(\sqrt{n}\). Exam tip: identify both the unit side and the right angle.

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

Explanation: ((\sqrt{30})^2+1^2=31), so the perpendicular is drawn on (\sqrt{30}). For the next square root, identify the previous inner value.

Correct answer: A. The difference of lengths is not (1), the difference of their squares is (1)

Explanation: ((\sqrt{11})^2-(\sqrt{10})^2=1), but (\sqrt{11}-\sqrt{10}\neq1). In exams, understand the difference between length and square.

Correct answer: A. (6\sqrt{2})

Explanation: The direct answer is A: \\(6\\sqrt{2}\\). Simplify the radical by looking for a perfect-square factor: \\(72=36\\times2\\). Therefore, \\(\\sqrt{72}=\\sqrt{36\\times2}=\\sqrt{36}\\sqrt{2}=6\\sqrt{2}\\). The square-root spiral does not change this algebraic simplification. A is correct. B, \\(8\\sqrt{2}\\), is wrong because its square is \\(128\\), not 72. C, \\(12\\sqrt{2}\\), is wrong because its square is \\(288\\). D, \\(36\\sqrt{2}\\), is wrong because its square is \\(2592\\). The error in the wrong choices is taking an incorrect multiplier outside the root. Memory cue: remove the largest perfect-square factor; \\(\\sqrt{36}=6\\).

Correct answer: B. (\sqrt{49})

Explanation: The direct answer is option B: \(\sqrt{49}=7\). A natural number is a positive whole number such as 1, 2, 3, and so on. In the square-root spiral, the successive hypotenuse lengths are square roots of positive integers. A square root is a natural number when its radicand is a perfect square. Check each option: \(\sqrt{45}=3\sqrt{5}\), not a natural number, because 45 is not a perfect square. \(\sqrt{49}=7\), and 7 is a natural number, so option B works. \(\sqrt{52}=2\sqrt{13}\), not a natural number. \(\sqrt{60}=2\sqrt{15}\), also not a natural number. The fact that these lengths can occur in the spiral does not make every one of them a whole number. Exam cue: test whether the number under the radical is a perfect square; 49 is \(7^2\).

Correct answer: C. (\sqrt{81})

Explanation: In the next step, the inner number increases by (1), so (\sqrt{81}) is formed. The spiral does not skip numbers in order.

Correct answer: C. \(\sqrt{a+1}\)

Explanation: In the standard square-root spiral construction, \(PQ\) is perpendicular to \(OP\). By Pythagoras’ theorem, \(OQ^2=OP^2+PQ^2=(\sqrt{a})^2+1^2=a+1\). Hence, \(OQ=\sqrt{a+1}\). The expression \(a+1\) is the value of \(OQ^2\), not the length of the hypotenuse itself. Exam tip: add the squares of the perpendicular sides first, then take the square root.

Correct answer: B. (\sqrt{98}) and (\sqrt{100})

Explanation: In order, the inner number decreases by (1) before and increases by (1) after. So (\sqrt{98}) comes before (\sqrt{99}), and (\sqrt{100}) comes after it.

Correct answer: B. 11

Explanation: \(121\) is a perfect square because \(11 \times 11 = 121\). Therefore, \(\sqrt{121}=11\), so its actual length in the spiral is 11 units. \(121\) is the number under the square root, not its square root. Exam tip: Memorising squares from 1 to 15 helps solve such questions quickly.

Correct answer: C. \(\sqrt{22}\)

Explanation: In a square root spiral, each new step increases the number under the square root of the hypotenuse by 1. Since 7 new steps lead to \(\sqrt{29}\), the starting number was \(29-7=22\). Hence, the starting hypotenuse was \(\sqrt{22}\). If it had been \(\sqrt{21}\), seven steps would give \(\sqrt{28}\), not \(\sqrt{29}\). Exam tip: Add or subtract the number of steps from the radicand, not from the square root itself.

Correct answer: B. \(3\sqrt{6}\)

Explanation: Since \(54=9\times6\), and \(9\) is the greatest perfect-square factor of 54, \(\sqrt{54}=\sqrt{9\times6}=\sqrt9\times\sqrt6=3\sqrt6\). The expression \(2\sqrt{27}\) is neither in simplest form nor equal to \(\sqrt{54}\). Exam tip: first identify the greatest perfect-square factor inside the radical.

Correct answer: B. Drawing a (1) unit perpendicular to the previous hypotenuse

Explanation: The main construction depends on drawing a (1) unit perpendicular to the previous hypotenuse. This creates the next right triangle.

Correct answer: B. No, the direction changes because each new perpendicular is drawn on the previous hypotenuse

Explanation: At each new step, the perpendicular is drawn on the previous hypotenuse, so the direction gradually changes. The spiral is formed by this turning.

Correct answer: B. No, because a right angle is necessary for Pythagoras theorem

Explanation: (\sqrt{16}) is certain only when the new segment is perpendicular to the previous hypotenuse. Without a right angle, Pythagoras theorem cannot be applied.

Correct answer: C. 4

Explanation: In a square root spiral, each new step represents the next number: \(\sqrt{1}, \sqrt{2}, \sqrt{3}, \ldots\). From \(\sqrt{3}\) to \(\sqrt{7}\), the radicand changes from 3 to 7, so the number of new steps is \(7-3=4\). The four steps correspond to \(\sqrt{4}, \sqrt{5}, \sqrt{6}\), and \(\sqrt{7}\). Exam tip: subtract the starting radicand from the ending radicand in such questions.

Correct answer: A. \(3\sqrt{10}\)

Explanation: Since \(90=9\times10\) and \(9\) is a perfect square, \(\sqrt{90}=\sqrt{9\times10}=\sqrt9\sqrt{10}=3\sqrt{10}\). \(9\sqrt{10}\) is incorrect because \(\sqrt9=3\), not 9. Exam tip: identify the greatest perfect-square factor when simplifying a surd.

Correct answer: C. (\sqrt{101})

Explanation: The next hypotenuse is (\sqrt{101}), which is not a simple integer. Even after (\sqrt{100}=10), the sequence continues with (\sqrt{101}).

Correct answer: A. The distance obtained from the spiral can be marked on the number line using a compass

Explanation: The spiral gives an exact distance that can be transferred to the number line with a compass. It is useful for showing irrational numbers.

Correct answer: B. \(\sqrt{62}\)

Explanation: In a square root spiral, the number under the radical increases by 1 at each successive step. Going 3 steps back from \(\sqrt{65}\) gives \(65-3=62\). Hence, the required hypotenuse is \(\sqrt{62}\). \(\sqrt{63}\) is only 2 steps before the final hypotenuse. Exam tip: subtract the number of backward steps from the radicand.

Correct answer: C. Length of every new perpendicular segment is (2)

Explanation: A square-root spiral is built by repeatedly drawing a new segment perpendicular to the previous segment. In the standard construction, each newly drawn perpendicular segment has a fixed length of 1 unit. The initial segment also has length 1 unit, and these two perpendicular unit segments form the first right triangle. Its hypotenuse is therefore
(1^2+1^2)=\sqrt{2}
, which begins the sequence of square roots.

Thus, the statements saying that the initial segment is 1 unit, every new perpendicular segment is 1 unit, and the first triangle uses two segments of length 1 are consistent with the standard construction. Choosing a new perpendicular segment of length 2 changes the Pythagorean calculation and produces a different construction, not the usual square-root spiral. Therefore option C is the incorrect fixed length. The supplied answer and explanation are accurate.

Correct answer: B. (4\sqrt{15})

Explanation: (\sqrt{120}=\sqrt{4\times30}=2\sqrt{30}) is simplest, while (4\sqrt{15}) squares to (240). Check each option carefully.

Correct answer: A. \(2\sqrt{30}\)

Explanation: Since \(120=4\times30\), and \(4\) is a perfect square, \(\sqrt{120}=\sqrt{4\times30}=2\sqrt{30}\). As 30 has no perfect-square factor greater than 1, this is the simplest form. The close distractor \(4\sqrt{15}\) is incorrect because its square is \(16\times15=240\), not 120. Exam tip: identify the greatest perfect-square factor inside the radical before simplifying.